Regular Local Rings

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Regular Local Rings

Recalling §System of Parameters, ⁋Proposition–Definition 3, it is obvious that in a regular local ring $(A, \mathfrak{m})$, elements $a_1,\ldots, a_d$ with $d=\dim A$ generating $\mathfrak{m}$ form a system of parameters of $A$. We call this a regular system of parameters.

Corollary 1 A regular local ring is an integral domain.

Proof

We prove this by induction on the dimension of $(A, \mathfrak{m})$. The case $d=0$ is trivial since $A$ is a field. Assume the claim holds for dimension $\dim A=d$, and consider the case $\dim A=d+1$. Then in particular $\mathfrak{m}\neq 0$, so by §Integral Extensions, ⁋Lemma 8 (Nakayama) we know that $\mathfrak{m}\neq \mathfrak{m}^2$. On the other hand, by §Associated Primes, ⁋Theorem 7, the minimal prime ideals of $A$ are finite in number. Let these be $\mathfrak{p}_1,\ldots, \mathfrak{p}_k$. If

\[\mathfrak{m}\subseteq \mathfrak{m}^2\cup \mathfrak{p}_1\cup\cdots\cup \mathfrak{p}_k\]

then by §Associated Primes, ⁋Lemma 2 (Prime avoidance lemma) and the above fact that $\mathfrak{m}\neq \mathfrak{m}^2$, we would have $\mathfrak{m}=\mathfrak{p}_i$ for some $i$, which would imply

\[d+1=\dim A=\codim \mathfrak{m}=\codim \mathfrak{p}_i=0\]

a contradiction. Therefore there must exist some $a\in \mathfrak{m}$ with $a\not\in \mathfrak{m}^2\cup \mathfrak{p}_1\cup\cdots\cup \mathfrak{p}_k$.

Now let $A'=A/(a)$, and consider the maximal ideal $\mathfrak{m}'=\mathfrak{m}A'$ of $A'$. By the choice of $a$, none of the prime ideals of $A'$ corresponds to any $\mathfrak{p}_i$, so necessarily $\dim A'<\dim A$; combining this with §System of Parameters, ⁋Corollary 7 we see that $\dim A'=d-1$. Thus from the identity

\[\mathfrak{m}'/(\mathfrak{m}')^2=\mathfrak{m}/(\mathfrak{m}^2+(a))\]

and §Integral Extensions, ⁋Lemma 8 (Nakayama), we know that $\mathfrak{m}'$ is generated by $(d-1)$ elements, and hence by the induction hypothesis $A'$ is an integral domain. That is, $(a)$ is a prime ideal, and therefore for some $i$ we have $\mathfrak{p}_i\subsetneq (a)$.

Now for any $x\in \mathfrak{p}_i$, choose $\alpha\in A$ such that $x=\alpha a$. Since $a\not\in \mathfrak{p}_i$, we have $\alpha\in \mathfrak{p}_i$, and thus $\mathfrak{p}_i=a \mathfrak{p}_i$, from which $\mathfrak{p}_i=\mathfrak{m}\mathfrak{p}_i$. Applying §Integral Extensions, ⁋Lemma 8 (Nakayama) again gives $\mathfrak{p}_i=0$, so $A$ is an integral domain.

Since we will use this corollary frequently in what follows, we introduce a new definition.

Definition 2 Elements $a_1,\ldots, a_d$ of a ring $A$ are called an $A$-regular sequence, or simply an $A$-sequence, if $(a_1,\ldots, a_d)$ is proper and for each $i$, the image of $a_{i+1}$ is a non-zerodivisor in $A/(a_1,\ldots, a_i)$.

Corollary 3 A regular system of parameters in a regular local Noetherian ring forms an $A$-sequence.

Proof

For each $i$, the quotient $A/(a_1,\ldots, a_i)$ is also a regular local ring, and by Corollary 1 it is an integral domain, so $x_{i+1}$ becomes a nonzero element of this ring.

Proposition 4 For a regular local Noetherian ring $(A, \mathfrak{m})$ and a prime ideal $\mathfrak{p}$ thereof,

\[\dim A/\mathfrak{p} + \codim \mathfrak{p} = \dim A\]

holds.

Proof

As we saw in §Dimension, in general we have $\dim A/\mathfrak{p} + \codim \mathfrak{p} \le \dim A$. Thus it suffices to prove the reverse inequality.

