## 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.

1. 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 $\leq k$.

2. We say $R$ has property $(S_ k)$ if for every prime $\mathfrak p$ the local ring $R_{\mathfrak p}$ has depth at least $\min \{ k, \dim (R_{\mathfrak p})\}$.

3. 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:

1. $M$ has no embedded associated prime, and

2. $M$ 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 $\text{depth}(M_{\mathfrak p}) = 0$. Then we see (arguing backwards using the lemmas cited above) that $\mathfrak p$ is an embedded associated prime. $\square$

Lemma 10.157.3. Let $R$ be a Noetherian ring. The following are equivalent:

1. $R$ is reduced, and

2. $R$ 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 a $x \in \mathfrak m$ 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. This implies that $R_{\mathfrak p}$ is a subring of localizations of $R$ at primes of smaller height. Thus by induction on the height we conclude that $R$ is reduced. $\square$

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. $\square$

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

1. for any nonzero $a \in R$ the quotient $R/aR$ has no embedded primes, and all its associated primes have height $1$

2. $R = \bigcap \nolimits _{\text{height}(\mathfrak p) = 1} R_{\mathfrak p}$
3. 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). $\square$

Comment #71 by hoxide on

In the proof of "lemma-criterion-no-embedded-primes", the relation between $\mathfrak p$ and $\mathfrak q$ seems should be $\mathfrak p \supset \mathfrak q$

Comment #4258 by Antoine Chambert-Loir on

In the definition of $S_k$, the convention that the dimension of the zero module can be made useless by requiring that the prime ideal $\mathfrak p$ belongs to the support of the module.

Comment #4428 by on

OK, the comment on the dimension of the empty set was there because in a previous iteration we had a different convention on the depth of the zero module. Currently the depth of the zero module is $+\infty$ and hence the definition of $(S_k)$ works as written no matter what your convention for $\dim(\emptyset)$ is. I changed the discussion following the definition slightly so that this becomes more clear, see here.

Of course, I don't know if it would be better to define $\dim(\emptyset) = -1$ rather than $\dim(\emptyset) = -\infty$. Do you?

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