Let $K$ be a field. Is $K[X,Y]$ a Dedekind domain?
2025-01-12 23:44:15.1736725455
Polynomial rings which are Dedekind domains.
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No, because a Dedekind domain is an integrally closed noetherian domain (this is true for $K[X,Y]$) of Krull dimension $1$, and $\dim K[X,Y]=2$.
$K[X,Y]$ is integrally closed, because a polynomial ring over an integrally closed domain is itself integrally closed.
As for the assertion about the dimension, the general fact is that for any $K$-algebra of finite type $A$ which is an integral domain with fraction field $L$, we have $$\dim A=\operatorname{deg\,tr} L $$ (Bourbaki, Algebra, Ch. 8 ‘Dimension’, § 2, n°4, theorem 3).