Regular Local Rings

Considering §System of Parameters, ⁋Proposition–Definition 3, it is obvious that in a regular local ring $(A, \mathfrak{m})$, the $d=\dim A$ elements $a_1,\ldots, a_d$ generating $\mathfrak{m}$ form a system of parameters of $A$. These are called 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 $(A, \mathfrak{m})$. When $d=0$, there is nothing to prove because $A$ is a field. Assume that the given claim holds up to $\dim A=d$, and let us show the case $\dim A=d+1$. Then, in particular, $\mathfrak{m}\neq 0$, so from §Integral Extensions, ⁋Lemma 8 (Nakayama) we know that $\mathfrak{m}\neq \mathfrak{m}^2$. Meanwhile, from §Associated Primes, ⁋Theorem 7, we know that there are only finitely many minimal prime ideals of $A$. 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 computation $\mathfrak{m}\neq \mathfrak{m}^2$, we must have $\mathfrak{m}=\mathfrak{p}_i$, and since this yields

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

which is a contradiction, we know that there must exist some $a\in \mathfrak{m}$ such that $a\not\in \mathfrak{m}^2\cup \mathfrak{p}_1\cup\cdots\cup \mathfrak{p}_k$.

Now let $A'=A/(a)$, and consider the maximal ideal $\mathfrak{m}'=\mathfrak{m}A'$ of $A'$. Then by the choice of $a$, no prime ideal of $A'$ corresponds to any $\mathfrak{p}_i$, so necessarily $\dim A'<\dim A$ holds, and combining this with §System of Parameters, ⁋Corollary 7, we know that $\dim A'=d$. Therefore, from the equation

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

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

Now, for any $x\in \mathfrak{p}_i$, choose $\alpha\in A$ such that $x=\alpha a$. Then 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, we obtain $\mathfrak{p}_i=0$, so $A$ is an integral domain.

Since this corollary will be used frequently in what follows, we make the following new definition.

Definition 2 Elements $a_1,\ldots, a_d$ of a ring $A$ are said to be an $A$-regular sequence, or simply an $A$-sequence, if $(a_1,\ldots, a_d)$ is proper and, for each $i=0,\ldots, d-1$, the image of $a_{i+1}$ is a non-zerodivisor in $A/(a_1,\ldots, a_i)$.

By convention, we set $(a_1,\ldots, a_0)=0$. Then in the case $i=0$, what the above definition requires is that $a_1$ is a non-zerodivisor in $A$.

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

Proof

Let the maximal ideal of $A$ be $\mathfrak{m}$, set $d=\dim A$, and let $a_1,\ldots, a_d$ be a regular system of parameters generating $\mathfrak{m}$. Then $(a_1,\ldots, a_d)=\mathfrak{m}$ is proper. Now, for each $i$, let $A_i=A/(a_1,\ldots, a_i)$; then $A_i$ is a Noetherian local ring, and its maximal ideal $\mathfrak{m}A_i$ is generated by the images of $a_{i+1},\ldots, a_d$, that is, by $d-i$ elements, so $\dim A_i\leq d-i$ by §Krull Dimension, ⁋Theorem 7 (Principal Ideal Theorem). Conversely, applying §System of Parameters, ⁋Corollary 7 $i$ times gives $\dim A_i\geq d-i$, so $\dim A_i=d-i$ and therefore $A_i$ is also a regular local ring.

Then by Corollary 1, each $A_i$ is an integral domain, so it suffices to show that the image of $a_{i+1}$ in $A_i$ is not $0$. If $a_{i+1}\in (a_1,\ldots, a_i)$, then $\mathfrak{m}$ is generated by the elements of $a_1,\ldots, a_d$ other than $a_{i+1}$, namely $d-1$ elements, which contradicts that the maximal ideal of a Noetherian local ring with $\dim A=d$ can only be generated by at least $d$ elements. (§Krull Dimension, ⁋Theorem 7 (Principal Ideal Theorem))

Proposition 4 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 some field, then $A\cong \kappa[[\x_1,\ldots, \x_d]]$, and this isomorphism sends the variables $\x_i$ to a regular system of parameters of $A$.

