Does anyone know of a domain in which
- every ideal is generated by at most three elements
- at least one ideal is generated by no less than three elements?
What if three is replaced by n: a positive integer bigger than 3?
$\begingroup$
$\endgroup$
3
-
3$\begingroup$ What about the subring $k[x^3,x^4,x^5]$ of $k[x]$? (I'm not sure whether it satisfies 1. If so, generalization for higher $n$ should work fine.) $\endgroup$– YCorJun 5, 2018 at 8:03
-
$\begingroup$ @YCor It feels like it should work. But I think $k[[x^3,x^4,x^5]]$ definitely does. $\endgroup$– Jeremy RickardJun 5, 2018 at 12:20
-
$\begingroup$ @JeremyRickard thanks, of course you're right, localizing clears up all the mess. $\endgroup$– YCorJun 5, 2018 at 12:31
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
This is too long to be a comment.
If we restrict our attention to the case of finitely generated ideals, then there are two classical results by Heitmann and Swan.
Theorem 1 (Heitmann 1976) A finitely generated ideal in a Prüfer domain of Krull dimension $n$ needs at most $n + 1$ generators.
Theorem 2 (Swan 1984) For every $n \in \mathbb{N}$, there exists a Prüfer domain of Krull dimension $n$ with an ideal $I$ that cannot be generated by less than $n+1$ elements.
Unfortunately, we cannot hope to use the above results in order to give an answer to the original question, since a Noetherian Prüfer domain is a Dedekind domain, and it is well-known that in such a domain every ideal is either principal or generated by two elements, see MSE question 597543.
References.
R. C. Heitmann: Generating ideals in Prüfer domains, Pacific J. Math., Volume 62, Number 1 (1976), 117-126.
R. G. Swan: $n$-generator ideals in Prüfer domains, Pacific J. Math., Volume 111, Number 2 (1984), 433-446.
answered Jun 5, 2018 at 8:53