# Integral domain that is integrally closed, one-dimensional and not noetherian

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.

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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.

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