I got four people that seemed more-or-less to want me to keep going with these, so I have decided to do it, and not just because it means I have less work to do regarding finding a topic for each post. Anyway, on my post which was an overview of resolution of singularities (which I intend to, at some point, go into more detail on the proof…just need to understand it properly first), John asked about the details of normalization. Right there, I did an example, and here I’m going to do a bit of theory for the process, and then repeat that example and add a second one.

As is often the case in algebraic geometry, we start with the algebra. We say that an integral domain R is normal if the roots in the field of fractions of R for every monic polynomial (polynomial with lead coefficient 1) in R[x] are already in R. We’ll start with two examples, and generalize to a theorem. First up, the integers are normal. The field of fractions there is the field \mathbb{Q}. Now, if you take a monic polynomial with integer coefficients, the only rational solutions you can get are integers again. Similarly, the ring k[x] is normal for any field k. The key thing these two rings have in common is unique factorization (that is, every element can be written uniquely as a product of finitely many elements which cannot be factored further, up to an invertible element). So we can actually prove the following:

Proposition: Any unique factorization domain is normal.

Proof: Any element of K, the field of fractions, can be written as a/b uniquely in lowest terms, because we have a UFD. Now we plug it into a general monic, and we obtain \left(\frac{a}{b}\right)^n+c_{n-1}\left(\frac{a}{b}\right)^{n-1}+\ldots+c_0=0. Clearing denominators, we get a^n+c_{n-1}a^{n-1}b+\ldots+c_0b^n=0. A short algebraic manipulation later and we get to -a^n=b(c_{n-1}a^{n-1}+\ldots+c_0b^{n-1}). So then we have b divides a^n. That would mean that a and b have a common factor, which is a contradiction. QED.

So now we have a class of normal rings. That’s a good place to start from, and seeing that k[x_1,\ldots,x_n] is normal hints that there might be some nice property of affine space that this generalizes to other varieties and schemes. This is, in fact, the case.

We say that a scheme is normal if every local ring is. But to be completely honest, I don’t care much for normal schemes, so we’re going to focus on varieties (in the abstract), which are what really matters. (yes, I’m letting my geometric bias show here, just as I will later if I start working over \mathbb{C} explicitly again).

An easier definition to check is that a variety is normal if it is covered by open affine varieties which are normal, and an affine variety is normal if and only if its coordinate ring is. So that tells us that \mathbb{P}^n is always normal, because it can be covered by \mathbb{A}^n‘s, which have coordinate ring k[x_1,\ldots,x_n].

Now, we want to define a normalization of a variety. First we’ll define it with abstract nonsense and a universal property. Let Y be any variety. A normalization of Y is a morphism \pi:\tilde{Y}\to Y such that \tilde{Y} is normal and for any other normal variety Z and map \phi:Z\to Y a dominant morphism (that is, the image is dense) we get a unique morphism \theta:Z\to\tilde{Y} with \phi=\pi\circ\theta.

Now, it’s not AT ALL obvious that this exists in general. For affine varieties, however, it’s not so hard. Take X to be an affine variety, and k[X] its coordinate ring. Then the normalization of X is the affine variety Y with k[Y] the normalization (in the sense of rings) of k[X]. All this is is that we take all the monics with coefficients in k[X] and then take the ring of solutions to them in k(X), the field of fractions. That this is a ring is a fairly standard commutative algebra problem, and is left as an exercise (hint: reformulate in terms of modules and note that an element is integral over R if and only if R[x] is a finitely generated R-module). The morphism for normalization of affine varieties is the one given by the inclusion k[X]\to k[Y].

Uniqueness of normalization is pretty easy from the universal property, and in fact, it’s unique up to unique isomorphism. Existence is the tricky part, which we prove below:

Proof of Existence: Let X be any variety and \cup U_i be an open affine cover. For each U_i, we have a normalization \tilde{U}_i by taking the normalization of the coordinate ring, and these are affine. Now, as normalizations are unique, we have that \pi^{-1}_i(U_i\cap U_j) and \pi_j^{-1}(U_j\cap U_i) must be isomorphic by a unique isomorphism making all the relevant diagrams commute. Thus, we can glue the \tilde{U}_i together by identifying them along these isomorphisms. This gives us a variety which we will call \tilde{X}. (It does require a check that \tilde{X} is separated, but I’m not going to do it here.) In fact, we get a map \tilde{X}\to X because the gluings were uniquely determined by the conditions of being locally normal, and so they agree on overlaps. Finally, we note that \tilde{X} is in fact normal, and as any dominant map Z\to X where Z is normal will locally factor through \tilde{X} in a nice way, the whole map factors through, and so we have that \tilde{X}\to X is the normalization. QED

