# Showing that something is a maximal ideal.

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?

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.

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.