So this is, okay, we go on with algebraic geometry. We were, last time we had talked about irreducible components and say the ideal of an algebraic set and finally we had started to talk about Hilbert's Nulsternsatz. So the theorem of zeros, Nulsternsatz. So which is, which tells us something about the relation between an algebraic set and the corresponding ideal. So first we have the weak version, which I stated the other time. So weak Nulsternsatz. Anyway, so as I said, Nulsternsatz means theorem of zeros. But for some reason it's always referred to by its German name. So this says that if I is an ideal in the polynomial ring, which is a strict ideal, so proper ideal. So an ideal which is not equal to the whole of the polynomial ring, then its zero set is non-empty. As I said, I will not prove this now. If you are interested, I will prove it later in the chapter on dimension. When it follows from the things that we do anyway. I will instead use it to prove the strong form of the Nulsternsatz and make some conclusions. So the way we usually want to use this theorem is in this kind of trivial reformulation, which is the following remark. Usually we will use this as follows. It's used in the following form. So if i is an ideal such that its zero set is empty, then it follows that one is contained in i. That's usually how one will use it. So sometimes you want to show that something is a polynomial. And then this ideal will somehow be the possible denominators. And then one of the denominators will be one. And so it's a polynomial or things like that. Anyway, this is how one usually uses it. And obviously this is equivalent. Because if an ideal contains one, it's equal to the whole of kx1 to xn. And then it's just a contraposition of the original statement. And one should also remark, we have assumed in the whole course that our ground field k is algebraically closed. Otherwise, this theorem is wrong. For instance, the most trivial example is if you take the ideal of x squared plus one in kx, in rx over the real numbers, then its zero set, if I call this i, zero set of i will be the set of all points a and k in r such that a squared is equal to minus one. And so this is the empty set. OK. And obviously this is not the whole of ix. This ideal doesn't contain the constants. So now we want to, so this is kind of the weak form. And the question is what's the precise relation between an ideal and its zero set. We had seen, so we know that if, say, j is an ideal, then we have that zero set of j is some fine algebraic set. And if we take its ideal, then this will contain j. Because just by definition, this ideal consists of all polynomials which vanish on the set. And these are among these polynomials. And conversely, if x is a subset, it's in a fine algebraic set, a n, a fine algebraic set, then we have that we can look at the zero set of the ideal of x. And we had seen that this is actually equal to x that we had seen. It's also easy. I mean, the obvious question is whether this is also true here, which means that we would have a bijection between ideals and zero sets. So is for all ideals. And so that, which would mean that we have a bijection between ideals and kx1 to xn and the fine algebraic sets. And the answer is no. And it's actually easy to see that this will not be the case because it comes from the fact that if you have a polynomial and the power of this polynomial, they have the same zero set. So for instance, we can just look in a1. We have that if we look at the zero set of the ideal generated by say x squared. This is obviously equal just to 0.0. But I mean the ideal of zero. So all the polynomials that vanish in one variable that vanish at zero are precisely the polynomials with zero constant term. So this is the ideal generated by x. So they are different. But the statement of the null stanzatz of the theorem of zeros is that in a suitable sense, this is precisely all that can go wrong. So that it is the difference between an ideal and the ideal of the zero set of the ideal is by being allowed to take powers of the elements. And so this we want to formalize by introducing this definition. So let I be an ideal in a ring R. So then I introduce the radical. And one can either write it like this, or I often will write it like square root. But it's obviously not the square root. So this radical of i is this which is the set of all elements in the ring such that some power of the element lies in i. So there exists an n bigger than zero such that r to the n is an element in i. It's an easy, maybe I don't do it. So exercise this radical of i is an ideal. If you just straightforwardly go through the definitions, we'll find that this satisfies the definition of an ideal. And so we call, so i is called a radical ideal if while one can say it in two words, so if it's equal to its own radical and you can easily see that this is equivalent to the fact that i is equal to the radical of some ideal. This comes from the fact that from the definition it easily follows that if you take the radical of the radical, this is equal to the radical. So now we see directly that this has something to do with this story here because we see that if x isn't a finite set, then its ideal is radical ideal. I mean to be a radical ideal, the reformulation is that if any power of an element lies in the ideal, then the element itself lies there. So let's see. I mean that's again trivial but at least it's... So let's, we take a polynomial such that some power of it lies in the ideal of x. Well, so in other words, we have that f vanishes on the whole of x. So then for all points p and x, we have that f to the n of p is equal to 0. But this is just the same as f of p to the power n. And if a power... So we have, this is some number, its nth power is 0, so the number is 0. So as it follows, f of p is equal to 0. So we see that, and thus this means precisely that f lies in the ideal of x. So trivially by the definition, the ideal of a fine algebraic set is a radical ideal. And now we come to the formulation of a strong form of the Nullstandsatz, which as I said is, which says theorem, which says that, so let i be an ideal in kx1 to xn. Then if I take the ideal of the zero set of i, this is equal to the radical of i. So maybe you mind this notation that the i has two different meanings. So this i is just an ideal, and this is this process of taking the ideal of a zero set. I excuse myself, but the point is I would like to use the j in the proof, so I don't want to call it j. But anyway, I think you can distinguish what these two i's mean something different. Okay, so this is not so easy theorem. I mean actually the first part, the Nullstandsatz is more difficult. This is in some sense a consequence of the weak one. So we'll prove this as a consequence of the weak Nullstandsatz, but even this proof is not so straightforward. So how does it go? So before doing this, I, so in the proof, I will talk about the fraction field or the quotient field of an integral domain. So I briefly will introduce it because I talk about the field of rational functions in several variables. I mean it's something you I think already had in the algebra course, but you know I just briefly recall what you are supposed to know. So this is the quotient field of an integral domain. So R should be an integral domain. So the quotient field Q of R is a set of some equivalence classes of pairs. So is the set of equivalence classes and of pairs f, g, where f and g are in R and g is different from 0. Under the equivalence relation that f, g is equivalent to f prime, g prime. Well, if and only if f, g prime is equal to f prime, g in R. So these equivalence classes I write, so equivalence class of f, g is denoted f divided by g, which also comes somehow explains this thing because if the ring for instance would happen to be the integers, then these equivalence classes correspond precisely to the rational numbers. And you can do this however for any integral domain. And you have again multiplication, addition and multiplication are defined by the usual formulas for fractions as if you were in elementary school.  