I'm overcoming my fear of tensor products, and the following exercise got me wondering:
Give an example of commutative rings $A$ and $B$ with $\operatorname{char}A=\operatorname{char}B$ such that the map $$A \longrightarrow\ A\otimes_{\Bbb{Z}}B:\ a\ \longmapsto\ a\otimes1,$$ is not injective.
Examples with $\operatorname{char}A\neq\operatorname{char}B$ are of course abundant, and examples with $\operatorname{char}A=\operatorname{char}B=0$ aren't too difficult either. But I am unable to find an example with $\operatorname{char}A=\operatorname{char}B>0$. An example would be very welcome, but a clue as to where to look for one would also be much (or even more?) appreciated.
Off the top of my head, I think this can't be done. Assuming there's nothing wrong with the following proof:
Suppose $A$ and $B$ are rings of characteristic $m$, and that $a \otimes 1 = 0$ in $A \otimes_\mathbb{Z} B$ for some $a \in A$.
This means that we must have $n \in \mathbb{Z}$ and $b \in B$ such that $1 = n b$ and $n a = 0$, so that $$a \otimes 1 = a \otimes (n b) = (n a) \otimes b = 0 \otimes b = 0.$$
Now we have both $m a = 0$ and $n a = 0$, so by the Euclidean algorithm we see that $$\gcd(m, n) a = 0.$$
On the other hand, we have $m b = 0$ and $n b = 1$. Write $m = \gcd(m, n) \, m'$. Then we have
$$m' (1) = m' (n b) = (m' n) b = 0$$
since $m = m' \gcd(m, n)$ divides $m' n$. Since $m'(1) = 0$ in $B$, $0 < m' \leq m$, and $B$ is of characteristic $m$, this means that $m'=m$, i.e. $\gcd(m,n)=1$. From above, this says that $1 \cdot a = 0$, i.e. $a = 0$.