OK. So this is the second lecture of algebraic geometry. So we introduced the fine algebraic sets. So these are subsets 0 set of s in a k, where s is some set of polynomials, the mimecate n in n variables. And z of s just means it's a set of all points p in a n, such that f of p is equal to 0 for all f in s. We had seen that these fine algebraic sets also can be just obtained as 0 sets of ideals. So of course, the 0 set of s is equal to the 0 set of the ideal generated by s. And we had also introduced the ideal of a fine algebraic set. So if x subset, we have the ideal of x, which is just a set of all polynomials f in k x1 to xn, such that f restricted to x is the 0 polynomial. And say, if we assume that x is in a fine algebraic set, we can consider this. So we have a way how to associate to an ideal. It's 0 set, which is a fine algebraic set. And we have also the possibility of associating to a fine algebraic set an ideal. And this ideal is somehow associated in a natural canonical way to x. So you can somehow hope that the properties of the ideal, I mean the algebraic properties of the ideal of x, are closely related to the geometric properties of x. Let us see that this is the case. And then what else do we have? We had the Hilbert-Base's theorem, which used in the form that we have that k x1 to xn is Neutering. So that means, in particular, one way of formulating is that every ideal in this ring is finitely generated. And as a corollary, we would get that every fine algebraic set is a 0 set of finitely many polynomials. So this is just clear. So we know that x is 0 set of some ideal. And this ideal is finitely generated because every ideal in k x1 to xn is finitely generated. So we can write i because the ideal is generated by some elements f1 to fr for some r. And then we have that x is 0 set of the ideal generated by the fi, which is the same as 0 set of the fi. So in our ritual definition, we allow 0 sets of arbitrary sets of polynomials. But in the end, one only needs finitely many, so it somehow feels a bit more comfortable. So now, who's that? Now I want to give an application of a geometric application in some sense of this Hilbert-Basey's theorem. I already hinted at it. So the statement is that every fine algebraic set is a finite union of a fine algebraic set which cannot be further decomposed. So we'll have to see precisely what I mean by that. So there is some kind of way how you can really say that in a fine algebraic set is a union of finitely many pieces. And these pieces will be called the irreducible components. So irreducible components. So you know, for instance, that if in A2, if you take the 0 set of x times y, where x and y are the coordinates, then we know this is the 0 set of z of x, union is 0 set of y. If you look at the real points, then it would look this one would be the union of two lines, the horizontal and the vertical line. And you can evidently see that this thing in some sense has two pieces. And somehow we want to see that it's always like that. And what one cannot see here immediately, but which turns out to be true, is that these pieces cannot be further decomposed. One cannot write it as a union of a fine algebraic sets. I mean, at least not a finitely many. So in order to make sense of this statement and prove it, we want to formulate it in a topological way. So we have this risky topology. In terms of this, we can say what we mean by something being irreducible. And then we want to show this statement about the irreducible components. So let's see. So formulate it more generally as a topological statement. Then later, such statements can be applied in slightly larger generality. So we say definition. A topological space x is called reducible if we can write it as a union of two closed subsets, none of which is equal to x. So if x is equal to x1 union x2, where x1 and x2 are closed subsets, and x1 and x2 are strictly contained in x. So to say it, so for instance here, this is evidently the case. This zxy is the union of z of x and z of y, and both of them are closed subsets. And none of them is equal to this union. And then what we really are interested in is irreducible, which is the opposite. So something is called irreducible if it's not reducible. So explicitly it says that, I mean, because we use it most like in that way. So if x is irreducible, if x is the union of x1 union x2 with xi in x closed subsets, then it follows that x is equal to x1, or x is equal to x2. So that is what it means to be irreducible. OK, that's just the definition. So maybe I should say that these topological statements, we will apply to find algebraic varieties with the Zariski topology. And if you look at it in the usual sense, if you look at subsets of Rn with the usual topology, these kind of concepts make essentially no sense. I mean, if for instance in a Hausdorff topological space, the only irreducible subsets are the points. So it's not such an exciting concept if you use this in the things you usually study. But the Zariski topology is very strange. This concept does make sense. So one thing is that, for instance, if x is irreducible and u subsets x an open subset, then it follows. So and u is non-empty. So any non-empty open subset of an irreducible space, then the closure of u, then u is dense in x. So if you have an irreducible topological space, then any non-empty open subset is dense. And this is easy because we can certainly write x equal to x minus u, which is a closed subset because u is open, union, the closure of u, which is also a closed subset. And as x is irreducible, it follows that x is equal to x minus u or x is equal to the closure of u. But u was non-empty. So this is obviously not true. This is not true. So it follows that x is equal to the closure of u and u is dense in x. And in a similar way, you can show it's an easy exercise. So under these assumptions, also u is irreducible. So if x is irreducible and u is a non-empty open subset, then u is also irreducible. And it's an easy exercise in the definitions to see that. So now we give some examples in the examples. So first, as I said, a point P in an is irreducible. Because if you write it as a union of two closed subsets, then one of them must be the point. And then we also see that the zero set of x, y in a2 is reducible because it's a union of z of x and z of y. And you can easily see that none of these two is equal to x to the zero set of these two. Take the point, whatever, 1, 0. It will lie in one. If you take the point 0, 1, it lies in the other. And so this is reducible. Now we want to show the statement that I made about irreducible components. We want to show that a fine algebraic sets can be written as a union of finitely many irreducible fine algebraic sets, where I call in the fine algebraic set irreducible as it's irreducible as a topological space. And again, to make it more generally applicable, we will prove a slightly more general statement in a topological context. So we first define something which will be true for the fine algebraic sets. And then we prove that the statement about decomposition into irreducible components holds for each such thing that we define. So this is the following statement. Definition, a topological space is called Neuterium if the following holds for if every descending chain, so x contains x1, contains x2. So an infinite chain like this of closed subsets becomes stationary. So that means from some point onwards, they are all equal for some n. So this is obviously very similar to the definition of a Neuterium ring. So there we also have some chain condition. But for a Neuterium ring, we had the ascending chain condition. So if you have a chain of ideals where they become larger and larger, then they will become stationary. And here we have it for closed subsets, but they become smaller and smaller. But if you think of it, this should somehow correspond. And we'll see that because for a fine algebraic sets, if we have an inclusion i1 is contained in i2, then the zero set of i2 is contained in i1. We have the inclusion gets reversed so that in ascending chain will become a descending chain. So its definition is precisely made in such a way to match, via this kind of duality, the Neuterium ring for rings. OK, so I make some mark. So first, any subspace of a Neuterium topological space topological is Neuterium. So we have a subspace, we have a topological space. We will call the subspace y of a Neuterium topological space x is Neuterium. So subspace mean is a subset with induced topology, as usual. So this is almost obvious. Let's see. So assume we have a descending chain. Take a descending chain in y. So y contains y1, contains y2, a descending chain of closed subsets. Well, the subspace topology means precisely that I can take these closed subsets as the intersection of y with closed subsets of x. So then we have that for all i, yi can be written as x intersected xi, where xi in x is a closed subset. So now one could be tempted to say that the xi form a descending chain in x, but this is actually not true. But with the tiny modification, they do. So we put xi prime to be the intersection for all j smaller equal to i. So one smaller equal to j smaller equal to the given i. So for all j from 1 to i of the xi. Then it is clear that if I take xi prime intersected y, this is the same as x intersected. It's the same as yi. Because you know that the intersections of the xi with y form a descending chain. So if we first intersect them in this way, we get the same intersection with y. And now these xi prime form a descending chain in x. So then x descending chain of closed subsets of x. So it becomes stationary. So for some n, we have that xn prime is equal to xn plus 1 prime. Well, and as the yi are just given as intersection of y with this thing, with these, so then they also become stationary. Thus also this chain becomes stationary. So these are some very easy observations. So I also need a second statement to now finally come back to our fine algebraic sets. So the claim is that an is in a terren topological space. And thus by 1, every a fine algebraic set, and in fact any subset of an with induced topology is in a terren. So we only have to prove this statement. The second follows by 1 because we have the statement that any subspace of a terren topological space is in a terren. So how is that? Well, and this is precisely the story with the chains of closed subsets corresponding to chains in the other direction of ideals. So let say x contains x1, contains x2 be a chain. So x is an, after all, of closed subsets in an. Well, then I can look at the ideals of these closed subsets. So the ideal of x1 will be contained in the ideal of x2 is an ascending chain of ideals in kx1, xn. So we know it becomes stationary. Thus we have that same, it becomes stationary. It means i xn is equal to i of xn plus 1. And but note that we have seen that for a close, for a fine algebraic set, the zero set of the ideal is equal to the set itself. So thus also xn, which is the zero set of i of xn, is equal to xn plus 1. So this proves that an is in a terren topological space. We have a chain, our given chain of ideals of closed subsets becomes stationary. And so whatever we have said about a new terren topological spaces applies to a fine algebraic sets. And also to open subsets of a fine algebraic sets, whatever we want. Now we want to make a statement about these irreducible components. This is theorem and also definition. So we prove that something exists and then give it a name. So let x be a terren topological space. Then there are two statements. First, x is a union of finitely many irreducible topological spaces. So I can let x is equal to x1 union. Maybe irreducible, I don't think I want like this, irreducible closed subsets. And the second statement is that in a suitable sense, this decomposition is unique.  You can always add to this decomposition some closed subset, which is contained in one of the xi's. You can add as many additional components as you want by just taking something very small, which fits into one of them.  So if we require that xi is not contained in xj for i different from j, then this decomposition is unique up to reordering. So let's see.  