And f divided by g times f prime divided by g prime is equal to f, f prime divided by g, g prime. So again, like you learn in elementary school, except that now we are in a general ring and not dealing with integers. The whole point of this is that this is all that happens. You can compute this if you had them. You can view r as a suppressing of qr. So first easy to see that this thing q of r is a field. And we have that r. So identifying an element r in r with the kind of fraction r is the one in q of r. So the map which sends such a thing to this is easy seen to be injective and you can just identify it if you want. Then we get that r is a suppressing of q of r. So this is all very standard. Obviously the inverse of f divided by g is g divided by f as in elementary school and so on. So in particular, we will only look at a very special case, namely that our ring is the polynomial ring in several variables. So in particular, the quotient field of the polynomial ring kx1 to xn is denoted usually like this with round brackets. It's called the field of rational functions in these variables. And you can see that these are just, so the elements are just of the form f divided by g with f and g polynomials and you compute with them in the obvious way. You just multiply them and add them and divide them really as fractions and this will be a field. So now after this thing we can, I will use this very briefly in a moment. Now we want to actually come to the proof of this null stands and we will use this thing. So we start with our idea. So we know that it's finitely generated. So we can as we write that, so proof, write i equal to f1 to fr for some polynomials fi, some polynomials in i. Okay. Now, I mean we know, we have seen that if we take the ideal of the zero set of i, this is a radical ideal and we also know it contains i. So it follows that the ideal of the zero set of i contains the radical of i. So we are supposed to prove the other conclusion. Well, and so basically now we just take an element here and we have to show it lies here. So let f be an element in the ideal of the zero set of i and we have to show there exists an n bigger than zero such that f to the n lies in i. So from the fact that it vanishes here we have to prove that. And that's actually, you know, if you think of it, that doesn't look so easy because, you know, for instance, it's not at all clear where you should get this n from. And so there must be some kind of trick and we somehow maybe also want to use the weak null stands, but it doesn't look very clear how one could do it. But the trick consists in using the weak null stands in one variable more. So we use weak null stands as in k x1 xn, t. It's not quite clear what this will do for us, but at least let's do that. And so for this we will have to produce an i, using i we will have to produce an ideal here whose zero set is empty. So let's write it down. J, so let J be now the ideal generated by these self-simple normials. And one more, we have here this f. So we take f times t minus one. So this is the ideal in k x1 xn, t. So we first have the ideal generated in this bigger ring by these polynomials which only depend on the first few and then we add one more polynomial which also depends on t. And so we take this ideal and we ask ourselves what is the zero set of J? So let's look at it. So we take a point, so let's say p, a be a point in an plus one. So where p is a point in an and a is a point in k. So then you know this is an n plus one pupil. And we ask ourselves what does it mean? So we want to know whether it can be that p, a is an element in the zero set of J. Well, when will this happen? So all of these polynomials must vanish. So these ones depend only on the first few coordinates. So they only give coordinates for p. So the first statement is that f1 of p until fr of p should be zero. But we know that the zero set of them is, well anyway, so we have that f1 of p is equal to fr of p equal to zero which in this itself is equivalent to the fact that p is an element in the zero set of i. And the second statement is that, well, f of p times a should be equal to y. But f lies in the ideal of the zero set of i. That is, if p lies in zero set of i then f of p is equal to zero. And so this cannot be true. So this implies that f of p is equal to zero and therefore this is impossible. So these two conditions can never be fulfilled at the same time. And so we deduce that indeed z of j is equal to the empty set. So and now I told you that we have this way how we usually use the weak null stanzatz. So this implies that j contains the element one. So by the weak null stanzatz we have that one is an element of j. Well, and we can write out what this means. No, just it's a linear combination with coefficients in kx1 and t of these elements. So that means we have thus we can write one is equal to say g zero times f t minus one plus sum i equals one to r g i f i. And this identity so where the g zero to g r are polynomials in x1 to xn and t. And this identity holds in kx1 to xn, t. So we have this identity. Now the obvious question that one can ask oneself here is why should this be helpful to us? We are interested in finding out something about i. Now we have made a different ideal in one more variable and we find that that contains one. So we have to have a way to get back to kx1 to xn. And how do we do this? We replace t by a polynomial. We express t in terms of something in x1 to xn. In fact, we replace t by one over f. But this only makes sense in the fractured field. So now we want to go back to kx1 to xn. But this is via the fraction field kx1 to xn. So we define a ring homomorphism say phi from this ring to the fraction field or the quotient field or the field of rational functions in x1 to xn which is done by taking any polynomial and sending it to this polynomial where we let the first two variables in the first variables we just take the same and for the last one we put one over f. Now this is clearly a homomorphism because if you take the sum of two polynomials you get here the sum of the rational functions. I mean it's clear what this notation means, I hope. And if you take the product and so on it will always be like this. But if the polynomial here contains some t you will get some power of f in the denominator. So you don't end up in the polynomial ring but in this fractured field. You have this homomorphism. So now you take this identity and want to see what happens to the identity by applying this homomorphism. So applying phi, so this may be a star or something. So apply phi on both sides to this identity star. We get one. Well, let's see. Before we do it we maybe have to look at it a little bit. So what happens to ft minus 1? So f is a polynomial in x1 to xn so nothing changes. It's sent to 1 over f. So we have f divided by f which is 1 minus 1. So this becomes 0. So this term is just not there. Then the fi again depend only on x1 to xn so they do not get changed. So what we find is that we get 1 is equal to sum i equals 1 to r. So phi of gi times fi. And this is an identity in this field of rational functions. Now we want to get rid of the denominator so that we get a statement in kx1 to xn. So we have to ask ourselves what are the denominators. So if you look at this, what have we done? So we have taken this polynomial. We have replaced t by 1 over f. So the only denominators that we can have are powers of f. Whenever we have some power of t, we get the same power of 1 over f. So these fi are anyway just polynomials in x1 to xn but it applies to this. So we have that there exist some numbers n i in non-negative integers such that phi of gi can be written as gi divided by f to the n i to the power n i where gi is a polynomial in kx1 to xn. And so now if we have that, we can clear the denominators. We can multiply with whatever the maximum of the f to the maximum of the n i. Then there will be no denominators here until we get the power of f which actually will turn out to prove the theorem. So let n be the maximum of the n i. So from i equals 1 to r. So we multiply by f to the n. Then we get the identity f to the n is equal to what is gi?  