I need to prove that the Krull dimension of a ring $R$ equals the dimension of the Spectrum. I started by showing that in $\operatorname{Spec}(R)$ the closed irreducible sets are the prime ideals and then the length of every chain $C$ in spectrum will be equal to the supremum of the length of prime chains. Is this enough? I can definitely understand why it stands but I’m not sure this is enough for a proof… I read something about geometric interpretation, is it maybe proven geometrically?
2026-03-28 08:09:53.1774685393
Proof that the Krull dimension of a ring R equals the dimension of Spec(R)
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