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If $f:A\rightarrow R$ be a ring homomorphism, where $A$ and $R$ are commutative rings. If $f$ is surjective and $P$ is a prime ideal in $A$, how to prove that $f(P)$ is a prime ideal in $R$?

This a an exercise I came across while self-studying joseph rotman's book advanced modern algebra. I searched this website and find similar questions but they have another condition: $P$ contains the kernel of $f$. However, I want to know how to prove this result without that condition. Give thanks to any useful help!

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  • $\begingroup$ The condition that $P$ contains $\ker f$ is necessary, otherwise the result is not true. $\endgroup$ – Crostul May 30 '16 at 9:47
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It's impossible: let $\mathfrak m$ be a maximal ideal of $A$ which doesn't contain $\ker f$. As $B\simeq A/\ker f$, we have $$f(\mathfrak m)=\mathfrak m \cdot A/\ker f=(\mathfrak m+\ker f)/\ker f=A/\ker f,$$ which is not a prime ideal in the quotient.

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  • $\begingroup$ Do you mean that if M is a maximal ideal of $A$ which doesn't contain $ker f$, then $f(M)$ has to be the whole ring $B$? $\endgroup$ – 王李远 May 30 '16 at 9:04
  • $\begingroup$ It's exactly that, since $\mathfrak m+\ker f =A$, by maximality of $\mathfrak m$. $\endgroup$ – Bernard May 30 '16 at 18:51
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Suppose $r,s\in R$ and $rs\in f(P)$. Since $f$ is surjective, you have $r=f(a)$ and $s=f(b)$ for some $a,b\in A$, so the condition is $$ f(a)f(b)=f(c) $$ for some $c\in P$. In turn this becomes $ab-c\in\ker f$. If $P\supseteq\ker f$, you can conclude $ab-c\in P$, so also $ab\in P$ and it's easy to finish.

Without the assumption $\ker f\subseteq P$ there's no way to prove the statement. For instance, consider $A=\mathbb{Z}$ and $R=\mathbb{Z}/2\mathbb{Z}$, with $f$ the canonical projection. Then $f(3\mathbb{Z})=R$, which is not a prime ideal.

If you think more about the problem, the assertion would be equivalent to the statement that, if $P$ is a prime ideal of $A$ and $I$ is any ideal of $A$, then $P+I$ is prime: just consider the projection $f\colon A\to A/I$. Saying that $f(P)$ is prime in $A/I$ implies that $f^{-1}(f(P))=P+I$ is prime in $A$.

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