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Home , Regular local ring, Total ring of fractions

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 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) = {f ∈ k[x1, ... , xn]: f(x) = 0 ∀x ∈ S}. Definition. For an A and its fraction field K, all integral elements, i.e. those x ∈ 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-).

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 = {y2 = x3} ⊂ 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 \{0} 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 = {y2 = x2(x + 1)}. 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, g) ∈ k[X] × k[Y]: f(0) = g(0)} (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 k[x, y]/(y − x )(x,y) where, again, y/x ∈ 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 ∈ 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 m of the point p is a 2 , 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 A is normal iff

1. Ap is a regular local ring for all prime p of height ≤1; and

2. depth Ap ≥ inf {2, ht(p)} 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= ∅ and the ideal of Y 0 = Y ∩ X0 in k[X0] is principal. Assuming this, there is at least one point y0 ∈ 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 . However the examples in section 1 provide some non-examples. In the case {y2 = x3}, 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 \ 0. And let integral closure we have constructed be n : C C, with C = C \ n−1(0). Analytically, we have this Riemann extension induced isomorphism:

0 → {holomorphic functions on C} =∼ {holomorphic functions on C that is bounded near n−1(0)} =∼ {holomorphic functions on C0 that is bounded near 0}

- 3 - 0 where the second isomorphism is induced the isomorphism between C0 and C . Analogously we would suspect that algebraically:

{regular functions on C} =∼ ? {regular functions on C0 that is bounded near 0}

Same with the first case, the RHS should be interpreted as rationals function that are regular on C0 and is bounded near 0. But how should we interpret bounded in the language of algebra, i.e. not of analysis? Let notations be as above, we know the map n is birational, thus the function fields of C and C are isomorphic. 0 If f is a regular function on C and is not bounded near 0, since it is, in particular, a rational function on C, it must have a pole at 0. Looking at the Laurent series around 0, we thus get a homomorphism:

{rational functions on C} =∼ {rational functions on C} {Laurent series around 0}

and a rational function is bounded iff its Laurent series has no→ pole near 0. Denote the ring of formal Laurent series L, and those with no poles Ls. Now what we want is that any “no pole Laurent series” extends to a regular function on C; i.e. we attempt to make the following equivalent definition of normalization: Definition. (*) Let A be the ring of regular functions on an affine curve with a single singularity at origin, and K the field of rational functions. The normalization of A in K is the unique largest subring A0 such 0 that the homomorphim φ : A Ls extends to φ : A Ls.

More generally, Ls is the discrete→ of→L, so we can replace Ls by any DVR R. And if we replace the single singularity with a set of singularities, then every homomorphism A R must extend 0 to A Ls. Let’s prove the equivalence in the easy setting with only one singularity at origin. → Proof. →(usual definition) (definition*): Pick any t satisfying a monic polynomial ⇒ n n−1 t + an−1t + ··· + a0 = 0, with ai ∈ A.

Let φ : A Ls be the Laurent series. Now if the Laurent series of t has a pole of order r > 0, then the first term tn has a pole of order nr, while the rest have a pole at most (n − 1)r. This is impossible. So φ extends to→ include t. (definition*) (definition): note that φ extends automatically to a partially defined homomorphism Φ : K Frac (Ls) by a1/a2 7 φ(a1)/φ(a2) whenever φ(a2) 6= 0. So in definition*, the normalization 0 −1 is the ring A =⇒Φ (Ls) (in the more general setting, the intersection of all such preimages, while 0 ranging→ all homomorphism A→ R). So take any element t = s1/s2 ∈ A , then the order of 0 of φ(s1) is larger or equal to φ(s2). This should remind us of the→ property that A = ∩V for all valuation rings A ⊂ V ⊂ K. Now we need to show t ∈ V. Let’s suppose t is not in V. By construction, we know V is in a maximal pair (V, f) for f : V C homomorphism. If t 6∈ V, we can extend V to V[t] C by sending t t(0) ∈ C, thus contradicting the maximality. → → → References

[1] Atiyah, Macdonald. Introduction to . Addison-Wesley; 1969. [2] Kollr J. Lectures on Resolution of Singularities. Annals of Mathematics Studies, Number 166. Princeton and Oxford: Princeton University Press; 2007. Pages 20-25.

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