Proof.
We may replace X and Y by suitable affine open neighbourhoods of x and y. Then Y is the spectrum of a domain A and X is the spectrum of a finite type A-algebra B. Let \mathfrak q \subset B be the prime ideal corresponding to x and \mathfrak p \subset A the prime ideal corresponding to y. The local ring A_\mathfrak p = \mathcal{O}_{Y, y} is geometrically unibranch. The ring map A \to B is unramified at \mathfrak q. Also, the point x' in (3) corresponds to a prime ideal \mathfrak q' \subset \mathfrak q such that A \cap \mathfrak q' = (0). It follows that A_\mathfrak p \to B_\mathfrak q is injective. We conclude by More on Algebra, Lemma 15.107.2.
\square
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