This talk is part of an online commutative algebra course and will be a little bit more about stably free modules except that, well, it will really be about the technique called the Islandberg Mesa Swindle and some applications of this to the structure of free modules. So the first example of the Islandberg Mesa Swindle is a proof that one is equal to zero and the proof goes like this. What we do is we look at the series one minus one plus one minus one plus one and so on. And we can write the series in two ways. We can either write as one minus one plus one minus one plus one minus one where we pair off elements like this. And you see this is zero and this is zero and this is zero. So the series is equal to zero. On the other hand, we can pair off the elements like this. So if we pair off these two elements, we get zero and these two elements, we get zero and so on. So this sum is equal to one. So we find one is equal to zero. Well, obviously this is not a valid proof. So there's a problem and it's not difficult to see what the problem is. The infinite sums are not well-defined. Now the Islandberg-Mazer swindle works in cases where you can get infinite sums that are well-defined. And the same argument shows that if A plus B is equal to zero and infinite sums are well-defined and behave well. So behave well means that there's some sort of notion of the infinite sums being associative and commutative which I'm not going to specify precisely. Anyway, it says that if A plus B is equal to zero, then A and B must both be zero. And you can see this by considering the infinite sum A plus B plus A, A plus B plus and so on. And this is equal to A plus B plus A plus B and so on. And you can bracket the terms in two different ways. So we can bracket them like this and we find the term is A and we can bracket them like this and we find the sum is zero. So we can apply this whenever we can take infinite sums of things. And the first example I'm going to give has nothing to do with commutative algebra and it's a problem about knots. So we can take a sum of knots as follows. Suppose I've got a couple of knots. Try and suppose I've got a knot here. So this is, I hope, a truffle knot. And suppose I've got another knot here which might look like this. Not very good at drawing knots. This is something algebraic topologists know how to do. So how do I take the sum of these two knots? Well, that's quite easy.  And that gives me a sum of two knots. And obviously you can take sums of any numbers of knots and we can also note the sum is commutative. And we can see this as follows. So suppose I've got a little knot here which might go like this and suppose I've got another knot let's try and take a slightly different knot so it's not quite trivial there, they're commutative. So there I think is a figure of eight knot if I've got it right. And what I want to do is to show that this is equal that I can exchange the order of these. And it's kind of obvious how to do this. What I do is I make the first knot really small and I make the second knot really big. So now I can take hold of the first knot and just slide it along through this big knot and move it to the other side. So the sum is commutative and it's also fairly obvious that it's in some sense associative. The other nice thing is that infinite sums are defined. So suppose I've got a knot and I'm just going to draw this as a box with the letter A on it because I'm getting tired of misdrawing knots. And suppose I've got another knot B and suppose I've got another knot C I can draw the knot C even smaller and I can have an even smaller knot and so on. And I can continue making these knots smaller and smaller each one might be half the size of the next one and I can collect them all up and get an infinite sum of knots. Notice that this is infinitely fine for continuous knots. This doesn't work for smooth knots because at the limit point here the knot is obviously not going to be smooth. But we won't worry too much about that. So this is one of the constructions why topological manifolds are kind of different from smooth manifolds. There are some constructions you can do for topological manifolds that just aren't smooth and this is one of them. And now let's show that knots cannot cancel. So suppose I've got two knots A and B like this. So suppose A and B cancel. It sort of seems pretty unlikely that by taking a knot and by adding another knot you can eliminate both knots but that's not the same as a mathematical proof. You want to understand why knots don't cancel. And what we do is we just sum A plus B plus A plus B and so we take an infinite sum of these two knots. And now on the one hand you can pair off the terms like this. So I can pair off these two and these two and these two and they all cancel out. So by doing this green pairing we say that this is just the sort of circle that the trivial knot. On the other hand I can pair them off like this just as in the side pair those two and those two and those two and we see everything pairs out except the knot A. So I'm left with just the knot A. So we see that if A and B cancel out then in fact the knot A is the trivial knot and of course the knot B is also trivial. And so that's a topological example of the Islandberg-Mazer swindle. The topological version is usually called the Mazer swindle because Barry Mazer introduced this idea in a slightly