I think, as R is an ideal of itself, and R is Noetherian thus R is finite generated, assume it is generated by N elements, then every ideal is generated by at most N elements. But my conclusion is wrong, could you tell me where I made mistakes? Thank you!
2026-03-27 16:27:48.1774628868
If R is a Noetherian ring, is it true that there exists an integer N such that every ideal of R is generated by at most N elements?
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The ideal $R$ is always generated by the element $1$ as an ideal, for any ring. This is no surprise and has nothing to do with the noetherian property.
Even if $R$ was a finitely generated $k$-algebra where $k$ is a field, it is not guaranteed that the number of generators for $R$ as a $k$-algebra is an upper bound on the number of generators of an ideal. The generators of your algebra do not have to lie in the ideal nor help to generate it.
There are even noetherian rings of infinite dimension (i.e. such that the set of the heights of its prime ideals is not bounded), and by Krull's height theorem, the height of a prime ideal $p$ of $R$ is less than $n$ if and only if $p$ is minimal above an ideal of $R$ generated by $n$ elements. So for some noetherian rings, the number of generators necessary to generate a prime ideal can be arbitrary large (although always finite for each prime ideal).
Hope that helps,