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Let $R$ be ring and let $I$ be a nonzero proper ideal of $R$. If $B$ is a field and $f:R\rightarrow B$ is a ring homomorphism, and $f(x)\in{I^e}$ (where ${I^e}$ is extension ideal of $I$), then $$x\in{rad(I)},$$ where $rad(I)$ is the radical of $I$.

It seems to me it is easy question but I just proved with assuming that $f$ is surjective.

Any insight is much appreciated.

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    If $B$ is a field then $I^e$ must be $B$ right ? – 2017-02-22
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    In general this is wrong: let $R$ be an integral domain, $B$ its field of fractions, and $f$ the inclusion. – 2017-02-22
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    @user26857 oh yeah, but you comment disposes, of the question. – 2017-02-22
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    Even for surjective homomorphisms this is wrong, so check your notes. – 2017-02-22
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    user26857 Thanks for comment but I still confused since ${I^e}$, is an ideal in ,$k$, which is field , it must be the equal to , $ k$ . Is it right ? – 2017-02-22
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    What about the ideal $I=6\mathbb Z$ and the canonical surjection $\mathbb Z\to\mathbb Z/3\mathbb Z$? – 2017-02-22

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