 This talk is part of an online course on commutative algebra and will be about the geometric meaning of finite extensions and integral elements and normal rings. So we recall that if we've got two rings R contained in S, then we say S is finite over R. If it's finite as an R module, an element S in S is integral over R if R of S is finite in this sense. And finally, we say R is normal if in the quotient field, in the field of quotients K of R, every element that is integral over R is already in R. So we discussed these last lecture and what we're going to do is discuss the geometric meaning of these. What this means is that if you've got a map from R to S, you've also got a map from the spectrum of S to the spectrum of R as topological spaces. And we want to describe what these concepts mean in terms of this geometry. So first of all, we discuss what it means for S to be finite. Well, there's an obvious notion for this to be finite, which means that if this map is F, we can have F the minus one of a point is a finite set. Turns out this is closely related to the condition that S is finite as an R module, but isn't exactly the same. So this condition here is called quasi-finiteness. And there are plenty of examples of extensions of rings that are quasi-finite, but not finite. For example, we could have the extension R of X is contained in R, X, X the minus one. So you can think of this as being the affine line over R and this has been the affine line over R minus a point. So the spectrum of these, so the spectrum of this looked like a line and the spectrum of this looked like a line minus the origin. And the map from this one to this one is finite because the inverse image of any point is either empty or a point, but it's certainly not a finite extension because this is not finitely generated as a module over this ring. So quasi-finiteness and finiteness aren't exactly the same. However, they are quite closely rated because if R contains an S is finite, this implies that it's quasi-finite. The question of whether if you've got something quasi-finite, how far is it from being finite? It's a subtle one relation to something called Zariski's main theorem. It turns out that under fairly general conditions, you can obtain any quasi-finite morphism by first taking a finite extension and then localizing it. But anyway, what I'm going to do is to show that finite implies quasi-finite. So we have spectrum of S, maps to spectrum of R and we're going to pick some prime ideal of R. So P is in the spectrum of R. And if we call this, we want to show that F the minus one of P is a finite set. In other words, there only a finite number of primes of S whose intersection with R is finite. So we can do this in several steps. First of all, we can quotient by P. So what we do is we just set everything in P equal to zero and you can see this doesn't really change anything. So we can assume P is equal to zero. So in particular, R is an integral domain because we've now got zero is no prime ideal. Secondly, we can simplify further by localizing at P, which is just zero in R. What this means is we just invert all elements, all non-zero elements of R. So this means we replace R by its field of quotients. And again, you can check that this preserves the inverse image of P. So it's sufficient to prove finiteness under this condition. So we've reduced the case when R is a field. Well, in this case, S is a finite dimensional vector space over R, so is Artinian. And we showed earlier that Artinian rings have only a finite number of prime ideals. So the theorem is obviously true in the case when R is a field and we can reduce this case by quotient again and localizing. So it's true in general. And in fact, you see that the spectrum that the inverse image of P is not just finite, but it's actually finite and discreet as a set, as a topological space. So that sort of indicates partly what the geometric meaning of finiteness is. It has something to do with a map of spectra such as the inverse image of every point is finite. Now I want to talk about the geometric meaning of normalization or normal rings. And this is a sort of mild resolution of singularities. It definitely doesn't resolve all singularities in general, but it's sort of a first step in getting rid of singularities. So let me explain what this means by an example. Suppose we take R to be the ring K T squared T cubed. So it has a basis one T squared T cubed, T to the four and some where we sort of forgotten to put in the term T. And if we put Y equals T cubed, X equals T squared, we see that R is equal to K XY, and you're quotient out by the ideal Y squared minus X cubed or possibly Y cubed minus X squared. It's easy to get these the wrong way around. And if we look at this, we can see it's the coordinate ring of a curve with a singularity. Now we haven't actually described what a singularity is in general, but it's kind of pretty obvious that this is a singularity, whatever a singularity means. So we can say this ring R has a singularity at this ideal. So this is of course just the ideal generated by X and Y. So there's something funny going on at this point of the spectrum of R. Well, now we can normalize R. Well, this is quite easy to do because the field of quotients is just the ring of all rational functions in T and it's pretty obvious that T is integral over R because T squared is in R. So T must be in the integral closure of R. So the ring R is contained in K of T, and since this is a unique factorization domain, it's automatically normal. So this is the integral closure of the ring R and you can call it the normalization of R. And now we can draw a picture of the spectrum of K of T. So here the spectrum of K of T is just an affine line which obviously doesn't have singularities, whatever singularities have. And the inclusion of R and K of T just means that we've got a map from the affine line to this singular curve here. In other words, we've sort of resolved the singularity of this curve by taking the normalization of its coordinate ring. In fact, it turns out this works for any curve. You can resolve all singularities of a curve in one step just by taking normalization of its coordinate ring. Geometrically, what's going on by the way is if we draw a line, if we take a point here and draw a line through the point then T is, turns out to be Y over X, which is the slope of the line. So here we've got a point T in the affine line and it maps the point on this curve such that the slope of the line through the origin is T. If you've done blowing up an algebraic geometry, we'll notice that this is also an example of blowing up this curve at the point. So let's look more generally what happens if you take normalization of a quadratic extension. So I'm going to suppose R is a unique factorization domain to make everything simpler. And