Does every prime ideal in a ring arise as kernel of a homomorphism into $\mathbb{Z}$? Let $R$ be a commutative ring. Clearly the kernel of $h$ is a prime ideal whenever $h : R \rightarrow                                                                                                                
\mathbb{Z}$ is a ring homomorphism.  But is the converse true: does every prime ideal arise
as kernel of a homomorphism into $\mathbb{Z}$? 
 A: Let $R$ be a field of uncountable cardinality. For a minimal counterexample, let $R = \mathbb{F}_2$. 
The correct salvage is that every prime ideal arises as the kernel of a homomorphism into some integral domain (in fact, into some field). It shouldn't be possible to say anything stronger than this. 
A: The converse is very, very false. For instance if $R$ is a ring of nonzero characteristic (think $\mathbb Z/n\mathbb Z$ with $n>0$), there aren't any homomorphismes  $R\to\mathbb Z$ at all, yet $R$ will always (assuming the axiom of choice) have at least one prime ideal. Simple counterexample $\mathbb Z/2\mathbb Z$ and its zero ideal, or (if you don't like zero ideals) $\mathbb Z/4\mathbb Z$ and its ideal $\{\overline0,\overline2\}$.
A: If $h: R \rightarrow \mathbb{Z}$ is a ring homomorphism, then $h$ is also a homomorphism of (additive) abelian groups.  If $P$ is the kernel of $h$, then $R/P$ is isomorphic to the image of $h$, which is a subgroup of $\mathbb{Z}$.  And a subgroup of $\mathbb{Z}$ is either trivial or isomorphic to $\mathbb{Z}$.  
So if $P$ is a prime ideal of $R$ (prime ideals are by definition proper, so $R/P$ is never trivial) such that $R/P$ is not isomorphic to $\mathbb{Z}$ as an abelian group, then $P$ cannot possibly be the kernel of a homomorphism $R \rightarrow \mathbb{Z}$.  There are plenty of prime ideals which fit this description.  If $R$ is $\mathbb{Z}$ itself, and $P = \mathbb{Z} p$ for some prime number $p$, then $R/P$ is finite!
