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Let $X$ be an irreducible affine variety. Let $U \subset X$ be a nonempty open subset. Show that dim $U=$ dim $X$.

Since $U \subset X$, dim $U \leq$ dim $X$ is immediate. I also know that the result is not true if $X$ is any irreducible topological space, so somehow the properties of an affine variety have to come in. I have tried assuming $U=X$ \ $V(f_1,...,f_k)$ but I don't know how to continue on.

Any help is appreciated!

Chi Cheuk Tsang
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2 Answers2

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Let $X=\operatorname{Spec} A$, and pick some nonzero $f\in A$ with $\operatorname{Spec} A_f = D(f) \subset U$. Then we can lift chains of closed sets to see that $\dim D(f) \leq \dim U \leq \dim X$.

But as shown here, we have $\dim D(f) = \dim A_f = \dim A = \dim X$. We can also see this quite quickly by noting that $A$ and $A_f$ have the same fraction fields, and therefore the same transcendence degree over $k$.

Note that this is false for general rings $A$, even if $A$ is assumed to be a noetherian domain. In fact, any DVR is a counterexample.

Community
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Andrew Dudzik
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You should note that any nonempty open subset of an irreducible variety (or topological space) is dense : https://en.wikipedia.org/wiki/Hyperconnected_space. Then you can use the definition of the dimension to conclude.

paf
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  • But isn't the whole question about the correct way to "use the definition of dimension"? As pointed out in the post, this is false for irreducible spaces in general. – Andrew Dudzik May 18 '16 at 18:10
  • To utilize the assumption that $X$ is an irreducible affine variety, we could relate the function fields of $U$ and $X$ or - using the "chains of subvarieties" definition - we could note that the first in a chain of subvarieties for $U$ must be special. – Tomi Tyrrell May 18 '16 at 19:22
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    This does in no way answer the question because there exist examples of irreducible topological spaces (even a two-point example!) for which an open subset has dimension smaller than that of the space. – Georges Elencwajg May 19 '16 at 07:34