The prime ideals of $\mathbb{Z}[X]$ are

  1. $0$
  2. $(p)$ for prime integer $p$
  3. $(f)$ for irreducible polynomial $f$
  4. $(p, f)$ for prime integer $p$ and irreducible polynomial $f$ that remains irreducible mod $p$.

The radical ideals are intersections of prime ideals. It is easy to compute the intersection when finitely many prime integers are involved using Chinese remainder theorem: they are of the form $(\prod^{n}_{i=1} p_i, f)$ where $f$ is zero or square-free mod $p_i$ for distinct prime integers $p_i$.

However, I could not compute the intersection when infinitely many prime integers are involved.

Related question: if there is no simple form for all the radical ideals, is there a simple description of the Zariski-closed sets of the prime spectrum (which correspond to radical ideals)?

  • 3
    $\begingroup$ What would an element of an infinite intersection look like? $\endgroup$ – Lubin Aug 19 '13 at 2:08
  • $\begingroup$ Special case of math.stackexchange.com/questions/300170 $\endgroup$ – Martin Brandenburg Aug 19 '13 at 7:45
  • $\begingroup$ I should remark that the question how the radical ideals look like has NOT been answered yet in the other thread. Instead I just wanted to suggest to continue this discussion there. $\endgroup$ – Martin Brandenburg Aug 19 '13 at 12:54
  • $\begingroup$ Following Lubin's hint, I believe any infinite intersection of prime ideals can be written as a finite intersection of prime ideals. $\endgroup$ – Tim Aug 21 '13 at 2:18