Let $d = \dim A$ and $h = \codim \mathfrak{p}$. By Corollary 3, a regular system of parameters $x_1, \ldots, x_d$ is an $A$-sequence contained in $\mathfrak{m}$. In general, when a $d$-dimensional Noetherian local ring $(A,\mathfrak{m})$ has an $A$-sequence of $d$ elements inside $\mathfrak{m}$, we call $A$ a Cohen–Macaulay local ring; thus every regular local ring is a Cohen–Macaulay local ring. Therefore it suffices to prove the following two claims by induction on $d$.

Any associated prime $\mathfrak{q}\in \Ass A$ of a $d$-dimensional Cohen–Macaulay local ring $A$ satisfies $\dim A/\mathfrak{q}=d$.
Any prime ideal $\mathfrak{p}$ of a $d$-dimensional Cohen–Macaulay local ring $A$ satisfies $\dim A/\mathfrak{p}+\codim \mathfrak{p}=d$.

The case $d=0$ is trivial for both claims, so assume $d\geq 1$ and that both claims hold for $(d-1)$-dimensional Cohen–Macaulay local rings.

To prove the first claim, let $\mathfrak{q}=\ann(a)\in\Ass A$. Since $x_1$ is a non-zerodivisor, by §Associated Primes, ⁋Theorem 7 we have $x_1\not\in \mathfrak{q}$. We may assume that $a\not\in x_1A$. Indeed, if $a=x_1b$, then since $x_1$ is a non-zerodivisor we have $\ann(b)=\ann(a)=\mathfrak{q}$, and $Aa\subsetneq Ab$. If $Aa=Ab$, then from $b=ca=cx_1b$ we get $(1-cx_1)b=0$; since $x_1\in\mathfrak{m}$, the element $1-cx_1$ is a unit and hence $b=0$, i.e. $\mathfrak{q}=A$, a contradiction. Therefore the process of replacing $a$ by $b$ terminates in finitely many steps by the ascending chain condition.

Now let $\overline{A}=A/x_1A$. Then by definition $\overline{A}$ contains an $\overline{A}$-sequence $\overline{x}_2,\ldots, \overline{x}_d$ in its maximal ideal, and $\dim \overline{A}=d-1$. Indeed, by §System of Parameters, ⁋Corollary 7 we have $\dim \overline{A}\geq d-1$, and since the non-zerodivisor $x_1$ belongs to no minimal prime ideal of $A$ by §Associated Primes, ⁋Theorem 7, any prime ideal $\mathfrak{q}_0$ containing $x_1$ satisfies $\codim \mathfrak{q}_0\geq 1$, hence $\dim A/\mathfrak{q}_0\leq d-1$. In particular, $\overline{A}$ is a $(d-1)$-dimensional Cohen–Macaulay local ring.

Since $a\not\in x_1A$, the element $\bar{a}\in \overline{A}$ is nonzero and $\ann_{\overline{A}}(\bar a)\supseteq (\mathfrak{q}+x_1A)/x_1A$. By §Associated Primes, ⁋Theorem 7, the set $\Ass(\overline{A}\bar a)$ is nonempty and its elements contain $\ann_{\overline{A}}(\bar a)$, and by §Associated Primes, ⁋Lemma 5 we have $\Ass(\overline{A}\bar{a})\subseteq \Ass \overline{A}$. Therefore there exists some $\overline{\mathfrak{q}}'\in \Ass\overline{A}$ whose preimage $\mathfrak{q}'\subseteq A$ satisfies $\mathfrak{q}\subsetneq \mathfrak{q}'$ (since $x_1\in\mathfrak{q}'\setminus\mathfrak{q}$). By the first claim of the induction hypothesis, $\dim \overline{A}/\overline{\mathfrak{q}}'=d-1$, so

\[\dim A/\mathfrak{q}\geq 1+\dim A/\mathfrak{q}'=1+\dim \overline{A}/\overline{\mathfrak{q}}'=d\]

and the reverse inequality is obvious, yielding the first claim.

For the second claim, if $h=0$ then $\mathfrak{p}$ is a minimal prime ideal, and by §Associated Primes, ⁋Theorem 7 we have $\mathfrak{p}\in\Ass A$, so the first claim gives $\dim A/\mathfrak{p}=d$. Now assume $h\geq 1$. If $\mathfrak{p}$ were contained in some $\mathfrak{q}\in \Ass A$, then from the first claim and the general inequality we would get $\codim \mathfrak{q}=0$, but then $\mathfrak{p}\subseteq \mathfrak{q}$ implies $h=0$, a contradiction. Since $\Ass A$ is finite (§Associated Primes, ⁋Theorem 7), by prime avoidance (§Associated Primes, ⁋Lemma 2 (Prime avoidance lemma)) we can choose $x\in \mathfrak{p}$ not belonging to any associated prime; in particular, $x$ is a non-zerodivisor.

