Let $A$ be a Noetherian domain and $B$ a finite $A$-algebra containing $A$ as a subring. Suppose there is a number $n$ such that for every maximal ideal $\mathfrak{m}$ of $A$, $$\dim_{k(\mathfrak{m})} B \mathbin{\otimes_A} k(\mathfrak{m}) = n$$ where $k(\mathfrak{m}) = A / \mathfrak{m}$. Why is $B$ flat in this case?

It's generally true that a finitely-generated module over a Noetherian ring is flat if and only if it is locally free (in the sense of the stalks being free), and if a finitely-generated module $M$ over a Noetherian domain $A$ is locally free, then its has constant local rank, in the sense that there is a number $n$ such that for all $\mathfrak{p} \in \operatorname{Spec} A$, $$\dim_{k(\mathfrak{p})} M \mathbin{\otimes_A} k(\mathfrak{p}) = n$$ where $k(\mathfrak{p}) = A_\mathfrak{p} / \mathfrak{p} A_\mathfrak{p}$. The converse is true if $A$ is reduced. (See, for example, Hartshorne [Algebraic Geometry, Ch. II, Exercise 5.8]) If we strengthen our hypotheses and demand that $A$ be a Noetherian Jacobson domain (or, at least, a Noetherian domain with trivial Jacobson radical), it is sufficient to assume that only the stalks over maximal ideals have dimension $n$ to prove that $M$ is locally free. But as far as I know Noetherian domains may have non-trivial Jacobson radical, e.g. Noetherian local rings. Am I supposed to use the hypothesis that $A$ is a subring of $B$ to prove the claim?

  • $\begingroup$ This is not true. Consider the case when $A$ is a local domain (localization at a non-normal point of an algebraic variety) and $B$ its normalization. $\endgroup$ – user18119 Nov 9 '11 at 17:57
  • $\begingroup$ @QiL: Thanks. I was wondering if the question was even correct. If you post more details about your counterexample I will be glad to accept it as an answer. $\endgroup$ – Zhen Lin Nov 9 '11 at 20:51

More details on the counterexample in the comments.

Consider the singular curve associated to $k[x,y]$, $y^2=x^3$ ($k$ is any field). Let $\mathfrak m$ be the maximal ideal of $k[x,y]$ generated by $x,y$ and let $A$ be the localisation $k[x,y]_{\mathfrak m}$. The normalization of $k[x,y]$ is $k[t]$ where $t=y/x$ and $x=t^2, y=t^3$. Then $k[t]$ is finite over $k[x,y]$ (generated by $1, t$). So the localization $B:=k[t]\otimes_{k[x,y]} A$ is finite over $A$. Now $B\otimes_A k(\mathfrak m)=k[t]/(t^2, t^3)$ has dimension $2$ over $k=k(\mathfrak m)$, while $B\otimes_A k(\mathfrak p)$, where $\mathfrak p=\{0\}$, has dimension $1$ over $k(\mathfrak p)$. So $B$ is not flat over $A$.

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