But now what do we see? So this is a linear combination of the fi's. So the fi generates the ideal i, large i. So it means that this thing is an element of i. And we have proven what we wanted to prove. We have found some power of f which lies in i. But it comes about in this very roundabout way as the maximal denominator we need in this expression. So the fact that before we couldn't have a guess for what power of n one should need to take is reflected in the complicateness in the proof in such a way that there's no way how you can find out. So anyway, so this proves this theorem. And you can see it's a bit tricky. Now the moment we will not really have very many applications of it. So we now find that we have indeed this bijection between, say, a fine algebraic set in an and the radical ideals in kx1 to xn. So we have a bijection in both directions. So in the one direction one goes with the ideal. The other one with the zero set. And these are two bijections in both directions which are inverse to each other. So precise correspondence between radical ideals and a fine algebraic sets. Okay, so I want to give some small corollary. I had, we had proven as a statement that and a fine algebraic set is irreducible if and only if its ideal is a prime ideal. That sounds like a nice statement, but in practice it's not, wouldn't be so nice because it's very difficult. It's not so easy to find out what the ideal of an affine algebraic set is. Usually you are given the affine algebraic set as a zero set of some, some polynomials. And so it's not so obvious. But in this case as a corollary you can get a more direct statement. So the first is so if say I in kx1 to xn is a prime ideal then its zero set is irreducible. And the second statement which is in some sense a corollary is that if f in kx1 to xn is an irreducible polynomial so it cannot be written as a product of two polynomials unless one of them is constant then its zero set is also irreducible. So the other statement was that we had before is that if the ideal of the zero set is a prime ideal then it's irreducible. So that's not a part of quite the same thing but it actually is almost trivially so because from the definition it follows that prime ideals are radical because what does it say? To be a prime ideal means that if you have f and g in the ring such that f times g lies in the ideal then it follows f lies in the ideal or g lies in the ideal. And now if you have instead a power of an element so you have an element in the ring so that the power lies in the ideal then it should follow that already the element lies there this is a special case where these f and g can be taken as one f and the other one some power of f. And then you can make induction. So thus if i is a prime ideal then it is a radical ideal so if I take the ideal of the zero set of i this is just i. And so the statement and this is still a prime ideal and therefore it's irreducible. And we know so if that is a prime ideal so if I write so if I write x equal to z of i we have that i of x is prime so thus x is irreducible by what we had seen before. And the second part is a special case if one so for two if f is irreducible so it is a fact which was maybe mentioned maybe also proven but I don't know in the algebra course that k x1 to xn is a unique factorization domain I will not recall what that means but one of the things that follow is that every irreducible element is a prime element so that means if f in k x1 to xn is irreducible then it follows that the ideal generated by f is a prime ideal so this is something which holds in unique factorization domain well and so and so thus if this irreducible then this is a prime ideal and thus the zero set of f which is the same as the zero set of the ideal generated by f is irreducible so that was for the moment so this is the as much as I want to say about the nulstern that's for the moment so now I want to come to a different topic which is projective varieties and projective algebraic sets so you know we here do everything over some algebraic very closed field but for the intuition it's in some sense best to think about the complex numbers and so if you for instance take some a fine algebraic set which is does not happen just to be a finite set of points but it has some is somehow bigger over the complex numbers you will find that it always somehow goes to infinity so it will always be non-compact and this somehow I mean somehow it's nice to also look at compact things I mean they have a they behave nicer in many ways and in some sense the more interesting and more useful things to study I mean also from the viewpoint of topology or whatever anyway so therefore we want to somehow you know do something like compactifying these fine varieties so we want to look at varieties which have some kind of which at least if we would look at complex numbers with the standard topology would be compact now and so this is done by adding some points at infinity so a fine variety is always you know they are not bounded somehow but you somehow add something at infinity to get some projective varieties and this will have many nice properties I mean one of so the simplest projective variety or whatever would be projective space or something and there you can you have this thing that if you are in the usual space if you have two lines then you know usually they will intersect but sometimes they are parallel and they don't but in projective space what you call lines there they will always intersect if they don't intersect they are parallel they intersect at infinity and so somehow looking at projective against instead of the fine varieties will also kind of remove very many case distinctions and things like that they are just kind of nice in many ways so anyway after regardless of that I mean they you know in some sense it turns out that projective varieties are more important to study than the fine varieties and fine varieties you know you would more think of them by thinking at you know how locally the projective variety looks like so to look at a piece of it but the total thing you want to study would be rather a projective variety and so we want to now introduce them so I first tell you what the projective space is and then projective varieties will be 0 sets of polynomials in projective space where we even have to worry what we mean by that so first projective space and projective algebraic sets so on so this is a definition on KN so KN plus 1 without the origin so the origin in KN plus 1 I have an equivalence relation so I say that an N plus 1 tuple A0 to AN is equivalent to B0 to BN if and only if they differ by multiplying all of them by the same non-zero constant there exists a lambda in K without 0 such that A0 to AN is equal to lambda B0 until lambda BN so just the whole vector is obtained by multiplying by the same constant and I so the quotient so maybe write first A0 for N for the equivalence class with a square bracket and the quotient I call projective space so n dimensional projecting so projective N space or n dimensional projective space is the quotient quotient set divided by this equivalence relation so an element in projective space is just an N plus 1 tuple of elements in K or a vector in KN plus 1 up to multiplying by non-zero constant so which is equivalent to saying it is a line so in a one dimensional subspace of KN plus 1 because such a line is determined by a non-zero vector on the line up to multiplication by a non-zero constant so you can also view projective space as a space of lines in KN plus 1 lines means linear subspaces so lines to the origin so now we want to somehow show that this is somehow some relation also with a fine space so we want to somehow find that there are N plus 1 subsets which look a little bit like AN and we use one of them to identify AN as a subset of this so