## 10.106 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 10.110.5, and trace back the references.

Lemma 10.106.1. 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 10.60.10. 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 10.60.9 we know that the numerical polynomial $n \mapsto \dim _\kappa \mathfrak m^ n/\mathfrak m^{n + 1}$ has degree $d - 1$. By Lemma 10.58.10 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 10.51.4. 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 10.106.1 that either $f \in \mathfrak m^{a + 1}$ or $g \in \mathfrak m^{b + 1}$. Contradiction. $\square$

Lemma 10.106.3. 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 10.60.13 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 10.106.2 $x_1$ is a nonzerodivisor. $\square$

Lemma 10.106.4. 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 10.106.3). 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 10.106.2 and 10.60.13. $\square$

Lemma 10.106.5. 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 10.20.1 $m_1, \ldots , m_ r$ 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 10.106.6. 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 10.103.4, 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 10.106.5. $\square$

Lemma 10.106.7. 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 10.60.13. The last statement follows from the first and induction. $\square$

Lemma 10.106.8. 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 10.106.3 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 10.106.1) we see that $R$ is a domain. Hence $x$ is a nonzerodivisor and we conclude that $R$ is a regular local ring by Lemma 10.106.7. $\square$

Comment #685 by Keenan Kidwell on

In the proof of 00NO, isn't $d$ the degree of the Hilbert polynomial plus one, i.e., the degree of $n\mapsto\dim_k(\mathfrak{m}^n/\mathfrak{m}^{n+1})$ should be $d-1$, not $d$? 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 $d-1$. But it's exactly $d-1$, so there's no kernel, and we have an isomorphism.

Comment #686 by Keenan Kidwell on

In the statement of 00NR, "the maximal $\mathfrak{m}$" should be "the maximal ideal $\mathfrak{m}$."

Comment #687 by Keenan Kidwell on

In the proof of 00NR, $I/\mathfrak{m}^2$ should be $I+\mathfrak{m}^2/\mathfrak{m}^2$...of course this is minor. Also, I'm assuming Nakayama's lemma is being used to (implicitly) conclude that $x_1,\ldots,x_c$ generate $I$ because their images generate $I/\mathfrak{m}^2$, but I don't quite see how this works because there's a possible discrepancy between $I/\mathfrak{m}I$ and $I/\mathfrak{m}^2$.

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