How do ring homomorphism R → ℤ correspond to prime ideals of R?

Question

In "Category Theory" (Oxford Logic Guides, 2010 by Steve Awodey), pg 35, Awodey makes an off-hand comment:

"Ring homomorphisms A → ℤ into the initial ring ℤ ... correspond to so-called prime ideals, which are the ring-theoretic generalizations of ultrafilters."

I can find other sources that claim this to be true, that "clearly" prime ideals of a ring correspond to its homomorphisms into ℤ. I understand both of these concepts individually and I'm trying to construct a proof connecting them using the homomorphism law alone, but I'm getting stuck.

How do you get from "homomorphism into ℤ" to "prime ideal"?

Answer 1

I'm not sure what additional assumptions you're working under, but here is something that you can say:

For a commutative ring $R$, the prime ideals are exactly the kernels of homomorphisms of $R$ into integral domains.

This is immediate if you now the First Isomorphism Theorem and the fact that $A/I$ is a domain iff $I$ is a prime ideal of $A$.

Now, the kernels of homomorphisms into $\mathbb Z$ certainly correspond to some prime ideals, but definitely not all, and that was not really 100% transparent in the passage you quoted. For example, the field of two elements has a prime ideal that does not correspond to any homomorphism into $\mathbb Z$, and the same can be said for $A=\mathbb C$, or any field for that matter.

Answer 2

The claim is nonsensical:
There are plenty of prime ideals in $A=\mathbb C[X]$ but no ring morphism at all $\mathbb C[X]\to \mathbb Z$.
Worse still: there is only one ring morphism $\mathbb Z\to \mathbb Z$ but infinitely many prime ideals $\mathbb Z$, as proved by Euclid.
The claim is also false for every field and even for every algebra over a field!
Actually, offhand I don't know a single ring for which the claim is true, but I'm not going to waste one nanosecond looking for one...