# finitely many minimal prime ideals in $\mathbb{Z}[x]$?

In my commutative algebra course we are asked to (in the end) construct all prime ideals in $S=\mathbb{Z}[X]$. We are already given the minimal non-zero prime ideals of the form $(f)$ with $f$ irreducible in $S$. Now there is a corollary in my book that says that if $R$ is a Noetherian ring, then there are only finitely many minimal prime ideals of $R$.

Since $\mathbb{Z}[x]$ is a Noetherian ring, it must have finitely many minimal prime ideals. But there are infinitely many irreducible polynomials in $S$. So do some of them coincide? Or am I missing something?

• "Now there is a corollary in my book that says that if RR is a Noetherian ring, then there are only finitely many prime ideals of RR." That's just not true at all. Oct 30, 2016 at 19:17
• It is only finitely many minimal prime ideals, see here. Oct 30, 2016 at 19:19
• Excuse me, I meant 'minimal' prime ideals Oct 30, 2016 at 19:19
• My assignment is saying: the minimal non-zero prime ideals of S are precisely those (f) with f irreducible in S. But you are saying that is not correct? Oct 30, 2016 at 19:21

That is false: if $R$ is noetherian, there is a finite number of minimal prime ideals only. $\mathbf Z$ itself is noetherian, and it's been known since Euclid there's an infinite number of primes.
To construct all prime ideals in $\mathbf Z[x]$, consider its intersection with $\mathbf Z$: either this intersection is $(0)$, and it corresponds to a prime ideal in $\mathbf Q[x]$, or it is $(p)$ ($p$ prime), and it corresponds to a prime ideal in $\mathbf Z/p\mathbf Z[x]$.
[T]here is a corollary in my book that says that if $R$ is a Noetherian ring, then there are only finitely many prime ideals of $R$.
The correct statement is that in a commutative Noetherian ring there are only finitely many primes minimal over a given ideal. In the ring $\mathbb Z[X]$, the only minimal prime over $(0)$ is $(0)$.