A simple example of a ring that is an $A$-module but not an $A$-algebra

Question

Let $A$ and $B$ be commutative rings, and suppose (the underlying group of) $B$ has a structure of $A$-module. "Obviously", that doesn't imply that $B$ gets a structure as an $A$-algebra, but I can't come up with a simple example.

Surely there must be some relatively down-to-earth example, though. Can you help me?

Answer 1

Let $B=\def\QQ{\mathbb Q}\QQ(t)$, which is among other things a rational vector space of countable dimension. In particular, it has a structure of a $\QQ(t_1,t_2)$-module of rank one. But $B$ is not an $\QQ(t_1,t_2)$-algebra in any way, as there are no ring maps $\QQ(t_1,t_2)\to B$.

Answer 2

I came up with a counterexample myself: let $A=k[\partial_t]$, and $B=k[t]$, with $A$ acting by differentiation; clearly there is no ring map $\phi:A\to B$ compatible with that action.