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I'm trying to understand the following proof.

Let $f : A \to B$ be a surjective ring homomorphism. If $P$ is a prime ideal of $A$ that contains $\ker f$, then $f(P)$ is a prime ideal in $B$.

Proof: We have $\ker f \subseteq P \subseteq A$. So by the third isomorphism theorem, we have that $P/\ker f$ is an ideal of $A/\ker(f)$. Furthermore, we have that

$$(A/\ker f)/(P/\ker f) \cong A/P.$$

The latter is an integral domain because $P$ is a prime ideal, this proves that $P/\ker f$ is a prime ideal in $A/\ker f$. Furthermore by the first isomorphism theorem you know that because $f$ is surjective,

$$A/\ker f \cong B.$$

It follows that because $P/\ker f$ is a prime ideal in $A/\ker f$ that $f(P)$ is a prime ideal in $B$.

I understand all of the proof except the final line. I'm not sure why $P/\ker f$ being a prime ideal in $A/\ker f$ implies that $f(P)$ is a prime ideal in $B$. I was wondering whether someone can explain why this final implication is true.

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  • $\begingroup$ What you need is that if $g\colon C\to D$ is a ring isomorphism and $P\subset C$ is a prime ideal, then $g(P)\subset D$ is a prime ideal. Apply to $C=A/\ker f$ and $D=B$ $\endgroup$ – Tashi Walde Apr 12 '18 at 1:24
  • $\begingroup$ Hint: Consider the restriction of the canonical isomorphism $\varphi\colon A/\ker f\to B$ to $\varphi\mid_P\colon P/\ker f\to f(P)$ $\endgroup$ – Prasun Biswas Apr 12 '18 at 1:27
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There is an isomorphism $\Phi:A/\ker f\to B$. This isomorphism maps $P/\ker f$ to $f(P)$, that is $\Phi(P/\ker f)=f(P)$. As $\Phi$ is an isomorphism and $P/\ker f$ is a prime ideal of $A/\ker f$ then $\Phi(P/\ker f)$ is a prime ideal of $B$, that is $f(P)$ is a prime ideal of $B$.

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  • $\begingroup$ Why does the isomorphism map $P/Ker(f)$ to $f(P)$? $\endgroup$ – user262291 Apr 12 '18 at 1:29
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    $\begingroup$ @user262291: The canonical isomorphism $\Phi\colon A/\ker f\to B$ is defined by $a(\ker f)\mapsto f(a)$, so its restriction to $P/\ker f$ would be $p(\ker f)\mapsto f(p)~\forall~p\in P$ which is equivalent to saying $\Phi(P/\ker f)=f(P)$ $\endgroup$ – Prasun Biswas Apr 12 '18 at 1:35
  • $\begingroup$ @PrasunBiswas Even though this seems intuitive why does an isomorphism map a prime ideal to a prime ideal. $\endgroup$ – user262291 Apr 12 '18 at 1:36
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    $\begingroup$ @user262291: To see that they carry prime ideals, just being an epimorphism suffices. Suppose $P$ is a prime ideal of $R$ and $f\colon R\to R'$ be a ring epimorphism. Now, we need to show that $f(P)$ is a prime ideal of $R'$. Suppose $xy\in f(P)$. By surjectivity, $xy=f(a)f(b)$ for some $a,b\in R$. Now, since $f$ is a homomorphism, $xy=f(a)f(b)=f(ab)$ which implies $ab\in P$ from which you can conclude $f(a)\in f(P)$ or $f(b)\in f(P)$ $\endgroup$ – Prasun Biswas Apr 12 '18 at 1:46

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