The Stacks project

10.105 Regular local rings

It is not that easy to show that all prime localizations of a regular local ring are regular. In fact, quite a bit of the material developed so far is geared towards a proof of this fact. See Proposition 10.109.5, and trace back the references.

Lemma 10.105.1. Let $(R, \mathfrak m, \kappa )$ be a regular local ring of dimension $d$. The graded ring $\bigoplus \mathfrak m^ n / \mathfrak m^{n + 1}$ is isomorphic to the graded polynomial algebra $\kappa [X_1, \ldots , X_ d]$.

Proof. Let $x_1, \ldots , x_ d$ be a minimal set of generators for the maximal ideal $\mathfrak m$, see Definition 10.59.9. There is a surjection $\kappa [X_1, \ldots , X_ d] \to \bigoplus \mathfrak m^ n/\mathfrak m^{n + 1}$, which maps $X_ i$ to the class of $x_ i$ in $\mathfrak m/\mathfrak m^2$. Since $d(R) = d$ by Proposition 10.59.8 we know that the numerical polynomial $n \mapsto \dim _\kappa \mathfrak m^ n/\mathfrak m^{n + 1}$ has degree $d - 1$. By Lemma 10.57.10 we conclude that the surjection $\kappa [X_1, \ldots , X_ d] \to \bigoplus \mathfrak m^ n/\mathfrak m^{n + 1}$ is an isomorphism. $\square$

Proof. We will use that $\bigcap \mathfrak m^ n = 0$ by Lemma 10.50.4. Let $f, g \in R$ such that $fg = 0$. Suppose that $f \in \mathfrak m^ a$ and $g \in \mathfrak m^ b$, with $a, b$ maximal. Since $fg = 0 \in \mathfrak m^{a + b + 1}$ we see from the result of Lemma 10.105.1 that either $f \in \mathfrak m^{a + 1}$ or $g \in \mathfrak m^{b + 1}$. Contradiction. $\square$

Lemma 10.105.3. Let $R$ be a regular local ring and let $x_1, \ldots , x_ d$ be a minimal set of generators for the maximal ideal $\mathfrak m$. Then $x_1, \ldots , x_ d$ is a regular sequence, and each $R/(x_1, \ldots , x_ c)$ is a regular local ring of dimension $d - c$. In particular $R$ is Cohen-Macaulay.

Proof. Note that $R/x_1R$ is a Noetherian local ring of dimension $\geq d - 1$ by Lemma 10.59.12 with $x_2, \ldots , x_ d$ generating the maximal ideal. Hence it is a regular local ring by definition. Since $R$ is a domain by Lemma 10.105.2 $x_1$ is a nonzerodivisor. $\square$

Lemma 10.105.4. Let $R$ be a regular local ring. Let $I \subset R$ be an ideal such that $R/I$ is a regular local ring as well. Then there exists a minimal set of generators $x_1, \ldots , x_ d$ for the maximal ideal $\mathfrak m$ of $R$ such that $I = (x_1, \ldots , x_ c)$ for some $0 \leq c \leq d$.

Proof. Say $\dim (R) = d$ and $\dim (R/I) = d - c$. Denote $\overline{\mathfrak m} = \mathfrak m/I$ the maximal ideal of $R/I$. Let $\kappa = R/\mathfrak m$. We have

\[ \dim _\kappa ((I + \mathfrak m^2)/\mathfrak m^2) = \dim _\kappa (\mathfrak m/\mathfrak m^2) - \dim (\overline{\mathfrak m}/\overline{\mathfrak m}^2) = d - (d - c) = c \]

by the definition of a regular local ring. Hence we can choose $x_1, \ldots , x_ c \in I$ whose images in $\mathfrak m/\mathfrak m^2$ are linearly independent and supplement with $x_{c + 1}, \ldots , x_ d$ to get a minimal system of generators of $\mathfrak m$. The induced map $R/(x_1, \ldots , x_ c) \to R/I$ is a surjection between regular local rings of the same dimension (Lemma 10.105.3). It follows that the kernel is zero, i.e., $I = (x_1, \ldots , x_ c)$. Namely, if not then we would have $\dim (R/I) < \dim (R/(x_1, \ldots , x_ c))$ by Lemmas 10.105.2 and 10.59.12. $\square$

Lemma 10.105.5. Let $R$ be a Noetherian local ring. Let $x \in \mathfrak m$. Let $M$ be a finite $R$-module such that $x$ is a nonzerodivisor on $M$ and $M/xM$ is free over $R/xR$. Then $M$ is free over $R$.

