Let $f$ be a surjective homomorphism. Prove that $\ker(f)$ is a maximal ideal

Question

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.

Answer 1

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.

Answer 2

Since $R/\ker(f)$ is a field, $\ker(f)$ is a maximal ideal.

Answer 3

As you mention, the quotient $R / \text{ker}(f)$ is isomorphic to $S$, hence a field.

It remains to show that if $I \subset R$ has the property that $R/I$ is a field, then $I$ must be a maximal ideal. To see this, argue by contradiction. Suppose $I$ is not maximal; then there exists an ideal $I \subset J \subset R$ with $J \neq I$ and $J \neq R$. Let $a \in J$, $a \not\in I$. Since $R/I$ is a field, there exists $b \in R$ with $ab = 1 \mod J$. Since $a \in J$ we find that $1 \in J$, contradicting $J \neq R$.

Answer 4

Let $\Phi \in \mathbb{R} \rightarrow \mathbb{R}^\'$ be a ring homomorphism. Then $R/Ker \Phi$ is isomorphic to $\Phi (R)$. Since $\Phi (R)$ is onto and $\mathbb{R}'$ is a field, then $\mathbb{R}/Ker \Phi$ is also a field. Let $\mathbb{R}$ be a commutative ring with unity. Since $\mathbb{R}/Ker \Phi$ is a field, then $Ker \Phi$ is a maximal ideal of $\mathbb{R}$.

Any comment on my proof? Thank you.