So we want to see. I mean, the most important thing is the first part. We want to see that we can write x as a union of finitely many irreducible closed subsets. So let's start with the proof. So we first show, in some sense, the existence. So basically 1. So for the existence of a decomposition like this, also if you want to have this property, it's enough to show the existence of any such decomposition without this property. Let me say it more precisely. So it's enough to show the existence of a decomposition. x is equal to x1 union xr with xi irreducible closed subsets. So for the existence part of 2, which means there exists such a decomposition where the xi are not contained in the xj, is trivial. If some xi is contained in another xj, we just throw it away. And the union of all the xi's is still x. Because after all, the thing we threw away was contained in another one. So by removing the xi with xi contained in xj for i different from j, we get a decomposition as in 2. So x equal to the union. And xi is not contained in xj for i different from j. So this is kind of clear. Now we want to prove just that we have a decomposition into finitely many irreducible closed subsets. And this we do by an indirect argument. We assume such a thing does not exist and we get to a contradiction. So assume x does not have a decomposition into finitely many irreducible closed subsets. So we make this assumption. We will want to come to a contradiction. So one thing, if x doesn't have such a decomposition into finitely many irreducible closed subsets, certainly x is not irreducible. Because otherwise we could take x equal to x as the decomposition. So in particular, x is reducible. So we can write x is union of x1 and y1, where x1 and y1 are closed subsets. And it can also not be that both x1 and y1 have a decomposition into finitely many irreducible closed subsets. Because otherwise we just take the finitely many for this plus the finitely many for this. And we still have finitely many for x. So one of the two does not have a decomposition into finitely many closed irreducible closed subsets. So one of the two and whose x1 and y1 is just names that we give. So say y1, say x1, does not have a decomposition into finitely many irreducible closed subsets. But if you look at it now, this is precisely the assumption we made about x. The only assumption we made about x is that it does not. So I wrote something else. I said x does not have a decomposition. And I wrote it does. But obviously the assumption was that. Otherwise, how do we bring it to a contradiction? So we assume that x does not have a decomposition into finitely many irreducible closed subsets. Then in particular, x is irreducible. And now we find that one of the two does not have a decomposition into finitely many irreducible closed subsets. This is precisely the assumption we made about x. So whatever we have deduced about x, we also deduce of x1. So we can again write x1 as a union of two things, one of which does not have a decomposition and so on. So we just repeat the argument with it. So this is the same assumption, same statement as the assumption about x. So we get, we can repeat the argument. So repeat the argument. Then we get that x1 can be written as x2 union y2, where x2 does not have a decomposition into finitely many irreducible closed subsets. So we get, we can again apply to this. So we get inductively that in descending chain, x contains x1, x2, and so on. And if you look at it here, I mean, I haven't been careful in writing that x is the union of x1 and y1. But part of the story is it's the union of two such things, none of which is equal to x. We have in particular also that x1 is different from x and y1 is different from x. Because being reducible means that you can write it as a union of closed subsets of two closed subsets, none of which is equal to x. And the same applies here. So this means here x1 is not equal to x, x2 is not equal to x1, and so on. So we get a descending chain where at each step it gets strictly smaller. So that means the chain is not stationary. And this is a contradiction to x be in the terrain. So we can conclude that we were wrong in the beginning. And x does indeed have a decomposition into finitely many irreducible closed subsets. So we get a statement of one. OK. So now the second is sometimes. So you see, that's somehow how one usually proves results using this notoriety that you somehow make such a chain and bring things to contradiction. It always leads to some indirect argument which is very non-constructive. I mean, it actually gives you no idea how you would find these components. It just is an existence result. But this is what this assumption will give you. So let's prove two, I mean, or rather the uniqueness. So we assume now we have a decomposition into x1 union xr such that no xi is contained in another xj. Now we have seen that we can get that. And actually we assume we have two such decomposition. ys, the two compositions as in two. So x is the union of the xi. It's also union of the yj. These are all irreducible closed subsets. And none of the xi is contained in another xj. And none of the yi is contained in another yj. And we want to show that these are the same things. r is equal to s. And the yi are permutation of the xj. So we can certainly write xi is the intersection of xi with x, which is the same as the intersection as union. i equals 1 to s of xi intersected j equals 1 to s yj. Because after all, the union of these is just x. And so this is certainly the case. But now xi is irreducible. And these are closed subsets. So it follows that if xi is equal to the union of these finitely, many it must be equal to one of them. It follows that xi is equal to xi intersected yj for sum j. But what does that mean? It means that xi is contained in yj. So we find that each xi is contained in yj. In the same, I mean, each xi is contained in sum yj. In the same way, we can exchange the role of the x and y here. So similarly, we have that yj is contained in xk for some k. So we have this chain that xi is contained in yj is contained in xk. But then obviously, xi is contained in xk. And we have excluded that. The only possibility how this can be the case is that xi is equal to xk. And actually, the k is equal to i. So it is only possible if i is equal to k. But in this case, you see that yj is squeezed here in between. It contains this. And it's contained in the same. So they are equal. And yj is equal to xi. So we have that each xi is equal to sum yj. In the same way, you can exchange the word each yj is equal to sum xi. And this just means that the yj's are just the permutation of the xi's. We have here r different things. We have here s different things. But each of them is equal to one of these. And each of them is equal to one of these. So the number of things is the same. And they are just permutations of each other. So r is equal to s. And the xi's, or yj's, are a permutation, the xi's. So this was quite, OK. So here, this is actually quite simple. So we just use this condition that we put here. If one is contained in the other, they must be equal. Then just uses this together with this condition of irreducibility here. And one gets a diminutive. So the reason why we introduce these irreducible algebraic sets is that it's, from some point onwards, we will mostly just consider those. So in the future, but not immediately in the future, but most of the time, we will mostly consider only irreducible algebraic sets. And that's OK, because we know that if we have a general defined algebraic set, it's just a finite union of these. So if we know all the pieces, we know the thing. It's not such a thing. And so we consider it important enough that we give it a name. So the definition and the fine variety is an irreducible defined algebraic set. And I might see in a moment. And now one thing I had mentioned when I talked about the ideal of an defined algebraic set that because the ideal is kind of associated in a canonical way to the defined algebraic set, one should have that the algebraic properties of the ideal should somehow reflect some geometric properties of the defined algebraic set. And the first instance we have here, an defined algebraic set is irreducible if and only if its ideal is a prime ideal. This is a proposition. x in a defined algebraic set. Then x is irreducible if and only if. The ideal of x is a prime ideal. So we have some, at least, we have a somewhat interesting concept in commutative algebra, which is related to this geometric property of being irreducible. And this is actually quite simple. If you think of it, so we have these two directions. I would want to prove that. So we assume, say, x is irreducible. And then we have to show it's prime ideal. So recall the definition of a prime ideal, so an ideal. So that means ix is an ideal. And in addition, if a product f times g, where f and g are elements of kx1 to xn, so prime ideal in kx1 to xn. If f and g are two polynomials that take their product, if that lies in ix, then it follows that f is in ix or g is in ix. This is the definition of a prime ideal, once it's an ideal. So let x be irreducible.  So then our task is to prove that one of the two already lies in the ideal of x. Well, that's very simple. So the ideal of x is a set of all polynomials which vanish on x. So it means that if I take f times g vanishes on the whole of x, or saying it in a different way, we have that x is contained in the zero set of f times g. And so in particular, we can write x to be in similar fashion as here, to be x intersected the zero set of f. And recall that this is just the zero set of f union zero set of g. And from this one can already see that these concepts should be related. So therefore, we can write x as the union of x intersected the zero set of f, union x intersected the zero set of g. Because union of c of f and g of g contains x. So x is union is equal to that. So these are two closed subsets of x. So two closed subsets of x. And x is irreducible, so one of them must be equal to x. So it follows x is equal to x intersected z of f, or x is equal to x intersected z of g. But what does it mean? If x is equal to x intersected z of f, it means that x is contained in the zero set of f, which means that f lies in the ideal of x. And this was precisely what we had to see. So this is the one direction. And now the other direction is not much different, except that we decide to do it indirect. So we assume that x is reducible, so not irreducible. And we have to show that ix is not a prime ideal. So assume x reducible, show ix is not a prime ideal. Well, so again, we just spell out the definitions. And then we cannot avoid proving the result. So we assume it's reducible. So x is equal to x1 union x2, where the xi are closed subsets of x, and none of them is equal to x. So we have the zero set of the ideal of x1. We know that this is equal to x1. And this is contained in x, which is the zero set of the ideal of x. And it's not only contained, it's strictly contained. So we have this and same for x2. So we have that as we have the strict inclusion like this, we must have the strict inclusion the other around between the ideals. Because if the ideals were equal, then the zero sets could not be different. So it follows that i of x1 contains strictly i of x. And in the same way, obviously, ix2 constrain strictly i of x. So we can find an element f in i of x1, which does not lie here. And we can find in g here, which does not lie here. So let f in i of x1 on i of x, in g in i of x2 without i of x. Well, so here we have found two elements which do not lie in i of x, but the claim is obviously that the product lies in i of x. And so it's not a prime ideal. So obviously, if I take f times g, so f vanishes on x1. g vanishes on x2. So f times g vanishes on x1 union x2. So z vanishes just as a union of the zero sets. And this is equal to x. And as the ideal of x is just the zero set is just the set of all functions which vanishes on x. It means that f of times g is an element in the ideal of x. So we have indeed found two elements, f and g, which do not lie in the ideal of x that the product does. So i of x is not a prime ideal. OK, and so this proves this statement. So in the moment, this doesn't help us so very much to decide whether a given fine algebraic set is irreducible. So one thing we can see as an example is that An is irreducible. That was maybe not so clear before, but now it's clear because we know that An, what is the ideal of An? So what are all the polynomials which vanish on the whole of