# S is a ring with identity $f(1)$

Let $f:A\to B$ be a ring homomorphism with $f(r)\not=0$ for some $r\in {A}$. $A$ has identity 1 and $B$ has no zero divisors. I need to show that $B$ is a ring with identity $f(1)$.

I haven't made any progress on this problem and would appreciate a hint.

I don't have to prove that $B$ is a ring, right?(since it is a ring homomorphism). Also, to show that $f(1)$ is the identity, I need to show that $bf(1)=f(1)b$ for each $b$ in $B$ but I am not able to.

## 1 Answer

You have to prove that $f(1)b=b=bf(1)$, for all $b\in B$.

First of all, $f(r)\ne0$ implies $f(1)\ne0$; indeed, $f(1)=0$ implies $f(a)=f(1a)=f(1)f(a)=0$ for all $a\in A$.

Set $e=f(1)$, for simplicity. Then $e^2=f(1)f(1)=f(1\cdot 1)=f(1)=e$, so $e^2=e$.

Let $b\in B$: then $eb=e^2b$, so $e(b-eb)=0$. Since $B$ has no zero divisors and $e\ne0$, we conclude $b-eb=0$.

Can you go on?

• Thank you, this is truly ingenious. – user96343 Oct 26 '15 at 14:02
• Similarly, we can look at $be=be^2$. $e(b-be)=0\implies b=be$ – user96343 Oct 26 '15 at 14:03