Image of ring hom is ideal, kernel is subring.

[Kreizhn]

Homework Statement

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?

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.

 

[micromass]

Homework Statement

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?

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.

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...

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.

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

 

[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.

 

[micromass]

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?

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

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.

Indeed!