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 10.135.6. From Lemma 10.135.4 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 10.126.7 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
k[y_1, \ldots , y_ m]_{\mathfrak q} \to R \to S.
We have to show that the kernel of R \to S is generated by a regular sequence. By Lemma 10.135.4 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 10.135.6.
\square
Comments (1)
Comment #2359 by Simon Pepin Lehalleur on
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