2
$\begingroup$

Is every radical ideal in the ring of algebraic integers (i.e. the integral closure of $\mathbb{Z}$ considered as a subring of $\mathbb{C}$ via the unique homomorphism of unital rings $\mathbb{Z}\rightarrow\mathbb{C}$) a finite intersection of prime ideals?

$\endgroup$
  • $\begingroup$ Welcome to MathOverflow! This is a website for questions about research level mathematics (as is explained in the help centre). For questions about basic commutative algebra, Math StackExchange might be more suitable. $\endgroup$ – R. van Dobben de Bruyn Jun 8 at 20:41
  • $\begingroup$ Seems related to mathoverflow.net/questions/333526. If I believe what I read there, this ring is not noetherian. For noetherian commutative rings every radical ideal is finite intersection of prime ideals, but this is not true in general and not obvious to me here. $\endgroup$ – YCor Jun 9 at 2:18
  • 2
    $\begingroup$ @R.vanDobbendeBruyn I think I can answer this, but it doesn't seem like basic commutative algebra to me. This is a ring which is non-noetherian, but in a very simple and structured way. $\endgroup$ – David E Speyer Jun 9 at 3:13
  • 1
    $\begingroup$ @DavidESpeyer: maybe you're right. Regardless, the question (like some other recent questions by unregistered users) does not seem to be research-related, with no mention of what the OP has tried so far, no motivation given, and the answer is essentially one line. I don't find it a very high quality question. $\endgroup$ – R. van Dobben de Bruyn Jun 9 at 15:56
7
$\begingroup$

The answer is no.

Let $R$ be the ring in question, with fraction field $K$ (which is the field of algebraic complex numbers). Fix a prime $p$, and let $I=\sqrt{pR}$. Thus, $V(I)=\mathrm{Spec}(R/I)$ is the reduced fiber at $p$ of $\mathrm{Spec}(R)\to \mathrm{Spec}(\mathbb{Z})$. This is a zero-dimensional affine scheme with infinitely many points (they correspond to the valuations of $K$ extending the $p$-adic valuation). For a subset $S$ of these, to say that $I=\bigcap_{\mathfrak{p}\in S}\mathfrak{p}$ just means that $S$ is Zariski-dense in $V(I)$, hence cannot be finite.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy