The Stacks project

Proof. It is clear that (2) implies (1) and (1) implies (3). It is also clear that (4) implies (5). Let us show that (3) implies (4). Thus we assume there exists a surjection $R \to S$ with $R$ a regular local ring essentially of finite presentation over $k$ such that the ideal $\mathop{\mathrm{Ker}}(R \to S)$ can be generated by $\dim (R) - \dim (S)$ elements. We may write $R = (k[x_1, \ldots , x_ n]/J)_{\mathfrak q}$ for some $J \subset k[x_1, \ldots , x_ n]$ and some prime $\mathfrak q \subset k[x_1, \ldots , x_ n]$ with $J \subset \mathfrak q$. Let $I \subset k[x_1, \ldots , x_ n]$ be the kernel of the map $k[x_1, \ldots , x_ n] \to S$ so that $S \cong (k[x_1, \ldots , x_ n]/I)_{\mathfrak q}$. By assumption $(I/J)_{\mathfrak q}$ is generated by $\dim (R) - \dim (S)$ elements. We conclude that $I_{\mathfrak q}$ can be generated by $\dim (k[x_1, \ldots , x_ n]_{\mathfrak q}) - \dim (S)$ elements by Lemma 10.133.6. From Lemma 10.133.4 we see that for some $g \in k[x_1, \ldots , x_ n]$, $g \not\in \mathfrak q$ the algebra $(k[x_1, \ldots , x_ n]/I)_ g$ is a global complete intersection and $S$ is isomorphic to a local ring of it.

To finish the proof of the lemma we have to show that (5) implies (2). Assume (5) and let $\pi : R \to S$ be a surjection with $R$ a regular local $k$-algebra essentially of finite type over $k$. By assumption we have $S = A_{\mathfrak a}$ for some local complete intersection $A$ over $k$. Choose a presentation $R = (k[y_1, \ldots , y_ m]/J)_{\mathfrak q}$ with $J \subset \mathfrak q \subset k[y_1, \ldots , y_ m]$. We may and do assume that $J$ is the kernel of the map $k[y_1, \ldots , y_ m] \to R$. Let $I \subset k[y_1, \ldots , y_ m]$ be the kernel of the map $k[y_1, \ldots , y_ m] \to S = A_{\mathfrak a}$. Then $J \subset I$ and $(I/J)_{\mathfrak q}$ is the kernel of the surjection $\pi : R \to S$. So $S = (k[y_1, \ldots , y_ m]/I)_{\mathfrak q}$.

By Lemma 10.125.7 we see that there exist $g \in A$, $g \not\in \mathfrak a$ and $g' \in k[y_1, \ldots , y_ m]$, $g' \not\in \mathfrak q$ such that $A_ g \cong (k[y_1, \ldots , y_ m]/I)_{g'}$. After replacing $A$ by $A_ g$ and $k[y_1, \ldots , y_ m]$ by $k[y_1, \ldots , y_{m + 1}]$ we may assume that $A \cong k[y_1, \ldots , y_ m]/I$. Consider the surjective maps of local rings

We have to show that the kernel of $R \to S$ is generated by a regular sequence. By Lemma 10.133.4 we know that $k[y_1, \ldots , y_ m]_{\mathfrak q} \to A_{\mathfrak a} = S$ has this property (as $A$ is a local complete intersection over $k$). We win by Lemma 10.133.6. $\square$