I claim that flat local maps $(A, m_A) \to (B, m_B)$ are faithfully flat injections. To show faithful flatness of $B$ it suffices to show that $B \otimes_A k(m_A) = B/m_AB \neq 0$. However, by the local morphism definition, we know that $m_AB \subset m_B$ so that there is an induced surjection $B/m_AB \to k(m_B)$. In particular, $B/m_AB \neq 0$ so the map is faithfully flat.
Now, we want to show the map is injective. Indeed, if $K$ is the kernel, then $K \otimes_A B = 0$ which implies $K = 0$ by faithful flatness, since $B$ is a nonzero $A$-module. (see Stacks Lemma 10.39.15).
Is this reasoning correct?
Thanks!
Best Answer
I'd say you need more justification why $K\otimes_A B=0$. You have an injection $K\hookrightarrow A$ which you're tensoring with $B$ to get an injection $K\otimes_A B\hookrightarrow B$ where injectivity comes from flatness. The image is $KB$ so you're getting $K\otimes_A B\simeq KB=0$. Now the rest of what you say works.
Inspecting your proof you see that any faithfully flat morphism is injective, so together with your first paragraph (which all looks good) you have separately (flat + local implies faithfully flat) and (faithfully flat implies injective).