different context and the algebraic version of the Islandberg-Mazer swindle that I'm now going to do is named after Islandberg who introduced it. So the first application is the following problem. When we defined M to be stably free we said this was equivalent to saying M plus R to the N is free for N finite. And one question to ask is why do we take N finite? What happens if we take N infinite? So suppose we allow N to be infinite. Well, if we allow N to be arbitrary then we find M plus R to the N free and anything turns out to be just equivalent to M being projective. So this isn't giving us a new concept. So we can see this in two steps. First of all, if M plus something free is free this implies M is projective. And the reason for this is that any direct summand of a projective module is automatically projective which is very easy to check. And free modules are projective so M is a direct summand of a free module so it's projective. The part that involves the Islandberg-Mazer swindle is we want to show that if M is projective then M plus some free module is free. And we can see this as follows. First of all, if M is projective this implies that M plus N is free the sum N. That's because we can just find a free module mapping onto M and then this splits because M is projective so M is a direct summand of a free module. And now we use the Islandberg-McLaine swindle and we look at M plus N plus M plus N plus M and so on. And we can divide this up in two ways. First of all, if we divide it up like this we see that this is free. On the other hand, we can divide it up like this and this green bit is free and we've got M left over so this is equal to M plus a free module. So we found a free module of infinite rank such that M plus this free module is free. So that's why we add the condition that this number N here is finite in the definition of stably free because if we don't it's just projective modules. Next, I want to use the Islandberg-McLaine swindle to show that stably free of modules of infinite rank are free. So I haven't actually quite defined what the rank of a stably free module is but what I mean is if M plus R to the N is isomorphic to some infinite sum of copies of R. If you're a set there is to know there are different sorts of infinity. I just don't care. This can be any sort of infinity. So suppose that M plus some finite rank module is an infinite sum of copies of R and we want to show that M is actually free. What we do is we look at the map from R to the infinity to M plus R to the N. And now we can find some finite number of copies of these that map onto this finite module R to the N. So we can think of this as being R to the infinity plus R to the M is isomorphic to M plus R to the N where this map here is now onto with M and N both finite. Next, we can take all the basis elements of this and just subtract suitable elements of this in order to make the image of this free module lie inside M. So we can now reduce this to R infinity plus R to the M goes to M plus R to the N where this is onto and this map here is into and there may also be a map from R to the M to M of course and so on. Well, we've now got a map from R to the M which maps onto R to the N. So because R to the N is free and therefore projected this split. So we can write R to the M is equal to X plus R to the N where we can take this sub module X is going to map into M. So what we do is we find that M is equal to R to the infinity plus X where X is stably free X plus R to the N is R to the M. So now let's use the Island Bergman Swindle to show this implies M is actually free.  And let's just add R to the infinity to this and on the one hand, and this is free because X plus R to the N is R to the M. So this is a free module. I mean, the whole lot is a free module not just the R to the infinity. And on the other hand, we can, that's a free module because we're pairing it off like that I guess. On the other hand, we can pair it off like this. And again, we know that R to the N plus X is just R to the infinity. So this is just X plus R N plus X to the infinity plus R to the infinity which is just X plus R to the infinity which is M. So we found M is this module here which is free. So next lecture, we will be looking at locally free modules which are more or less the analog of vector bundles in topology.


This study found that residents in Japan who chose general medicine had higher levels of basic skill achievement than those who chose other specialties. Those who chose general medicine also had higher scores than those who chose highly specialized careers. This suggests that residents in training programs without special to specific competition may not have the same motivation as those in competitive systems. This article was authored by Takashi Watari, Yuji Nishizaki, Nathan Houchens, and others.


This study examined the motivations for complying with COVID-19 related policies in Germany and Switzerland. The results showed that social cohesion, consequences, and personal risk assessments were all important factors in determining whether someone would comply with the policies. Additionally, the study found that while some people felt it was important to follow the rules out of principle, others needed more evidence before they could be convinced. This article was authored by Bettina Zimmerman, Amelia Fisk, Stuart MacLennan, and others.


