**Proof.**
If $x$ is a closed point of $X$, then $y$ is too (look at residue fields). But conversely, this need not be the case, i.e., it can happen that a closed point of $Y$ maps to a nonclosed point of $X$. However, in this case $y$ cannot be a node. Namely, then $X$ would be geometrically unibranch at $x$ (because $x$ would be a generic point of $X$ and $\mathcal{O}_{X, x}$ would be Artinian and any Artinian local ring is geometrically unibranch), hence $Y$ is geometrically unibranch at $y$ (Varieties, Lemma 33.40.3), which means that $y$ cannot be a node by Lemma 53.19.7. Thus we may and do assume that both $x$ and $y$ are closed points.

Choose algebraic closures $\overline{k}$, $\overline{K}$ and a map $\overline{k} \to \overline{K}$ extending the given map $k \to K$. Using the equivalence of (1) and (3) in Lemma 53.19.7 we reduce to the case where $k$ and $K$ are algebraically closed. In this case we can argue as in the proof of Lemma 53.19.7 that the geometric number of branches and $\delta $-invariants of $X$ at $x$ and $Y$ at $y$ are the same. Another argument can be given by choosing an isomorphism $k[[x_1, \ldots , x_ n]]/(g_1, \ldots , g_ m) \to \mathcal{O}_{X, x}^\wedge $ of $k$-algebras as in Varieties, Lemma 33.21.1. By Varieties, Lemma 33.21.2 this gives an isomorphism $K[[x_1, \ldots , x_ n]]/(g_1, \ldots , g_ m) \to \mathcal{O}_{Y, y}^\wedge $ of $K$-algebras. By definition we have to show that

\[ k[[x_1, \ldots , x_ n]]/(g_1, \ldots , g_ m) \cong k[[s, t]]/(st) \]

if and only if

\[ K[[x_1, \ldots , x_ n]]/(g_1, \ldots , g_ m) \cong K[[s, t]]/(st) \]

We encourage the reader to prove this for themselves. Since $k$ and $K$ are algebraically closed fields, this is the same as asking these rings to be as in Lemma 53.19.3. Via Lemma 53.19.6 this translates into a statement about the $(n - 1) \times (n - 1)$-minors of the matrix $(\partial g_ j/\partial x_ i)$ which is clearly independent of the field used. We omit the details.
$\square$

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