 this electron algebraic geometry will cover the relation between affine and projective varieties. So we recall we have affine space, which corresponds to its coordinate ring kx1 up to xn if we're living in n dimensional space. And an affine algebraic subset corresponds to a radical ideal of this ring. So remember, a radical ideal is one, which is equal to its radical. Then we have projective space. So projective space sort of corresponds to the ring kx naught up to xn where we consider this to be a graded ring. It's graded in the obvious way where we give each generator degree one. And a projective algebraic subset will be a subset of projective space given by various equations. And polynomials defining this subset should also be graded so that if some point is zero, this polynomial and so is any multiple of these points. So these projective algebraic subsets correspond to graded radical ideals. So a graded ideal means one where every element is a sum of homogenous elements in the ideal. So homogenous polynomial p has the probability that px naught up to xn is equal to naught if and only if p of lambda xx naught up to lambda xn equals naught whenever lambda is a non-zero element. Well, that's not quite true because there's one graded radical ideal that doesn't quite correspond to a projective algebraic subset. So this is except for the ideal generated by x naught up to xn. So the problem is that this corresponds to the point of an plus one which doesn't contain any lines at all. So this ideal here kind of corresponds to a completely empty subset. So what is going on is that a cone over a projective variety in pn sort of corresponds to an affine variety in an plus one. I haven't defined exactly what a cone over a projective variety is precisely but if you've got a projective variety defined by various equations px naught up to xn then you can consider the affine variety in an plus one defined by exactly the same equations and informally this is something like a cone over the projective variety. This is not just any old affine variety it's an affine variety that is invariant under rescaling. So we've got a point x naught up to xn we can rescale it as lambda x naught up to lambda xn. So projective varieties kind of correspond affine varieties invariant under these transformations. Next we will give some examples of converting an affine variety into a projective variety. So we recall projective space is just affine space together with some pointed infinity added. So if we've got a variety in affine space we can just sort of add pointed infinity in its closure and see what we get. So let's see some examples of this. So let's start with the very simple example. Let's just take the affine variety y equals x cubed. So this obviously isn't going to cause any great problems. So this lies in the xy plane. Now to make it into a projective variety we should add an extra variable. So we're going to add an extra variable z and look at points x, y, z in the projected plane. So this is a point x, y in affine space and we're now embedding affine space into projective space. So x, y in affine space corresponds to the point x, y, one in projective space. And in order to define a closed subset of projective space we need a homogenous polynomial. So we just make this homogenous by adding in powers of z. So we get yz squared equals x cubed. So here's an affine variety and here is a corresponding projective variety. And we want to know what does this projective variety look like? Well, to do this we notice that p2 is covered by three copies of affine space. So we can either take x not equal naught or y not equal naught or z not equal naught. It's also got lots of other copies of affine space covering it. These are just the easiest to work with. Well, if x is not zero, we may as well rescale to make x equal to one. y equals one and z equals one. And so if x is one, we've got a sort of yz plane. And if y equals one, we've got an xz plane. And if z equals one, we've got an xy plane. So p2 is covered by three copies of a2. You can think of it as being something like, in form is something like this. So here we've got three copies of a2 and altogether they cover p2. And we can look at this and we will have some sort of, here we've got some sort of curve living inside projective space. And we can look at the restriction of this curve to each of these three copies of affine space and see what it looks like. Well, first of all, we can look at the plane z equals one. Well, this is just the curve we started with. So we're guessing the curve y equals x cubed and it looks something like this. As everybody knows. Or we can look at the xz plane. And if we set y equals one, we find we get the equation z squared equals x cubed. So this plane, the coordinate looks, the curve looks like this. And finally, we may as well look at the x plane and in the, sorry, not the x plane, the plane where we fix x to be one. So this is the yz plane. And here if we set x equals one, we find that yz squared is equal to one. So this time the curve looks like this. So these are pictures of three pieces of this projective curve. And we now want to try and understand how they fit together. So here, if we take a point such as the point one one, this corresponds to the point one, colon one, colon one in projective space, which corresponds to this point here. So this is again, one, colon one, colon one. And again, it corresponds to this point one, one. It's one, one, colon one. So these three points are all really the same point on the curve. So let's try a different point because that really wasn't terribly exciting. What about this point here? So what about this point here? Let's see what it corresponds to. Well, this is the point where z equals one and y equals x equals naught. So it's naught, colon, naught, colon one. And so what point does it correspond to in this picture? Well, here we would have to rescale it to be y equals one, but we can't do this. So the green point becomes a point at infinity in some sense. It doesn't really appear on this plane. It's outside this plane. And again, here the green one is again the point at infinity. Except this time there are two points at infinity. We have to kind of figure out which it is. Well, here the point where z is very large and y is very small. So if z, sorry, I meant