Proof

By §Properties of Completions, ⁋Theorem 8 (Cohen structure theorem), we know from the given assumption that $A$ must contain $\kappa$. Now, by the first result of §Properties of Completions, ⁋Theorem 5, we obtain a $\kappa$-algebra homomorphism $\phi:\kappa[[\x_1,\ldots, \x_d]]\rightarrow A$, and by the second result, $\phi$ is surjective. Meanwhile, $\kappa[[\x_1,\ldots, \x_d]]$ is the formal power series ring of §Completion, ⁋Example 4, that is, the completion of $\kappa[\x_1,\ldots, \x_d]$ with respect to $(\x_1,\ldots, \x_d)$, so it is Noetherian by the first result of §Properties of Completions, ⁋Theorem 1, and by §Completion, ⁋Corollary 6 it is a local ring with $(\x_1,\ldots, \x_d)$ as its unique maximal ideal. Then by §Krull Dimension, ⁋Theorem 7 (Principal Ideal Theorem), we have $\dim \kappa[[\x_1,\ldots, \x_d]]\leq d$, and since $\phi$ is surjective, $\dim \kappa[[\x_1,\ldots, \x_d]]\geq \dim A=d$. Therefore $\kappa[[\x_1,\ldots, \x_d]]$ is a regular local ring of dimension $d$, and from this

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

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

Discrete Valuation Rings

Now we examine a $1$-dimensional regular local ring $(A,\mathfrak{m})$. Then by definition, $\mathfrak{m}$ must be generated by a single element $m$, which we call a regular parameter or uniformizing parameter of $A$.

Proposition 5 Suppose a 1-dimensional regular local ring $(A, \mathfrak{m})$ is given, and let $m$ be a regular parameter of $A$. Then in $\Frac(A)$, any non-$0$ element $x$ can be uniquely written 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), since $\bigcap \mathfrak{m}^i=0$, for any non-$0$ $x\in A$, $x\in \mathfrak{m}^i$ holds for only finitely many indices $i$. If we let $k$ be the largest among these, then from $x\in \mathfrak{m}^k=(m^k)$, we have $x=am^k$ for some $a\in A$. Then by the maximality of $k$, $a$ is a unit of $A$.

Now let an arbitrary element of $\Frac(A)$, $x$, be given. If we write $x=x_1/x_2$, then by the argument above we can write

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

Here $a=a_1a_2^{-1}$ is a unit, and if $am^k=bm^l$ and $k<l$, then $a=bm^{l-k}\in \mathfrak{m}$, which contradicts that $a$ is a unit; hence this expression is unique.

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

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

More generally, we define the following.

Definition 6 For an integral domain $A$ and a totally ordered abelian group $G$, if a group homomorphism $\nu:\Frac(A)^\times \rightarrow G$ satisfies, whenever $x+y\neq 0$, the inequality

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

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

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

is called the valuation ring of $\nu$.

In particular, if $G=\mathbb{Z}$ and $\nu$ is surjective, we call it a discrete valuation, and call the valuation ring of $\nu$ a discrete valuation ring.

Then that the map $\nu:\Frac(A)^\times \rightarrow \mathbb{Z}$ defined above is a discrete valuation is clear from the equation

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

Then by Proposition 4, we know that if two complete discrete valuation rings each contain a field and have isomorphic residue fields, they are isomorphic to each other. In general, however, no classification of this kind exists among discrete valuation rings that are not complete.

Serre’s Normality Criterion

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

Proposition 7 Let a reduced Noetherian ring $A$ and the total ring of fractions of $A$, $K$, be given. Then an element $x\in K$ belongs to $A$ if and only if, for every prime ideal $\mathfrak{p}$ associated to a non-zerodivisor, the image of $x$ in $K_\mathfrak{p}$ belongs to $A_\mathfrak{p}$.

Proof

By definition, an element of $K$ is, for arbitrary $a\in A$ and a non-zerodivisor $u\in A$, of the form $a/u$. 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 due to §Associated Primes, ⁋Corollary 4. Then for every prime ideal $\mathfrak{p}$ associated to a non-zerodivisor,

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

holds. Meanwhile, 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 can identify the total ring of fractions of $A_\mathfrak{p}$ with $K_\mathfrak{p}$. Examining the above inclusion again through this identification yields the desired result.

Through this, we can show the following.

Theorem 8 A Noetherian integral domain $A$ is a normal domain if and only if the following condition holds.

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

Proof

First assume ($\ast$); we show that $A$ is a normal domain. However, given normal domains sharing a common quotient field, it is clear that their intersection is also a normal domain. Here, for each prime ideal $\mathfrak{p}$ associated to a principal ideal, the fact that $A_\mathfrak{p}$ is a normal domain follows from ($\ast$) as follows: since $\mathfrak{p}A_\mathfrak{p}$ is principal by assumption, by §Krull Dimension, ⁋Theorem 7 (Principal Ideal Theorem) we have $\dim A_\mathfrak{p}\leq 1$, and therefore $A_\mathfrak{p}$ is either a field or a $1$-dimensional regular local ring whose maximal ideal is principal, that is, a discrete valuation ring. Since discrete valuation rings and fields are normal domains, it suffices to show the equality

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

where $A_\mathfrak{p}$ is viewed as a subset of $A$’s quotient field $K$. Now, the claim to be shown was treated in greater generality in Proposition 7.

