Lemma 10.135.11. 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 10.135.9 and 10.135.10. 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 10.135.4 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 10.135.4 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 10.18.5 the primes of $S'$ lying over $\mathfrak m$ correspond to primes in $K \otimes _ k \kappa$. By the Hilbert-Nullstellensatz Theorem 10.34.1 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 10.53). 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 10.112.7. (This can also be seen using Lemma 10.116.6.) 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 10.134.8 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 10.135.4 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. $\square$

Comment #3205 by Dario Weißmann on

Typo: In the direct proof the regular sequence $f_1,\dots, f_c$ is also mentioned as $f_1,\dots,f_e$.

In the second paragraph the inequality $\text{dim}S_{\mathfrak{m}}\geq \text{dim}S'_{\mathfrak{n}}$ does not imply $n-\text{dim}S_{\mathfrak{m}}\geq n - \text{dim}S'_{\mathfrak{n}}$. But as in reality the dimensions are equal it still works. The equality follows from Lemma 10.111.7.

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