# 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 $f(A)$. Mar 23, 2017 at 19:58
• @Bernard that was the statement I was setting out to prove. My question is why is proof wrong?
– GuPe
Mar 23, 2017 at 19:59
• Because every element in $B$ is not necessarily some $f(y)$. It works only if $f$ is surjective. Mar 23, 2017 at 20:13

No, it is not. Let $$i : \mathbb Z \to \mathbb Q$$ be the natural injection given by $$i(n)=n$$. Since $$\mathbb Q$$ is a field, its only ideals are $$0$$ and $$\mathbb Q$$. Take any ideal $$(n) \subseteq \mathbb Z$$ with $$n \ne 0$$; is its image $$i((n)) = n \mathbb Z$$ any of those two ideals of $$\Bbb Q$$? No, of course, so the image of an ideal is not necessarily an ideal.

• What does it mean natural injection? @Alex M. Dec 11, 2021 at 19:14
• @Bestmat: In general, if $A \subseteq B$, then the natural injection $i : A \to B$ is the map given by $i(a)=a$ for all $a \in A$. Dec 11, 2021 at 19:50

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?
– GuPe
Mar 23, 2017 at 20:01
• @GuachoPerez Yes, exactly. Mar 23, 2017 at 20:04

If it were so, then every unital ring homomorphism between fields would be an isomorphism.* That is not the case; consider $$\mathbb{R} \hookrightarrow \mathbb{C}$$.

*For let $$\phi:F \to K$$ be a unital ring homomorphism, $$F,K$$ fields. Where does the ideal $$F$$ go? The only ideals of a field are $$(0)$$ and $$(1)=$$ the field; therefore if homomorphic images of ideals were ideals, then $$\phi(F) = (0)$$ or $$(1)$$. Since $$\phi(1) \neq 0$$, we would have $$\phi(F) = K$$ if your claim were true. So $$\phi$$ would be surjective, and we already know that homomorphisms of fields are injective.