Lemma 33.5.1. Let $K/k$ be an extension of fields. Let $X$ be scheme over $k$ and set $Y = X_ K$. If $y \in Y$ with image $x \in X$, then

1. $\mathcal{O}_{X, x} \to \mathcal{O}_{Y, y}$ is a faithfully flat local ring homomorphism,

2. with $\mathfrak p_0 = \mathop{\mathrm{Ker}}(\kappa (x) \otimes _ k K \to \kappa (y))$ we have $\kappa (y) = \kappa (\mathfrak p_0)$,

3. $\mathcal{O}_{Y, y} = (\mathcal{O}_{X, x} \otimes _ k K)_\mathfrak p$ where $\mathfrak p \subset \mathcal{O}_{X, x} \otimes _ k K$ is the inverse image of $\mathfrak p_0$.

4. we have $\mathcal{O}_{Y, y}/\mathfrak m_ x\mathcal{O}_{Y, y} = (\kappa (x) \otimes _ k K)_{\mathfrak p_0}$

Proof. We may assume $X = \mathop{\mathrm{Spec}}(A)$ is affine. Then $Y = \mathop{\mathrm{Spec}}(A \otimes _ k K)$. Since $K$ is flat over $k$, we see that $A \to A \otimes _ k K$ is flat. Hence $Y \to X$ is flat and we get the first statement if we also use Algebra, Lemma 10.39.17. The second statement follows from Schemes, Lemma 26.17.5. Now $y$ corresponds to a prime ideal $\mathfrak q \subset A \otimes _ k K$ and $x$ to $\mathfrak r = A \cap \mathfrak q$. Then $\mathfrak p_0$ is the kernel of the induced map $\kappa (\mathfrak r) \otimes _ k K \to \kappa (\mathfrak q)$. The map on local rings is

$A_\mathfrak r \longrightarrow (A \otimes _ k K)_\mathfrak q$

We can factor this map through $A_\mathfrak r \otimes _ k K = (A \otimes _ k K)_{\mathfrak r}$ to get

$A_\mathfrak r \longrightarrow A_\mathfrak r \otimes _ k K \longrightarrow (A \otimes _ k K)_\mathfrak q$

and then the second arrow is a localization at some prime. This prime ideal is the inverse image of $\mathfrak p_0$ (details omitted) and this proves (3). To see (4) use (3) and that localization and $- \otimes _ k K$ are exact functors. $\square$

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