If $A$ is normal and $\Omega^1_{B/A}=0$ then $B$ is normal

Let $A\subseteq B$ be two noetherian domains with fraction fields $k$ and $L$, respectively. Assume that $A$ is normal and $B$ is finite as $A$-module. I'm asking myself if $B$ is also normal if $\Omega^1_{B/A}=0$?

Any suggestion or reference in the literature is welcome.

EDIT: I'm trying to show the claim above, the idea is the following.

IDEA: Let $\mathfrak{q}\subset B$ be a maximal ideal of $B$ and set $\mathfrak{p}=\mathfrak{q}\cap A$, where $\mathfrak{p}$ is maximal since $B$ is integral over $A$. Consider the map $k:=A/\mathfrak p\longrightarrow B/\mathfrak p B:=B'$ induced by the inclusion $A\subseteq B$. We get a finite $k$-algebra $B'$, with module of differentials $\Omega^1_{B'/k}\simeq B'\otimes_{B}\Omega^1_{B/A}=0$. Now, since $\Omega^1_{B'/k}=0$, then $B'\simeq K_1\times\cdots\times K_n$, where any $K_i$ is a finite (and separable) field extension of $k$, hence we get $$B'_{\mathfrak q}\simeq B_\mathfrak q/\mathfrak pB_\mathfrak q,$$ but the localization at a prime of a finite product of fields (which is $B'$) is a field, hence $\mathfrak p B_\mathfrak q$ is maximal, i.e. $\mathfrak p B_\mathfrak q=\mathfrak q B_\mathfrak q$. Now let $\mathfrak q'$ a prime strictly contained in $\mathfrak q$. We have $\mathfrak q'B_\mathfrak q\subset \mathfrak pB_\mathfrak q$, then $(\mathfrak q'B_\mathfrak q )\cap A=\mathfrak p'\subset \mathfrak p$. This shows (localizing at $\mathfrak q'$) that $\mathfrak q'B_{\mathfrak q'}= \mathfrak p'B_{\mathfrak q'}$, where $\mathfrak q'\cap A=\mathfrak p'$. We showed that for any prime $\mathfrak q\subset B$ there is a prime $\mathfrak p\subset A$ (with $\mathfrak q\cap A=\mathfrak p$), such that $$\mathfrak p B_\mathfrak q=\mathfrak q B_\mathfrak q \, \, \, (*).$$ By Serre's Normality Criterion (SNC) the ring $B$ is normal if and only if for every prime $\mathfrak q$ associated to a principal ideal the ring $B_\mathfrak q$ is a DVR, i.e. $\mathfrak q B_\mathfrak q$ is principal. If we are able to show (I don't know if it is true) that for every prime $\mathfrak q$ associated to a principal ideal we have $\mathrm{ht}(\mathfrak q)=1$, since $\mathrm{ht}(\mathfrak q)=\mathrm{ht}(\mathfrak q\cap A)$ ($B$ is integral over $A$) and $(\mathfrak pA_\mathfrak p)(B_\mathfrak q)=\mathfrak q B_\mathfrak q$, being $\mathfrak p A_\mathfrak p$ principal, we obtain that $\mathfrak qB_\mathfrak q$ is principal, hence by SNC the ring $B$ is normal.

• SGA 1, Corollaire I.9.11: "Let $f \colon X \to Y$ be a dominant morphism [of schemes], $Y$ being normal and $X$ connected. If $f$ is unramified, then $f$ is étale, and therefore $X$ is normal." – Martin Bright Jan 31 '18 at 15:32
• This result is also proved as Lemma 1.5 in Chapter I of the book "Etale Cohomology and the Weil Conjectures" by Freitag & Kiehl in much more robust generality: you can relax "finite as $A$-module" to "finitely generated as $A$-algebra" (they state it for localization of such). It ultimately rests on the etale-local structure of unramified maps. – nfdc23 Jan 31 '18 at 15:46
• Just for completeness, here's a counterexample if $B/A$ is not flat; take $A=R[x,y]$ and $B=R[x,y]/(y^2-x^3)$, for $R$ any field. – Daniel Litt Jan 31 '18 at 16:31
• @DanielLitt The map $A\longrightarrow B$ is not injective. – Vincenzo Zaccaro Jan 31 '18 at 16:37
• The whole point of the result in SGA1 and the Freitag-Kiehl book is that flatness does not need to be assumed (only injectivity/dominance), and the conclusion gives that flatness is therefore a consequence of the assumptions. Please remove the "EDIT" about the flat case; there is nothing interesting being asserted if one allows to assume flatness. – nfdc23 Jan 31 '18 at 17:10