The Stacks project

Proof. Set $R = k[x_1, \ldots , x_ n]_{\mathfrak q'}$. This is a Cohen-Macaulay local ring of dimension $\text{height}(\mathfrak q')$, see for example Lemma 10.135.3. Moreover, $\overline{R} = R/IR = R/I_{\mathfrak q'} = S_{\mathfrak q}$ is a quotient of dimension $\text{height}(\mathfrak q)$. Let $f_1, \ldots , f_ c \in I_{\mathfrak q'}$ be elements which generate $(I/I^2)_{\mathfrak q}$. By Lemma 10.20.1 we see that $f_1, \ldots , f_ c$ generate $I_{\mathfrak q'}$. Since the dimensions work out, we conclude by Proposition 10.103.4 that $f_1, \ldots , f_ c$ is a regular sequence in $R$. By Lemma 10.69.2 we see that $(I/I^2)_{\mathfrak q}$ is free. These arguments show that (2), (3), (4) are equivalent and that they imply the last statement of the lemma, and therefore they imply (5).

If (5) holds, say $I_{\mathfrak q'}$ is generated by a regular sequence of length $e$, then $\text{height}(\mathfrak q) = \dim (S_{\mathfrak q}) = \dim (k[x_1, \ldots , x_ n]_{\mathfrak q'}) - e = \text{height}(\mathfrak q') - e$ by dimension theory, see Section 10.60. We conclude that $e = c$. Thus (5) implies (2).

We continue with the notation introduced in the first paragraph. For each $f_ i$ we may find $d_ i \in k[x_1, \ldots , x_ n]$, $d_ i \not\in \mathfrak q'$ such that $f_ i' = d_ i f_ i \in k[x_1, \ldots , x_ n]$. Then it is still true that $I_{\mathfrak q'} = (f_1', \ldots , f_ c')R$. Hence there exists a $g' \in k[x_1, \ldots , x_ n]$, $g' \not\in \mathfrak q'$ such that $I_{g'} = (f_1', \ldots , f_ c')$. Moreover, pick $g'' \in k[x_1, \ldots , x_ n]$, $g'' \not\in \mathfrak q'$ such that $\dim (S_{g''}) = \dim _{\mathfrak q} \mathop{\mathrm{Spec}}(S)$. By Lemma 10.116.4 this dimension is equal to $n - c$. Finally, set $g$ equal to the image of $g'g''$ in $S$. Then we see that

and by our choice of $g''$ this ring has dimension $n - c$. Therefore it is a global complete intersection. Thus each of (2), (3), and (4) implies (1).

Assume (1). Let $S_ g \cong k[y_1, \ldots , y_ m]/(f_1, \ldots , f_ t)$ be a presentation of $S_ g$ as a global complete intersection. Write $J = (f_1, \ldots , f_ t)$. Let $\mathfrak q'' \subset k[y_1, \ldots , y_ m]$ be the prime corresponding to $\mathfrak qS_ g$. Note that $t = m - \dim (S_ g) = \text{height}(\mathfrak q'') - \text{height}(\mathfrak q)$, see Lemma 10.116.4 for the last equality. As seen in the proof of Lemma 10.135.3 (and also above) the elements $f_1, \ldots , f_ t$ form a regular sequence in the local ring $k[y_1, \ldots , y_ m]_{\mathfrak q''}$. By Lemma 10.69.2 we see that $(J/J^2)_{\mathfrak q}$ is free of rank $t$. By Lemma 10.134.16 we have

Thus $(I/I^2)_{\mathfrak q}$ is free of rank $t + n - m = m - \dim (S_ g) + n - m = n - \dim (S_ g) = \text{height}(\mathfrak q') - \text{height}(\mathfrak q) = c$. Thus we obtain (4). $\square$