- Define the nilradical \(\operatorname{Nil}(R)\) of a commutative ring \(R\), and state the characterization theorem relating it to the intersection of all prime ideals of \(R\).
- Prove that the polynomial ring \(F[x]\) over a field \(F\) contains infinitely many prime ideals.
Solution
The nilradical is \[ \operatorname{Nil}(R)=\sqrt{(0)} =\{a\in R: a^m=0\text{ for some }m\ge1\}. \] The standard characterization is \[ \boxed{\operatorname{Nil}(R)=\bigcap_{\mathfrak p\in\operatorname{Spec}R}\mathfrak p.} \] In particular, because \(F[x]\) is a domain, \[ \operatorname{Nil}(F[x])=(0). \]
To prove that \(F[x]\) has infinitely many prime ideals, suppose instead that its nonzero prime ideals are \[ (p_1),\dots,(p_m), \] where each \(p_i\) is irreducible. Set \[ h=p_1\cdots p_m+1. \] Then \(h\) is nonconstant, so it has an irreducible divisor \(q\in F[x]\). Hence \((q)\) is a nonzero prime ideal. By the assumed list, \(q\) is associate to some \(p_j\). But then \(p_j\) divides both \(p_1\cdots p_m\) and \(h\), hence divides \[ h-p_1\cdots p_m=1, \] a contradiction.
Therefore \(F[x]\) has infinitely many nonzero prime ideals, and hence infinitely many prime ideals.
The nilradical theorem is compatible with this conclusion, but it does not by itself force infinitude here: \((0)\) is already a prime ideal of the domain \(F[x]\), so the intersection of any family of primes containing \((0)\) is automatically \((0)\).