let's look at R where we adjoin the root of something in R, so we're taking R of T over T squared minus R. So I'm just taking a quadratic extension of R. And I want to ask what is the normalization of this quadratic extension? Well, what we have to ask is when is P times root of R plus Q and P times root of R plus Q integral over R? Well, if so, then P of root R minus Q is also integral. So two Q is integral. So it must be in R because it's in the quotient fields of R and R as a unique factorization domain. So it's integrally closed. Now, you can get cases when two Q is an R, but Q is not an R. So in fact, we've had this example several times before. We can just take Z and adjoin root of minus three. And then you see that root of minus three plus one over two is in the integral closure. So this factor of two really does turn up sometime. However, I don't want to worry about it for the moment. So from now on, we'll assume that a half is contained in R. So this means Q must be in R. So we're going to assume R is a unique factorization domain. And we're also going to assume that half is in R except for this example to show sometimes it isn't. So now we find that P root R is integral. So P root R squared is integral over R. And this is just equal to P squared R, which is in the K, which is the field of quotients. So this means that P squared times little R must actually be in R. And we can now see some examples of this. So here's the first example. Suppose we look at a cubic curve, Y squared equals X cubed plus A, X squared plus B, X plus C. And we can take its coordinate ring. Let's take K of X, Y, modulo Y squared minus X cubed minus A, X squared minus B, X minus C. And we can ask what is the integral closure of R, which is this ring here in the field of quotients. And what we see is that we must have here P squared times R has to be in R and we're taking R to be this element X cubed minus plus A, X squared plus B, X plus C. And let's factorize this as X minus alpha squared times X minus, so suppose this factorizes X minus alpha squared times X minus beta. Well, then we see that we can take P to be one over alpha, one over X minus alpha. So if this polynomial has a repeated root, then we can find there's an element of the integral closure given by, that this will be Y over X minus alpha. So this is in the integral closure. And furthermore, this generates the integral closure. On the other hand, if this factorizes X minus alpha, X minus beta, X minus gamma for alpha, beta and gamma different. This means the integral closure is R. In other words, R is just normal. And these two cases correspond to the cases when the curve has a singularity or doesn't have a singularity. So if we look at the equation Y squared equals X minus alpha squared X minus beta, the graph of this is going to look, might look something like this. And the curve has a singularity at X equals alpha, Y equals zero. And when we take the ring generated by, say K, X, Y over Y squared minus, that's stuff in X. When we adjoin the element Y over X minus alpha, this is now just K of T, where T is equal to Y over X minus alpha. So this is now the coordinate ring of an affine line. And as before, we've really found a resolution of the singularities of this curve because we've now got a map from an affine line going to this curve and it's non-singular. And in fact, as before, we can interpret the coordinate T on this affine line as being the slope of a line through a point and through the singular point. So if I just draw a straight line through here, slope T, then geometrically what's happening is a point on this curve with a singularity just correspond to a point on the affine line with coordinate T. On the other hand, if you look at Y squared equals X minus alpha, X minus beta, X minus gamma, this is just one of the celebrated elliptic curves or rather an affine piece of it. And it has no singular points and its coordinate ring is already normal. So normalizing does absolutely nothing. Another example is an example of a ring with a singular point which is still normal. So this time we're going to look at a double cone. So a double cone obviously has a singularity at this point. And this is going to be the graph of Z squared equals X squared plus Y squared graph of a cone. So our coordinate ring R is going to be just KXYZ modulo Z squared minus X squared minus Y squared. And we want to show R is normal even though it's got a singular point. So how do we show it's normal? When we notice R is a quadratic extension of KXY it's just equal to KXY and then we adjoin the square root of X squared plus Y squared. So this would be the elemental R in the previous discussion. And now we notice that X squared plus Y squared is not divisible by a square. Here we're assuming that the characteristic of K is not equal to two because if the character is two everything always goes wrong. So all we're doing is we're adjoining the square root of an element which is not divisible by any square of an irreducible element. So KXY root X squared plus Y squared is normal by what we said earlier. So a variety can have a normal coordinate ring even if it has a singular point. There's a criterion of SIR that I will just mention which says that an integral domain R is normal if and only if all singularities of the spectrum of R of co-dimension at least two and if the localization to a prime has depth, if RP has co-dimension has dimension at least two it has depth at least two. So this is a slightly technical criterion. The depth of a ring is something we'll be discussing later when we discuss cone Macaulay rings. And it turns out this criterion here is nearly always satisfied for any reasonable ring more precisely for cone Macaulay rings. So in practice it often turns out that ring R is normal provided all singularities of co-dimension at least two. Incidentally normalization is sometimes used in resolution of singularities of a surface. What we do to resolve singularities of a surface is we normalize and this gets rid of all singularities of co-dimension one. So we find all singularities of points. And next we blow up all the points. So blowing up is an operation algebraic geometry that allows you to get rid of singularities on points or for that matter more general varieties. So you may think you can now get rid of the singularities of points by blowing up but it turns out that when you blow up a point you may get a new singularity that is co-dimension one. So you go back to step one and normalize to get rid of the co-dimension one singularities and you keep repeating this and hope you can show that this process eventually terminates which I think Zariski and Walker proved back in the early 20th century. This process does actually resolve all singularities of a surface. So next lecture we'll be talking about filtrations of rings.