1
$\begingroup$

Let $R$ and $R'$ be commutative rings with unity $1$ and $1'$. Let $\phi: R\to R'$ be a surjective ring homomorphism with $\phi(1) = 1'$. I'm to show that when $M$ is a maximal ideal in $R$ such that ker $\phi \subseteq M$, then $\phi[M]$ will be a maximal ideal in $R'$.

Now I was thinking that since $R$ is a commutative ring with unity and $M$ is a maximal ideal, then $R/M$ is a field. Then $\psi: R/M \to R'/\phi[M]$ given by $\psi(a+M) = \phi(a)+\phi[M]$ is an isomorphism.

But I think I have to show that every non-zero element in $R'/\phi[M]$ is a unit, in order to say that it is a field. And I'm not sure how to do that.

Once that is done I can say that $R'/\phi[M]$ is a field and so $\phi[M]$ is a maximal ideal in $R'$, right?

$\endgroup$
2
$\begingroup$

You've already shown that $\psi:R/M \to R'/\phi[M]$ is an isomorphism. Since $R/M$ is a field and $R'/\phi[M]$ is isomorphic to $R/M$, $R'/\phi[M]$ is also a field.

Note: when you state that $\psi$ is an isomorphism, it bears mentioning (as justification) that $\psi$ is a surjective ring homomorphism from a field.

$\endgroup$
1
$\begingroup$

Suppose $\varphi(M)$ is not a maximal ideal, suppose $N'$ is an ideal containing $\varphi(M)$ and consider $\varphi^{-1}(N')$.

Hopefully we already know that the preimage of ideals is an ideal. To show that is is bigger than $\varphi(M)$ you should use that $\varphi^{-1}(m')$ is already contained in $M$ for all $m'\in \varphi(M)$ (to do this use that $\ker\varphi$ is contained in $M$).

You could also just use the correspondence theorem.

$\endgroup$

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.