0
$\begingroup$

Let $f$ be a surjective ring homomorphism from $R$ to $S$ and $P$ be a prime ideal of $R$ such that $\ker(f)<P$; then $f(P)$ is a prime ideal of $S$.

Any hints? I don't know how to use the property $\ker(f)<P$

$\endgroup$
0
$\begingroup$

Since $f$ is surjective, the ideals of $S$ have the form $f(I)$, for a unique ideal $I\supseteq\ker(f)$.

Suppose $f(P)\supseteq f(I)f(J)=f(IJ)$. Since both $P$ and $IJ$ contain $\ker(f)$ this is the same as $IJ\subseteq P$. Can you finish?

$\endgroup$
  • $\begingroup$ @NourAlnajjarine Changed without the assumption of commutativity. $\endgroup$ – egreg Apr 16 '17 at 14:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.