definition so let we take UI for I is an element from 0N we define this to be the set of all A0 to AN in PN we denote this as PN projective space in PN such that AI is non-zero and now I define a bijection of this to AN so I take phi I from UI to AN which sends a point to AN to a point here in AN so I take E0 divided by AI and then it goes on A1 divided by AN at some point I arrive at AI divided by AI and this I leave out I want only N coordinates and then it goes on until AN divided by AI so this will be something in AN and I am allowed to make this quotient because I have assumed that AI is non-zero so this is certainly a well defined map and I claim it's a bijection because I can write down the inverse so say UI from AN to UI which sends where I write like this I want to have to write down an N tuple here I do this by writing an N plus 1 tuple of which I leave one out so I take say B0 and Bi is not there until BN we send this to B0 and in the ith position I put 1 I mean the ith position is not here but you know just where and BN and you can check that these two maps are well defined and inverse to each other I mean I don't know whether you might have already seen something like this in different geometry or something if you define many folds in a projective space as a many fold by giving local charts to it and this would somehow be such a local would be such a set of local charts anyway we have this it's very easy to check that this is a bijection with this property so usually we fix we concentrate on the and and we want to we want to view AN as a subset of PN by identifying the point say A1 to AN in AN with its image under this A1 AN so I mean later we will actually do this I mean obviously they are not precisely the same thing but in some sense you can see that precisely the same information is contained it's just a notation issue whether you write this or whether you add the 1 in the beginning and put the brackets around it contains the same information so but I mean strictly speaking the statement would be so okay I don't say that but given this so with this identification we have PN is equal to the union of AN and H infinity where H infinity defined to be well whatever in PN does not lie in the image of this U0 namely this would be the set of all A0 well whatever I can write this is 0 to N in PN such that A0 is equal to 0 this would be called the hyperplane at infinity so when I write this notation strictly speaking what it really means is that PN is equal to U0 of AN union H infinity but I kind of use this U0 you know as it's a bijection and it even I mean not so much happens but I kind of can pretend that this is really just that I identify the image with what I started it so anyway but if we have this picture we can really see that PN is obtained from AN by adding something namely this so it is you know AN plus something added at infinity yeah in some that is well it is somehow in some sense it is PN minus 1 I mean we don't yet know what it means to be isomorphic but it is in some sense it will be isomorphic to PN minus 1 it's very close to PN minus 1 because it's just you know if you have 0 and then A1 to AN so that you know A1 to AN so in some sense this will be this is essentially PN minus 1 okay so yeah you could see it is okay now that we have the projective space we want to talk about projective algebraic sets so we want to define projective algebraic sets as 0 sets of polynomials in this projective space and we immediately run into some small problem that it doesn't completely make sense so we want to define projective algebraic sets as 0 sets of polynomials so of sets of polynomials in so so if we are projective algebraic sets in PN this will be of polynomials in k x0 to xn and now there is a problem namely it is not true that the polynomial in k x0 to xn defines a function on PN so but f in k x0 to xn does not define a function on PN no if we have no obvious it is just not true actually will never be true that I mean unless f is constant or something that f of A0 to AN is always different from f lambda A0 lambda PN we don't have that is a function so it's not really clear how we could talk about so we cannot say what f of P is for a point P in PN because it depends on the representative but it still sometimes makes sense about a 0 of a function namely if it's 0 for all representatives and this makes sense if f is homogeneous can still say whether P in PN is a 0 of f or not and so why is that well this is the you know so I hope you remember what it means for polynomial to be homogeneous so you have that it's a linear combination of monomials so some product of the XIs with some powers and it's homogeneous if the degree of all monomials is the same and that would then be called the degree of this polynomial so this is the following remark if f is homogeneous is homogeneous of degree D then we have what if we take f of lambda a 0 until lambda a n then this will be equal to lambda to the d times f of a 0 a n and this is actually kind of obvious because if you take any monomial of this polynomial it has degree D so it's the product of d of this XI counted with that power so if you put it into you get d of these lambda AIs so out of each factor you get out of each of these you get one factor lambda so you get lambda to the d comes out of every monomial we have that so in particular so thus whether f of a 0 to a n is equal to 0 depends only on the class of a 0 on the point in projective space because lambda is a non-zero if you multiply by non-zero constant then the question whether or not it is 0 is not changed so with this we can define what we mean by a point in projective space to be a 0 of a homogeneous polynomial it is a 0 if for any representative or equivalently for all representatives it is a 0 so let G be a homogeneous polynomial so a point p equal to a 0 a n is a 0 of G and we will write this just we will just write this as G of p is equal to 0 although I mean in general G of p doesn't make sense but the question whether G of p is 0 makes sense if well G of a 0 to a n is equal to 0 for one and equivalently all representatives of this point so if we have we can p is the equivalence class of some n plus 1 tuple and we just take you know a representative and put into the polynomial if we get 0 we say it's this and it doesn't depend on which representative we take because of this statement now we can talk about 0 sets of polynomials so let S in K X 0 X n is set of polynomials set of homogeneous polynomials so then the 0 set of S is defined to be so Z of S the set of all points p in p n such that f of p is equal to 0 for all f in S as before in the affine case with this definition of what f of p is equal to 0 means so a subset of p n of the form C of S is called a projective algebraic set so now we have defined what projective algebraic set is so we only had this tiny difficulty with saying what the 0 of a polynomial is but for homogeneous polynomial sets okay and now we want to also I mean you want to basically repeat what we did for the affine algebraic sets so for instance you want to show that every projective algebraic set is a 0 set of an ideal but one has to be a little bit more careful because we have somehow have been forced to talk only about homogeneous polynomial so somehow that will also play a role here and so we will have to deal with so-called homogeneous ideals before that maybe I can give a couple of trivial examples which anyway use so for instance we have that the empty set is a 0 set of 1 and p n p n is say the 0 set of the empty set of polynomials so we again have the most trivial sets then we also can the points also find algebraic sets so if p equal to a 0 to a n is a point in p n then we have that where p can be written as the 0 set of what so we say for instance we take x 1 minus a 1 x 0 minus a 0 x 1 a 2 x 0 minus a 0 x 2 and so on a n x 0 minus a 0 x n so if I'm not mistaken you find that this precisely will determine this equivalence class I mean so precisely that you know that up to multiplying by a non-zero