7
$\begingroup$

Let $R$ and $S$ be nonzero rings with identity and denote their respective identities by $1_R$ and $1_S$. Let $\varphi: R \to S$ be a nonzero homomorphism of rings.

Prove that if $\varphi(1_R) \ne 1_s$ then $\varphi(1_R)$ is a zero divisor in $S$.

I've seen many variations of this problem elsewhere on MSE, but none in this particular form. I've got the following:

Since $\varphi$ is a homomorphism, it follows that $\varphi(1_R) = \varphi(1_R) \varphi(1_R)$; and so

$$\varphi(1_R)1_S = \varphi(1_R) \varphi(1_R) \implies \varphi(1_R)[1_S - \varphi(1_R)] = 0. $$

Since $\varphi(1_R) \ne 1_S$ by assumption, either $\varphi(1_R)$ is a zero divisor (in which case, we're done), or $\varphi(1_R) = 0$.

Here's where I'm stuck because I don't see why $\varphi(1_R)$ can't be equal to $0$. Indeed, the statement of the problem says that $\varphi$ is a nonzero homomorphism of rings, but is there some reason why just $1_R$ can't be in ker$(\varphi)$?

$\endgroup$
7
$\begingroup$

If $\phi(1_R)=0$, then we have for all $r\in R$ $$ \phi(r)=\phi(1_Rr)=\phi(1_R)\phi(r)=0, $$ hence $\phi=0$, which was excluded.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.