10.157 Serre's criterion for normality
We introduce the following properties of Noetherian rings.
Definition 10.157.1. Let R be a Noetherian ring. Let k \geq 0 be an integer.
We say R has property (R_ k) if for every prime \mathfrak p of height \leq k the local ring R_{\mathfrak p} is regular. We also say that R is regular in codimension \leq k.
We say R has property (S_ k) if for every prime \mathfrak p the local ring R_{\mathfrak p} has depth at least \min \{ k, \dim (R_{\mathfrak p})\} .
Let M be a finite R-module. We say M has property (S_ k) if for every prime \mathfrak p the module M_{\mathfrak p} has depth at least \min \{ k, \dim (\text{Supp}(M_{\mathfrak p}))\} .
Any Noetherian ring has property (S_0) and so does any finite module over it. Our convention that the depth of the zero module is \infty (see Section 10.72) and the dimension of the empty set is -\infty (see Topology, Section 5.10) guarantees that the zero module has property (S_ k) for all k.
Lemma 10.157.2. Let R be a Noetherian ring. Let M be a finite R-module. The following are equivalent:
M has no embedded associated prime, and
M has property (S_1).
Proof.
Let \mathfrak p be an embedded associated prime of M. Then there exists another associated prime \mathfrak q of M such that \mathfrak p \supset \mathfrak q. In particular this implies that \dim (\text{Supp}(M_{\mathfrak p})) \geq 1 (since \mathfrak q is in the support as well). On the other hand \mathfrak pR_{\mathfrak p} is associated to M_{\mathfrak p} (Lemma 10.63.15) and hence \text{depth}(M_{\mathfrak p}) = 0 (see Lemma 10.63.18). In other words (S_1) does not hold. Conversely, if (S_1) does not hold then there exists a prime \mathfrak p such that \dim (\text{Supp}(M_{\mathfrak p})) \geq 1 and \text{depth}(M_{\mathfrak p}) = 0. Since \text{depth}(M_{\mathfrak p}) = 0, we see that \mathfrak p \in \text{Ass}(M) by the two Lemmas 10.63.15 and 10.63.18. Since \dim (\text{Supp}(M_{\mathfrak p})) \geq 1, there is a prime \mathfrak q \in \text{Supp}(M) with \mathfrak q \subset \mathfrak p, \mathfrak q \not= \mathfrak p. We can take such a \mathfrak q that is minimal in \text{Supp}(M). Then by Proposition 10.63.6 we have \mathfrak q \in \text{Ass}(M) and hence \mathfrak p is an embedded associated prime.
\square
Lemma 10.157.3.slogan Let R be a Noetherian ring. The following are equivalent:
R is reduced, and
R has properties (R_0) and (S_1).
Proof.
Suppose that R is reduced. Then R_{\mathfrak p} is a field for every minimal prime \mathfrak p of R, according to Lemma 10.25.1. Hence we have (R_0). Let \mathfrak p be a prime of height \geq 1. Then A = R_{\mathfrak p} is a reduced local ring of dimension \geq 1. Hence its maximal ideal \mathfrak m is not an associated prime since this would mean there exists an x \in \mathfrak m with annihilator \mathfrak m so x^2 = 0. Hence the depth of A = R_{\mathfrak p} is at least one, by Lemma 10.63.9. This shows that (S_1) holds.
Conversely, assume that R satisfies (R_0) and (S_1). If \mathfrak p is a minimal prime of R, then R_{\mathfrak p} is a field by (R_0), and hence is reduced. If \mathfrak p is not minimal, then we see that R_{\mathfrak p} has depth \geq 1 by (S_1) and we conclude there exists an element t \in \mathfrak pR_{\mathfrak p} such that R_{\mathfrak p} \to R_{\mathfrak p}[1/t] is injective. Now R_\mathfrak p[1/t] is contained in the product of its localizations at prime ideals, see Lemma 10.23.1. This implies that R_{\mathfrak p} is a subring of a product of localizations of R at \mathfrak p \supset \mathfrak q with t \not\in \mathfrak q. Since these primes have smaller height by induction on the height we conclude that R is reduced.
\square
Lemma 10.157.4 (Serre's criterion for normality).referenceslogan Let R be a Noetherian ring. The following are equivalent:
R is a normal ring, and
R has properties (R_1) and (S_2).
Proof.
Proof of (1) \Rightarrow (2). Assume R is normal, i.e., all localizations R_{\mathfrak p} at primes are normal domains. In particular we see that R has (R_0) and (S_1) by Lemma 10.157.3. Hence it suffices to show that a local Noetherian normal domain R of dimension d has depth \geq \min (2, d) and is regular if d = 1. The assertion if d = 1 follows from Lemma 10.119.7.
