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Let $A,B$ be rings. If $f:A\rightarrow B$ is a homomorphism from $A$ onto $B$ with kernel $K$, and $J$ is an ideal of $A$ such that $K\subseteq J$, then $f(J)$ is an ideal of $B$:

My solution: Let $a,b\in J$. Then $a+b\in J$, so $$f(a)+f(b)=f(a+b)\in f(J).$$ Also, $-a\in J$, so $$-f(a)=f(-a)\in f(J).$$ Finally, if $c\in A$, then $$f(a)f(c)=f(ac)\in f(J).$$

We don't need the fact that $K\subseteq J$, or am I mistaken somewhere?


marked as duplicate by rschwieb, Davide Giraudo, Ayman Hourieh, TZakrevskiy, TMM Jan 14 '14 at 20:59

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    $\begingroup$ Long story short: no, it isn't necessary. No matter what ideal you pick in $A$, $f(A)\lhd B$, and $f(A+K)=f(A)$. If you look at preimages of ideals in $B$, though, these preimages are all ideals containing $K$. $\endgroup$ – rschwieb Jan 14 '14 at 20:22
  • $\begingroup$ Yes, you’re mistaken: take the homomorphism $f\colon k[x]\to k$ by $f(p)=p(0)$, the constant term of the polynomial $p$. Here, $k$ may be taken to be a field. The kernel is $(x)$. Now let $J$ be the set of all multiples of $x+1$. $\endgroup$ – Lubin Jan 14 '14 at 20:27
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    $\begingroup$ @Lubin: in your case, $f(J) = k$ is still an ideal (the unit ideal) $\endgroup$ – zcn Jan 14 '14 at 20:34
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    $\begingroup$ @Lubin: In this case the image of $J$ is all of $k[y]$, again the unit ideal. If you're trying to give an example where the OP's reasoning fails, necessarily one must try a homomorphism that is not surjective $\endgroup$ – zcn Jan 14 '14 at 20:54
  • $\begingroup$ @JJBeck Were you implicitly assuming a commutative ring? $\endgroup$ – rschwieb Jan 14 '14 at 21:13

The proof is correct, since $f$ is onto and for every $c \in A$ we have that $f(c)f(a) \in f(J)$ this implies that for every $c \in B$ we have $cf(a) \in f(J)$. To be exact you should complete the proof proving also that for every $c \in B$ and every $a \in A$ you have that $f(a)c \in f(J)$ but this proof is symmetrical to the other one.

So the result holds even if $J$ doesn't contain $K$.

Edit: Thanks to rschwieb I could correct a mistake in the previous answer.

  • $\begingroup$ Thanks, Giorgio. Isn't that just one mistake, not two? :) $\endgroup$ – JJ Beck Jan 14 '14 at 20:57
  • $\begingroup$ The second being the requirement that $J$ have to contain $K$ (as you to supposed in your question). :) $\endgroup$ – Giorgio Mossa Jan 14 '14 at 21:00
  • $\begingroup$ Dear @GiorgioMossa : Actually, the "onto" the user used in the problem statement does seem to indicate that they have been assuming surjectivity from the beginning. And I don't see any error in the $f(a)f(c)\in f(J)$ argument, except perhaps omitting that they were thinking of commutative rings, or else forgetting to do the "other side "$f(c)f(a)\in f(J)$". Otherwise the provided solution does not really have any flaw... $\endgroup$ – rschwieb Jan 14 '14 at 21:13
  • $\begingroup$ @rschwieb Thank you. Yes, surjectivity is assumed, except that I didn't make it explicit where I used it in my solution. $\endgroup$ – JJ Beck Jan 14 '14 at 21:36
  • $\begingroup$ @rschwieb my bad I'll edit the answer $\endgroup$ – Giorgio Mossa Jan 14 '14 at 21:37

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