Conversely, suppose $A$ 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 about localization anyway, and by §Integral Extensions, ⁋Proposition 12 the localization of a normal domain is again a normal domain, we may assume that $(A,\mathfrak{p})$ was a local ring. Here, letting $K$ be the field of fractions of $A$, if we 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 $A$. Now, from the maximality of $\mathfrak{p}$, either $\mathfrak{p}^{-1}\mathfrak{p}=\mathfrak{p}$ or $\mathfrak{p}^{-1}\mathfrak{p}=A$ must hold. However, if $\mathfrak{p}^{-1}\mathfrak{p}=\mathfrak{p}$, then by §Integral Extensions, ⁋Lemma 5, every element of $\mathfrak{p}^{-1}$ is integral, and thus $\mathfrak{p}^{-1}\subseteq A$. But since $\mathfrak{p}b\subseteq (a)$, we have $b/a\in \mathfrak{p}^{-1}$, and from this $b\in (a)$, which is a contradiction.

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

This theorem can be generalized further. First, let us define the following.

Definition 9 Fix a ring $A$ and $A$’s total ring of fractions $K$.

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

Then the following holds.

Theorem 10 (Serre) A Noetherian ring $A$ is a (finite) direct product of normal domains if and only if both of the following conditions hold.

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

Proof

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

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

of other rings, then every 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 an associated prime ideal of $0$ is obtained by taking $\mathfrak{p}_k$ in the above form to be an associated prime of $0$ (in $A_k$). Similarly, an associated prime of a non-zerodivisor

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

of $A$ is 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 Theorem 8, and condition (R1) holds because, through the description of $\mathfrak{p}$ above, the localization of $A$ at a prime ideal $\mathfrak{p}$ of codimension $c$ can be viewed as the localization of $A_k$ at a prime $\mathfrak{p}_k$ of codimension $c$.

Conversely, suppose conditions R1 and S2 hold. Then first, $A$ is a reduced ring. This is because 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 all have codimension $0$ by condition S2, and by condition R1 we know that the localization here is a field. Then for each $i$, we have $\mathfrak{q}_i=\mathfrak{p}_i$, because since $\mathfrak{p}_iA_{\mathfrak{p}_i}=0$, for any $a\in \mathfrak{p}_i$ there exists $s\not\in \mathfrak{p}_i$ such that $sa=0\in \mathfrak{q}_i$, and since $\mathfrak{q}_i$ is $\mathfrak{p}_i$-primary and $s\not\in \sqrt{\mathfrak{q}_i}=\mathfrak{p}_i$, we have $a\in \mathfrak{q}_i$. Therefore $0=\bigcap \mathfrak{p}_i$ is an intersection of prime ideals, and $A$ is reduced.

Now we can apply Proposition 7. By condition S2, every prime ideal $\mathfrak{p}$ associated to a non-zerodivisor has codimension $1$, and by condition R1, $A_\mathfrak{p}$ is a discrete valuation ring. Then, if an element $x$ of the total ring of fractions $K$ of $A$ is integral over $A$, the image of $x$ in $K_\mathfrak{p}$ is integral over $A_\mathfrak{p}$, and since a discrete valuation ring is a normal domain, this belongs to $A_\mathfrak{p}$. Therefore, by Proposition 7, $x\in A$, and so $A$ is integrally closed in $K$.

Meanwhile, since $A$ is reduced, $K$ is a finite direct product $K_1\times\cdots\times K_n$ of fields (§Associated Primes, ⁋Corollary 8), and the idempotent $e_i\in K$ corresponding to each factor is a root of the monic polynomial $\x^2-\x$, so it is integral over $A$, and therefore $e_i\in A$. Then, for any maximal ideal $\mathfrak{m}$ of $A$, since $A_\mathfrak{m}$ is a local ring its only idempotents are $0$ and $1$, and since $\sum_i e_i=1$, exactly one of the images of the $e_i$ is $1$ and the rest are $0$; it follows that $A_\mathfrak{m}$, being a localization of the subring $Ae_i$ of the field $K_i$, is a domain. Now, by §The Jordan-Hölder Theorem, ⁋Theorem 5, $A$ is a finite direct product of domains, and since $K$ is the product of the total rings of fractions of the factors, the fact that $A$ is integrally closed in $K$ means that each factor is integrally closed in its own total ring of fractions. That is, $A$ is a finite direct product of normal domains.

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