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If $X$ is an irreducible (quasi-)affine variety it is well known that each maximal sequence $C_0 \subsetneq C_1 \subsetneq \dots \subsetneq C_d$ of irreducible closed subsets has the same length $d = \dim(X)$. In particular, $X$ has the same local dimension at each point. Is the same true for arbitrary irreducible varieties (locally ringed spaces which can be covered by finitely many open affine sub-varieties)? What can we say about (quasi-)projective varieties for example?

Thank you in advance!

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Yes, the local dimensions of closed points are all equal (I assume that the word "variety" includes being reduced.): for every open affine $U\subseteq X$ the ring of sections $O_X(U)$ is a finitely generated $K$-algebra without zero-divisors, $K$ the base field. Since $X$ is irreducible the rings $O_X(U)$ lie in the function field $K(X)$, which is their field of fractions (*). By Noether normalization the Krull dimension of $O_X(U)$ equals the transcendence degree of $K(X)|K$, hence the Krull dimensions $\dim (O_X(U))$ are all equal. A closed point $p\in X$ corresponds to a maximal ideal $M_p$ in at least one of the $O_X(U)$. The lengths of maximal chains of prime ideals in $O_X(U)$ -- in particular those ending with $M_p$ -- are all equal to $\dim (O_X(U))$, since finitely generated $K$-algebras are catenary.

(*) Two non-empty open sets of $X$ always have a non-empty intersection due to the irreducibility of $X$. Hence for a given open set $U$ we have $K(X)=K(U)$. On the other hand for affine $U$ we have $K(U)={\mathrm Frac}(O_X(U))$.

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