We now use the fact that $\overline{A}=A/xA$ is again a $(d-1)$-dimensional Cohen–Macaulay local ring. This is a standard result from depth theory — quotienting by a non-zerodivisor reduces both depth and dimension by exactly $1$ — and since we have not yet covered depth theory on this blog, we defer its proof to [Stacks, Lemma 10.104.2]. Applying the second claim of the induction hypothesis to $\overline{\mathfrak{p}}=\mathfrak{p}/xA$ gives

\[\dim \overline{A}/\overline{\mathfrak{p}}+\codim \overline{\mathfrak{p}}=d-1\]

But $\overline{A}/\overline{\mathfrak{p}}\cong A/\mathfrak{p}$, and $\codim \overline{\mathfrak{p}}=\dim \overline{A}_{\overline{\mathfrak{p}}}=\dim A_\mathfrak{p}/xA_\mathfrak{p}$. On the other hand, $x$ is also a non-zerodivisor in $A_\mathfrak{p}$: if $(x/1)(b/s)=0$, then for some $u\not\in\mathfrak{p}$ we have $uxb=0$, and since $x$ is a non-zerodivisor we get $ub=0$, i.e. $b/s=0$. Then since $x/1\in \mathfrak{p}A_\mathfrak{p}$, applying the same argument as in the proof of $\dim\overline{A}=d-1$ to the local ring $A_\mathfrak{p}$ yields $\dim A_\mathfrak{p}/xA_\mathfrak{p}=\dim A_\mathfrak{p}-1=h-1$. Putting everything together,

\[\dim A/\mathfrak{p}+(h-1)=d-1\]

and we obtain the desired result.

Proposition 5 Let $(A, \mathfrak{m})$ be a complete regular local Noetherian ring of dimension $d$, and let $\kappa=A/\mathfrak{m}$ be its residue field. If $A$ contains a field, then $A\cong \kappa[[\x_1,\ldots, \x_d]]$, and this isomorphism identifies each variable $\x_i$ with an element of a regular system of parameters of $A$.

Proof

By §Properties of Completion, ⁋Theorem 8 (Cohen structure theorem), the given hypothesis implies that $A$ must contain $\kappa$. Now by the first part of §Properties of Completion, ⁋Theorem 5, we obtain a $\kappa$-algebra homomorphism $\phi:\kappa[[\x_1,\ldots, \x_d]]\rightarrow A$, and by the second part $\phi$ is surjective. On the other hand, $\kappa[[\x_1,\ldots, \x_d]]$ has dimension $d$ by Corollary 1, so

\[d=\dim A=\dim \im(\phi)=\dim \kappa[[\x_1,\ldots,\x_d]]/\ker\phi\leq \dim \kappa[[\x_1,\ldots, \x_d]]-\codim \ker\phi=d-\codim\ker\phi\]

and for this inequality to hold we must have $\codim\ker\phi=0$. But $\kappa[[\x_1,\ldots, \x_d]]$ is an integral domain by Corollary 1, so this means $\ker\phi=0$.

Discrete Valuation Rings

We now examine a one-dimensional regular local ring $(A,\mathfrak{m})$. By definition, $\mathfrak{m}$ must be generated by a single element $m$, which we call a regular parameter or uniformizing parameter of $A$.

Proposition 6 Let $(A, \mathfrak{m})$ be a one-dimensional regular local ring, and let $m$ be a regular parameter of $A$. Then any element $x$ of $\Frac(A)$ can be written uniquely in the form

\[x=a m^k\qquad \text{$k\in \mathbb{Z}$, $a$ a unit of $A$}\]

Proof

First, $A$ is an integral domain by Corollary 1. Now by §Blowup Algebras, ⁋Corollary 8 (Krull intersection theorem), we have $\bigcap \mathfrak{m}^i=0$, so for any nonzero $x\in A$ there are only finitely many indices $i$ with $x\in \mathfrak{m}^i$. Let $k$ be the largest of these; then from $x\in \mathfrak{m}^k=(m^k)$ there exists $a\in A$ such that $x=am^k$. By the maximality of $k$, the element $a$ is a unit of $A$.

