An integral domain that is Noetherian, integrally closed, but not one-dimensional

Question

A Dededind domain is defined as an integral domain which is integrally closed, one-dimensional, and Noetherian.

Also I know an equivalent characterization that a domain is Dedekidn if and only if every nonzero ideals can factor into prime ideals.

Thus, can anyone provide me an integral domain that is Noetherian, integrally closed, but not one-dimensional, and show an ideal that cannot factor into primes and/or does not own unique factorization?

Answer 1

As suggested in the comments, let's take $k[x, y]$ for $k$ a field, algebraically closed for simplicity. This is Noetherian and integrally closed but $2$-dimensional. The prime ideals of this ring are (reference):

• $(0)$,
• $(f(x, y))$ where $f$ is irreducible, and
• $(x - a, y - b)$ where $a, b \in k$.

I claim that the ideal $(x^2, y)$ does not possess a prime factorization. Note that any prime ideal occurring in such a prime factorization must contain $(x^2, y)$, but from an inspection of the above list the only prime ideal containing $(x^2, y)$ is $P = (x, y)$, and $(x^2, y)$ lies strictly between $P$ and $P^2$.

In this setting a common salvage of prime factorization is primary decomposition.

Answer 2

Here is a geometric example. We consider the affine quadric hypersurface $ \Bbb{C}^3$ corresponding to $\Bbb{C}[x,y,z]/(xy - z^2)$. By Hartshorne II Ex. 6.4 the ring is integrally closed. If you really want to show unique factorization, follow the method of proof of Alex's comment above and show $(z^2,x,y)$ does not have unique factorization.