## 33.5 Change of fields and local rings

Some preliminary results on what happens to local rings under an extension of ground fields.

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

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

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

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

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.
$\square$

Lemma 33.5.2. Notation as in Lemma 33.5.1. Assume $X$ is locally of finite type over $k$. Then

\[ \dim (\mathcal{O}_{Y, y}/\mathfrak m_ x\mathcal{O}_{Y, y}) = \text{trdeg}_ k(\kappa (x)) - \text{trdeg}_ K(\kappa (y)) = \dim (\mathcal{O}_{Y, y}) - \dim (\mathcal{O}_{X, x}) \]

**Proof.**
This is a restatement of Algebra, Lemma 10.116.7.
$\square$

Lemma 33.5.3. Notation as in Lemma 33.5.1. Assume $X$ is locally of finite type over $k$, that $\dim (\mathcal{O}_{X, x}) = \dim (\mathcal{O}_{Y, y})$ and that $\kappa (x) \otimes _ k K$ is reduced (for example if $\kappa (x)/k$ is separable or $K/k$ is separable). Then $\mathfrak m_ x \mathcal{O}_{Y, y} = \mathfrak m_ y$.

**Proof.**
(The parenthetical statement follows from Algebra, Lemma 10.43.6.) Combining Lemmas 33.5.1 and 33.5.2 we see that $\mathcal{O}_{Y, y}/\mathfrak m_ x \mathcal{O}_{Y, y}$ has dimension $0$ and is reduced. Hence it is a field.
$\square$

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