Now given any element $x$ of $\Frac(A)$, write $x=x_1/x_2$. By the above argument,

\[x=\frac{x_1}{x_2}=\frac{a_1m^{k_1}}{a_2m^{k_2}}=a_1a_2^{-1}m^{k_1-k_2}=am^k\]

where $a=a_1a_2^{-1}$ is a unit, and the uniqueness of this expression is almost obvious.

From the uniqueness of the expression proved above, we can define a group homomorphism from the multiplicative group $\Frac(A)^\times$ to $\mathbb{Z}$ by

\[\nu:\Frac(A)^\times \rightarrow \mathbb{Z};\qquad am^k\mapsto k\]

More generally, we define the following.

Definition 7 For an integral domain $A$ and a totally ordered abelian group $G$, a group homomorphism $\nu:\Frac(A)^\times \rightarrow G$ satisfying the inequality

\[\nu(x+y)\geq \min(\nu(x), \nu(y))\]

is called a valuation. For a valuation $\nu$, the ring

\[S=\nu^{-1}\left(\{g\in G\mid g\geq 0\}\right)\]

is called the valuation ring of $\nu$.

In particular, when $G=\mathbb{Z}$ we call this a discrete valuation, and the valuation ring of $\nu$ is called a discrete valuation ring.

That the $\nu:\Frac(A)^\times \rightarrow \mathbb{Z}$ defined above is a discrete valuation follows from the identity

\[am^k+bm^l=(am^{k-\min(k,l)}+bm^{l-\min(k,l)})m^{\min(k,l)}\]

Then by Proposition 5, we know that if two complete discrete valuation rings each contain a field and have isomorphic residue fields, then they are isomorphic. However, in general no such classification exists among discrete valuation rings that are not complete.

Serre’s Normality Criterion

For convenience, for a non-zerodivisor $u$ in a ring $A$, we shall call an associated prime ideal $\mathfrak{p}$ of $A/(u)$ associated to a non-zerodivisor $u$. This is the same exception as in §Associated Primes, ⁋Definition 1.

Proposition 8 Let $A$ be a reduced Noetherian ring and let $K$ be its total ring of fractions. Then an element $x\in K$ belongs to $A$ if and only if its image in $K_\mathfrak{p}$ belongs to $A_\mathfrak{p}$ for every prime ideal $\mathfrak{p}$ associated to a non-zerodivisor.

Proof

By definition, elements of $K$ are of the form $a/u$ for arbitrary $a\in A$ and non-zerodivisor $u\in A$. Now

\[\frac{a}{u}\in A\iff a\in (u)\iff a=0\mod{(u)}\iff \epsilon_\mathfrak{p}(a)= 0\text{ in $(A/(u))_\mathfrak{p}=A_\mathfrak{p}/(u)A_\mathfrak{p}$ for all $\mathfrak{p}$ associated prime of $A/(u)$}\]

holds. Here $\epsilon_\mathfrak{p}: A \rightarrow A_\mathfrak{p}$ is the canonical morphism, and the last equivalence is by §Associated Primes, ⁋Corollary 4. Then for any prime ideal $\mathfrak{p}$ associated to a non-zerodivisor,

\[\epsilon_\mathfrak{p}(a)\in(u)A_\mathfrak{p}\]

On the other hand, since $A$ is reduced, $K$ is a finite direct product of fields (§Associated Primes, ⁋Corollary 8), and since localization commutes with finite direct products we may identify the total ring of fractions of $A_\mathfrak{p}$ with $K_\mathfrak{p}$. Re-examining the above inclusion through this identification yields the desired result.

From this we can prove the following.

Theorem 9 A Noetherian integral domain $A$ is a normal domain if and only if the following condition holds.

($\ast$) For any prime ideal $\mathfrak{p}$ associated to a principal ideal, $\mathfrak{p}A_\mathfrak{p}$ is a principal ideal of $A_\mathfrak{p}$.

Proof

First, assuming ($\ast$), we show that $A$ is a normal domain. Now it is obvious that if normal domains with a common quotient field are given, their intersection is again a normal domain. Thus it suffices to show the identity

\[A=\bigcap_\text{\scriptsize$\mathfrak{p}$ associated to a principal ideal}A_\mathfrak{p}\]

where each $A_\mathfrak{p}$ is viewed as a subset of the quotient field $K$ of $A$. The claim we wish to prove was treated in the more general case in Proposition 8.

