Image of ring hom is ideal, kernel is subring.

[Kreizhn]

Homework Statement

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

The Attempt at a Solution

I think the second one is easy. Since [itex] \ker \phi [/itex] is a subring with unity, we have that [itex] \phi(1_R) = 0_S [/itex]. But it is necessary that [itex] \phi(1_R) = 1_S [/itex] since [itex] \phi [/itex] is a homomorphism. Thus [itex] \phi \equiv 0 [/itex] 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 [itex] \phi [/itex] is surjective, and here is my reasoning. If [itex] \text{im}\phi [/itex] is an ideal, then for all [itex] s \in S, t \in \text{im}\phi [/itex] we get that [itex] st \in \text{im}\phi [/itex]. Let [itex] u, v \in R [/itex] be such that [itex] \phi(u) = s, \phi(v) = st [/itex], then
[tex] \phi(v) = st = s \phi(u) [/tex]
So now I want to be able to say that there is necessarily a [itex] w \in R [/itex] such that [itex] \phi(w) = s [/itex], but I can't quite see how to get there.

 

[micromass]

Homework Statement

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

The Attempt at a Solution

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

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

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

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

 

[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!