# The Stacks Project

## Tag: 06DQ

This tag has label spaces-descent-lemma-descend-unibranch and it points to

The corresponding content:

Lemma 51.8.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $x \in |X|$. If $f$ is flat at $x$ and $X$ is geometrically unibranch at $x$, then $Y$ is geometrically unibranch at $f(x)$.

Proof. Consider the map of étale local rings $\mathcal{O}_{Y, f(\overline{x})} \to \mathcal{O}_{X, \overline{x}}$. By Morphisms of Spaces, Lemma 45.27.8 this is flat. Hence if $\mathcal{O}_{X, \overline{x}}$ has a unique minimal prime, so does $\mathcal{O}_{Y, f(\overline{x})}$ (by going down, see Algebra, Lemma 9.36.17). $\square$

\begin{lemma}
\label{lemma-descend-unibranch}
Let $S$ be a scheme.
Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.
Let $x \in |X|$.
If $f$ is flat at $x$ and $X$ is geometrically unibranch at $x$, then $Y$ is
geometrically unibranch at $f(x)$.
\end{lemma}

\begin{proof}
Consider the map of \'etale local rings
$\mathcal{O}_{Y, f(\overline{x})} \to \mathcal{O}_{X, \overline{x}}$.
By
Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-flat-at-point-etale-local-rings}
this is flat. Hence if $\mathcal{O}_{X, \overline{x}}$ has a unique minimal
prime, so does $\mathcal{O}_{Y, f(\overline{x})}$ (by going down, see
Algebra, Lemma \ref{algebra-lemma-flat-going-down}).
\end{proof}


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