Lemma 10.135.10. Let K/k be a field extension. Let S be a finite type algebra over k. Let \mathfrak q_ K be a prime of S_ K = K \otimes _ k S and let \mathfrak q be the corresponding prime of S. Then S_{\mathfrak q} is a complete intersection over k (Definition 10.135.5) if and only if (S_ K)_{\mathfrak q_ K} is a complete intersection over K.
Proof. Choose a presentation S = k[x_1, \ldots , x_ n]/I. This gives a presentation S_ K = K[x_1, \ldots , x_ n]/I_ K where I_ K = K \otimes _ k I. Let \mathfrak q_ K' \subset K[x_1, \ldots , x_ n], resp. \mathfrak q' \subset k[x_1, \ldots , x_ n] be the corresponding prime. We will show that the equivalent conditions of Lemma 10.135.4 hold for the pair (S = k[x_1, \ldots , x_ n]/I, \mathfrak q) if and only if they hold for the pair (S_ K = K[x_1, \ldots , x_ n]/I_ K, \mathfrak q_ K). The lemma will follow from this (see Lemma 10.135.8).
By Lemma 10.116.6 we have \dim _{\mathfrak q} S = \dim _{\mathfrak q_ K} S_ K. Hence the integer c occurring in Lemma 10.135.4 is the same for the pair (S = k[x_1, \ldots , x_ n]/I, \mathfrak q) as for the pair (S_ K = K[x_1, \ldots , x_ n]/I_ K, \mathfrak q_ K). On the other hand we have
\begin{eqnarray*} I \otimes _{k[x_1, \ldots , x_ n]} \kappa (\mathfrak q') \otimes _{\kappa (\mathfrak q')} \kappa (\mathfrak q_ K') & = & I \otimes _{k[x_1, \ldots , x_ n]} \kappa (\mathfrak q_ K') \\ & = & I \otimes _{k[x_1, \ldots , x_ n]} K[x_1, \ldots , x_ n] \otimes _{K[x_1, \ldots , x_ n]} \kappa (\mathfrak q_ K') \\ & = & (K \otimes _ k I) \otimes _{K[x_1, \ldots , x_ n]} \kappa (\mathfrak q_ K') \\ & = & I_ K \otimes _{K[x_1, \ldots , x_ n]} \kappa (\mathfrak q'_ K). \end{eqnarray*}
Therefore, \dim _{\kappa (\mathfrak q')} I \otimes _{k[x_1, \ldots , x_ n]} \kappa (\mathfrak q') = \dim _{\kappa (\mathfrak q'_ K)} I_ K \otimes _{K[x_1, \ldots , x_ n]} \kappa (\mathfrak q_ K'). Thus it follows from Nakayama's Lemma 10.20.1 that the minimal number of generators of I_{\mathfrak q'} is the same as the minimal number of generators of (I_ K)_{\mathfrak q'_ K}. Thus the lemma follows from characterization (2) of Lemma 10.135.4. \square
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