k to the n? I mean, a polynomial vanishes on the whole of k to the n precisely when it's 0. And we know that kx1 to xn doesn't have zero revives or anything, so 0 is a prime ideal, which is a prime ideal. So we see that An is indeed irreducible. I should maybe say that this fact that if x is irreducible, then its ideal is a prime ideal is actually the main reason why we will want later to restrict attention to irreducible algebraic sets to a fine varieties. Because it is very useful for us that the ideal is a prime ideal. The main reason is the following. So this is a remark. So if x is in a finite algebraic set, so we will consider the following, the so-called a fine coordinate ring, A of x. So I assume that x is maybe contained in kx1 in An. So Ax will be just defined to be all polynomials divided by the ideal of x. And we will use this ring very much because we can view this as somehow functions on x. And we use this to make maps between such things. So these are somehow our functions on x. And so we will make a lot of use of this. And the point is that if x is irreducible, if and only if ix is a prime ideal, if and only if this Ax is an integral domain. And it's very useful for us that this thing is an integral domain because it's much easier to work with integral domains than with rings which have zero devices. In particular, as we will want to consider the so-called the field of fractions of A of x, which doesn't even exist if it's not an integral domain. OK. Yeah, well, these are polynomial functions somehow. But I mean, we will come back to that. So you find, if you think of it, that these are all the functions from x to k which can be obtained as restrictions of polynomials to x. OK. And these are the kind of natural functions that one might want to consider. But we will not, in the moment, do this. But this is the main reason why we later will restrict our attention to irreducible algebraic sets and call them varieties. Yeah, I want to very briefly also touch some other subject which we will not give us. So just that you have seen it when we finally do it. So I want to just say two words about dimension. So there's a concept of dimension for a fine algebraic set and which surprisingly turns out to work rather nicely and similarly to what one would expect. It's part of the fact that the definition will look rather strange to you. And so again, we introduce the dimension more generally for so-called, for neutral anthropological spaces. So we have a, but you think of a fine algebraic set. And the idea, the basic idea is very simple. So see if x is irreducible and y is also irreducible and y is a close subset in x, then somehow y is smaller than x. And we would really feel that y is given by some close condition in x. It should really have a smaller dimension. So then we want the dimension of y is strictly smaller than the dimension of x. And somehow if we cannot fit another irreducible close subset between y and x, we would want that the difference of dimension is precisely y. This somehow should define the dimension. So one can actually turn this into a definition. As I said, we are not going to work with this now. We will introduce it once. And then in the second half of this course, we will actually work with it and show that one actually can prove something about it. But in the moment, we don't have the tools. So let x be a, say, non-empty irreducible topological space. The dimension, which we call dmx of x, is the largest n so that we have some chain of close subsets. So it's the largest n is an ascending chain, empty set, x0 strictly contained in x1, strictly contained in x2, and so on. And finally, we come to xn, which is equal to x, of irreducible close subsets. And if x is not irreducible, if just in the terrarium space, not irreducible, we take the maximum of the dimensions of the irreducible components. So if x is in the terrarium topological space, then the dimension of x equal to the maximum of the dimensions of the irreducible components of x. Now this, I think, reminds me of the fact that I did not, or that I at least don't recall, but I don't think I properly did it. When we have had this theorem and definition, I somehow forgot the definition part. So we had the statement that if, so I just make this as a side remark to have it once completely on the blackboard. So we had x in a terrarium topological space. Then we had that x can be written as x1 union xr, finite decomposition with the xi irreducible close subsets. And this was unique if we assumed that xi is not contained in xj for i different from j. And then under these assumptions, so when we have this thing, this unique decomposition, the xi are called the irreducible components of x. So now this definition here makes sense if we have an irreducible topological space. The dimension is the length of the longest chain of the largest n for the length of a chain of irreducible close subsets. And if x is not irreducible, we just do this for each irreducible component of x. And we take the maximum of the dimensions. And that's the dimension of the whole thing. So this definition in the moment is not very practical because we don't have any idea how to find out something about all close subsets of an affine algebraic set. So there are very few things that we can say. We can say that, first, the points p in n have dimension 0 because the chain will only consist of the point itself. And then that's it. As it also should be, a point has dimension 0. And it's also comforting that a1 has dimension 1. And this we actually proved the other time without seeing it. Namely, we had seen that the only irreducible close subsets of a1 are the points and a1 itself. Because we have seen that the close subsets are the empty set, the finite sets, and a1. And so if you have a finite set of points, it's not irreducible, unless it's just one point. So we only have this. So the maximum length of a chain is the empty set contained in a point, contained in the whole of a1. So the dimension is 1. And something which we cannot see, which obviously is necessary for the whole thing to make any kind of sense, is that in the moment we cannot prove, but we will later, but it's true that an has dimension n. If you look at this definition, this is not so clear. But it's evident that the dimension of n is at least n. So it's easy. It's bigger equal to n. I mean, if, for instance, you are in two-dimensional space, what kind of chain can you take? You can start with a point. Then you take a line through the point. And then you take the whole thing. So this is a chain of length 2. You can do the same thing in an. You take a point. You take a one-dimensional linear subspace. Then a two-dimensional linear subspace. And each of these are fine algebraic sets. I mean, you can see that they are irreducible. Maybe that is not so evident. But it is quite evident. But anyway, as we are not proving it, so let me just write it down. So we have an ascending chain. So 0.0 is contained, say, in the 0 set of the variables x2 to xn. This is contained in and always strictly 0 set of x3 to xn and so on. 0 set of xn contained in an. And this is a chain of length n. It's not completely evident that these are all irreducible. We will see this later soon anyway. But if you believe that for the moment, we have an ascending chain of length n of irreducible closed subsets. And so we get the dimension of n is at least n. And later, we will prove really something about the theory of dimensions. We'll see that there's actually a lot that one can say. But for this, we need to develop quite a few tools. So now we just go. So this was just some kind of preview for what we want to do. So now we come to another big theorem, which is the Hilbert-Nulsteinsatz. So let me, this is another section kind of. So this is again after the well-known German mathematician David Hilbert. And Nulsteinsatz is a German word, which is how everybody calls it in spite of just because that's how Hilbert called it. And this means theorem of zeros. But if you look in a book, it's called like that, not like that, even if it's an English book. I mean Nulsteinsatz means zero. Nulsteinsatz means the place where the zero is. And Zatz means theorem. So it's all easy. Now, so this is about the relations of ideals and their zero sets. So remember, if you have an ideal in the polynomial ring, we can associate to it, we have its zero set, which is in a fine algebraic set. On the other hand, if x is in a fine algebraic set, we can associate to it the ideal of x, which is an ideal. So if you want, you can say we have two maps between the ideals and the fine algebraic sets in both directions. So we have here, say, which we do want it, the fine algebraic sets in the N. And here, we have the ideals in kx1 to xn. And we have two maps. We can associate to a fine algebraic set its ideal. And we have a map in the other direction, which associates to an ideal its zero set. Now, in the best of all worlds, you could hope that, for instance, these are inverse bijections. So that we somehow talking about the fine algebraic sets is the same as talking about ideals. Just you can go backwards and forwards. We have to figure out whether that's true. And it actually turns out to be not quite true. So are these inverse to each other? If not, what is the precise relation? So we had already seen, this was an exercise we know, that if we take the zero, so if x is in a fine algebraic set, and we take the zero set of the ideal of x, this is equal to x. More generally, if x is any subset of an, then the zero set of the ideal of x is the closure of x. So we know, at least, that if we start from here, we go here, and here we get the identity, but it turns out, we'll see it's not true, that in the other way, we get the identity, that the ideal of the zero set of an ideal j, the j ideal, is equal to j. It is clear again, whether it's again obvious, but clearly we have that the ideal of the zero set of j contains j. This is just by definition, because what do we have here? We have the zero set of j is a set of all points, where all point on what? We will see, I write it a little bit larger, we will see. This is not true. In one direction, it's true. The ideal of the zero set of an ideal is that ideal. And I also, even in the other direction, it's almost, there's some part of it, too. If we take, so it is not, this will not be true, that this is equal, but we will see that it's almost, that at least this is true, this we see immediately, and I was just explaining it. So the zero set of j is a set of all points, where all elements of j vanish. And the ideal of the zero set is a set of all polynomials, which vanish on the set. As the elements of j vanish on the set, j belongs to it. So it's just by pure elementary logic, this is true. But as I said, the other inclusion will generally not be true, and the answer will be a little bit more interesting. But first, before one does that, our time is almost up. Let me see. But we can ask a simpler question first. First, we ask a simpler question. Namely, when is it true that the zero set of an ideal is the empty set, or when it's not the empty set? So we know that the zero set of the whole polynomial ring is the empty set, because in particular, this polynomial contains the constant polynomial 1, which vanishes nowhere. And the first theorem, which is the weak form of the Nolsternsatz, says that this is the only ideal for which the zero set is the empty set. So let i, strictly contain in x1, px1, xn, your proper ideal. So if I have an ideal which is not equal to the whole of the polynomial ring, then the zero set of i is non-empty. So this is a surprisingly difficult theorem. It's so difficult that I don't want to prove it now. So it's not outrageously difficult, but the proof is very long. I mean, it would take me maybe, I don't know, two lectures or something to prove it. So in the notes, there is actually a proof, but it's in the chapter about dimension. So if you have time and you are interested, I will prove it then, because then we have developed the tools to do this. If now we have to do it from scratch, it would take a very long time. So we, for the moment, assume this result. And then we will prove as a consequence the strong form of the nullstellensatz, which tells you about the precise relation between ideals and their zero sets. And OK, and then we will do next time. OK, thanks.


