Lemma 10.116.7. Let k be a field. Let S be a finite type k-algebra. Let K/k be a field extension. Set S_ K = K \otimes _ k S. Let \mathfrak q \subset S be a prime and let \mathfrak q_ K \subset S_ K be a prime lying over \mathfrak q. Then
\dim (S_ K \otimes _ S \kappa (\mathfrak q))_{\mathfrak q_ K} = \dim (S_ K)_{\mathfrak q_ K} - \dim S_\mathfrak q = \text{trdeg}_ k \kappa (\mathfrak q) - \text{trdeg}_ K \kappa (\mathfrak q_ K)
Moreover, given \mathfrak q we can always choose \mathfrak q_ K such that the number above is zero.
Proof.
Observe that S_\mathfrak q \to (S_ K)_{\mathfrak q_ K} is a flat local homomorphism of local Noetherian rings with special fibre (S_ K \otimes _ S \kappa (\mathfrak q))_{\mathfrak q_ K}. Hence the first equality by Lemma 10.112.7. The second equality follows from the fact that we have \dim _ x X = \dim _{x_ K} X_ K with notation as in Lemma 10.116.6 and we have \dim _ x X = \dim S_\mathfrak q + \text{trdeg}_ k \kappa (\mathfrak q) by Lemma 10.116.3 and similarly for \dim _{x_ K} X_ K. If we choose \mathfrak q_ K minimal over \mathfrak q S_ K, then the dimension of the fibre ring will be zero.
\square
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