Let R be a local Noetherian normal domain with maximal ideal \mathfrak m and dimension d \geq 2. Apply Lemma 10.119.2 to R. It is clear that R does not fall into cases (1) or (2) of the lemma. Let R \to R' as in (4) of the lemma. Since R is a domain we have R \subset R'. Since \mathfrak m is not an associated prime of R' there exists an x \in \mathfrak m which is a nonzerodivisor on R'. Then R_ x = R'_ x so R and R' are domains with the same fraction field. But finiteness of R \subset R' implies every element of R' is integral over R (Lemma 10.36.3) and we conclude that R = R' as R is normal. This means (4) does not happen. Thus we get the remaining possibility (3), i.e., \text{depth}(R) \geq 2 as desired.
Proof of (2) \Rightarrow (1). Assume R satisfies (R_1) and (S_2). By Lemma 10.157.3 we conclude that R is reduced. Hence it suffices to show that if R is a reduced local Noetherian ring of dimension d satisfying (S_2) and (R_1) then R is a normal domain. If d = 0, the result is clear. If d = 1, then the result follows from Lemma 10.119.7.
Let R be a reduced local Noetherian ring with maximal ideal \mathfrak m and dimension d \geq 2 which satisfies (R_1) and (S_2). By Lemma 10.37.16 it suffices to show that R is integrally closed in its total ring of fractions Q(R). Pick x \in Q(R) which is integral over R. Then R' = R[x] is a finite ring extension of R (Lemma 10.36.5). Because \dim (R_\mathfrak p) < d for every nonmaximal prime \mathfrak p \subset R we have R_\mathfrak p = R'_\mathfrak p by induction. Hence the support of R'/R is \{ \mathfrak m\} . It follows that R'/R is annihilated by a power of \mathfrak m (Lemma 10.62.4). By Lemma 10.119.2 this contradicts the assumption that the depth of R is \geq 2 = \min (2, d) and the proof is complete.
\square
Lemma 10.157.5. A regular ring is normal.
Proof.
Let R be a regular ring. By Lemma 10.157.4 it suffices to prove that R is (R_1) and (S_2). As a regular local ring is Cohen-Macaulay, see Lemma 10.106.3, it is clear that R is (S_2). Property (R_1) is immediate.
\square
Lemma 10.157.6. Let R be a Noetherian normal domain with fraction field K. Then
for any nonzero a \in R the quotient R/aR has no embedded primes, and all its associated primes have height 1
R = \bigcap \nolimits _{\text{height}(\mathfrak p) = 1} R_{\mathfrak p}
For any nonzero x \in K the quotient R/(R \cap xR) has no embedded primes, and all its associates primes have height 1.
Proof.
By Lemma 10.157.4 we see that R has (S_2). Hence for any nonzero element a \in R we see that R/aR has (S_1) (use Lemma 10.72.6 for example) Hence R/aR has no embedded primes (Lemma 10.157.2). We conclude the associated primes of R/aR are exactly the minimal primes \mathfrak p over (a), which have height 1 as a is not zero (Lemma 10.60.11). This proves (1).
Thus, given b \in R we have b \in aR if and only if b \in aR_{\mathfrak p} for every minimal prime \mathfrak p over (a) (see Lemma 10.63.19). These primes all have height 1 as seen above so b/a \in R if and only if b/a \in R_{\mathfrak p} for all height 1 primes. Hence (2) holds.
For (3) write x = a/b. Let \mathfrak p_1, \ldots , \mathfrak p_ r be the minimal primes over (ab). These all have height 1 by the above. Then we see that R \cap xR = \bigcap _{i = 1, \ldots , r} (R \cap xR_{\mathfrak p_ i}) by part (2) of the lemma. Hence R/(R \cap xR) is a submodule of \bigoplus R/(R \cap xR_{\mathfrak p_ i}). As R_{\mathfrak p_ i} is a discrete valuation ring (by property (R_1) for the Noetherian normal domain R, see Lemma 10.157.4) we have xR_{\mathfrak p_ i} = \mathfrak p_ i^{e_ i}R_{\mathfrak p_ i} for some e_ i \in \mathbf{Z}. Hence the direct sum is equal to \bigoplus _{e_ i > 0} R/\mathfrak p_ i^{(e_ i)}, see Definition 10.64.1. By Lemma 10.64.2 the only associated prime of the module R/\mathfrak p^{(n)} is \mathfrak p. Hence the set of associate primes of R/(R \cap xR) is a subset of \{ \mathfrak p_ i\} and there are no inclusion relations among them. This proves (3).
\square
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