and today we're going to do something special I'm going to answer that question I keep getting on the Facebook group what is better pre-smoking or post-smoking I'm going to actually take two pork butts I'm going to pre-smoke one before I serve sous vide and then I'm also going to take one and I'm going to sous vide it first and post-smoke it so we're going to try it out this one here I'm going to do probably two and a half to three hours smoke on a big Joe at about 250 with some pecan wood to see if it's going to be better than if the post-smoke one this one I'm going to sous vide it for probably about 38 36 hours or so and then I'm going to put it on the big Joe for some smoke after the fact so we're going to answer that age all question that I keep getting on the Facebook group what's better pre-smoking or post-smoking so we're going to try that out so follow along with us when we do that and I'm going to get these rubbed down and we're going to get cooking so I'll be right back as we're back and I guess that I got these two pork butts they're about the same size I went ahead and I trimmed some of the fat cap off because when you sous vide you're not really going to render as much fat as you would if you had them in the smoker the whole time so I did trim a good portion of the big fat cap off these pork butts got a lot of fat marble in them so they're going to have plenty of fat in there it's going to be still nice and juicy it's not going to hurt anything to trim a lot of the fat cap off so I'm going to be rubbing these down with the pork candy every little while pork button ribs so we're going to go ahead and get these all okay all got these all rubbed down I got the one that I'm going to sous vide first I got it all in the vacuum bag the expandable vacuum bag I was showing you got the pork candy all rubbed down in there and the one that I'm going to smoke first and just put it in the regular two gallon zip lock bag and I'm going to let the smoker so when it gets closer time we're going to kind of coordinate the two try to get them done close to the same time to get these going we're going to cook this pork butt here when we're going to sous vide first at 155 for 48 hours so we're going to throw that in but I do have my smoker coming up the tent so I just started the fire on that just close it down put some pecan wood so as soon as this gets up to around 250 I'm going to go ahead and throw my other pork butt in and we're going to cook that for about two and a half to three hours so it gets a nice bark on it before we throw that in the same sous vide cooler so I'm going to go ahead and throw this one in and then as soon as I throw the other one on the cooler around the smoker I'll be back and show you that guys the sous vide is telling me that the water's reached 155 as soon as I hit set it's going to go ahead and the timer will start ticking down 48 hours so I got my sous vide pretty sous vide you know the non-pre-smoked in there got my pre-smoked one put on smoke's rolling I'm going to go ahead and close it up and see the temp to drop a little bit but that will get right back up to 250 and we're locking in that's going to roll for about a little over two hours I'm going to see if we can get a good bark on it not going to go too much longer than that okay guys I'm back it's been about two and a half hours since I threw this pork butt on the grill been sitting right at about 275 it's got plenty of smoke to it got some nice color to it I'm going to go ahead and take this off and I'm going to put it on my pan and we're going to go ahead and throw throw it in the vacuum seal bag and we're going to put it in the sous vide with the other guys just to show you these bags again these expandable bags from we'll use with the vacuum sealer so big they get so I'm going to go ahead and slide this right on in there and you can just kind of see I gotta fold it over just like you're right in a little warm so slide it right in the bag okay guys I have the one that was not pre-smoked still in there and now I'm putting the pre-smoked one in right next to it and what I'm going to do is because I know that pork butt will float sorry about the camera I know that float pork butt will float so I am going to put a couple of these racks on top and it will keep it under the water I know when I put the top back on it's going to hold that underneath the water there so they may float a little bit but that rack's going to kind of hold it still underneath the water so I don't have to worry about them not cooking fully so we're going to let these go it's been like right at a little over 46 hours and it's just cooking away I'm going to go ahead and get the fire going and then we're going to throw the one that was not pre-smoked welcome back on I just wanted to show you I just took the pork butt that was not pre-smoked out of the sous vide man just kind of wanted to show you what it looks like and pick up this color just from being in the sous vide and with the rub I had in there but as you can see there's a lot of juice in there I'm going to go ahead and dump that out we're not going to use that today but I'm going to go ahead and dump that out take this out of the bag put it on here we're going to dry it off a little bit and then we're going to put some more of the pork candy on there and then we're going to throw it on the smoker for a couple hours the smoker is just about up the temp it's right at about 250 so once I get this all set up and ready to go back on it'll be ready to perfect temp we'll throw that on for two and a half hours and then we'll see how it turns out hey guys I don't know if you can see this but it pretty much fell out I fell apart when I was taking it out of the bag so it's been cooking over 46 hours at 155 so I guess that's a good amount of cooking so it's pretty much cooked all the way through so I'm just going to go ahead and put some more of my rub on and I'm probably going to have to leave it on this pan while it's in the smoker because it's falling apart so I will leave it on the pan I'm going to put some more of this rub on and then we're going to throw it on there so I can get some smoke to it okay oh it's been on there a little over two hours it's got some color to it got some smoke I'm going to go ahead and pull the other one out that was pre-smoked and then I'm going to I'm going to hit that one with the sears off I'm going to pull that out of the sous vide and put on a pan and we'll be right back this one's going to stay in here for another 10 minutes or so and then we'll be back and pull it out and do our taste test all right guys I just got the pre-smoked one out of the sous vide bath and still kind of wet and juicy and fatty you can tell though it's got that smoke on here I'm going to go ahead and so everything's fair since I did put some more rub on the one that is being post-smoked I'm going to put a little bit more rub on this one what I'm also going to do since it's been sitting in the bath for you know 46 hours since the bark was on it I'm going to go ahead and hit it with the sears all for a second not too long just enough to kind of get a little crisp to it because that's what we're getting on the one in the uh in the smoker right now so I'm just going to go ahead and hit this real quick have to kind of crisp it up in some spots because we are going to get that one that's in the smoker we're not really cooking it anymore just kind of getting some smoky crispy chunks to it the rub we just put on is going to get crunch to it as well all right guys you can look at them side by side okay can you tell which one is which which one was pre-smoked which one was post-smoked probably can because of the pans I used that and the one on your left is the pre-smoked and the one on your right is the post-smoked okay post-smoked pre-smoked all right they both look pretty good you know you can tell that the one uh on the right there kind of fell apart a little bit more I'm going to go ahead and get these all washed up and we're going to have my son Spencer who's going to do a taste test A or B and see which one he actually likes better so I'm going to go ahead and he doesn't know he hasn't watched me do this at all all day so he's not going to know which one's which so I'm just going to go ahead and plate that up and we're going to do it as I got my 16 year old son Spencer here loves to eat he doesn't know which one's which and he's going to do the blind taste test to see which one he prefers there's no sauce really I just took them right off the pan put them on the plate okay Spencer go ahead and try A first okay how's the taste pretty good is it juicy yeah tender is it falling apart mm-hmm is it too tender no I'd say just tender enough okay all right well kind of swish your mouth out a little bit and then see if you can try B could it be does it taste different than the other one yeah I'd say this one's a bit more tender what about anything else any other flavor difference I can kind of taste more fat more flavor and it tastes more smoky yeah it could be once you try A again I prefer B you prefer B mm-hmm try B again this one's a little bit more smoky I think this one's more tender this one's more smoky okay well A was the one that was smoked afterwards okay so I didn't smoke it first I put it in the sous vide then we smoked it afterwards and then the first one was smoked on the on the smoker first and then sous vide so you're saying the A is more smoky yeah I would say A is more so the one that was smoked afterward was more smoky than the one that was smoked beforehand okay but you'd like B better because it tastes a little bit more tender yeah it was a little bit more tender it could just be the cup but it could be it's more tender as they both cooked about the same in in the smoker and in the sous vide all right well there you have it guys that's the you know from the judgment from Spencer so I hope you guys enjoyed this video we will be back for another one go ahead make some comments down below make sure you subscribe follow us if you want us to do any other taste test videos like this I'm sure what we do on them so thanks again and then Spencer's gonna eat some more of that pork so all right thanks again guys yeah I just want to give you my opinion on that cook I actually like both of them they both were pretty good the post smoke one was probably a little bit more mushy right by what even what Spencer was saying it was a little more smoky usually what I would do is do a little bit of both and last one I did before this cook I did a pre-smoke and I put it back on the smoker for about an hour after the sous vide so and I think that one would produce an actual you know a better result and maybe I'll do a cook like that soon down the road here but I just wanted to kind of do this one now to show you but they both turned out great experiment if you want to both of them were good I mean you just can't complain about smoky pulled pork so I'd probably go a little bit less time next time I think 46 hours made them both a little too tender so I think you know 36 is probably perfect at that temperature so I'd either go a little bit less temperature or a little bit less time so but thanks for watching I appreciate it make sure you follow us on Facebook we have a group and a page make sure you subscribe and like this video and we will be seeing you again soon


