**Proof.**
The equivalence of (1) and (2) is immediate from the definitions. Let $x \in X$ with $y = f(x)$. By definition $f$ is flat at $x$ if and only if $\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}$ is a flat ring map, and $X_ y$ is geometrically regular at $x$ over $\kappa (y)$ if and only if $\mathcal{O}_{X_ y, x} = \mathcal{O}_{X, x}/\mathfrak m_ y\mathcal{O}_{X, x}$ is a geometrically regular algebra over $\kappa (y)$. Hence Whether or not $f$ is regular at $x$ depends only on the local homomorphism of local rings $\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}$. Thus the equivalence of (1) and (4) is clear.

Recall (More on Algebra, Definition 15.41.1) that a ring map $A \to B$ is regular if and only if it is flat and the fibre rings $B \otimes _ A \kappa (\mathfrak p)$ are Noetherian and geometrically regular for all primes $\mathfrak p \subset A$. By Varieties, Lemma 33.12.3 this is equivalent to $\mathop{\mathrm{Spec}}(B \otimes _ A \kappa (\mathfrak p))$ being a geometrically regular scheme over $\kappa (\mathfrak p)$. Thus we see that (2) implies (3). It is clear that (3) implies (5). Finally, assume (5). This implies that $f$ is flat (see Morphisms, Lemma 29.25.3). Moreover, if $y \in Y$, then $y \in V_ j$ for some $j$ and we see that $X_ y = \bigcup _{i \in I_ j} U_{i, y}$ with each $U_{i, y}$ geometrically regular over $\kappa (y)$ by Varieties, Lemma 33.12.3. Another application of Varieties, Lemma 33.12.3 shows that $X_ y$ is geometrically regular. Hence (2) holds and the proof of the lemma is finished.
$\square$

## Comments (0)