Switch over to part of the new Highway 23 now in effect. Proveya's Sheboygan testing site location to change, U.S. Supreme Court refuses to extend Wisconsin's vote count. These and other local stories are coming up on this edition of Community News Review. This is Community News Review, service WSES-TV News content, provided by WHBL. I'm Maddie Fister and this is Community News Review for Tuesday, October 27, 2020. The Sheboygan County Division of Public Health issued its first report of the week and added 388 new cases of COVID-19 between 8 a.m. Friday through 8 a.m. Monday, October 26. Only two of those were K-M-C-I inmates, where 858 cases of the disease have been recorded. Sheboygan County now has 1,464 active cases, a number now revised per the definition of laboratory-confirmed PCR tests administered within the last 30 days. Positive tests are counted one time per person unless at least 90 days had passed since the first positive test. Recovered cases will now be defined as confirmed cases, where 30 days have passed since testing positive and has not resulted in death. That number is now 2,244. Both figures were brought into conformity with Wisconsin DHS standards. Of the active cases, 23 are hospitalized. Within 958 negatives returned the positive rate as of Monday was the 26th was 28.8%. Wisconsin received 12,749 test results since Sunday. With 2,883 returning positive, that return rate was 22.6%. 10 more deaths were recorded, raising the death toll to 1,788. Another 84 persons were hospitalized for COVID-19, and as of Sunday 1,293 persons in the state were hospitalized for COVID-19, with 320 of those in the ICU, and another 115 patients awaited their test results while hospitalized, and 445 persons were on ventilators. Prevea Health has been operating a COVID-19 testing site outside of their Sheboygan Medical Office building at Seiman Avenue and Taylor Drive, and will continue for another few weeks, but as of November 16th, that site will shift from the north to the south side. The new location will be in Southtown Mall at 3335 South Business Drive. Prevea Health offers testing at no charge to anyone experiencing one or more COVID-19 symptoms, and many in which are also occur in the common cold or flu, such as cough, fever, chills, sore throat, and runny nose. But Prevea Health president and CEO, Dr. Ashok Rye, said that one should not assume that it is just a cold. Rye said that COVID-19 can be spread even when symptoms are mild, and appointments for the free testing can be made in the same day. Travelers on Highway 23 between Sheboygan and Fond du Lac are using some of the newly constructed lanes as this past Saturday. The Wisconsin DOT made the move a few days earlier than anticipated, and drivers are being reminded to use caution and be aware of the traffic pattern change. The traffic will remain two lanes for now, as contractors still have plenty of work to do before the expansion to four lanes and completed in 2022. This week, the contractor will work on pavement marking, shouldering, topsoil, and landscaping. Work continues at the county tea cul-de-sac branch road and the old Plank Road recreational trail. Signing and delineator work also continues. A hoped for return to daily meal delivery in November has been put on hold for now by Fresh Meals on Wheels of Sheboygan County. Kelly Anderson, the CEO of Meals on Wheels of Sheboygan County, said that the rising number of COVID-19 cases in Wisconsin caused the organization to choose continuation of weekly bulk deliveries, and the decision was made to protect the safety of our meal recipients, volunteers, and staff. Those meals made by culinary staff and then flash frozen are delivered in sanitized vehicles by volunteers that also serve to check on the safety and welfare of their clients. With only weekly deliveries for now, those more frequently, those checks are made by so-called caring callers by phone, and there is room for more volunteers to do just that. Volunteers must pass a background check, and anyone interested in becoming a caring caller can email client services at freshmealsonwheels.org. Meanwhile, anyone needing a meal delivered should contact Meals on Wheels at 451-7011. There are no waiting lists, and there are no income restrictions. And finally, absentee ballots in Wisconsin are due back by 8 p.m. on election day. No exceptions. The U.S. Supreme Court on Monday ruled against a Madison-based federal judge who wanted to allow local election managers to count ballots that same day that came in as late as six days after the election day. The court ruled five to three that state lawmakers, not federal judges, are the ones who write the election laws in Wisconsin. The decision means late absentee ballots will not be counted this year. And that is all we have for today. Join me again next time for another recap of our local news on Community News Review.


