# For a maximal left ideal $M$ of $S$, is $f^{-1}(M)$ a maximal left ideal of $R$ when $f$ is surjective?

Let $f:R \longrightarrow S$ a surjective ring homomorphism. Is the inverse image $f^{-1}(M)$ a maximal left ideal of $R$ for any maximal left ideal $M$ of $S$?

Comments: I tied something like this: if $M$ is maximal then

$M \neq S$ and if $J$ is a left ideal such that $M \varsubsetneq J \subseteq S \Rightarrow J = S$.

I want to show that: $f^{-1}(M) \neq R$ and if $I$ is a left ideal such that $f^{-1}(M) \varsubsetneq I \subseteq R \Rightarrow I = R$.

The first statement follows from the fact that $f$ is surjective. I am unable to prove the second.

If $I$ is a left ideal such that $f^{-1}(M) \varsubsetneq I \subseteq R \Rightarrow I = R$.
Let $a\in I$, $a\notin f^{-1}(M)$. It follows $f(a)\notin M$. Since $f^{-1}(M)\subset I$ we get $f(f^{-1}(M))\subset f(I)$. $f$ surjective gives you $M=f(f^{-1}(M))$, so $M\subset f(I)$. But $f(a)\in f(I)-M$. Now note that $f$ surjective implies that $J=f(I)$ is a left ideal of $S$, an thus you get $M\subsetneq J\subseteq S\Rightarrow J=S$, so $f(I)=S$ hence $I=R$ (why?).
Edit. Let's prove the last claim. If $r\in R$ then $f(r)\in S=f(I)$, so there is $i\in I$ such that $f(r)=f(i)$. It follows $f(r-i)=0$, so $r-i\in\ker f$. But $\ker f=f^{-1}(0)\subseteq f^{-1}(M)$, so $\ker f\subset I$ hence $r-i\in I\Rightarrow r\in I$.
• I did a similar analysis with her before posting the question, and I got $f(I) = S$. However, I can not conclude that $I=R$, I tried to assume what's different but I get no nonsense. – Croos May 9 '15 at 18:11
Let $K:=ker f$. Then $R/ K \cong S$. So maximal ideals of $S$ correspond to maximal ideals of $R/ K$. On the other hand maximal ideals of $R/K$ correspond to those maximal ideals of $R$ which contain $K$.