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?