Lemma 33.39.8. 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 $X$ is geometrically reduced at $x$ and $\dim (\mathcal{O}_{X, x}) = \dim (\mathcal{O}_{Y, y}) = 1$. Then
Proof. Set $A = \mathcal{O}_{X, x}$ and $B = \mathcal{O}_{Y, y}$. By Lemma 33.6.2 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. Then
is 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 $A_ f = A'_ f$. Let $B''$ be the integral closure of $B$ in its total ring of fractions as in Lemma 33.39.2. Then $B' \subset B''$. Thus the $\delta $-invariant of $Y$ at $y$ is $\text{length}_ B(B''/B)$ and
by Algebra, Lemma 10.52.13 since $A \to B$ is flat (as a localization of $A \to A \otimes _ k K$). Since $\text{length}_ A(A'/A)$ is the $\delta $-invariant of $X$ at $x$ and since $\text{length}_ B(B/\mathfrak m_ A B) \geq 1$ the lemma is proved. $\square$
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