What's good y'all Chris from justice case breaks comm doing a 2022 panini immaculate WWE one box break number seven Joe That's nice enough to let me grab one of these and do this with y'all since I'm a big wrestling fan I'm your resident wrestling mark here at the shop, but R&B style one break zero to nine We've got our friends here Ed Joe Anthony Patrick Peter Chris and Brandon got those two last spots the last spot Mojo, thanks again everyone again that I'll quit out bro ricochet and brawn is an excellent example Oliver that's my guy And our spots one down two zero List right here. Let's randomize with our dice. Good luck everyone. Thanks again We'll roll both two and one three times. So let's do it one two and three ding ding ding ring the bell Patrick down to Chris and Then three times here good luck one two and last one three Five down to one Mini Cooper and locks of love. I like that. I'm all right with that He's here And let's take a look real quick. So Patrick spot five Brandon It's spots nine and three Ed P with two Chris would zero It's any live any redemptions even if they are numbered one-on-one in the checklist Peter with four Joe C with six Anthony with seven Joe C with eight and Chris with one any active live one-on-ones Go to Chris. All right. Oh the trade open open up for like a second or two We'll pause the video when we come back. We'll get into this one box break of Immaculate WWE. Thanks again y'all All right, so no trade chatter. So let's just get right into the break. Shall we some fun stuff coming up? What a one box break One of the you know, I was a actually a Dudley boys guy So we're talking Hardee's edging Christian in the Dudley's Those are my dudes All right, well, we got our break sheet printed. We got four boxes here from a case Right here. So we'll roll Joe's calling them the raw and smack down dice Let's go smack down since tomorrow is Friday Right here. So let's roll our dice. Good luck everyone One two three and four five and six will be re-rolls And it's three box three Not Brian Alice deal spike Dudley What a gimmick all you needed was a pair of glasses and a tie-dye shared and you were a Dudley boy All right Here's our list Kristen spot zero and one all the way to down to Brennan with nine. Thanks again y'all always fun doing the WWE like I said, so Let's get to it Yeah, North Carolina royalty if wasn't for Rick flair, you know, North Carolina would be hardy country 3d the Dudley deaf device Bushwhacker Luke butch whacker butch. I'll be lucky. I'll be Luke and Joe can be butch No redemption all live and active cards. Let's get it. Good luck everyone Starting off. Whoo Charlotte flair the current Smackdown women's champion 65 out of 60 spot spot spot five Going to Patrick it might I might be thinking there's another that's so many I think the the Dudley death device is the the doomsday device like a legion of doom used to do Next up Patch of Ridge Holland 39 out of 99 spot 9 going to Brendan Perfect time to get it. She'll be a I don't know if she'll be made of any one of the WrestleMania nights, but her and Oscar This would be a heck of a match next up Immaculate standards 73 out of 99 of the phenomenal a jay style spot 3 going to Brendan They I don't know what they're gonna be doing with Bray I hope it's not omas and Brock at mania like actually Don't want all that. It's good to our autos here first up 95 of 99 Simon Dean the fitness guru spot 5 going to Patrick Murphy Now that's a name. I have not thought about in a long time long time Simon Dean With the milkshakes and the Dean system that is crazy. What a pull penultimate card Very nice on card marks of greatness 43 out of 49 for Raquel Rod Regas Spot 3 going to Brendan and the last one. Thanks again y'all. You got more immaculate loaded up on the site right after this break Let's get a long boy here to end it. Oh Hell yeah 49 of 49 stone cold Steve Austin Nice one there. Oh, Steve awesome. Hall fame induction induction awesome 316 Austin 316 That's 49 of 49 spot 9 those two last spot mojo's If this wasn't a family show, I would give people the double bird and then shotgun of beer beer Yeah, I don't right here and double bird the crowd then wow some of the stone cold center Very very nice everyone nice box there nice stone cold to pull it up Thanks again y'all more immaculate put up on the site very very soon jazby's case breaks calm. Thanks again. Bye