When dealing with commutative algebras, a usefull criterion for faithfull 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 naive 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 falt?

$\DeclareMathOperator\Spec{Spec}$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?