Existence of a ring homomorphism implies characteristic divisibility

Question

Let $A$ and $B$ be rings having respective characteristics $n$ and $k.$ Show that if there exists a ring homomorphism $f : A \to B$, then $k$ divides $n$.

I don't know how to start this problem. If someone has tips, I would appreciate it.

Answer 1

Let's assume for now that the characteristics are both positive. As suggested by Anurag, note that $f(1_A)=1_B$, so that $n\cdot 1_A=0$ implies that $f(n\cdot 1_A)=n\cdot 1_B=0$. However, the characteristic $k$ of $B$ is the least number so that $k\cdot 1_B=0$. Hence, $k\le n$. Write $n=mk+r$ with $0\le r<k$. Then: $$ 0=(mk+r)1_B=mk\cdot1_B+r\cdot 1_B=r\cdot 1_B$$ whence $r=0$. So, $k\mid n$. In the case where $n=0$, then $k\mid n$ automatically. If $k=0$, then $n\cdot 1_B=0$ implies that $n=0$, whence we are done, since $0\mid 0$.