# Ring homomorphism is surjective iff image of ideal is ideal

I am searching for requirements that a ring homomorphism $f: A \to B$ must meet in order that it maps arbitrary ideals $I \subset A$ to ideals $f(I) \subset B$.

My thought was that the requirement is that $f$ must be surjective. That is

$f$ surjective $\Longleftrightarrow f(I)$ ideal in $B$

The $\Rightarrow$ direction was easy and straightforward (in fact there are multiple questions here that deal with this direction).

I'm struggling with showing that $f$ is surjective if we require $f(I)$ ideal in $B$. Especially for non-trivial elements in $B$ that are not contained in $f(I)$.

My initial assumption that $f$ should be surjective might also be wrong, I'm not sure since I am unable to prove $\Leftarrow$. Can you hint me into the right direction?

• Your rings have a unit element? Are ring homomorphisms required to map $1$ to $1$? – Mariano Suárez-Álvarez Nov 27 '17 at 20:13
• @MarianoSuárez-Álvarez Yes and yes. The rings contain $1$ as neutral element for $\cdot$ and $f(1) = 1$. And also $1 \ne 0$. – Zabuza Nov 27 '17 at 20:19
• Are ideals required to be proper? Or is $A$ an ideal of $A$? – user14972 Nov 27 '17 at 20:20
• @Zabuza: As an aside, it's a little weird to include $1 \neq 0$ in the definition of ring. – user14972 Nov 27 '17 at 20:21
• It's good to have a little bit of skepticism for the problem you want to prove. Think about what this is saying for fields, where the only ideals are zero and the whole ring. – D_S Nov 27 '17 at 22:33

Since $A$ is an ideal of $A$, the requirement that a map $f : A \to B$ sends ideals to ideals implies that $f(A)$ is an ideal of $B$.
However, $f(A)$ is an ideal of $B$ containing $1_B$, and thus $f(A) = B$. So $f$ is surjective, as required.
The theorem is no longer true if we merely require $f$ to map proper ideals to ideals. For example if $A$ is a field, then its only proper ideal is the zero ideal, and thus every ring homomorphism $f : A \to B$ has the property that it sends proper ideals to ideals.
• Why is $1_B \in f(A)$? – Zabuza Nov 27 '17 at 20:26
• @Zabusa: $1_B = f(1_A)$ and $1_A \in A$. – user14972 Nov 27 '17 at 20:30
• Sorry, what confused me in the first place is why $1_A \in A$. I mean ideals aren't required to contain $1$, are they? – Zabuza Nov 27 '17 at 20:31
• @Zabuza: Ideals of $A$ aren't required to contain $1_A$. $A$ is the only ideal of $A$ that contains every element of $A$. – user14972 Nov 27 '17 at 20:37
• Ah now I see. Sorry I got confused by $A$. I thought it would be $I$ that may be less than $A$. But you are actually using $A$ itself, which obviously contains $1_A$ and thus gets mapped to an ideal that contains $1_B$ and thus $A$ gets mapped to $B$. – Zabuza Nov 27 '17 at 20:40