So now normalizations exist for any variety, which means that we can attempt to do things like work on normal varieties and pushforward whatever properties we get to see if we can get something on arbitrary varieties. But we’re left wondering what nice properties do normal varieties have? Well, here’s the big one:

Theorem: The set of singular points of a normal variety has codimension \geq 2.

Proof: Suppose that X is normal of dimension n and let S be the set of singular points. Suppose that it contains a component Y of dimension n-1. Then (and this is nontrivial, but not something I intend to prove, though it is where we actually use the fact that we have a normal variety) there exists an affine open set X' such that Y'=Y\cap X' is cut out by a single equation. There exists y\in Y' which is a nonsingular point of Y' but not of X'. Look at \mathscr{O}_{Y',y}, the local ring of Y' at y, and u_1,\ldots,u_{n-1} local parameters (that is, their images in \mathfrak{m}_y/\mathfrak{m}_y^2 form a basis).

Now, I_{Y'} is a principal ideal in k[X'], so k[Y']=k[X']/(u) for some u. Similarly, we must have \mathscr{O}_{Y',y}=\mathscr{O}_{X',y}/(u) and \mathfrak{m}_{X',y} is the inverse image of \mathfrak{m}_{Y',y} under the natural quotient map of local rings. Choose v_1,\ldots,v_{n-1} to be inverse images of u_1,\ldots,u_{n-1}. Then \mathfrak{m}_{X',y}=(v_1,\ldots,v_{n-1},u), and so \dim \mathfrak{m}_{X',y}/\mathfrak{m}_{X',y}^2\leq n, because it is generated by at most n elements. So then, y is a nonsingular point of X', contradicting our assumption that it was. So S has no codimension 1 components. QED.

The value here is that this shows that normal varieties are regular in codimension one, and so we can define Weil divisors for them, and as a geometer, I consider Weil divisor to be better whenever we can use them. Now, if the local rings are all UFDs (which is stronger than normal), Weil divisors and Cartier divisors are the same. However, for normal varieties, Cartier divisors form a subground of Weil divisors. The other nice thing this does is prove that curves admit resolutions, that is, every curve is birational to a nonsingular curve. This is simply the normalization of a curve, as the singularities are in codimension 2, and that’s the empty set, there can be no singularities on a normal curve.

So now, as promised, some examples of normalization. Both will be simple normalizations, in fact, both are rational curves, and I’m choosing the simplest singularities out there, the cuspidal cubic and the nodal cubic.

Example 1: The cuspidal cubic curve is defined by the equation y^2=x^3 in the plane. It thus has coordinate ring k[x,y]/(y^2-x^3), which is isomorphic to k[t^2,t^3] by x\mapsto t^2 and y\mapsto t^3. Now, once we find a normal ring that contains k[t^2,t^3] and is contained in the normalization, we’re done. Such a ring is k[t], because x^2-t^2=0 is a monic polynomial with coefficients in k[t^2,t^3] and it has t as a root in the field of fractions, which is k(t). So then, k[t] must be the normalization, and so the normalization of the cuspidal cubic curve X is \mathbb{A}^1\to X given by taking the point t to (t^2,t^3) in the plane.

Example 2: The nodal cubic Y is defined by y^2=x^2+x^3. Here we don’t have as nice a characterization of the coordinate ring, so we need to work a bit less directly. Look at the parameter t=y/x on the curve. So then in the field of fractions, the equation can be reduced to t^2=1+x by dividing by x^2. The point of that is that now we have that t^2-x-1, which is monic with coefficients in k[Y] is zero, and so t is in the normalization. Now, if we take any element integral over k[Y], it can be written as a polynomial in t, as any instance of x becomes t^2-1 and any instance of y becomes xy/x=(t^2-1)t=t^3-t. Thus, k[t] is the integral closure, so again \mathbb{A}^1\to Y is the normalization map. Note that along the way we showed that k[Y]=k[t^2-1,t^3-t], and from there it is a bit clearer that k[t] is the normalization.

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