Lemma 10.133.10. Let $k \subset K$ 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 10.133.5) 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 10.133.4 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 10.133.8).

By Lemma 10.115.6 we have $\dim _{\mathfrak q} S = \dim _{\mathfrak q_ K} S_ K$. Hence the integer $c$ occurring in Lemma 10.133.4 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

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 10.19.1 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 10.133.4. $\square$

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