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\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.133.11. Let $k \to K$ be a field extension. Let $S$ be a finite type $k$-algebra. Then $S$ is a local complete intersection over $k$ if and only if $S \otimes _ k K$ is a local complete intersection over $K$.

Proof. This follows from a combination of Lemmas 10.133.9 and 10.133.10. But we also give a different proof here (based on the same principles).

Set $S' = S \otimes _ k K$. Let $\alpha : k[x_1, \ldots , x_ n] \to S$ be a presentation with kernel $I$. Let $\alpha ' : K[x_1, \ldots , x_ n] \to S'$ be the induced presentation with kernel $I'$.

Suppose that $S$ is a local complete intersection. Pick a prime $\mathfrak q \subset S'$. Denote $\mathfrak q'$ the corresponding prime of $K[x_1, \ldots , x_ n]$, $\mathfrak p$ the corresponding prime of $S$, and $\mathfrak p'$ the corresponding prime of $k[x_1, \ldots , x_ n]$. Consider the following diagram of Noetherian local rings

\[ \xymatrix{ S'_{\mathfrak q} & K[x_1, \ldots , x_ n]_{\mathfrak q'} \ar[l] \\ S_{\mathfrak p}\ar[u] & k[x_1, \ldots , x_ n]_{\mathfrak p'} \ar[u] \ar[l] } \]

By Lemma 10.133.4 we know that $S_{\mathfrak p}$ is cut out by some regular sequence $f_1, \ldots , f_ c$ in $k[x_1, \ldots , x_ n]_{\mathfrak p'}$. Since the right vertical arrow is flat we see that the images of $f_1, \ldots , f_ c$ form a regular sequence in $K[x_1, \ldots , x_ n]_{\mathfrak q'}$. Because tensoring with $K$ over $k$ is an exact functor we have $S'_{\mathfrak q} = K[x_1, \ldots , x_ n]_{\mathfrak q'}/(f_1, \ldots , f_ c)$. Hence by Lemma 10.133.4 again we see that $S'$ is a local complete intersection in a neighbourhood of $\mathfrak q$. Since $\mathfrak q$ was arbitrary we see that $S'$ is a local complete intersection over $K$.

Suppose that $S'$ is a local complete intersection. Pick a maximal ideal $\mathfrak m$ of $S$. Let $\mathfrak m'$ denote the corresponding maximal ideal of $k[x_1, \ldots , x_ n]$. Denote $\kappa = \kappa (\mathfrak m)$ the residue field. By Remark 10.16.8 the primes of $S'$ lying over $\mathfrak m$ correspond to primes in $K \otimes _ k \kappa $. By the Hilbert-Nullstellensatz Theorem 10.33.1 we have $[\kappa : k] < \infty $. Hence $K \otimes _ k \kappa $ is finite nonzero over $K$. Hence $K \otimes _ k \kappa $ has a finite number $> 0$ of primes which are all maximal, each of which has a residue field finite over $K$ (see Section 10.52). Hence there are finitely many $> 0$ prime ideals $\mathfrak n \subset S'$ lying over $\mathfrak m$, each of which is maximal and has a residue field which is finite over $K$. Pick one, say $\mathfrak n \subset S'$, and let $\mathfrak n' \subset K[x_1, \ldots , x_ n]$ denote the corresponding prime ideal of $K[x_1, \ldots , x_ n]$. Note that since $V(\mathfrak mS')$ is finite, we see that $\mathfrak n$ is an isolated closed point of it, and we deduce that $\mathfrak mS'_{\mathfrak n}$ is an ideal of definition of $S'_{\mathfrak n}$. This implies that $\dim (S_{\mathfrak m}) = \dim (S'_{\mathfrak n})$ for example by Lemma 10.111.7. (This can also be seen using Lemma 10.115.6.) Consider the corresponding diagram of Noetherian local rings

\[ \xymatrix{ S'_{\mathfrak n} & K[x_1, \ldots , x_ n]_{\mathfrak n'} \ar[l] \\ S_{\mathfrak m}\ar[u] & k[x_1, \ldots , x_ n]_{\mathfrak m'} \ar[u] \ar[l] } \]

According to Lemma 10.132.8 we have $\mathop{N\! L}\nolimits (\alpha ) \otimes _ S S' = \mathop{N\! L}\nolimits (\alpha ')$, in particular $I'/(I')^2 = I/I^2 \otimes _ S S'$. Thus $(I/I^2)_{\mathfrak m} \otimes _{S_{\mathfrak m}} \kappa $ and $(I'/(I')^2)_{\mathfrak n} \otimes _{S'_{\mathfrak n}} \kappa (\mathfrak n)$ have the same dimension. Since $(I'/(I')^2)_{\mathfrak n}$ is free of rank $n - \dim S'_{\mathfrak n}$ we deduce that $(I/I^2)_{\mathfrak m}$ can be generated by $n - \dim S'_{\mathfrak n} = n - \dim S_{\mathfrak m}$ elements. By Lemma 10.133.4 we see that $S$ is a local complete intersection in a neighbourhood of $\mathfrak m$. Since $\mathfrak m$ was any maximal ideal we conclude that $S$ is a local complete intersection. $\square$


Comments (2)

Comment #3205 by Dario WeiƟmann on

Typo: In the direct proof the regular sequence is also mentioned as .

In the second paragraph the inequality does not imply . But as in reality the dimensions are equal it still works. The equality follows from Lemma 10.111.7.

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