to say z is this axis and y is this axis, I hope I've got that the right way around. So it's going to be the point at infinity that's kind of in this direction here. So let's look at another point. What about this point here? So this is the point where y is one and x and z are zero. So it's this point here. On the projective curve. So what's it going to be here? Well, now we have to set z equals to one, which again, we can't do. So this red point is going to be the point at infinity. And similarly here, the red point would be the point at infinity. Only it will be the point in this direction here. So we see that there are some points which don't appear on all three of these diagrams. There's a green point, which only appears in this picture and a red point, which only appears in this picture. All other points will appear on all three diagrams. For example, suppose we take this point here, which is the point. Let's take x equals two, y equals eight and z equals one. So which point will this correspond to? Well, in this diagram, we have to set y equals one, we can rescale by dividing by eight. So we get the point one quarter one and one over eight. And in this picture here, we have to set x equals one. So if we rescale this so x is one, we see that y is four and z is a half. So it's a point up here somewhere. So this would be the point a half four. So that point, that point and that point are all really the same point on this projective curve here. Incidentally, we see that the point, that this point at infinity is singular. Well, we haven't quite defined singular points here, but it's quite obvious that something is going wrong because there's a sort of spike or cusp there. So even though this curve in the xy plane look completely harmless, we've got the curve y equals x cubed, which doesn't seem to have anything wrong with it. We see this affine curve really picks up a singular point at infinity if we sort of compactify it by extending it to a projective curve. So for the next example, let's look at an elliptic curve, y squared equals x cubed plus bx plus c. So this is a famous example of a so-called elliptic curve. At least it will be an elliptic curve when we make it into a projective curve. So this is the affine version of it. Now let's look at the projective version. So here we've got an xy plane. The projective version, we're going to three coordinates, x, y and z. And we need to make this homogenous, which we do just by sticking in factors of z just to make everything of degree three. So it will look like this. So now we want to know what does this curve look like? Well, let's find all the curves on it. First of all, when z equals one or z is none zero, we just get the curve y, I'm sorry, that should be a y squared and that should be a z. We just get the curve y squared equals x cubed plus bx plus c, which is the curve we started with. Typically looks something like this. Some sort of cubic curve. Well, if z is none zero, we just get the curve we started with. What about if z is equal to zero? Or in this case, we get the curve y squared times naught equals x cubed plus naught plus naught. In other words, we find x must be equal to zero. So the only point at infinity is the point where x is zero. So y must be none zero and z is zero because it's a point at infinity. So we have just one infinite point. And now we can ask what does the curve look like at this near this infinite point? We can ask, for example, is this infinite point singular or non-singular? Well, to do that, we look at the xz plane with y equals one. And if we set y equals one, we find z is equal to x cubed plus bxc squared plus cz cubed. And this curve will now look something like this. Well, the point at infinity of this curve will be the point where x and z are zero. So the point at infinity in this picture will be the point zero zero in this picture. Now, if x and z are both small, then we're saying z is approximately zero or slightly better approximation bz equals x cubed plus something very small. So the curve is going to look a bit like this. I don't know what it does elsewhere. Anyway, we definitely see that it is non-singular. In fact, any curve where they're given by an equation in x and z where there is a non-zero linear term in x or z will always be non-singular at the origin because it looks roughly like a line. We'll see more about that later. So this elliptic curve, if you turn it into a projective curve is non-singular at infinity. So, so far all the examples have just had a one point at infinity. Let's just look at a simple example when there is more than one singular point at infinity. So if you look at y squared equals x squared plus one, it's going to look something like this. It's some sort of hyperbola. So this is the x-axis. This is the y-axis. Now I want to write down its points at infinity. Well, you can sometimes do this without really calculating because it's kind of pretty obvious that this part of the hyperbola is sort of tending to a point that looks like something very large where x and y are both very large and approximately equal. So this will obviously correspond to an infinite point one, one, zero. And similarly, this asymptote here will kind of tend to an infinite point of the form one minus one, zero. So without doing any calculations, we can sort of guess that there are going to be two infinite points that look like this. If we want to check by calculation, that's quite easy. We homogenize it by multiplying by powers of z. So we're now looking at x, y, z. And now we want points where z is equal to naught. This just gives us y squared equals x squared. So y is equal to plus or minus x. And that gives us, if y is plus x, it gives us this point here. And if y is minus x, it gives us this point there. So it does indeed have two points at infinity. So far, all the examples we've looked at have just been curves in the plane. Let's look at an example in slightly high dimensions. This is the famous example of the twisted cubic. So in affine space, it's a set of points of the form t, t squared, t cubed. Here, t is a parameter, and this is going to be the point x, y, z. So t is equal to x for the moment. We'll see why we're not calling t x a little bit later. Well, what we want to do is to look at the ideal that defines this curve. And it's pretty obvious what