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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 $\text{depth}(M_{\mathfrak p}) = 0$. 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. 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 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 $\mathfrak q \supset \mathfrak p$ with $t \not\in \mathfrak q$. Since theses primes have smaller height 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). $\square$

Comments (4)

Comment #71 by hoxide on

In the proof of "lemma-criterion-no-embedded-primes", the relation between and seems should be

Comment #4258 by Antoine Chambert-Loir on

In the definition of , the convention that the dimension of the zero module can be made useless by requiring that the prime ideal 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 and hence the definition of works as written no matter what your convention for 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 rather than . Do you?

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