The Stacks project

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. $\square$

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. $\square$

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

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. $\square$

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. $\square$

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. $\square$

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. $\square$

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. $\square$