constant the coordinates have to be these anyway you can check that so now we want to come to the story with the ideals so if we have a polynomial we can always write it as a sum of homogenous polynomials so the polynomial f in k x 0 x n can be written uniquely as f equal to say f 0 plus 1 plus whatever f d where d is the degree of the polynomial where the fi homogenous of degree i so we just take all the monomials of degree i together with their coefficients and we take the sum of those and put them together so these fi are called homogenous components of f now an ideal is called homogenous if with every polynomial it contains all its homogenous all its homogenous components so an ideal i is called homogenous, so homogenous ideal if for every f in i also all homogenous components fi i in i so this is like the strange definition but there's a you can formulate it in another way namely have the following proposition some more remark an ideal is homogenous if it's generated by homogenous polynomials if and only if it is generated by homogenous polynomials so if and only if there's a set of generators so that all the generators are homogenous i don't know whether i can also prove it although it's not very exciting but so we assume we assume that i is homogenous and so we take some system of generators well, so then for each of the generators we can take the homogenous components if we take all the homogenous components of all the generators they will certainly generate because we can write the generators as a combination of them so then the f i for i for i and i are a set of homogenous generators because after all f alpha is the sum of the f alpha i so whatever you can express in terms of the f alpha you can do in terms of the f and the other direction is ever so slightly more complicated so let i be generated by homogenous polynomials generated where these are the i are homogenous okay so I have to show that for every f and i I can write it as a we have that all the homogenous components of f also in i so we take an f in i so then thus we can write f is equal to sum over i a finite sum and a i g i because after all the g i are generators with a i sum elements in k x0 to xn not necessarily homogenous and now but notice but g i is homogenous so the homogenous part of a certain degree of this product is just this g i multiplied by the corresponding homogenous part of a i so thus the homogenous part of a i g i of degree d is just g i times the homogenous part of degree sorry the homogenous part of degree d minus the degree of g i you get something of degree d by taking here something of degree d minus the degree of this that's the way how you get the polynomials the sum of all the monomials of a i which have the degree that adds up to d well but if that is so then we know how to get the homogenous part of degree d of f we just take this part so thus it follows that fd is the sum of all i of a i to the d minus the degree so the homogenous part of degree d minus g i times g i and so the g i so it is a linear combination of the generator so it follows that this thing is indeed an element in i so I'm essentially up but I can at least write down the definition we were in for and we come back at the next time so we want to say what the zero set of an ideal is so according to our and we want to say this for homogenous idea but we cannot say it's the zero set of all the elements in the idea because we have only talked about zero sets of homogenous polynomial so we only talk it's the zero set of all the homogenous elements in the idea definition that i in k x0 to xn in ideal so so in some sense projective but we will not write it zero set of i is denoted again z of i which is the set of all points p in pn such that f of p is equal to 0 for all homogenous elements I do I want an ideal or I want a homogenous I want a homogenous idea for all homogenous elements f in i so the zero set of i is not the zero set of all the common zero sets of all elements in i because we have only said what it means to be zero for homogenous polynomial and conversely we can talk about the ideal of a projective algebra set this is just the ideal which is generated by all homogenous polynomials which vanish on that set or just for a subset x in pn the homogenous but again we will not say ideal of x is i of x which is defined to be a set of all f in k x0 to xn such that f is homogenous and f vanishes on the whole of x f of p is equal to 0 for all p in x and by definition this is a homogenous ideal so we have this definition now next time we will work a little bit with it so we have seen we have kind of made the same I mean the analogous definitions to the affine case we want to then prove that most of the results we proved in the affine case also hold in a suitable way in the projective case and then ok that we will do the next time so thank you and see you next week


It's the mat work. What's one? What do you say is the best thing for artists to do that has no budget? That's for no budget. Yeah. So we always break down everything you do in this is either you're gonna have to spend money into something or you're gonna have to spend a lot of time in the figuring something out. So usually what I would say with that is pick like one or two platforms that could be platform that you personally like to use this on your day-to-day basis or just platform that you feel like would really be beneficial for you to just like you know learn how to master and then really dive into and study how those platforms work. Look at how they move content out you know what I'm saying like how what different types of things trigger the algorithm on these platforms. What type of content moves the most the best on these platforms like as far as like natively like what content you make natively for the platform that moves really well.  Dive into all of that shit for the platform that you want to go with and then just get into the the game of like content creation for like consistent content creation because that's something that you should be figuring out anyway.   So like I'm really big on even before you start to put money into yourself figure out that content foundation that we're talking about anyway pick your two platforms that you're going to go hard on because you're not going to be able to handle trying to do everything it's going to immensely exhaust you it's going to exhaust you from time and content for it and it's going to exhaust you monetarily.  Right right so that's what I've been doing on pretty much YouTube right now just like one video a week every sunday just starting to build it up I think it's definitely slow right now but no I'm definitely I'm sure things are going to pick up in this time. Yeah bro slow is bad slow is bad and nothing man. That's facts.  YouTube really takes a lot of a lot of like consistent postings. I read something where it's like the algorithm doesn't really start the public awareness you get close to like 100 posts or so. I don't know how accurate that number is but I know like all these platforms pretty much like categorize you and they look for certain things that you do to let them know like okay this is a person that's actively trying to grow on the platform or this is a person that just like has to be on the platform. So the more of the ways that you play the game on the platforms the way that they want you to play it the more seriously they start to view your account and they go like okay cool this person wants to grow on Instagram but this person wants to grow on YouTube. Now let's see if their content measures up to whatever quality is put in. Instead of putting in front of a certain group of people see how those people react to it and if those people put with it then they go like all right let's keep pushing more of their content out. So like that's the game you have to get into playing.