the ideal is. The ideal is just given by y is equal to x squared. So we have y minus x squared and z is equal to x cubed. So let's add in a generator for this. So we've got two equations defining this curve here. And now we can attempt to find out what this curve looks like in projective space. So here we've worked out what is an affine space A3. What we want to do is to find its closure in projective space P3. And let's first of all do the wrong answer. So the wrong answer, well, how about we just take generators for the ideal and homogenize them? So let's homogenize some generators. So here we've got generators y minus x squared z minus x cubed. So let's homogenize them just by adding a point w. So we get w, y minus x squared, w, z squared, sorry, w squared z minus x cubed. So this is a graded ideal in the ring k, w, x, y, z. And defines, does it define a curve? Well, we'll see, it defines something in P cubed. And if we restrict it to the set of points where w is not equal to zero, it certainly becomes this twisted curve. It certainly becomes this twisted qubit we had earlier. If you look at it, there's something a bit fishy going on because this ideal also contains things like y squared minus cx. And the trouble is y squared minus cx is not contained in this ideal here. So this does not contain y squared minus cx. And not only does it not contain y squared minus cx, this ideal also contains z minus xy. And if we homogenize this, we get w, z minus xy. And this ideal doesn't contain w, z minus xy either. So we can't homogenize an ideal just by homogenizing a set of generators. We need to homogenize all elements of the ideal. Because some of them, the point is that this polynomial here is indeed in the ideal generated by these two, but its homogenized version of it is not in the homogenized, is not in the ideal generated by the homogenized generators. So in projective space, the twisted cubic, its closure is in fact given by s cubed, s squared t, st squared t cubed. So if s is equal to one, it gives this point here. And if s is zero, then it gives one extra point. So it's the curve above where s is equal to one plus one extra point, s equals zero. And the ideal spanned by, sorry, the ideal of polynomials that vanish on this, it's not too difficult to check. It is actually the ideal w y minus x squared, y squared minus cx, w z minus xy. So it contains this element here and also these two that were accidentally admitted by our rather careless attempt at homogenizing an ideal. Well, what I would like to do is to ask the following question, what exactly happens is if you take this incorrect set of generators for the closure of the twisted cubic, let's see what happens. So we've got the, we're looking at what happens if you take the equations w y minus x squared and w squared z minus x cubed. So this is an ideal in four variables. And we ask what algebraic set does this define in projective space consisting of points w, x, y. z. Well, projective space is of course covered by four copies, very not equal, it's covered by four copies of affine three-dimensional space, which of the four copies to the point where we can take w equals one or x equals one or y equals one or z equals one. So what we're going to do is to look at each of these four copies of affine space and see what this algebraic set looks like. First of all, w equals one is easy. Here we just get the equations y minus x squared equals naught and z minus x cubed equals naught. So it's the twisted cubic we started with. So this is just the twisted cubic in a three, nothing very exciting happens there. x equals one, nothing very exciting happens either. We find w, y equals one, w squared z equals one. And again, we just get one curve. It's really a copy of the affine line. So it's a copy of the affine z line minus a point because z can be any non-zero number and then w squared is one over z squared and y is then one over w. So this looks just like the twisted cubic with one point missing. If we take y equals one, something a bit stranger happens. So with y equals one, here we get the curve w equals x squared. And we get w squared z equals x cubed from this equation here. Well, so if we eliminate w because w is equal, we can just set w equal to x squared. So we get x to the four z equals x cubed. And this sort of splits as x cubed times zx minus one. Sorry, I should say that equals zero. So we actually get two things going on. First of all, we've got a curve zx equals one, which turns out to be again another view of this curve here. But then we've got a strange sort of triple line here. So we get a sort of triple line where x is equal to nought equals w. And in some sense, we get not just one copy of this line, but three copies of it. Because instead of saying x equals nought, we've only got a much weaker equation, x cubed equals nought. So what is going on is we can take our original cubic in affine space. And if we try and make it into a curve in projective space in this rather careless way, we find we also pick up a sort of triple line at infinity, which is definitely not what we wanted. If c is equal to one, this is kind of similar to what happens if y is equal to one. We again pick up a curve plus a sort of strange triple line. So you have to be a bit careful when homogenizing affine varieties. If you're careless about just homogenizing generators of the ideal, you might find you're getting not just the closure of the affine variety you started with, but you might also pick up a lot of strange junk at infinity that shouldn't be there. Incidentally, the fact that this contains two components, the curve you started with, plus a triple line corresponds to the following Lasker-Nurter style decomposition of this ideal. So we can see that the ideal w y minus x squared, w squared z minus x cubed is actually an intersection of two ideals. So we can take the ideal w y minus x squared, y squared minus x z, w z minus x y, and take the intersection of that with w y minus x squared, w squared, w x. So this ideal corresponds to the line we actually wanted. This ideal corresponds to this weird triple line infinity. You can see there's something fishy about this ideal here. It's not reduced. For instance, it contains w squared, but doesn't contain w, which kind of accounts for the fact this is a multiple component.