Welcome to the CUBE's coverage of KubeCon EU 2024, live from Paris, France. Join hosts Savannah Peterson, Dustin Kirkland, and Rob Strecce, as they interview some of the brightest minds in cloud-native computing. Coverage of KubeCon cloud-native con is brought to you by Red Hat, CNCF, and its ecosystem partners. The CUBE's coverage of KubeCon EU 2024 begins right now. And welcome back to Paris. We're here at KubeCon cloud-native con, EU really digging deep into what's going on in a lot of different projects and foundations around Kubernetes and cloud-native technology. I'm here joined by Dustin Kirkland, helping me out with analyzing things, and Amkar, who's the GM of OpenSSF, has joined us as well, you were on a little bit ago when it was your second day as GM. I was shocked, there was a big conference, you guys were very accommodating. Lovely to be back. You feel more settled in now? A little, a little. It's great to have you here, because I know we were just talking beforehand, and you're running all over the place, advocating for what has to happen, and regulations, and where things are going, and not only working with the groups here, but groups outside, like small groups like the UN, and others, and things of that nature. Little groups like that. I mean, when we think about this mission of securing open-source software, which, you know, it's easy, we're going to be done by the end of the year, there's three main stakeholders, right? Which is private sector, so the larger organizations that make up our premier membership and our board. The public sector, so US government, the EU, the UN, all these organizations that are accountable for setting policy and providing direction and protecting their citizens, and then the community. So how do we help the community who, you know, you may not be a security expert, but gosh, you really want to make sure that your code is secure, or the provenance is good. So those are our stakeholders, and by engaging with those three stakeholders, we believe we can help improve the security of open-source software. Yeah, I think for us, we've been talking about S-bombs and attestation and signing of containers with the Notary Project who was just on. And I think it's, people are looking to get beyond S-bombs. And where do you see us in that journey? To be, you know, because again, you had the federal directive and things of that nature, but that only gets you so far. Where do you see the state of things? S-bombs are great, but S-bombs on their own aren't a security property per se. They allow you to do a lot of interesting security things, but on their own, they don't provide security utility. So one of the things that we're going to be doing, we're going to be at open-source summit North America, we're having our source, securing open-source software community day on the 15th of April, just before that. And in addition to all the wonderful talks that we normally have, we're going to be having three workshops. The first workshop is on Scorecard. So if you always wanted to contribute to Scorecard, but didn't know how, come in, learn about the code base, raise your first PR, do great things. The second is a Salsa workshop. So if you want to adopt Salsa and you want to be Salsa two or three level, a lot of people don't know how to do that and the documentation is extensive. The Salsa team will be there and they will help contributors and maintainers achieve Salsa levels within that workshop. And then the third, the one I'm most excited about, I was on Wall Street for about 10 years and our regulators always made us do these simulation exercises called tabletops.  You're not reading the process book for the first time when things are melting down. We're going to do a tabletop exercise with maintainers, cloud providers and the community at Open Source Summit North America to really exercise that, okay, so I have an S-bomb, security incident happened, what the heck am I going to do? How am I going to use Sigstore in an incident? So all of that's going to be exercised and the output of that is going to be a guide that we're going to provide to the community to help us whenever the next log for J should occur, to have a playbook by which we can operate. Amazing, so you're going to help develop and share that process for organizations that may not have a tabletop exercise process in place today. Absolutely, and this isn't a one and done, right? So this is going to be the base artifact. From there, we're going to continue to iterate at our events and explore one of the things I'd love to see after each of these kind of exercises, we typically, when I was on Wall Street, do a hot wash where we sit back and we reflect what went well, what didn't. I really want to hear from the community where they feel we're a little bit soft in our incident response so that we as OpenSSF can start advocating for whether it be technology, education, additional guidance to help make that better next time. Where do you see kind of the scorecard playing out and then kind of describe the scorecard because there's a lot of organizations who don't know about it and I know you've released it and you actually had some findings that you published at the end of the year or the beginning of this year about what had happened last year and the scorecard is something that I think organizations should be looking at to your point about, hey, they should also be doing tabletops and we advocate for tabletops all the time but I think so that's awesome because I think getting involved in understanding that process because having been someplace, we're logged for J and we had to figure out, okay, where is this in all of our software, which was not a trivial thing to put in my whole play. How does the scorecard really help organizations and what really is it? Absolutely, so scorecard as the name implies is literally a report card that you can run against the open source dependencies that you have in your project and it will take a set of known-to-be-good security properties and test the repo either in GitHub or GitLab to validate that they're present. Could be things like branch protection, well, I guess two-factor authentication's a default now but it checks all these security properties and pops out a score for you. As a consumer of open source and there's two different perspectives here but as a consumer of open source that allows you to reason over, I need to use an SSL library. Should I use open SSL, polar SSL, boring SSL and it allows you to make a risk-based assessment based on the criteria as to which one you should take on. The second perspective is that of a maintainer. All maintainers want their projects to be adopted by the world. What better way to do that than to have this scorecard badge that says, hey, open SSF has checked my security properties and... I'm safe to adopt. Exactly, yeah, exactly. You mentioned in your sort of list of three, the scorecard of getting more people involved and contributing to that. Do you have a list of GitHub issues? Oh, yeah. There's a longer list of GitHub issues that are open. Our TPM's going to have a burn-down list ahead of our contributor workshop. And we're looking forward to getting everybody out there and helping build scorecard out. How do you see the community kind of coming together because there's a lot of projects out there aimed at different aspects of security? Yeah. How do you see that all coming together? Again, looking at there's 114 or so sandboxed, there's like another 50 or so in incubation and 27 graduated, how does the community, how do you see them all working together?  We find that a lot of the most avid users of open SSF technologies are CNCF projects. So I believe there was a talk just earlier, it was either today or yesterday on how one of the projects was using SigStore in order to attest provenance of their lineage. The other opportunity that we have or the other thing that we saw as an opportunity, and I'm speaking as somebody has been doing both open source and security for a really long time. The open source community has their events and their watering holes. The security community has theirs, right? So on the open source side, it's Fozdom, it's CNCF, KubeCon, it's open source North America and EU. And similarly on the security side, we have the trade shows, right? RSA, Black Hat, we have the hacker community at DEF CON, we have the academics at Usenix and Real World Crypto and stuff like that. And one of the safe spaces we want to make for the intermingling of the open source and security communities is a conference we're going to be hosting this year in Atlanta in October called Sauce Fusion. And it is focused on that cross pollination between our communities to make sure the security folks and the open source folks are working hand in hand. So we're really looking forward to that. I think that's a great idea. Wow, OpenSSF is the organizers of that? We are the organizers, it is called Sauce Securing Open Source Software Fusion. Really cool, I think that, and also we'll be at Open Source Summit in Seattle. So be great to have you on after you've done the tabletop and have some, you know, just, you know, I would say qualitative output that you found because I think that organizations need to see this is how they have to do it. I mean, we go to a lot of the different security conferences and I think one of the things is that if you haven't done a tabletop in two years, the last two years, you're pretty screwed. And I mean, because, and having run DR for a major financial company, I know where you're coming from. I know where you're coming from because I had the book. And you didn't want it to be the first day.  