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Let $f:A\to B $ be a homomorphism between the rings $A$ and $B$ and let $J$ be an ideal of $A$, then $f(J)$ is an ideal of $B$.
If $f(x)\in f(J)$ and $f(y)\in B$ then $f(x)f(y)=f(xy)$ and since $x\in J$ then $xy\in J$ and therefore $f(x)f(y)\in f(J)$. For $f(x),f(y) \in f(J)$ then $f(x)-f(y)=f(x-y)\in f(J)$ since $x,y \in J$ and $J$ is closed under subtraction.
However, nowhere in this proof did I use the fact that $\text {ker} f \subset J$, which was a part of the problem statement, leading me to believe that my proof is wrong or incomplete. I cannot see my mistake, are there hidden assumptions I am making?

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    $\begingroup$ You're wrong from the starting point: $f(J)$ is not an ideal of $B$. It's only an ideal of the subring $f(A)$. $\endgroup$ – Bernard Mar 23 '17 at 19:58
  • $\begingroup$ @Bernard that was the statement I was setting out to prove. My question is why is proof wrong? $\endgroup$ – Guacho Perez Mar 23 '17 at 19:59
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    $\begingroup$ Because every element in $B$ is not necessarily some $f(y)$. It works only if $f$ is surjective. $\endgroup$ – Bernard Mar 23 '17 at 20:13
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Since a ring homomorphism is a homomorphism of the underlying abelian groups under addition, $f(J)$ is an additive subgroup of $B$ (maps of abelian groups send subgroups to subgroups).

However, in general, the multiplicative absorption property of ideals need not be transferred: not every element of $B$ need be in the image of $f$, so showing that $f(x)f(y)\in f(J)$ for $x\in A$, $y\in J$ is not enough. You would need to show that for any $b\in\color{red}{B}$ and $x\in J$, $b f(x)\in f(J)$, and this is not generally true.

As others note, the image $f(J)$ is only an ideal of the subring $f(A)$ of $B$, not an ideal of $B$ (this is what you actually proved). It is not hard to find examples of morphisms of rings sending ideals to sets which are not ideals, as in Alex M.'s answer.

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It is not true that the homomorphic image of an ideal is an ideal in general. Take any ring $R$ that can be embedded into a field $F$ (such as $\mathbb{Z}$ in $\mathbb{R}$). Then the image of a proper ideal $J \subset R$ under the embedding map will not be an ideal of $F$, as there are no proper nonzero ideals in a field.

In your proof, you never checked that if $a \in A$ and $b \in B$, then $bf(a) \in f(A)$, which is also necessary for an ideal.

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  • $\begingroup$ So my mistake lies in assuming that $f$ is surjective? $\endgroup$ – Guacho Perez Mar 23 '17 at 20:01
  • $\begingroup$ @GuachoPerez Yes, exactly. $\endgroup$ – Stahl Mar 23 '17 at 20:04
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No, it is not: let $i : \Bbb Z \to \Bbb Q$ be the natural injection. The only ideals of $\Bbb Q$ are $0$ and $\Bbb Q$. Take any ideal $(n)$ with $n \ne 0$ - is its image any of those two ideals of $\Bbb Q$? No, of course, so the image of an ideal is not necessarily ideal.

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