Why doesn't $0$ being a prime ideal in $\mathbb Z$ imply that $0$ is a prime number? I know that $1$ is not a prime number because $1\cdot\mathbb Z=\mathbb Z$ is, by convention, not a prime ideal in the ring $\mathbb Z$.
However, since $\mathbb Z$ is a domain, $0\cdot\mathbb Z=0$ is a prime ideal in $\mathbb Z$. Isn't $(p)$ being a prime ideal the very definition of $p$ being a prime element?
(I know that this would violate the Fundamental Theorem of Arithmetic.)

Edit:
Apparently the answer is that a prime element in a ring is, by convention a non-zero non-unit (see wikipedia).
This is strange because a prime ideal of a ring is, by convention, a proper ideal but not necessarily non-zero (see wikipedia).
So, my question is now: Why do we make this awkward convention?
 A: Generally we make nice conventions because they make the statements of theorems nice.  The theorem relevant to prime ideals is that $P$ is a prime ideal of $R$ if and only if $R/P$ is an integral domain.  The theorem relevant to prime elements is prime factorization (when it holds).  
These two concepts almost coincide for principal ideals, but we must distinguish between the generic point $(0)$ and closed points, and there are good reasons for doing this.  (The zero ideal, for example, can't occur in the factorization of a nonzero ideal in a Dedekind domain.)
A: You have a point here: absolutely we want to count $(0)$ as a prime ideal in $\mathbb{Z}$ -- because $\mathbb{Z}$ is an integral domain -- whereas we do not want to count $(1)$ as being a prime ideal -- because the zero ring is not an integral domain (which, to me, is much more a true fact than a convention: e.g., every integral domain has a field of fractions, and the zero ring does not).
I think we do not want to call $0$ a prime element because, in practice, we never want to include $0$ in divisibility arguments.  Another way to say this is that we generally want to study factorization in integral domains, but once we have specified that a commutative ring $R$ is a domain, we know all there is to know about factoring $0$: $0 = x_1 \cdots x_n$ iff at least one $x_i = 0$.
Here is one way to make this "ignoring $0$" convention look more natural: the notions of factorization, prime element, irreducible element, and so forth in an integral domain $R$ depend entirely on the multiplicative structure of $R$.  Thus we can think of factorization questions as taking place in the cancellative monoid $(R \setminus 0,\cdot)$.  (Cancellative means: if $x \cdot y = x \cdot z$, then $y = z$.)  In this context it is natural to exclude zero, because otherwise the monoid would not be cancellative.  Contemporary algebraists often think about factorization as a property of monoids rather than integral domains per se.  For a little more information about this, see e.g. Section 4.1 of http://alpha.math.uga.edu/~pete/factorization2010.pdf.
A: There are good reasons behind the convention of including $(0)$ as a prime ideal
but excluding $(1).\ $ First, we include zero as a prime ideal because it facilitates many useful reductions. For example, in many ring theoretic problems involving an ideal $\, I\,$, one can reduce to the case $\,I = P\,$ prime, then reduce to $\,R/P,\,$ thus reducing to the case when the ring is a domain. In this case one simply says that we can factor out by the prime $ P\,$, so w.l.o.g. assume $\, P = 0\,$ is prime, so $\,R\,$ is a domain. For example, I've appended to the end of this post an excerpt from Kaplansky's classic textbook Commutative Rings, section $1\!\!-\!\!3\!:\,G$-Ideals, Hilbert Rings, and the Nullstellensatz.
Thus we have solid evidence for the utility of the convention that the zero ideal is prime. So why don't we adopt the same convention for the unit ideal $(1)$ or, equivalently, why don't we permit the zero ring as a domain?  There are a number of reasons. First, in domains and fields it often proves very convenient to assume that one has a nonzero element available. This permits proofs by contradiction to conclude by deducing $\,1 = 0.\ $ More importantly, it implies that the  unit group is nonempty, so unit groups always exist. It'd be very inconvenient to have to always add the proviso (except if $\, R = 0)\,$ to the many arguments involving units and unit groups. For a more general perspective it's worth emphasizing that the usual rules for equational logic are not complete for empty structures so that is why groups and other algebraic structures are always axiomatized to prevent nonempty structures (see this thread for details).
Below is the promised Kaplansky excerpt on reduction to domains by factoring out prime ideals. I've explicitly emphasized the reductions e.g. reduce to....

Let $\, I\,$ be any ideal in a ring $\, R.\,$ We write $\, R^{*}\,$ for the quotient ring $\, R/I.\,$ In the polynomial ring $\, R[x]\,$ there is a smallest extension $\, IR[x]\,$ of $\, I.\,$ The quotient ring $\, R[x]/IR[x]\,$ is in a natural way isomorphic to $\, R^*[x].\,$ In treating many problems, we can in this way reduce to the case $\, I = 0,\,$
and we shall often do so.
THEOREM $28$. $\,$ Let $\, M\,$ be a maximal ideal in $\, R[x]\,$ and suppose that the contraction $\, M \cap R = N\,$ is maximal in $\, R.\ $ Then $\, M\,$ can be generated by $\, N\,$ and one more element $\, f.\ $ We can select $\, f\,$ to be a monic polynomial which maps $\!\bmod N\,$ into an irreducible polynomial over the field $\, R/N.\ $
Proof. $\,$ We can reduce to the case $\, N = 0,\,$ i. e., $\, R\,$ a field, and then
the statement is immediate.
THEOREM $31$. $\,$ A commutative ring $\, R\,\,$ is a Hilbert ring if and only if the polynomial ring $\, R[x] \,\,$ is a Hilbert ring.
Proof. $\,$ If $\, R[x]\,$ is a Hilbert ring, so is its homomorphic image $\, R\,$.
Conversely, assume that $\, R\,$ is a Hilbert ring. Take a G-ideal $\, Q\,$ in
$\, R[x]\,$; we must prove that $\, Q\,$ is maximal. Let $\, P = Q \cap R\,$; we can reduce the problem to the case $\, P = 0,\,$ which, incidentally, makes $\, R\,$ a domain.
Let $\, u\,$ be the image of $\, x\,$ in the natural homomorphism $\, R[x] \to R[x]/Q.\,$
Then $\, R[u]\,$ is a G-domain. By Theorem $23$, $\,u\,$ is algebraic over $\,R\,$ and $\,R\,$ is a G-domain. Since $\,R\,$ is both a G-domain and a Hilbert ring, $\,R\,$ is a field. But this makes $\, R[u] = R[x]/Q\,$ a field, proving $\, Q\,$ to be maximal.