Yeah, when the fiber got backcoded by people, it's like, you're sitting there like, okay, now what? It's like, okay, I'm getting the page at two in the morning, now what do I do to shut down all these servers, stuff like that? Well rehearsed incident response. I mean, you can see the results when that's well rehearsed and everyone knows what they're doing. It's still a tough situation, but at least people knowing, you know, how to address and what their responsibility is. It's going to be a tough situation, but it shouldn't be a panic. Right, right. It should be not routine, but you should not be worried about going into executing. One of my neighbors is a paramedic and there's an analog with how paramedics operate. You'll never see a paramedic running through an accident scene. They will move deliberately, they'll move quickly, but they will always follow their steps because rush and panic in an incident can actually cause more damage to what you're trying to rectify. Absolutely. Where do you see OpenSSF beyond, okay, so we get to the Sauce Fusion. Where do you see by next year this time when we're sitting somewhere up north of here? So, I mean, I already told you we're going to fix all the open source problems in security by the end of the year, so we're going to be done. There we go, awesome. Mission accomplished. Yes. So, a lot of the stuff we're working on really hard this year, focuses in three main areas. One, convening. We talked a lot about bringing the communities together, talking with public sector and ensuring that they're advocating for things that'll support better security and open source. We're also really strong believers in education. We're right now in the middle of revamping our secure developer education. There's a survey that Linux Foundation Research is putting out right now where we've asked the community, hey, based on your incidents, based on the issues you have in your code base, where could we be doing better in terms of providing you instruction? And in addition to that, if we think about the watering holes of our security community, it's really the security repos, or the package repos. It's PyPI, it's Rust Cargo, it's NPM. We recently put out some guidance along with CISA talking about how to secure repositories because we believe that's a great choke point in which we can positively affect the security properties of open source software that everyone consumes. We're going to be putting a lot more effort behind that over the course of the year. And to the point that we're going to develop some much more specific security standards that repos owners and maintainers should be adhering to. So that there's an even bar that when you choose to enter the business of being a repository, you know that there's good provenance, you know that there's good production quality control and access control, all of these different properties. So we're looking forward to doing that as well. Reproducibility maybe, and being able to be able to verify those attestations. Verify the attestations, a well-formed S-bomb perhaps, yeah. So I think it's, you also get to see this from a different level because you are interacting with governments and things of that nature. Kind of give us an idea of what you're seeing. Are there differences between what's happening in the US and what's happening over here in Europe or in Asia? What are you seeing from a, how people are putting themselves out there and really trying to work always in a secure manner? So from our perspective as the foundation, we're here to ensure technical correctness. And the reason that I start there is I'm a software engineer, and most of the folks that work on OpenSSF are technical. We're not lawyers and we're not politicians. And we are quite happy for those folks to invoke the appropriate legislative processes that are the cultural norms of where they are. We're here for technical correctness. So what we see in the US is the US politically has generally let innovation occur. And then when things start trending badly, come in with a bit of a heavy hammer around legislation and regulation. In Europe, culturally, it seems like the EU wants to get ahead of that. So they're trying to look beyond the next horizon and figure out what's coming up. Examples being things like the CRA, as well as the AI Act that they recently passed. So culturally, I think the top down versus bottoms up approach to legislation is different, but the intent's the same. Like everybody recognizes security is incredibly important and we can't keep getting burned with supply chain issues and ransomware attacks on such a regular basis. Otherwise, not only will our economies collapse, but our citizens won't be safe anyway. So you hit on it there and we'll kind of end on this question, but AI, I mean, you know, signing models, trying to understand where models came from in their lineage and the data, securing the data, making sure that PII is protected. Where do you see open SSF playing a role from an open source perspective with that? Two perspectives on that. One, I think AI itself can be a great enabler of security for open source. So we're partnered with DARPA on the AI cyber challenge in which DARPA has challenged the community and researchers to, I think that's up to $18 million now in prize money. The open competition is occurring right now, midpoint check-in at DEF CON 2024, finals DEF CON 2025, and we challenged the community to say, hey, how can you apply large language models to solve entire class of vulnerabilities in open source? As for the security of LLMs, a lot of the existing open source, open SSF security software can be used. So if you're talking about model provenance, you can sign using Sigstore and assure the provenance of the model. If you're talking about some of the basic open source components that go into the LLM, again, you can use security, or you can use open SF scorecard and tools such as that. We have an AI and ML workgroup that's currently working on some of the higher order challenges around AI and machine learning, but I'd love to see us produce something like a responsible disclosure guidance for large language models and things of that nature. Well, Omkar, thank you for coming on. I think, again, even two days in, you were great, but I always learn something when I talk to you and I really appreciate you coming on board today and being with Dustin and I to talk about this. Thank you. It's a pleasure, look forward to seeing you in Seattle and talking about the tabletop. Absolutely, we'll see you there. And we'll see you here, so keep it tuned from KubeCon, CloudNativeCon, EU, live from Paris, here on theCUBE, the leader in high tech news and analysis. Stay tuned.


Okay, just wanted to say something quick about a US-China war, you know, first of all, there's more and more talk about this It's almost like the more people talk about it. The more it becomes a self-fulfilling prophecy This seems to be both among Democrats and among Republicans a real desire As I've said this before and I'll say this again and maybe do a whole show about this real desire for a new Cold War and And and but what is starting to really sink in is the possibility of a hot war a possible real possibility of a war over Taiwan and The United States is trying to shift resources to Pacific we've talked about that we've talked about the Navy We've talked about submarines and aircraft carriers But I did see a story in foreign affairs yesterday or today that basically documents the war games that the United States has played You know assuming a war in Taiwan and assuming the United States gets involved and assuming even that South Korea and Japan and Maybe Australia Get involved the reality is that The United States and this is a problem that has become evident with the war in Ukraine The United States no longer has the manufacturing capacity weapons manufacturing capacity To meet the demand of a war In the war games the United States runs out of Smart ballistic missiles Anti-ship missiles which are going to be crucial because one of the key aims of the United States is to knock out the ships that are moving troops and and Supplies and ammunition and tanks and whatever from mainland China to Taiwan and and the goal of the US Navy will be To knock out as many of those ships before they reach Taiwan as possible. Well, the United States runs out of ammunition for anti-ship smart ballistic missiles within a week and given that the the Arena of war is 5,000 miles away from California So maybe 7,000 miles away from California. Anyway, some ridiculous amount of miles away from California Not only with these new bombs have to be built Very very fast They then would have to be shipped to I guess Hawaii or move to Hawaii very very quickly loaded up again on submarines on on Destroyers aircraft carriers or the kind of ships that carry these kind of missiles and then and then move closer to the arena and deployed Just saying that takes a long time I mean imagine what actually doing it and of course the bottleneck right now beyond the distance is manufacturing capabilities so And this is what hooks me and again. This is where I'll go after Congress instead of talking about industrial policy and micro managing and Why doesn't the United States focus on it national truly national defense and and and build industrial capacity build the kind of capacity to Build the weapon systems that we actually need to defend ourselves The Chinese have and and are doing so even they're the defense budget even though the defense budget is something like 20% of what ours is of the United States is They have the capacity to build large numbers of missiles they have the capacity to build weapons very very fast It seems that we don't now. I suspect that in actual war It'll turn out that we have a massive technological advantage. Then maybe we don't have the numbers, but we have the effectiveness There's a lot of talk about the fact that you know, China Has a lot of a defense systems and I'll prevent American airplanes from from being it but from you being useful in the in the Taiwan Straits, I think I don't believe that I think the United States probably has the capabilities To dismantle and to make it effective the Chinese defense systems and I think very quickly the United States could command The air above the Taiwanese Straits I think it's the Chinese technology is overrated But I do worry just about sheer quality quantity of weaponry sheer quality quantity of of missiles and and and other weapons and I do worry about Chinese hypersonic missiles I do worry about the fact that this is very close to China and very far from the US a Delivery of weapons supply chain management all of that is very very going to be very very difficult And I worry about the fact that I'm primary worry my number one worry Is the fact that I don't get a sense that there is real strategic thinking going on and that there is a commitment To acting on that strategic thinking.  We spend a lot of money on all kind of political correct BS Of bringing woke into the military We need to get rid of all of that. We need to focus our efforts on what is actually Absolutely unequivocally necessary for national defense We need to get troops home from the hundreds of from the dozens and dozens and dozens of countries We have troops in we need to refocus our efforts to what a war over Taiwan would look like I think that a war over Taiwan is for now, you know, you know strategic international interests to defend Taiwan not to go to war for Taiwan, but to defend Taiwan Because of chip technology So we need to be able to do that if that's our strategic national interest We need to be able to do that to do that. We need to orient all of our efforts around that We need to be able to contain Russia In Europe and I think we do that by shifting the burden to our European allies and demanding that they bear most of the burden of the Ukraine war and by shifting our resources of thinking our efforts on Navy and and and our weapons production to the Pacific to the to the Pacific area and and boosting our productive capabilities for weapons that we believe would be necessary not for the last war But for the next war for for war in the Pacific and that would require a robust Navy and it's going to require long-distance ballistic missiles and it's going to require Both hypersonic weapons, but even more important than hypersonic weapons. It's going to require Weapons that they could defend against hypersonic Missiles that both the Russians and the and the Chinese have Thank you for listening or watching the Iran book show if you'd like to support the show We make it as easy as possible for you to trade with me You get value from listening. You get value from watching Show your appreciation. You can do that by going to your on book show dot com slash support by going to patreon subscribe star locals and just making a appropriate contribution On any one of those any one of those channels also if you'd like to see the Iran book show grow Please consider sharing our content and of course subscribe Press that little bell button right down there on youtube so that you get announcement when we go live and for those of you Who already subscribers and those of you who already supporters of the show? Thank you. I very much appreciate it


