Let $f:R\to S$ be a surjective homomorphism, where $R$ is a commutative ring and $S$ is a field. Prove that $\ker(f)$ is a maximal ideal.
I already know that $\ker(f)$ is an ideal of $R$. I tried to consider some ideal $J$ of $R$ such that $\ker(f) \subset J$. If we can show that for arbitrary $y\in J$, $f(y)=0$ then we are good. But I don't know how to show that. Specifically, how does being surjective come into play?
Another theorem I know is that if $f$ is a surjective homomorphism, then quotient ring $R/\ker(f)$ is isomorphic to $S$. Don't know if that's gonna help.