# Is the homomorphic image of an ideal an ideal?

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?

• You're wrong from the starting point: $f(J)$ is not an ideal of $B$. It's only an ideal of the subring $f(A)$. – Bernard Mar 23 '17 at 19:58
• @Bernard that was the statement I was setting out to prove. My question is why is proof wrong? – Guacho Perez Mar 23 '17 at 19:59
• Because every element in $B$ is not necessarily some $f(y)$. It works only if $f$ is surjective. – Bernard Mar 23 '17 at 20:13

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.

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.

• So my mistake lies in assuming that $f$ is surjective? – Guacho Perez Mar 23 '17 at 20:01
• @GuachoPerez Yes, exactly. – Stahl Mar 23 '17 at 20:04

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.