How many minimal prime Ideals in an integral domain? I am wondering right now whether every ring, which is commutative, has a 1, and is reduced (i.e. no nilpotent elements) always has just one minimal prime Ideal.
Does anyone have a proof or a counterexample? The motivation for this question is that said rings correspond to coordinate rings $k[X]$ for irreducible algebraic sets $X$. And the minimal prime ideals should correspond to the irreducible components of $X$, which in case that $X$ itself is irreducible should just be $X$. Possible it is necessary to assume that $k$ is also an integral domain. 
As there was some confusion, because I did not phrase the question clearly (sorry for that) and  the answer is so simple, here is the short solution:
If $R$ is an integral domain then $0$ is a prime ideal, and clearly it is also the unique minimal prime ideal. If $R=k[X]$ (i.e. $R$ is the coordinate ring of $X$), then $0$ corresponds to $X$ and $X$ is irreducible.
 A: If you're allowing $(0)$ as a minimal ideal, then every ring has $(0)$ as a unique minimal ideal.

So I'll assume that by "minimal ideal", you mean an ideal which is minimal among nonzero ideals.

But then minimal ideals need not exist. For example, the ring $\mathbb{Z}$ has no minimal ideals.

In your latest edit, you changed "minimal ideal" to "minimal prime ideal".

In this context, "minimal" presumably means truly minimal in the set of prime ideals.

If the ring $R$ is required to be an integral domain, then the zero ideal is a prime ideal, and in that case, the zero ideal is the unique minimal prime ideal.

If $R$ is required to be reduced, but not necessarily an integral domain, then 
there need not be a unique minimal prime ideal.

For example, consider the ring $R=Z_6$.

It's easily verified that $R$ is reduced.

The prime ideals of $R$ are the principal ideals $(2)$ and $(3)$, which are distinct, and which are each both minimal and maximal.

Thus, $R$ does not have a unique minimal prime ideal.