Hello, welcome to Clock Talk, I'm Crystal here. Culture of women today, what are we gonna talk about? Well, we're gonna introduce a different lens on women, a hidden history of women warriors. Cool, what does that mean? Well, what about piracy? Nobody ever thinks that they were female pirates. Well, indeed there were, and we're gonna unravel that one with a famous historian we have here. I'm so excited to have him to present his knowledge and to share the space with me to talk about women pirates, so join us. So, let me introduce our distinguished guest. He is the Dan and Maggie Distinguished Chair of Democratic Ideas at UH right now. In fact, I'm taking a course with him and that's why I pulled him over. Let me introduce the world-renowned visiting professor, Dr. Professor Marcus Redaker. Marcus, welcome. Hello Crystal, I'm very pleased to be with you, thank you. Thank you for coming all the way from Pittsburgh. That's right. So, you are the acting distinguished professor of Atlantic history at University of Pittsburgh. That's right. This is your background.  So, you grew up in the South. I did. And so, you have a lot of influence growing up from a working class family.  That's right. Can we talk a little bit about that before we go into the minutes? Sure, sure. Dropped out of college, I guess I was 19 years old, and went to work. Because I wasn't a serious student, I had been a basketball player, I had had a severe injury, but what I really wanted was real world experience. And so, I went to work in a factory, which is what all the members of my family had always done. This was a factory of about 3,500 workers in Richmond, Virginia. And as it happens, a lot of my most serious education began there. I began to read much more seriously. And then finally, I went to night school and finished up my university degree and then went on to graduate study. Now, were these factories with women and men separated? Were you worked together? We worked together. This was a factory owned by DuPont. And there were two parts to it. On one side, they made textile fibers like nylon and Teflon. And on the side that I was on, we made cellophane. Huh, wow. And I worked shift work around the clock. Wow, can I ask you what the wage was at the time? Do you remember? It was actually very good because I had one of those jobs that has now really disappeared. I was an American industrial worker. That species is almost extinct. So I made quite good money and got a lot of overtime. Okay, oh, you got more than you bargained for maybe. Now, when I take your class, the first thing you like to ask us to unpack each book is, you know, what are the backgrounds of these authors? How were they influenced before they wrote this?  Yes, well, as you said, I grew up in a working class family and that had a big impact on my decision to study history from below, as we call it, meaning emphasizing the history of ordinary working people who are usually left out of the national stories. Another big influence on me were the movements, the social movements of the 60s and 70s, the civil rights movement, the anti-war movement and those kinds of movements also demanded a new kind of history. For example, the civil rights movement wanted a history that took race and slavery seriously and the women's movement is another major impact. Of course, that has led to a lot of history of the more numerous part of humanity. Right, what about personally?  Yes, there are a lot of strong women in my life that's always been very important but in terms of my specific family, I was the first one to have the opportunity to go to college. So no one really knew quite what lay beyond. When I became a professor in 1982, it was something that my family actually had a hard time understanding. I couldn't quite figure out what I actually did. But I think my interest in women's issues is very strongly influenced by the rise of women's movement in this country. Okay, so let's go into the world of piracy because a lot of people don't know and you're saying that there is a fascination based on some books that you've written that have inspired people to even think about writing screenplays, which unfortunately my ignorance to the world of piracy is reduced to films like the Pirates of the Caribbean. And maybe some other people feel that way too. So let's go into unpacking that world and what time are we talking about that you focus on? Yes, I write about what's called the Golden Age of Piracy and that's basically from the 1650s to about 1730. But I more specifically have written a book about the third generation of these pirates. They were active in the 17 teens and 1720s and they were especially important because they created a crisis in the Atlantic trading system by capturing the ships of Great Britain, France, Spain, Portugal. There were several thousand of these pirates. But what I found to be most interesting was that it wasn't only about robbery at sea. It was unbelievably fascinating to see how pirates organize themselves in their daily activities. For example, in a time when poor people really know where in the world had any democratic rights, pirates always elected their captain and they could depose their captain if they saw fit. They also divided up the plunder that they took in very equal ways in contrast to how the merchant ships and naval ships of that day were organized where the captain would make 60 or 80 or 100 times more than the common sailor. Sailors, pirates divided up their plunder in shares. The captain got two share, the common sailor got one. So it was very egalitarian. And I think this is one of the things that actually attracted a lot of common sailors to join the pirates because they saw that they were really gonna live in a very different sort of social world and a world that was really all about liberty. In other words, another impact on the pirate ship was that captains didn't have the power to flog a sailor with a cat and nine tails the way it was so common on other kinds of ships. So they limited the disciplinary violence of the ship captain. So in some ways the pirate ship was like an antithesis of the normal ship of the day in which the lives of the common sailor were very, very hard. Low wages, very poor quality food and extremely violent discipline. On a pirate ship, it was the opposite. Really? We have a photo, I think it was a painting, a rendition of a pirate ship back in those days. Is that a typical portrayal of what it was?  And this sort of sums up the courage that these people had in taking on what were really the most powerful empires of their day, the Spanish empire, the British empire. And they captured literally thousands of ships like that and they plundered them of all their gold and silver and other commodities. And so how did the women come on? I mean, that's not a representing of the entire typical pirate ship but now it's quite small.  Well, that wasn't really a pirate ship. That was a boat. What they would do is they would lower a small boat from their much larger ship and they would then make the attack that way. But basically what pirates did was an average pirate ship would have 80 to 90, maybe 100 sailors on board. Some of those people became pirates through the act of mutiny. Okay. If we could have the next image. This is a very famous pirate captain named Bartholomew Roberts who is probably the most successful pirate of the golden age. You can see he had a whole flotilla of ships. He's actually here on the west coast of Africa where he's disrupted the slave trade. Where was he from originally? He was originally from Wales in Great Britain. He had been a mate on a ship and he rose to be probably the most powerful pirate of them all. Now, the British government wasn't gonna stand for this. So if we could have the next image, we'll see that they ran a concerted campaign to annihilate pirates. They hanged pirates by the hundreds. This is the hanging of a man named Steve Bonnet in Charleston, South Carolina in 1719. And my argument in this book, Villains of All Nations, is that pirates were annihilated and hanged in such large numbers, not only because they were attacking the property of ship owners, but also because their democratic and egalitarian ships posed a