Proof. Let $m_1, \ldots , m_ r$ be elements of $M$ which map to a $R/xR$-basis of $M/xM$. By Nakayama's Lemma 10.19.1 $m_1, \ldots , m_ r$ generate $M$. If $\sum a_ i m_ i = 0$ is a relation, then $a_ i \in xR$ for all $i$. Hence $a_ i = b_ i x$ for some $b_ i \in R$. Hence the kernel $K$ of $R^ r \to M$ satisfies $xK = K$ and hence is zero by Nakayama's lemma. $\square$

Lemma 10.105.6. Let $R$ be a regular local ring. Any maximal Cohen-Macaulay module over $R$ is free.

Proof. Let $M$ be a maximal Cohen-Macaulay module over $R$. Let $x \in \mathfrak m$ be part of a regular sequence generating $\mathfrak m$. Then $x$ is a nonzerodivisor on $M$ by Proposition 10.102.4, and $M/xM$ is a maximal Cohen-Macaulay module over $R/xR$. By induction on $\dim (R)$ we see that $M/xM$ is free. We win by Lemma 10.105.5. $\square$

Lemma 10.105.7. Suppose $R$ is a Noetherian local ring. Let $x \in \mathfrak m$ be a nonzerodivisor such that $R/xR$ is a regular local ring. Then $R$ is a regular local ring. More generally, if $x_1, \ldots , x_ r$ is a regular sequence in $R$ such that $R/(x_1, \ldots , x_ r)$ is a regular local ring, then $R$ is a regular local ring.

Proof. This is true because $x$ together with the lifts of a system of minimal generators of the maximal ideal of $R/xR$ will give $\dim (R)$ generators of $\mathfrak m$. Use Lemma 10.59.12. The last statement follows from the first and induction. $\square$

Lemma 10.105.8. Let $(R_ i, \varphi _{ii'})$ be a directed system of local rings whose transition maps are local ring maps. If each $R_ i$ is a regular local ring and $R = \mathop{\mathrm{colim}}\nolimits R_ i$ is Noetherian, then $R$ is a regular local ring.

Proof. Let $\mathfrak m \subset R$ be the maximal ideal; it is the colimit of the maximal ideal $\mathfrak m_ i \subset R_ i$. We prove the lemma by induction on $d = \dim \mathfrak m/\mathfrak m^2$. If $d = 0$, then $R = R/\mathfrak m$ is a field and $R$ is a regular local ring. If $d > 0$ pick an $x \in \mathfrak m$, $x \not\in \mathfrak m^2$. For some $i$ we can find an $x_ i \in \mathfrak m_ i$ mapping to $x$. Note that $R/xR = \mathop{\mathrm{colim}}\nolimits _{i' \geq i} R_{i'}/x_ iR_{i'}$ is a Noetherian local ring. By Lemma 10.105.3 we see that $R_{i'}/x_ iR_{i'}$ is a regular local ring. Hence by induction we see that $R/xR$ is a regular local ring. Since each $R_ i$ is a domain (Lemma 10.105.1) we see that $R$ is a domain. Hence $x$ is a nonzerodivisor and we conclude that $R$ is a regular local ring by Lemma 10.105.7. $\square$


Comments (4)

Comment #685 by Keenan Kidwell on

In the proof of 00NO, isn't the degree of the Hilbert polynomial plus one, i.e., the degree of should be , not ? That's consistent with the lemma that's invoked, which says that, if there were a non-zero kernel to the surjection, then the degree of the numerical polynomial under consideration would be strictly less than . But it's exactly , so there's no kernel, and we have an isomorphism.

Comment #686 by Keenan Kidwell on

In the statement of 00NR, "the maximal " should be "the maximal ideal ."

Comment #687 by Keenan Kidwell on

In the proof of 00NR, should be ...of course this is minor. Also, I'm assuming Nakayama's lemma is being used to (implicitly) conclude that generate because their images generate , but I don't quite see how this works because there's a possible discrepancy between and .


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