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Suppose $R$ is a commutative ring with identity and let $I$ be a nonzero proper ideal of $R$. Prove that if $f: R \to k$ is a ring map ($k$ is field) and $f(x)\in\left$, then $x\in\mathrm{rad}(I)$.

I believe this is not correct unless $f$ is surjective.

My attempted. I assumed that $f$ is surjective.

$f(x)\in\left$, and we will get $x\in\mathrm{rad}(I)$ since $I\subseteq\mathrm{rad}(I)$.

My question is this question can be true without assumption that $f$ must be surjective, if not, could give a counterexample. By the way, ring map is homomorphism.

Any help will appreciated .

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    you should try and pick much more descriptive titles to your posts – 2017-02-07
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    I edited my title, I got confused, since I spent much time to prove it without $f$ surjective but I couldn't find.so, do you believe that question is right? – 2017-02-07
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    Your claim is wrong and your attempted proof has no meaning for me. Try $\mathbb Z\to\mathbb Z/2\mathbb Z$ and $I=3\mathbb Z$. – 2017-02-07
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    Since $K$ is a field, $\langle f(I) \rangle$ is either $0$ or $K$. In both cases, you may find easy examples which show that your claim is false. – 2017-02-12

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