challenge to the way ships were normally organized. So a threat to the way the system was, as always. Exactly, exactly. It's interesting that you, because your other very well-known book on slavery, the slave ship compared to the pirate ship were so different. Completely different. So they're different microcomps. Perfect, I'm so glad you see that because the pirate ship was a space of freedom, a space of liberty. So some people began as mutineers, but the vast majority of people who joined the pirate ship do so from a vessel that has been captured. So the pirates would take a ship and before they would even begin to plunder it, they would call all the sailors and the captain up on board on the main deck and they'd say to the sailors, how does your captain treat you? And if they said, if the sailor said our captain doesn't treat us well, he'd beat us, that captain was in big trouble. Oh wow. Well, he was in big trouble. Some were executed, some were flogged. They would tie up the captain in the same place where he would flog his own sailors and then they give that captain a beating. So the pirate was kind of an avenger of the common sailor. Yes. And he represents that rebellious, kind of self-empowering position of the underdogs having a chance to overthrow. Precisely, precisely. And they caused a major crisis and believe me, the authorities of these various empires wanted to take them out as quickly as they could. Hence all these bodies hanging in chains at the entrance to every harbor. Anytime a merchant ship came in, you would see the remains of pirates there as a warning from the government, don't you dare turn pirate because this is what we'll do to you. So what about the glorifying of pirates? Like the images of the skull and the flags of the black and... Exactly. Well, the famous black flag, the skull and crossbones, comes from this generation of pirates and that is called the banner of king death.  So the flag is kind of an instrument of terror to force people to give up and people did overwhelmingly give up because you don't want to fight these people. They have very fast ships, they're very skilled sailors and they're going to capture you sooner and later, you better just give up early. So don't fight it.  Well, the image of a pirate, we don't want to mess with them. So where do the women come in? Why are these two famous women? Yes, these two women, their names are Mary Reed and Anne Bonny. They are actually quite famous in the history of piracy. Mary Reed was one of these people who was aboard a captured ship. The pirates came on board. Now she was dressed as a sailor. Okay. She dressed as a man. She had actually had a lot of experience fighting in wars in Europe. She had been a soldier, Mary Reed. Very tough, very expert on weapons, the use of weapons. So she was on board this vessel that was captured and the pirates did what they usually did at the end of their asking the sailors how the captain treat them. They would say, who wants to come with us? And so a group of people would join the pirates. And if you spoke badly about your captain, you had better go with the pirates. So Anne. And Mary Reed joined up from a captured vessel. Okay. And dressed as a man. And when she went on board, people thought she was a man. Until boom, boom, boom, boom. Okay, wait, this is all, we hope the audience can tell you they're captured and they go on to the pirate ship. What happens to these women? We're gonna take a break, but if you wanna know about the lives of female pirates aboard these ships, don't go away. Hello, I'm Dave Stevens, host of the Cyber Underground. This is where we discuss everything that relates to computers that's just gonna scare you out of your mind. So come join us every week here on thinktecawaii.com 1 PM on Friday afternoons. And then you can go see all our episodes on YouTube. Just look up the Cyber Underground on YouTube. All our shows will show up. And please follow us. We're always giving you current, relevant information to protect you. Keepin' you safe. Aloha. Aloha, I'm Wendy Lo. And I'm coming to you every other Tuesday at two o'clock, live from thinktecawaii. And on our show, we talk about taking your health back. And what does that mean? It means mind, body, and soul. Anything you can do that makes your body healthier and happier is what we're gonna be talking about. Whether it's spiritual health, mental health, fascia health, beautiful smile health, whatever it means, let's take healthy back. Aloha. Back on Quark Talk, I'm Crystal here with Professor Marcus Rattaker, talking about the fiercest sea robbers. And we're talking women here. Now we just mentioned before the break, Anne Bonney and Mary Reed, two well-known female pirates from the Anglo-American piracies from the 18th century. Now Marcus, let's get right into these characters because I did a little research and apparently, so they dressed as men, but they were born illegitimate and they were raised as boys, or at least to appear as boys. So that influenced that. Yes, so these women pirates, Anne Bonney and Mary Reed, were like almost all other pirates in that they came from very poor backgrounds, poor working-class people. And in both their cases, Anne Bonney, originally in Ireland, and then in Charleston, South Carolina and Mary Reed in London, they had both been dressed as boys. And I guess they liked it because they went on to become soldiers and sailors and really fearsome pirates. Now let me tell you the story of Anne Bonney. It turns out, even though she was an illegitimate child, her father was quite a wealthy man and she was living with him in Charleston, South Carolina. And she didn't like that wealthy lifestyle one bit. She was a rebel. Okay.  And what happened to him was that he was given such a beating by Anne Bonney that apparently he couldn't walk for many days. So this was a woman very skilled in the arts of self-defense from an early age. And in defiance of her father, she ran away with a poor seaman, went off to sea, went to Jamaica and there she joined a pirate vessel captained by a man with the great name Calico Jack Rackham. That's not the same Jack in Pirates of the Caribbean. It might be based on him because of the style of dress because Calico Jack was a very flamboyant in the way that he dressed. And I think Jack Sparrow might have been based on him. So anyway, Anne Bonney discards that other sailor she had followed and she gets on board Jack Rackham's ship. She's dressed as a man. And he doesn't know she's a woman. He does know, but nobody else does. Then they take on board Mary Reed. So now by a wild coincidence, you have two women dressed as men on the same ship and they become very interested in each other and then they finally let each the other know their secret. Now there's a lot of gender play here. Not only are we talking about the superficial cross-dressing, but we're talking about the interesting human relations of being attracted to somebody, not knowing what they are or thinking. This is like Tame of the Shrew. I mean, this is. Exactly, exactly. And you want, we need to remember too that the ship in this day and age, the deep sea sailing ship, was a hyper masculine environment. So these women were really in a tough spot. So how did they gain the respect of their fellow pirates? First of all, by being very good at the work they did, they were strong, they were bold, they were the first to volunteer to attack another vessel. And so this gained them a lot of credibility with their fellow pirates. And I'll give you one example of just how tough they were. They were in the Caribbean and the crew were drinking. Pirates liked to drink. So a lot of the crew were really drunk. Anne Bonney and Mary Reed were not. And then they see a British warship nearby. The captain, Calico Jack, and most of the rest of the pirates run down into the hold of the vessel to hide. Whereas Anne Bonney, Mary Reed, and one male pirate start working the cannon to fire at that vessel to try to keep them off. Well, it didn't work. They were captured and the whole