Lemma 33.12.5. Let $k$ be a field. Let $f : X \to Y$ be a morphism of locally Noetherian schemes over $k$. Let $x \in X$ be a point and set $y = f(x)$. If $X$ is geometrically regular at $x$ and $f$ is flat at $x$ then $Y$ is geometrically regular at $y$. In particular, if $X$ is geometrically regular over $k$ and $f$ is flat and surjective, then $Y$ is geometrically regular over $k$.

Proof. Let $k'$ be finite purely inseparable extension of $k$. Let $f' : X_{k'} \to Y_{k'}$ be the base change of $f$. Let $x' \in X_{k'}$ be the unique point lying over $x$. If we show that $Y_{k'}$ is regular at $y' = f'(x')$, then $Y$ is geometrically regular over $k$ at $y'$, see Lemma 33.12.3. By Morphisms, Lemma 29.25.7 the morphism $X_{k'} \to Y_{k'}$ is flat at $x'$. Hence the ring map

$\mathcal{O}_{Y_{k'}, y'} \longrightarrow \mathcal{O}_{X_{k'}, x'}$

is a flat local homomorphism of local Noetherian rings with right hand side regular by assumption. Hence the left hand side is a regular local ring by Algebra, Lemma 10.110.9. $\square$

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