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Integral Closure and Normalization

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Integral closure and normalization Wanlong Zheng 1 Introduction This article, in a rather informal manner, will present a few interesting connections between certain algebraic operations such as taking integral closure or completion, and their geometric interpretations. We will be dealing with affine varieties (affine algebraic sets) mostly. One can rather easily generalize to schemes by the glueing property. We first recall some basic definitions. Definition. The coordinate ring of an affine variety V is the ring of regular functions A = k[x1, ::: , xn]/I(V), where the operation I on any set S ⊂ k[x1, ::: , xn] is defined to be I(S) = ff 2 k[x1, ::: , xn]: f(x) = 0 8x 2 Sg. Definition. For an integral domain A and its fraction field K, all integral elements, i.e. those x 2 K such that x satisfies some monic polynomial with coefficients in A, form a new ring A ⊂ A ⊂ K, called the integral closure. Proposition. The integral closure is the intersection of all valuation rings V sit between A and K. This result is nontrivial, and it depends on Zorn’s lemma to come up with valuation rings. See [1] for a complete proof. There are a few basic algebraic properties related to this operation. For example, the integral closure of A will still be A, and therefore we can say a ring is integrally closed (c.f. completion of rings/ring is complete). In particular, we will assume the following result. Lemma. If A is a reduced, finitely generated k-algebra, then so is A (in fact it is a finitely generated A-module). This is a rather irrelevant lemma, but it shows the integral closure is the coordinate ring of some variety. For proof, see Local Algebra by Serre. The questions then become what is the new variety and how do we find out what the integral closure is. Let’s look at a few examples. 2 Examples Example. V = fy2 = x3g ⊂ C2. Assuming V is irreducible, we have the coordinate ring A = k[X, Y]/(Y2 - X3) an integral domain, and we can take the integral closure. Upon substituting X = t2, Y = t3, A is isomorphic to k[t2, t3], with fraction field k(t). This is because 1/t = t2/t3 = X/Y. But then k[t] is a UFD, thus is already integrally closed. So the integral closure will be k[t] = k[Y/X], which simply corresponds to a straight line. Observe that we defined a map of k-algebras (rings) φ∗ : k[X, Y]/(Y2 - X3) k[t] X 7 t2 Y 7 t3 by sending , , which is also injective. This happens! iff the map 2 3 ! ! φ : V(t) V(y - x ) !1 has dense image. φ also restricts to V n f0g an isomorphism. So somehow, the fact that the ring is not integrally closed has something to do with the singularity of the curve at the origin. There is a different way of interpreting this: if we consider the rational map 1 f : V 99K P by sending (X, Y) to [Y : X] = [Y/X : 1], as a map between quasi-projective varieties, then this map is not regular at (0, 0) (can be easily verified). However, this map can be extended over the origin: we write the image as [t3 : t2] = [t : 1], and then at origin this map has image [0 : 1]. Example. A slightly more subtle case is V = fy2 = x2(x + 1)g. A similar argument shows t = Y/X is integral. However, in this case, the rational function cannot be extended over origin to give a continuous function on V. Instead, we have limits ±1 as we approach the origin. In particular, it is bounded near 0. Example. We first slightly generalize the definition to include non-integral domains. Instead of working in fraction field, we work in the total ring of fractions, i.e. invert those elements that are not zero-divisors. Now consider the cross defined by XY = 0, with coordinate ring A = k[X, Y]/(XY) = f(f, g) 2 k[X] × k[Y]: f(0) = g(0)g (should be easy to verify). We can find a rational function, for example f = X/(X + Y), that takes value 0 on the y-axis, and 1 on the x-axis (in particular, it stays bounded near 0). This rational function is clearly in the integral closure: f2 = X2/(X2 + Y2), i.e. it defines the same equation as f: f2 - f = 0. Or similarly, if f = X2/(X + Y), then f2 -(X + Y)f = 0, where f takes X when Y = 0, and 0 otherwise (it extends continuously over 0). Generalize this idea, we see the integral closure is actually k[X] + k[Y], which corresponds to the disjoint union of two lines. 