Hi everyone, Joe for jaspyscasebreaks.com. Happy Friday. We got our jaspys reigning threes. Basketball mixer coming at you featuring flawless collegiate basketball from 2019, 2020. 2021 prison basketball and a blast from the past. 13, 14 timeless treasures basketball. Pretty nice stuff. 30 spots, obviously all cardship. Everyone gets a team in the NBA. We're chasing a lot of great players here. Very big thanks to this group for making it happen. Congrats again to the people who won spots in those fillers and thanks again to the people who bought their spots straight up, I appreciate it. There's all the 30 teams right here, let's do it. Let's roll it, let's randomize names and teams three times for the reigning threes. Two and a one, three times. One, two, and three. After three in the reigning threes basketball mixer, Nicholas down to Christopher. Two and a one and and one three times for the teams. One, two, and third and final time. After three we got the Nets down to the 76ers.  Can I get this all on one screen? Here we go. Let's sort by column B by team. Jose has the Trailblazers up. Ryan very happy with his Knicks draw. There you go, I've never been more pleased than I am right now. See, took a little time off, got right back into it. Got a team that you like. Now here's the box, here's the hobby box of 2021 Prison Basketball. And back here is the 2019 flawless collegiate basketball, right there, collegiate. And the 13, 14 timeless treasures is from a fresh inner case. So let's pop this open and let's see which box we're gonna do while you're considering trade. I'll go back to that other screen in a second. All right, so there's 10 boxes here. There's five on each side.  We're gonna select that die right there. And we'll go one, two, three for the left side, four, five, six for the right side, and then we'll pick an individual box. Six, one, two, three, four, five, six. We're gonna save this for some other time. And now from here, we'll just roll it one, two, three, four, five, right? And if I roll a six, I'll just roll again. And it's five, bottom box right there. And we'll save these for some other time. Maybe a future mixer, who knows? Probably a future mixer. So there's the timeless treasures. There's the prism. There's the flawless collegiate. This is the order that we're gonna go in. We'll do this one first, then that, then we'll close out with that. And we got the nets up for trade as well. We're gonna pause the video when we come back. We're gonna see if any of those trades pan out and then we'll have the break. Stick around, BRB. All right, welcome back, folks. Little bit of trade chatter, but in the end, no deals were done. Thanks, everyone. All right, now for flawless collegiate basketball, let me just double check to see if we need to go by, if there's a checklist that we go by, or, I should have done this during the trade window, sorry, you can fast forward through this part. See if there's a checklist I could go by or if we're going by our non-pro uniform rules. Well, I think there is a checklist. Good, that's helpful. Aha, boom. All right, so there's a checklist for flawless collegiate basketball. We're going strictly by that checklist. Well, let's see what happens here. Two autographs and one memorabilia card per box on average, a blast from the past. Some nice stuff in here. I think this still might be the era where, where we have to say that Charlotte stuff will go to Charlotte and New Orleans stuff will go to New Orleans. Are these hits on the bottom? No, that's a hit on the bottom, what is that? Is that a base card on the bottom? All right. It's been a while since I've done these. All right, we got timeless teams. The 94, 95, Vernon Maxwell Rockets. We got a Kyrie Irving. I'm pretty sure that's rookie. No, is Kyrie Irving's rookie in 12, 13 set? Or is this the year after the double draft class? Yeah, I think the double draft class was 2011 and then 2012, right? So this is the year after that draft class. This would be the honest draft class actually. That's what we're looking for. So not the 20, so this is 2013, 14 timeless treasures. So yeah, this is the, this is the Ola Depot. Yannis draft class. Rudy Gaubert is in this one too. All right, so that's what we're looking for. All right, this still goes to the Cavs, Benjamin. And then we got Kyrie Irving promising pros, Relic. I don't think it's numbered. He is a promising pro. And we've got Solomon Hill. Two color jersey and autograph. That is nine out of 49. Pacers, that's gonna be for Ben, Ben H. Ben C and then Ben H. Kind of like that design. All right, and the last autograph here is validating marks, Marshawn Brooks. Sean O with the Celtics, validating marks. All right, that was a little blast from the past. 13, 14 timeless treasures. Let's flash forward to the future. The way of the future. The way of the future. The way of the future. And let's do 2020, 2021, a Panini prison basketball. Hobby edition. I know we've been working on some first off the lines. Those boxes look exactly the same except for, this one doesn't have the first off the line sticker. Also, we had saved this. This was from an old random team, I guess. We do have a box of first off the line on the website right now. If you wanna get into that, ladies and gentlemen, we can run that tonight if you'd like. Jaspy's casebreaks.com. Same year, same product, prison. Except first off the line. Now, prison basketball is already nice, right? Just hobby edition. But there's all sorts of numbered cards, low numbered cards in that first off the line. If you wanna get that going. So think about your current draft class, of course, and all this. And around all cards shipped too. So you'll get all these prism cards. Good luck, everybody. We've got Ishmith, Red Wave, Wide Screen Kauai, Zignagi, Robert Woodard, Mitchell Robinson. We've got Tim Hardaway Jr., 24 out of 25. Dallas Mavericks, that's gonna be for Burke. And we've got Bucks edition of Wesley Matthews, that's your autograph. Milwaukee Bucks, Sean Oh, for the Bucks. He won that team from one of our Mosaic filler breaks. Bucks looked really great tonight, blowing out the Atlanta Hawks in game two. That series is tied. 1-1. Grant Riller, let's look for his teammates. Here's Achillian Hayes, Hyper Rookie for Detroit Keith with the Pistons. Detroit getting the number one overall pick. These Pistons rookies and youngsters from previous years can be running alongside a number one overall pick. That could increase their stock. There's Skyler Mays, Red Wave for the ATL, Brian Ferguson with the Hawks. Rookie Red Wave, Aaron Naismith, Mike Conley-Silver, Jalen Brown, Red Wave. Brooke Lopez, and here's an Emanuel quickly. This guy was a second round pick, and then has been playing meaningful minutes off the bench all season long. He could be their sixth or seventh man, maybe their sixth or seventh man for that team right there. Yeah, that's your guy, Ryan Harreld. Yeah, he's got a lot of upside. There's Isaiah Stewart for Detroit. Be for Keith Black. Second year Zion. That'll be for the Pelicans. That's gonna be Alec Johnson with that one. Jordan Kerry Jr., Cole Anthony. Kevin Porter Jr., and there's a red RJ Barrett. 271 out of 299, that's Ryan H. with the Knicks. Getting the quickly rookie card and an RJ Barrett. Second year RJ Barrett. LeBron James. I know the silver versions of these can still do pretty well on the secondary market. A little bit of a bend to these prism cards. Let's look at straightened out inside the top loader here. That'll be for Tyler and the Lakers. Yeah, Knicks look really good in the regular season. They look like they're ahead of schedule. Good for them. The second year John Morant. I don't know if these Clay Thompson Silver, what kind of secondary market value these have, but I don't know, it's silver. It's pretty nice. Here's John Morant. Yeah, free Caruso. He should Caruso, Alex Caruso should know better. Here's John Morant. You're not in California anymore, Alex. There's Killian Hayes and Kyrie Irving, Cracked Ice and Jay Sean Tate. Sensational signatures. He had some moments during the season. This could be someone to keep an eye on Paul and the Houston Rockets. And there's the Kyrie 110 out of 125. That'll be for Brooklyn. That's gonna go to Nicholas. Udoka, Sexton, Donovan Mitchell, Purple, Cracked Ice, Purple Ice, 76 out of 175. And then we've got LeBron James downtown, downtown insert for Tyler and the Lakers. Ooh, and look at this. We just looked this up the other day. LeBron James Prism Silver, Retro Prism Silver, I think still goes for a few hundred bucks. Tyler with the Lakers. There you go, Tyler. Won that spot in one of our Mosaic sell-o-pack rakes. Got lucky enough to get randomized by Lakers. And I'm sure if this grades out nicely, it could grades out at a nine or a 10, I'm sure it'll go for even more. Nice. And that's the last time LeBron's gonna be in that number for the Lakers. Caliobre Jr., Silver, Andrew Wiggins, Red Wave, Chris Middleton, Shaq, Kenyon Martin Jr. showed some moments too. That's him upside. Jamal Murray, Tyler Hero, 114 out of 175 for Miami. That'll be for Kyman. Kevin Durant, Covington, and Patty Mills. All right, just a couple boxes there. Let's end with that flawless collegiate. Could be some real monsters in there. Good luck. All right, so we'll do the card in the slider box first. I forget how many encased cards there are. I think three. I'm trying not to look, so I wanna be as surprised as you are when we take a look at those. All right, once again, if you're watching live, I did drop the checklist that we're gonna use for this case. I'll drop that link again if you wanna follow along. If you're re-watching the video, we're using the checklist on groupbreakchecklist.com, 2019, 2020, Miami Flaws Collegiate Basketball Checklist by MBAT. Let's start off with 25 out of 25 Bob McAdoo, premium ink autograph, 180 top loader. Starting off with old school McAdoo, and the premium ink autograph will go to the Clippers. Gale with the Clip Show. Clipper is staying alive and snagging a game, crucial game for them. Next up is, wow, five out of five Grant Hill. Now look how young he looks there, Grant Hill. Five out of five Grant Hill, and that'll be signatures and parallels for the Orlando Magic. And that goes to Alec, Alec Johnson, and the Orlando Magic. Nice Grant Hill, all aboard the Big Hit Express. Woo-hoo, back here we've got Sir Charles, another three out of five, Charles Barkley. Charles Barkley, 76ers, nice Sir Charles for Christopher Gerrard. And out of fives and under, get the train whistle. All aboard the Big Hit Express. Woo-hoo, next is, Romeo Langford, one out of 15. Romeo Langford would be Celtics on this checklist, I believe. Yes, Celtics. I remember someone coming to the shop not too long ago who was from Indiana and was really high on Romeo Langford. It was just like, it could be a late bloomer, but legend out in those parts, apparently. So someone to watch for Sean O, couple pieces of his jersey. We've got the Admiral, Admiral Schofield, not that Admiral, the other one, 10 out of 15. Two-color jersey and on-card autograph. And that'll be for Sean O and the Washington Wizards.  