Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've tried to construct examples of rings that match all except one of the properties in the definition of a Dedekind domain. (This is an old number theory qual question from Berkeleys MGSA website).

The only starting point that I can think of would be

$R:=K[X_1,X_2,\ldots]$

for $K$ a field. This is clearly integrally closed as any polynomial relation is contained in $K(X_1,\ldots,X_n)$ for some large enough $n$ and $K[X_1,\ldots,X_n]$ is integrally closed being a UFD.

The problem is then to get rid of enough ideals from this ring $R$, so that its dimension would be 1, but I can't seem to figure out what to do.

Anyone have any other ideas for rings that could work as a starting point? The polynomial ring with an infinite number of variables is pretty much the only example that I know of for how to construct a non-noetherian ring that's still a domain.

share|improve this question
add comment

1 Answer

The ring of all algebraic integers is integrally closed, one-dimensional and not Noetherian.

It is integrally closed since it is defined as the integral closure of $\mathbb{Z}$ in $\overline{\mathbb{Q}}$.

It is one-dimensional since it is an integral extension of $\mathbb{Z}$, and if $S/R$ is an integral extension of domains then $S$ and $R$ have the same Krull dimension.

It is not Noetherian since, for instance, $\{(2^{\frac{1}{2^n}})\}_{n=1}^{\infty}$ is an infinite strictly ascending chain of ideals.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.