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## Tag: 00SC

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Lemma 9.127.4. Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Let $\mathfrak q$ be a prime of $S$. Choose any presentation $S = k[x_1, \ldots, x_n]/I$. Let $\mathfrak q'$ be the prime of $k[x_1, \ldots, x_n]$ corresponding to $\mathfrak q$. Set $c = \text{height}(\mathfrak q') - \text{height}(\mathfrak q)$, in other words $\dim_{\mathfrak q}(S) = n - c$ (see Lemma 9.110.4). The following are equivalent
1. There exists a $g \in S$, $g \not \in \mathfrak q$ such that $S_g$ is a global complete intersection over $k$.
2. The ideal $I_{\mathfrak q'} \subset k[x_1, \ldots, x_n]_{\mathfrak q'}$ can be generated by $c$ elements.
3. The conormal module $(I/I^2)_{\mathfrak q}$ can be generated by $c$ elements over $S_{\mathfrak q}$.
4. The conormal module $(I/I^2)_{\mathfrak q}$ is a free $S_{\mathfrak q}$-module of rank $c$.
5. The ideal $I_{\mathfrak q'}$ can be generated by a regular sequence in the regular local ring $k[x_1, \ldots, x_n]_{\mathfrak q'}$.
In this case any $c$ elements of $I_{\mathfrak q'}$ which generate $I_{\mathfrak q'}/\mathfrak q'I_{\mathfrak q'}$ form a regular sequence in the local ring $k[x_1, \ldots, x_n]_{\mathfrak q'}$.

Proof. Set $R = k[x_1, \ldots, x_n]_{\mathfrak q'}$. This is a regular local ring of dimension $\text{height}(\mathfrak q')$. 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 9.18.1 we see that $f_1, \ldots, f_c$ generate $I_{\mathfrak q'}$. Since the dimensions work out, we conclude by Proposition 9.98.4 that $f_1, \ldots, f_c$ is a regular sequence in $R$. By Lemma 9.68.2 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 9.59. 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 $\dim(S_{g''}) = \dim_{\mathfrak q} \mathop{\rm Spec}(S)$. By Lemma 9.110.4 this dimension is equal to $n - c$. Finally, set $g$ equal to the image of $g'g''$ in $S$. Then we see that $$S_g \cong k[x_1, \ldots, x_n, x_{n + 1}] / (f_1', \ldots, f_c', x_{n + 1}g'g'' - 1)$$ 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 9.110.4 for the last equality. As seen in the proof of Lemma 9.127.3 (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 9.68.2 we see that $(J/J^2)_{\mathfrak q}$ is free of rank $t$. By Lemma 9.126.16 we have $$J/J^2 \oplus S_g^n \cong (I/I^2)_g \oplus S_g^m$$ 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). $\square$

\begin{lemma}
\label{lemma-lci}
Let $k$ be a field.
Let $S$ be a finite type $k$-algebra.
Let $\mathfrak q$ be a prime of $S$.
Choose any presentation $S = k[x_1, \ldots, x_n]/I$.
Let $\mathfrak q'$ be the prime of $k[x_1, \ldots, x_n]$ corresponding
to $\mathfrak q$. Set
$c = \text{height}(\mathfrak q') - \text{height}(\mathfrak q)$,
in other words $\dim_{\mathfrak q}(S) = n - c$
(see Lemma \ref{lemma-codimension}). The following are equivalent
\begin{enumerate}
\item There exists a $g \in S$, $g \not \in \mathfrak q$
such that $S_g$ is a global complete intersection over $k$.
\item The ideal $I_{\mathfrak q'} \subset k[x_1, \ldots, x_n]_{\mathfrak q'}$
can be generated by $c$ elements.
\item The conormal module $(I/I^2)_{\mathfrak q}$ can be generated by
$c$ elements over $S_{\mathfrak q}$.
\item The conormal module $(I/I^2)_{\mathfrak q}$ is a free
$S_{\mathfrak q}$-module of rank $c$.
\item The ideal $I_{\mathfrak q'}$ can be generated by a regular sequence
in the regular local ring $k[x_1, \ldots, x_n]_{\mathfrak q'}$.
\end{enumerate}
In this case any $c$ elements of $I_{\mathfrak q'}$
which generate $I_{\mathfrak q'}/\mathfrak q'I_{\mathfrak q'}$
form a regular sequence in the local
ring $k[x_1, \ldots, x_n]_{\mathfrak q'}$.
\end{lemma}

\begin{proof}
Set $R = k[x_1, \ldots, x_n]_{\mathfrak q'}$. This is a regular local
ring of dimension $\text{height}(\mathfrak q')$. 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 \ref{lemma-NAK}
we see that $f_1, \ldots, f_c$ generate $I_{\mathfrak q'}$.
Since the dimensions work out, we conclude
by Proposition \ref{proposition-CM-module} that
$f_1, \ldots, f_c$ is a regular sequence in $R$.
By Lemma \ref{lemma-regular-quasi-regular} 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).

\medskip\noindent
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 \ref{section-dimension}. We conclude that $e = c$.
Thus (5) implies (2).

\medskip\noindent
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 $\dim(S_{g''}) = \dim_{\mathfrak q} \Spec(S)$.
By Lemma \ref{lemma-codimension} this dimension is equal to $n - c$.
Finally, set $g$ equal to the image of $g'g''$ in $S$.
Then we see that
$$S_g \cong k[x_1, \ldots, x_n, x_{n + 1}] / (f_1', \ldots, f_c', x_{n + 1}g'g'' - 1)$$
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).

\medskip\noindent
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 \ref{lemma-codimension} for the last equality.
As seen in the proof of Lemma \ref{lemma-lci-CM} (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 \ref{lemma-regular-quasi-regular} we see that
$(J/J^2)_{\mathfrak q}$ is free of rank $t$.
By Lemma \ref{lemma-conormal-module-localize} we have
$$J/J^2 \oplus S_g^n \cong (I/I^2)_g \oplus S_g^m$$
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).
\end{proof}


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