TITLE: Descent of functions along finite birational morphisms
QUESTION [13 upvotes]: Let $A\to B$ be a morphism of (unitary commutative) rings such that $B$ is module-finite over $A$ and there exists $f\in A$ which is a nonzerodivisor in $A$ and in $B$, with $A[1/f]\to B[1/f]$ an isomorphism.
My question is: is the diagram $A\to B \rightrightarrows B\otimes_A B$ exact ?
In geometric terms, this is more or less equivalent to the question whether descent of functions along a finite, scheme-theoretically birational morphism is possible.
Here are some comments:

Of course, I'm mainly interested in the non-flat case (otherwise it's just faithfully flat descent).
I computed several examples, including the normalization of the scheme formed by $n$ lines meeting in the origin in $\mathbf{A}^2$ (i.e. the scheme $V(x^n-y^n)\subset\mathbf{A}^2$); and the resolution $k[x^n,y^n,xy]\to k[x,y]$ of a plane quotient singularity. Any time, the answer is yes.
I suspect that the result might be known and maybe present somewhere in the literature. The question is related also to descent of modules for universally injective morphisms, as exposed in the Stacks Project here but I could not find there something really close to the statement I'm after.

Thanks for any help!
EDIT (Jan 1st, 2016): the claim that the normalization of the scheme formed by $n$ lines meeting in the origin in $\mathbf{A}^2$ is an example is not true if $n>2$. Explanations are given in this MO answer.

REPLY [8 votes]: Apologies for a long comment masquerading as an answer.
I think you were on the right track when you mentioned universally injective ring homomorphisms, which are also often called pure ring homomorphisms.
You can find a well developed geometric theory of such homomorphisms in papers of Mesablishvili. The theory includes a generalization of the result that you and user52824 worked out in your answer. See, in particular, Theorem 6.5 of "More on descent theory for schemes," which states
"A quasi-compact morphism of schemes is stable schematically dominant if and only if it is pure if and only if it is a stable regular epimorphism."
Note that there is no "finite" hypothesis and no Noetherian hypothesis in the theorem.

The easiest way to get a pure ring homomorphism $R\to S$ is to have an $R$-module retraction $S\to R$. One often calls such a retraction a ``Reynolds operator'' because of the representation-theory situation (invariants of a linearly reductive group) that gives rise to such retractions. In fact, an elementary argument shows that if $S$ is finitely presented as an $R$-module (which is the situation you consider) and pure, then there is such a retraction.
For a proof that an $R$-algebra finitely presented as an $R$-module is pure if and only if it admits an $R$-module retraction onto $R$, one can see Corollary 5.3 of Hochster/Roberts "The purity of the Frobenius and local cohomology." The paper is now freely available online thanks to Elsevier's recent opening of access to older math articles. Although the corollary is in the middle of the paper, it belongs to a section about the basics of pure morphisms, and its proof is entirely contained in Section 5.
To get endless examples of pure ring homomorphisms, you can thus use representation theory: let $k$ be a field, let $G$ be a finite group whose order is invertible in $k$, and let $V$ be a finite-dimensional representation of $G$ over $k$. The inclusion $k[V]^G \to k[V]$ is then a pure ring homomorphism, since we have a $k[V]^G$-module retraction (the Reyonlds operator mentioned earlier) $k[V]\to k[V]^G$ defined by averaging over $G$. This sort of example generalizes your quotient-singularity computation.