3 Deeper theory 3.1 Normalization The fact is that taking integral closure always normalizes a variety. Definition. A variety is normal if the coordinate ring localizing at every point is integrally closed. Example. A line with coordinate ring k[t] is normal, since k[t](t) is obviously integrally closed (its field of fraction is also k(t) which is already integral over the UFD k[t]). 2 3 The cuspidal cubic has local ring k[x, y]/(y - x )(x,y) where, again, y/x 2 k(x, y) is not integral. Definition. A normalization of a variety X is a normal variety X˜ with a map X˜ X such that if Y X is a dominant map from a normal variety, then it factors uniquely through X˜. ! ! Uniqueness follows from the universal property. Here we prove the existence. Proposition. The normalization of a variety X with coordinate ring A is V(A). Proof. Inclusion of rings gives the map on varieties. Now suppose given f : A C, and let s 2 A be any element. Since s integral over A, A[s] can be generated by r1, ::: , rn as an A-module. Then it is fairly 0 easy to set f : A C by sending s = λiri to λif(ri). ! This result can be extended to schemes.P A schemeP X is normal if all its localizations are integrally closed domains. If!X is an integral scheme, then for each affine open U = Spec A, we can form U = Spec A. Then glue them together to get a normal variety X˜. - 2 - Formally we just proved to “take integral closure” is to “normalize” the variety, although this gives no extra geometric information at all... 3.2 Normal variety Let us build some intuition about normal varieties. Definition. A variety is regular at a point p if the localization at the maximal ideal m of the point p is a 2 regular local ring, i.e. dimkm/m = dim A. You can check this is equivalent to the usual definition we see. We state without proof the following fact: Theorem. (Serre’s criterion for normality) A Noetherian ring A is normal iff 1. Ap is a regular local ring for all prime p of height ≤1; and 2. depth Ap ≥ inf f2, ht(p)g for all prime p. Proposition. Smooth varieties are normal. Proof. (Sketch) Smooth varieties have regular coordinate rings. STP regular rings are normal. This follows from the fact that local regular rings are integral domains and Cohen-Macaulay, hence satisfy the theorem above. For more detail, see Bruns and Herzog. Another proof is to use that the local ring of a smooth point is a UFD, which, again, is not so trivial. Proposition. Normal varieties have singular points of codimension at least 2. Proof. Let X be the variety, and S its singular locus. We know S is closed in X. Now suppose S contains a component Y of dimension n - 1. We claim that there exists an affine open X0 ⊂ X such that X0 \ Y 6= Æ and the ideal of Y 0 = Y \ X0 in k[X0] is principal. Assuming this, there is at least one point y0 2 Y 0 that is non-singular as a point in Y 0, but is singular in X0. Now combine these two pieces together, we get something extraordinary. Theorem. For affine curves, smooth is the same as normal. This explains our intuition in the previous examples, where normalization of a curve is to simply “unwind” the singularity. To take integral closure always corresponds to normalize the variety. However, it is not necessarily the case we always resolute the singularities in higher dimensions. 3.3 Analytic properties of normalization of curves In complex analysis, we know there is this Riemann extension theorem, which says if a bounded meromorphic function is holomorphic at the complement of one point, then it is actually holomorphic on that point (and thus the whole open set). We would like something similar to hold in algebraic geometry. However the examples in section 1 provide some non-examples. In the case fy2 = x3g, consider the function f = y/x. This is regular away from 0, and the only sensible thing to do is to set f(0) = 0. However, this does not make f regular. By considering another representation f = a(x, y)/b(x, y), we quickly obtain a contradiction. This prompts us to ask (following ideas from [2]): what would be the criterion for a rational function to be integral? For notation, let our complex affine irreducible curve be C, with the unique singular point at origin, 0 and let C0 = C n 0. And let integral closure we have constructed be n : C C, with C = C n n-1(0).
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