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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.

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    $\begingroup$ This is false if $S$ is not a field: take the natural projection $\mathbb Z\to \mathbb Z/6\mathbb Z$. The kernel is $(6)$, but this ideal is not maximal, since it's contained in $(2)$, a proper ideal. $\endgroup$ – Nishant Dec 11 '14 at 22:00
  • $\begingroup$ Is $S$ a field? $\endgroup$ – Kevin Carlson Dec 11 '14 at 22:00
  • $\begingroup$ Indeed this is not true if $S$ is not a field, for example $k[x,y]\rightarrow k[x,y]/(xy)$ is surjective, but $(xy)\subset k[x,y]$ is not even prime, so cannot be maximal. $\endgroup$ – Moss Dec 11 '14 at 22:01
  • $\begingroup$ @KevinCarlson Yes it is. $\endgroup$ – 3x89g2 Dec 11 '14 at 22:03
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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.

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  • $\begingroup$ Hi @egreg, just wondering where I might find a proof of this: "if f:R→S is a surjective homomorphism, then f induces a bijection between the ideals of R containing kerf and the ideals of S"? $\endgroup$ – jimbobeanz Feb 8 '17 at 10:32
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    $\begingroup$ @jimbobeanz If $\mathcal{R}_f$ is the set of ideals of $R$ containing $\ker f$ and $\mathcal{S}$ is the set of ideals of $S$, the bijections are defined by $I\in\mathcal{R}_f\mapsto f(I)$ and $J\in\mathcal{S}\mapsto f^{-1}(J)$. It's in every algebra book. $\endgroup$ – egreg Feb 8 '17 at 10:36
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Since $R/\ker(f)$ is a field, $\ker(f)$ is a maximal ideal.

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  • $\begingroup$ I think this statement probably also had to be proved. $\endgroup$ – Kevin Carlson Dec 11 '14 at 22:55
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    $\begingroup$ Fields have exactly two ideals. QED $\endgroup$ – Martin Brandenburg Dec 11 '14 at 22:59
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    $\begingroup$ That's probably not clear to someone who's just been introduced to rings. $\endgroup$ – Kevin Carlson Dec 11 '14 at 23:49
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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 $J \neq R$.

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  • $\begingroup$ Why contradiction? Direct proof is better and faster. $\endgroup$ – Martin Brandenburg Dec 11 '14 at 23:17

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