Let Φ: R⇒S be a ring homomorphism. Prove....

1. If ker(Φ) is a subring of R, then ker(Φ) is an ideal of R
2. If S is an integral domain, then ker(Φ) is a prime ideal of R.


sooo confused

1.
A subring K of R is also an ideal if:
kr is in K, and rk is in K, for all k in K and for all r in R. In other words, multiply a subring element by anything in the ring, and you're still inside the subring.

Now let K be ker(phi). This means for any element k in K, phi(k) = 0 (where 0 is the additive identity of S). So: for any r in R, you need to prove that phi(kr) = phi(rk) = 0.
To do this, use the ring homomorphism formula:
phi(kr) = phi(k)phi(r)
and you know what phi(k) is, so... can you see it from there?

2.
For a prime ideal, we need to additionally prove that, if r and q are in R but not in K, then rq is not in K.
One of the extra properties of an integral domain is that it has are no zero divisors (ie. you can't multiply two nonzero elements get 0).
So think about two non-kernel elements r and q.
If rq is in the kernel
then: phi(rq) = 0
then: phi(r)phi(q) = 0
But if you think about what I just wrote about zero divisors, you'll see why this is not possible...