Suppose that $S'$ is a faithfully flat $S$-algebra, and let $T$ be a $S'$-algebra.
My question: is there some theorem that says that the canonical morphism $$T\underset{S}{\otimes}T\rightarrow T\underset{S'}{\otimes}T$$
is also faithfully flat?
Thanks in advance.
(All rings are supposed to be commutative and unitaries)
Let us suppose $f:=T\otimes _S T \to T\otimes _{S’}T$ is faithfully flat. Since $f$ is surjective ring homomorphism, it is a quotient by some ideal. Bur for a general ring $A$ and its ideal $I$, there is an equivalence such that $A/I$ is faithfully flat over $A \iff $ $I=0$, namely, $f$ must be an isomorphism.
Proof of the equivalence: suppose $A/I$ is faithfully flat over $A$. Tensoring exact sequence $$0\to I \to A\to A/I \to 0$$ with $A/I$ over $A$, we obtain exact sequence $$ 0\to I\otimes _A A/I \to A/I \overset{\sim}{\to} A/I \to 0$$ showing $I\otimes_A A/I =0$. If $I\neq0$, this contradicts to the faithfully flat assumption.