Faithful flatness and non-commutative algebras

Question

$\DeclareMathOperator\Spec{Spec}$When dealing with commutative algebras, a usefull criterion for faithful flatness is the following:

Let $f:A\rightarrow B$ be a morphism of commutative algebras. Then $f$ is faithfully flat if and only if $f$ is flat and the associated map $f^*:\Spec(B)\rightarrow \Spec(A)$ is surjective.

This criterion follows from the fact that a flat map between local rings is faithfully flat, and that a module is zero if and only if its localizations at prime ideals are zero.

Now, what happens if $A$ and $B$ are not necessarily commutatitve? In this situation, localization does not work as easily, and one cannot define the spectrum of a non-commutative algebra (at least in a naïve way). Hence, the criterion above will not hold in this setting.

So, apart from checking the definition, are there any general criteria to show that a flat map of non-commutative algebras is faithfully flat?

Answer 1

There is a simple criterion. I learned from Jacob Lurie's paper Tannaka Duality for Geometric Stacks, Lemma 5.5. His argument works, but since his statement does not cover the current situation, let me phrase an adapted statement and transcribe his proof, hopefully correct.

Let $f\colon A\to B$ be a map of associative rings. Let $A''$ denote the cokernel of $f$ as an abelian group, which carries a canonical $A$-$A$-bimodule structure. Then the map $f$ exhibits $B$ as a faithfully flat left $A$-module if and only if

1. the map $f$ is injective, and
2. the left $A$-module $A''$ is flat.

Proof. We start with assuming the injectivity of $f$ and the flatness of $A''$. Then the left $A$-module $B$ is an extension of $A''$ by $A$. Since flat modules are stable under extensions, the left $A$-module $B$ is flat. For every right $A$-module $N$, we have an exact sequence $$\DeclareMathOperator\Tor{Tor} 0=\Tor^A_1(N,A'')\to N\to N\otimes_AB\to N\otimes_AA''\to0. $$ If $N\otimes_AB=0$, then $N=0$. It follows that the left $A$-module $B$ is faithfully flat.

Now we assume that the map $f$ exhibits $B$ as a faithfully flat left $A$-module. Then the map $B=A\otimes_AB\xrightarrow{f\otimes1}B\otimes_AB$ of abelian groups is injective, since it admits a retract given by the multiplication map $B\otimes_AB\to B$. Since the left $A$-module $B$ is faithfully flat, it follows that the map $f$ of right $A$-modules is injective.

It remains to see that the left $A$-module $A''$ is flat. Let $N'\to N$ be any injective map of right $A$-modules. We have to see that the map $N'\otimes_AA''\to N\otimes_AA''$ of right $A$-modules is injective, which is equivalent to the injectivity of $N'\otimes_AA''\otimes_AB\to N\otimes_AA''\otimes_AB$ by faithful flatness of the left $A$-module $B$. We now examine the short exact sequence $$ 0\to B=A\otimes_AB\xrightarrow{f\otimes1}B\otimes_AB\to A''\otimes_AB\to0 $$ of left $A$-modules (by left flatness of $B$). As previously, the map $f\otimes1$ of left $A$-module admits a retract given by the multiplication $B\otimes_AB\to B$ which is also left $A$-linear. It follows that the left $A$-module $A''\otimes_AB$ is a direct summand of the left $A$-module $B\otimes_AB$. Since the left $A$-module $B\otimes_AB$ is flat, so is the left $A$-module $A''\otimes_AB$, and the result follows.

Answer 2

If $A$ happens to be (left) Noetherian then to show $f$ makes $B$ a faithfully flat right $A$-module it is enough to check that $B\otimes_A S\neq 0$ for every simple (left) $A$-module $S$ since, in that case, any $A$-module $M$ has a simple subquotient $S$ and $B\otimes_A S$ will be a subquotient of $B\otimes_A M$. This is because if $N$ is a submodule of $M$ then $B\otimes_A N \to B\otimes_A M$ is an injection and if $S$ is a simple quotient of $N$ then $B\otimes_A N\to B\otimes_A S$ is a surjection.