Conversely, assume $A$ is a normal domain, and let $\mathfrak{p}$ be an associated prime of a principal ideal $\mathfrak{a}=(a)$. That is,

\[\mathfrak{p}=\ann(b+\mathfrak{a})\]

and we must show that $\mathfrak{p}A_\mathfrak{p}$ is a principal ideal of $A_\mathfrak{p}$. Since this is a matter of localization, we may assume that $(A,\mathfrak{p})$ is a local ring; let $K$ be the field of fractions of $A$, and consider the inverse of $\mathfrak{p}$

\[\mathfrak{p}^{-1}=\{x\in K\mid x \mathfrak{p}\subseteq A\}\]

Then $\mathfrak{p}^{-1}\mathfrak{p}$ is an ideal between $\mathfrak{p}$ and $A$. By the maximality of $\mathfrak{p}$, we must have either $\mathfrak{p}^{-1}\mathfrak{p}=\mathfrak{p}$ or $\mathfrak{p}^{-1}\mathfrak{p}=A$. But if $\mathfrak{p}^{-1}\mathfrak{p}=\mathfrak{p}$, then by §Integral Extensions, ⁋Lemma 5 every element of $\mathfrak{p}^{-1}$ is integral, and hence $\mathfrak{p}^{-1}\subseteq A$. But since $\mathfrak{p}b\subseteq (a)$, we have $b/a\in \mathfrak{p}^{-1}$, which implies $b\in (a)$, a contradiction.

Therefore we must have $\mathfrak{p}\mathfrak{p}^{-1}=A$. Moreover, since $(A, \mathfrak{p})$ is local, combining these two conditions we know that there must exist some $x\in \mathfrak{p}^{-1}$ with $x \mathfrak{p}=A$. Hence $\mathfrak{p}=A x^{-1}$ is principal.

This theorem admits a further generalization. First we define the following.

Definition 10 Fix a ring $A$ and its total ring of fractions $K$.

A ring $A$ is called normal if $A$ is reduced and integrally closed in $K$.
The normalization of a reduced ring $A$ is defined as the integral closure of $A$ in $K$.

Then the following holds.

Theorem 11 (Serre) A Noetherian ring $A$ is a (finite) direct product of normal domains if and only if the following two conditions both hold.

(R1) The localization of $A$ at a codimension $1$ prime is a DVR, and the localization at a codimension $0$ prime is a field.
(S2) The associated primes of a principal ideal generated by a non-zerodivisor in $A$ have codimension $1$. Also, the associated primes of $0$ all have codimension $0$.

Proof

First, in general, if a Noetherian ring $A$ is a direct product of other rings

\[A=A_1\times\cdots A_n\]

then any prime ideal of this ring is of the form

\[A_1\times\cdots\times A_{k-1}\times \mathfrak{p}_k\times A_{k+1}\times\cdots\times A_n\]

for some prime ideal $\mathfrak{p}_k\subseteq A_k$, and the associated primes of $0$ are obtained by taking $\mathfrak{p}_k$ to be an associated prime of $0$ in $A_k$. Similarly, the associated primes of a non-zerodivisor

\[a=(a_1,\ldots, a_n),\qquad\text{$a_i$ a non-zerodivisor of $A_i$}\]

are the same as those obtained by taking $\mathfrak{p}_k$ to be an associated prime of $a_k$.

Now let us prove the claim. First, if each $A_i$ is normal, then condition (S2) follows from the result of Theorem 9, and condition (R1) holds because the localization of $A$ at a codimension $c$ prime ideal $\mathfrak{p}$ can be viewed, via the above description of $\mathfrak{p}$, as the localization of $A_k$ at a codimension $c$ prime $\mathfrak{p}_k$.

Conversely, assume that conditions R1 and S2 hold. Then first, $R$ is a reduced ring. Indeed, if

\[0=\bigcap \mathfrak{q}_i,\qquad\text{$\mathfrak{q}_i$ a $\mathfrak{p}_i$-primary ideal}\]

is a minimal primary decomposition of $0$, then the $\mathfrak{p}_i$ appearing here are all codimension $0$ ideals by condition S2, and their localizations are fields by condition R1. Therefore we can apply Proposition 8, and applying §The Jordan-Hölder Theorem, ⁋Theorem 5 yields the desired result.

References

[Eis] David Eisenbud. Commutative Algebra: with a view toward algebraic geometry. Springer, 1995.

[Stacks] The Stacks Project Authors. The Stacks Project. https://stacks.math.columbia.edu.

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Regular Local Rings

Discrete Valuation Rings

Serre’s Normality Criterion

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