crew of them taken into Port Royal Jamaica to stand trial for piracy. And they are convicted and sentenced to death. Isn't this where they find out one of them is pregnant? Actually, they were both pregnant. They were both pregnant by Jack? No, not by Jack. Anne Bonney was pregnant by Jack. Mary Reed had her own lover on board the pirate ship. See, they still don't make do. They do. And there's a story about her too. Pirates had a tradition that if two people got into a scrap to prevent violence from breaking out on the ship, one of them would challenge the other to a duel and then they would go on shore and have a duel with pistols. Okay, so Mary Reed's lover got into a scrap with a really rugged pirate. And the other pirate challenged him to a duel and Mary was sure that her lover was gonna be killed. So what did she do? She then picked a scrap with the same rugged pirate, challenged him to a duel, and scheduled it on shore one hour earlier than the duel that was supposed to take place with her lover. She then went ashore with that other pirate in the duel, shot him dead, and she saved her lover's life in that way. So anyway, they're both captured. They're both pregnant and that means they can't be hanged. Oh, so that's the main reason. Under British law, that's why they were not hanged with Calico Jack and all the rest. So the time comes when Calico Jack and his shipmates have all about the rope around their neck. Like that photo we saw earlier. Yes, exactly. Actually, is there a photo of the women that you have in the image? Yes, let's see the women. This is not a very realistic picture, but we've got Anne Bonney and Mary Reed and you see the ships in the background and notice that they're armed. They've got their cutlasses. And then we can see a more romanticized image next. The next one? Yeah. Okay, here we are. I don't think they went around bare-breasted. But they didn't have bras, I'm sure. No, they didn't. But most sailors never really undressed, so you wouldn't know. That's a sexual rendition of this. Exactly, it's sexualized in an effort to sell copies of the book. Interesting. Okay, so at the hanging of Calico Jack Rackham, who we will remember went down into the hold of the ship to avoid fighting, Mary Reed and Anne Bonney are standing there to watch the hanging and Calico Jack looks over at Mary for a look of sympathy.  I love it. You are humanized. You are recreating this in a very filmic portrayal. How do you even get their resources to find out about this information? Well, it turns out these pirates were very popular in their own day, and several people wrote books about them. And one man in particular, a man called Captain Charles Johnson, wrote a history of the pirates based on interviews, among other things, with people who had been pirates. So the stories of Anne Bonney and Mary Reed were told to him, he wrote them down, and now they exist in books that we can go and read. But a lot of times, the legend of these women are preserved in ballads and photos like that. But after a while, when things changed, when the history of women changed, and there was more celebration of the feminine, then they turned into a very ugly portrayal of the pirates. Exactly. And it is very important that these were, as you said at the outset, women warriors. And there is a tradition in Great Britain of ballads celebrating women warriors. And it turns out historians are discovering that a lot more women dressed as men and went to sea, went to war, did all kinds of things that people had never before known them to have done. So in my view, Anne Bonney and Mary Reed are heroes. Yes, more than heroes though. I think about all the implications in the layers because of their gender on the ship as cross-jessing women warriors who were lovers and mothers and pirates. And I don't know what your speculation is, but I read somewhere where they implied that there was some kind of monage or twang going on between these two women. And Jack was, what do you think? What's your take on that? I think probably not because Jack was a jealous person. He was also afraid of inspiring jealousy in his fellow pirates. So if he had not one but two women and they didn't, that wouldn't have sat well. But it is important that in the end, Anne Bonney and Mary Reed came out as women on board the ship. And when they did, there was kind of a scandal, but in the end, their fellow pirates accepted them because they were just as tough as they were. But that's impressive, but they had to prove themselves at whatever toughness meant. I mean, they had to swear, they had to be able to kill, they had to lift hard things. Whatever it meant to be tough in those days, in those terms. But so what I wanna know is how sexuality played out in this piracy world and whether it was a tool for them or whether it was a curse. Well, for Anne Bonney and Mary Reed, do you mean? Well, particularly. I think the main thing that, the main lesson we can draw from their lives is that like these other poor sailors who were overwhelmingly male, they managed to create a life of genuine freedom on the high seas in a way that very few women were able to do in that time. But it was also true that very few poor people could elect their own leaders. See, the other thing is, Anne Bonney and Mary Reed were real leaders on board the ship. They were always in the boarding party, they were always leading by example. And I think this is one of the things that makes them really interesting for our own time. I think somebody should make a film about them. I think so too, and I'm surprised there haven't been anything done on that. I wanted to draw back to the larger picture of the history of women and what context. Did you have a slide or an image of something? Yes, there's another image that appeared in a book of pirates published in 1725. And this is an allegory of piracy, meaning a sort of symbolic rendering of pirates in which you have a woman, an armed woman beneath the Jolly Roger, the black flag that was frequently not just a skull and crossbones, but an entire skeleton. At the left, you see bodies hanging. Those are the pirates who had been killed. At the right in the background, you see some burning ships. And you see this woman pirate striving up from below, over these figures below. In other words, she is power. She has got a sword in one hand and a torch in the other. But she's obviously female. That's the striking thing. She is very, very female. Well, Mark, with our small limited time left, there's one more image I wanted to leave with everyone. And as people see that, I just want to remind people of the beauty of the types of books you've written. Now, what is the significance of this painting? Yeah, this is a very famous painting by Eugene Delacroix, the great romantic artist in Paris. And I am convinced that he based this great painting on the previous painting of a woman holding a flag above her head and that that was actually a model. So my point is that Ann Bonny and Mary Reed inspired that allegory of piracy and that allegory of piracy inspired one of the greatest artists in all of world history. And I hope that you were inspired by all this information that you've just taken from Marcus. And if you want to know more about this, this is all extracted. There is a chapter in this book, Villains of All Nations, The Atlantic Pirates of the Golden Age, correct? Right. By Marcus Reddaker. And I just wanted to give a shout out to Marcus. You should Google him, really, because all of his books are so rich. You've got the slave ship, the famous Many-Headed Hydra that is co-written by Peter Lindbergh. And he's gonna be here too to talk about it. And we're talking about it in school. And just so many, so many rich details on history from the Atlantic. And Marcus, I really appreciate you coming and sharing this. This is just the tip of the iceberg for people who want to know more about women in history. And we appreciate that. And thank you for sharing these stories. That's right, thank you for having me on. Thank you. Crystal.