 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. What I do is I cut here and I cut there and I join them up like this. 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. So what we do is we look at X plus R to the N plus X plus R to the N plus X plus R to the N and so on. 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.