Lemma 33.39.9. Let $k$ be a field. Let $X$ be a locally algebraic $k$-scheme. Let $K/k$ be a field extension and set $Y = X_ K$. Let $y \in Y$ with image $x \in X$. Assume assumptions (a), (b), (c) of Lemma 33.27.6 hold for $x \in X$ and that $\dim (\mathcal{O}_{Y, y}) = 1$. Then the $\delta $-invariant of $X$ at $x$ is $\delta $-invariant of $Y$ at $y$.

**Proof.**
Set $A = \mathcal{O}_{X, x}$ and $B = \mathcal{O}_{Y, y}$. By Lemma 33.27.6 we see that $A$ is geometrically reduced. Hence $B$ is a localization of $A \otimes _ k K$. Let $A \to A'$ be as in Lemma 33.39.2. By Lemma 33.27.6 we see that $A' \otimes _ k K$ is normal. Hence

is normal, finite over $B$, and $B \to B'$ induces an isomorphism on total rings of fractions. Namely, pick $f \in \mathfrak m_ A$ satisfying (1) – (6) of Lemma 33.39.1; since $\dim (B) = 1$ we see that $f \in \mathfrak m_ B$ playes the same role for $B$ and we see that $B_ f = B'_ f$ because $A_ f = A'_ f$. It follows that $B \to B'$ is as in Lemma 33.39.2 for $B$. Thus we have to show that $\text{length}_ A(A'/A) = \text{length}_ B(B'/B) = \text{length}_ B((A'/A) \otimes _ A B)$. Since $A \to B$ is flat (as a localization of $A \to A \otimes _ k K$) and since $\mathfrak m_ B = \mathfrak m_ A B$ (because $B/\mathfrak m_ A B$ is zero dimensional by the remarks above and a localization of $K \otimes _ k \kappa (x)$ which is reduced as $\kappa (x)$ is separable over $k$) we conclude by Algebra, Lemma 10.52.13. $\square$

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