# Image of ring hom is ideal, kernel is subring.

1. Jun 15, 2011

### Kreizhn

1. The problem statement, all variables and given/known data

Let $\phi: R \to S$ be a ring homomorphism from R to S. What can you say about $\phi$ if its image $\text{im}\phi$ is an ideal of S? What can you say about $\phi$ if its kernel $\ker \phi$ is a subring (w unity) of R?

3. The attempt at a solution

I think the second one is easy. Since $\ker \phi$ is a subring with unity, we have that $\phi(1_R) = 0_S$. But it is necessary that $\phi(1_R) = 1_S$ since $\phi$ is a homomorphism. Thus $\phi \equiv 0$ the trivial homomorphism. I think this is all I can say about this.

The second one is a bit trickier. I want to say that $\phi$ is surjective, and here is my reasoning. If $\text{im}\phi$ is an ideal, then for all $s \in S, t \in \text{im}\phi$ we get that $st \in \text{im}\phi$. Let $u, v \in R$ be such that $\phi(u) = s, \phi(v) = st$, then
$$\phi(v) = st = s \phi(u)$$
So now I want to be able to say that there is necessarily a $w \in R$ such that $\phi(w) = s$, but I can't quite see how to get there.

2. Jun 15, 2011

### micromass

Staff Emeritus
That's not all you can say about this. How can $\phi$ ever be the trivial homomorphism 0 if $\phi(1_R)=1_S$. The only way phi can ever be trivial is if $\phi(1_R)=0$. Thus...

Can you do something with the information that $1_S\in im(\phi)$?

3. Jun 15, 2011

### Kreizhn

I had thought that first part was a bit fishy. My first inclination was to declare that S had to be the zero-ring, though the question asks what we can say about the homomorphism not what we can say about the codomain. Is this what you were going for?

Haha, I just figured out what you meant about 1 in the image. Obviously, if unity is in the ideal, the ideal is the whole ring, so the function is surjective.

Thanks.

4. Jun 15, 2011

### micromass

Staff Emeritus
Yes, this was what I meant. S has to be the zero ring, and phi has to be trivial!

Indeed!