Let $R$ and $S$ be commutative rings with $1$ and $\phi: R\rightarrow S$ be a surjective ring homomorphism. Then for an arbitrary maximal ideal $I$ of $R$, does $\phi(I)$ have to be maximal in $S$?
My idea: from the isomorphism theorem, maximal ideals of $R$ containing $Ker\phi$ are in 1-1 correspondence with maximal ideals of $S$. Hence there may exist some maximal ideals $I$ of $R$ such that $Ker \phi$ not contained in $I$, thus $\phi(I)$ is not a maximal ideal of $S$. But this is not valid proof (this is because that if we replace maximal with prime, we can use the same argument. However, the surjective homomorphism image of a prime ideal is prime, which I have proved using the definition). What should I do?
Best Answer
The answer is NO! take $R=k[x]$, $S=k[x]/x$, and $I=(x-1)$.