Few more out of here. Go-ho, there's Nikhil Alexander Walker, one out of 25. It's for the Pelicans. That's gonna be Alec with the Pels. That's Shae Gilgis Alexander's cousin. The Alexander part is the connector, their cousins. Couple more and then the encased cards. Oh, a dual autograph, uh-oh, randomizer? Deandre Hunter is a Atlanta Hawk, Brian Ferguson. And the Tijeroam, I'm almost certain, is a son. Yeah, Phoenix son. So once teammates no longer anymore, that'll be a randomizer between those two teams. Brian and Sean O, Sean O'Leary with the Phoenix sons. We'll do that at the end. And we got a Kobe White Tar Heel Nation, three out of 15. Oh yeah, Ryan Harrod would be interested in that Nikhil Alexander Walker. Kobe White, Michael Robinson with the Chicago Bulls. I feel like didn't quite get it going in 2020, 2021. Still young, keep an eye on him for next season. All right, let's see what we got in here. The encased cards, pretty solid box here. We'll go left, right and center. Left side, we've got, nice. Sapphire and on-card autograph, two out of 15, Rui Hachimura for Sean O'Leary and the Wizards. That's where the spot he bought straight up. Brian, and on the other side, oh no. Let's show you the other one first. Oh gosh. All right, here we've got Grant Williams, nine out of 25 diamond. Oh no. And these gems, right there, all naturally mine, not lab grown. Celtics for the Grant Williams. So there's a couple for Sean O'Leary. Oh no. All right, so empty. It's a dual autograph, two out of five. RJ Barrett and Zion, Duke teammates. Yeah, both on-card as well. And a train whistle out of fives and under. That's why I said, oh no, Ryan Harreld, because you're involved in a randomizer, right? You have the Knicks and oh no, because now Alec Johnson's involved in the randomizer. One person happy, one person very, very happy, and the other person very, very sad. Yeah, maybe not even watch Ryan Harreld. You don't want to go to bed angry or too happy. You don't want to be too jacked up or too depressed. I don't know, oh boy. All right, there's your break, ladies and gentlemen. All right, should I just cut the card in half, Gilo says? All right, let's go back to random.org. There's some new dice and there's two lists right there. First one was the dual auto, Deandre Hunter Atlanta and Ty Jerome Phoenix. And then the RJ Barrett, New York Knicks, and New Orleans Pelicans. Rex is like, not sure why you're freaking out. It's like a $100 card at best. Rex doesn't follow a lot of basketball. You have to excuse him, ladies and gentlemen. All right, let's roll it and randomize it three and a two, five times for each list. We're gonna randomize it five times for each list. Name on top or the team on top will be the winner of the card. Three and a two, five times. Let's do this one first. One, two, three, four, and once more. Fifth and final time goes to Atlanta. After five times, that'll be for Brian Ferguson with that dual autograph. Apologies to Sean and the Suns. Rex will gladly pay 150 bucks for this card right here.  That's being generous, he says. All right, let's go back to the screen again. Once again, five times, team on top. After five gets the RJ Barrett, Zion Williamsman, dual, on-card auto and train whistle. It's out of five. Five times for the out of five. One, two, three, four, and good luck. Fifth and final time stays with the Pelicans. After five times, that's for Alec Johnson. Alec won that team with that mosaic spot in that mosaic filler pack break. Got randomized the Pelicans and ends up with the RJ Barrett, Zion Williamson, dual on-card autograph out of five and out of fives and under. Get the train whistle. All aboard the Big Hit Express. Woo-hoo. Boom. There you go, boys and girls. I'm Joe for jazby'scasebrakes.com. Thanks for getting in on the Jasby's Reigning Threes basketball mixer. I'll see you next time for the next break. Bye-bye. What a break.


Hello, and welcome to the screencast, which is a sort of part two of finding limits using tables. In this video, we're going to do a very quick introduction to using spreadsheets in the context of limit calculations. The motivation here is that when determining whether a function has a limit, it's often simple to tell if the limit exists, but not so easy to tell what the value of the limit is with any degree of accuracy. Tables of data give us more accuracy than graphs many times, but we will also sometimes need very large tables to get really good accuracy. And since this is the 21st century, we don't fill out long tables by hand using a calculator, but rather we enlist a computer tool to automate that process, and that would be the spreadsheet. Now, a spreadsheet is a computer tool that allows us to make very quick computations on large sets of numerical data. It is an essential piece of software for those in science, engineering, and business, and so we're going to be using spreadsheets quite a lot here in calculus as well. Here I have open Microsoft Excel, which is a standard spreadsheet you can find on almost any computer. If you don't have Microsoft Excel, or if you prefer not to use it, there are other alternatives. For example, Google Docs has an online cloud-based spreadsheet that links with Google Drive, and there is a free open-source spreadsheet called OpenOfficeCalc, which you can download for free at openoffice.org. Even Geogebra has a spreadsheet built into it, but we'll stick to Excel for now since it's an industry standard. So I'm not going to go into depth here about how to use spreadsheets, but rather we're just going to see how to replicate the table calculations that we saw in the handwritten examples that were in the previous screencasts. If you want to know more about Excel, there are many, many places to go and learn about it online, and we'll be doing some of that in class. First, I'm going to set up my table here for x approaching 0 from the left.  Those are my x values. Notice they are all in column A starting row 1. Now each cell in a spreadsheet has an address that we can use to refer to it. For example, this cell is A2, that is column A row 2. Now let's go up to cell B1 and do the following. First, type in an equal sign, but do not press enter yet. If you make a mistake or if you happen to press enter, just click on the cell to highlight it and then hit the delete key on your keyboard to wipe out what's in the cell. Again, after the equal sign, type parenthesis 2, then the caret key for an exponent. Don't hit enter yet, and then click on the cell A1 or type in A1. Notice that when you click on that cell, Excel will put in A1 into the formula that we're creating. Now continuing, type minus 1, close the parenthesis, then the division symbol, and then click A1 again or type A1 and do not hit enter just yet. Notice what we have here is our formula for f of x, except instead of using a variable x, I have a cell value that I'm substituting in for x. Now we will hit enter and notice that Excel does the calculation that we did by hand in the previous video. But what's really the killer feature of spreadsheets is this. Click on the cell B1 that you just calculated. Notice there's a little square at the bottom right of the cell. Hover your cursor over that square so that the cursor turns into a plus sign. Click and hold on that plus sign, and while you're holding, drag down through the entire column and then let go. Notice that Excel applies the formula that you typed into cell B1 throughout all the cells you just dragged over and intelligently substitutes the new values of x into column A. If you click on cell B4, for example, notice that it's the same formula, but it's using cell A4 for the input. So now we very quickly created a big table of values to help determine our limit. Let's do this again for the values of x approaching zero from the right. Manually enter in the x values you want. Then in the cell next to the top entry, type in the formula for f, making sure to begin with an equal sign. And don't hit enter until you're done and make sure your parentheses match. Hit enter to calculate the value of that cell. Then drag the formula down the column to create the formula for f over all the cells. From these two tables, we get a very accurate read on the value of the limit of f of x as x approaches zero. Both tables have stabilized out to the 10 to the negative fourth place. So we can say confidently that at least a four decimal places of accuracy, the limit of 2 to the x minus 1 over x, exists and is equal to 0.6931. If we wanted more accuracy than just four decimal places worth, we could just simply build a larger table, and this is very easy to do. In fact, the value of this limit is exactly equal to the natural logarithm of 2, which equals this number out to 10 decimal places. But we have a little ways to go before we can figure out why that's the case. Thanks for watching.


Now, India has reported a record 200,000 new COVID-19 cases and first day, as many hospitals treating coronavirus patients reported severe shortages of beds and oxygen supplies. The surge was the seventh record daily increase in the last eight days and comes as India battles a massive second wave of infections in its epicenter, Mumbai. The western state accounts for about a quarter of the country's total cases. Two patients to one bed, gasping for air and wearing oxygen masks. This is the reality inside a government hospital in India's capital New Delhi, battling the country's growing COVID-19 crisis. Daily infections in India crossed 200,000 on Thursday, according to official data, making it the highest anywhere in the world. At Lok Nayak Jay Prakash Narayan Hospital, one of India's largest COVID-only facilities with more than 1,500 beds, a stream of ambulances ferried patients to the overflowing casualty ward. Dr. Suresh Kumar is the hospital's medical director. We are definitely overburdened. We are already working at the full capacity, rather double of the capacity. Our initial issue beds were less, but now we have increased the just double the number of issue beds. Initially it was only 54 issue beds. Now we have 300 plus issue beds. After imposing one of the world's strictest lockdowns for nearly three months last year, India's government relaxed almost all curbs by the beginning of 2021. Now, many regions including Mumbai and the rest of Maharashtra state are introducing localised restrictions. The government has blamed a widespread failure to practice social distancing and wear face masks. Experts have blamed everything from official complacency to aggressive variants. So you get notified about fresh news updates.