Flat ring homomorphism but not injective.

Question

Let $A\to B$ be flat ring homomorphism(i.e. $B$ is flat $A$ module.)

If $B$ is faithfully flat, then $A\to B$ is injection.

$\textbf{Q:}$ What is the example of flat but not injective ring homomorphism?(i.e. I want to fail faithfully flat but remain flat.) I think I need some ring $B$ as projective which realizes $B=F/N$ where $F$ is free $A-$module and this has to be compatible with ring structure as well. Clearly, I could not get this work over $A$ being a field.

Answer 1

Let $A$ be a nonzero and absolutely flat ring, let $\mathfrak{m}\subseteq A$ be a maximal ideal, and let $B=A/\mathfrak{m}$. Then, the canonical morphism $A\rightarrow B$ is flat but not injective.

Answer 2

As mentioned in the comments, one type of example is to take $A = B \times C$ for arbitrary non-zero $B$ and $C$, and let $A \to B$ be the projection. As a lazy way to see that this is flat, let $e = (1,0) \in B \times C$. Then the projection $A \to B$ can be identified with the localization map $A \to A[1/e]$, and localization is always flat.

For an example that is not of this form, take $k$ an arbitrary non-zero base ring and set $A = k[x,y] / (xy)$, $B = k[x, 1/x]$, and let $A \to B$ be the map $x \mapsto x$ and $y \mapsto 0$. You can check that $B$ is flat over $A$ by showing that $B = A[1/x]$ and again using that localization is always flat.

Note that if $A$ is an integral domain and $B$ is non-zero, then any flat ring map $\varphi \colon A \to B$ is injective. To see this, suppose towards a contradiction that there is some $f \in \ker(\varphi)$ with $f \neq 0$. Then $fA \cong A$ as $A$-modules, and in particular $fA \otimes_A B \cong B$ is non-zero. On the other hand, the injective map $fA \to A$ gives after tensoring the map $fA \otimes_A B \to B$ sending $a\otimes b$ to $\varphi(a)\cdot b$. But this is the zero map since $fA \subset \ker(\varphi)$. In particular, this map is not injective, contradicting the flatness of $A \to B$.