- #1

Kreizhn

- 743

- 1

## 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.