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.
Best Answer
An “elementwise” proof. Let $J$ be an ideal properly containing $\ker f$ and take $x\in J$, $x\notin \ker f$.
Then $f(x)\ne0$, so it is invertible, because $S$ is a field. Since $f$ is surjective, there exists $y\in R$ such that $f(y)=(f(x))^{-1}$ or, in other words, $$ f(xy)=1=f(1). $$ Therefore $t=xy-1\in\ker f$ and so $$ 1=xy-t\in J $$ (because $x\in J$ and $t\in\ker f\subset J$) which means $J=R$.
A more conceptual proof uses the homomorphism theorems: if $f\colon R\to S$ is a surjective homomorphism, then $f$ induces a bijection between the ideals of $R$ containing $\ker f$ and the ideals of $S$; such a bijection preserves inclusion (as it is defined by means of the direct image under $f$). If $S$ is a field, it has just the trivial ideals $\{0\}$ and $S$, so there is only one ideal in $R$ properly containing $\ker f$ and so $\ker f$ is maximal.