5
$\begingroup$

I'm currently trying to solve this problem.

Let $f: R \rightarrow S$ be a surjective ring homomorphism. Let $K = \ker(f)$. Assume $P$ is a prime ideal s.t. $K \subset P$. Show $f(P)$ is a prime ideal in $S$.

I solved the ideal part.

Let $y \in f(P)$, by definition then $y = f(x)$ for some $x \in P$. Now let $s \in S$. Then since $f$ is surjective $\exists r \in R$ s.t. $f(r) = s$. So then since $f$ is a homomorphism we have $f(x)\cdot s = f(x) \cdot f(r) = f(x\cdot r)$. Then since $P$ is ideal and $x \in P$, $x\cdot r \in P$, so $f(x \cdot r) = f(x)\cdot s \in f(P)$ so $f(P)$ is closed under multiplication by an element in $S$.

Now let $y, y' \in f(P)$, again by definition we have $x, x' \in R$ s.t. $f(x) = y, f(x') = y'$. By $f$ homomorphism we have $y - y' = f(x) - f(x') = f(x-x')$. Then by $P$ ideal we have $x-x' \in P$, thus $y-y' \in f(P)$.

Therefore $f(P)$ is an ideal.

The part I'm having trouble with is showing that $P$ prime implies $f(P)$ prime.

I started off let $AB \subset f(P)$, by definition we know we have $C \subset P$ s.t. $f(C) = AB$. From here I need to arrive at showing that either $A \subset f(P)$ or $B \subset f(P)$. I'm not really sure how to move forward from here but I know I somehow need to involve the fact that $K \subset P$ since I haven't used that yet.

Any help would be appreciated.

Note: I'm not assuming that $R$ or $S$ is commutative, I already know how to solve it if they are using the commutative definition of prime.

$\endgroup$
3
  • $\begingroup$ I think you are very close! Note that since $f$ is a surjection we have elements $x,y\in R$ such that $f(x)=A, f(y)=B$. What can you say about $f(C-xy)$? $\endgroup$
    – user214321
    Feb 12, 2015 at 6:27
  • 1
    $\begingroup$ @JRP I think that the prime ideals here are completely prime ideals in a non-commutative setting. The basic approach still works, but $A$ and $B$ are not elements, but ideals. $\endgroup$
    – Slade
    Feb 12, 2015 at 6:30
  • $\begingroup$ @Slade Ah, thank you. I've been working in a commutative setting for so long I had not considered an alternative. I was wondering why EgoKilla had suddenly switched to uppercase names. $\endgroup$
    – user214321
    Feb 12, 2015 at 6:32

2 Answers 2

Reset to default
3
$\begingroup$

$f:R\to S$ induces an isomorphism $\overline{f}:R/K \to S$. So it is enough to show that $P/K$ is a (completely) prime ideal of $R/K$.

Every ideal of $R/K$ can be written uniquely in the form $I/K$ for some ideal $I\supset K$ of $R$ (take the preimage under the projection $R\to R/K$). But if $A/K\cdot B/K \subset P/K$, then $AB \subset P+K = P$, and the conclusion follows shortly thereafter.

$\endgroup$
2
  • $\begingroup$ Is there a way to prove it more directly? $\endgroup$
    – EgoKilla
    Feb 12, 2015 at 16:59
  • $\begingroup$ @EgoKilla If you fill in the details, this is direct. Concretely, if $AB\subset f(P)$, then $f^{-1}(A)f^{-1}(B)\subset f^{-1}(f(P)) = P$, with the latter equality proved from $K\subset P$. $\endgroup$
    – Slade
    Feb 12, 2015 at 17:49
0
$\begingroup$

You should show $R/P \cong S/f(P)$. To that extend, consider the composition $$R \to S \to S/f(P).$$

$\endgroup$
2
  • $\begingroup$ Note that the setting is non-commutative. I am not sure whether we can conclude that $f(P)$ is (completely) prime directly from this isomorphism. $\endgroup$
    – Slade
    Feb 12, 2015 at 6:34
  • $\begingroup$ Hmm, I guess that it still works. $\endgroup$
    – Slade
    Feb 12, 2015 at 6:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .