# A small question about dimension of varieties

I am a bit puzzled while reading Period 11.2.1 of Vakil's $$The\space Rising\space Sea$$. A theorem was stated that for a finitely generated domain over a field $$k$$, the dimension is equal to the transcendence degree. Vakil then wrote for a irreducible $$k$$ variety, the dimension is just the transcendence degree of the function field $$\mathcal{O}_{X,\eta}$$ and therefore it can be calculated on any open set.

But he mentioned that this is false without the finite type hypothesis. I know that the above does not hold for domains not finitely generated. And my question is that is the finiteness criterion used in other parts of the above theorem? Like, if we just know that the function field of this integral scheme(which may not be a variety) is a finite extension of $$k$$, can we reach the same result?

• 1. In what sense do you mean "not a variety"? 2. If you have an irreducible variety over $k$ with function field $F$ a finite extension of $k$, then that variety is a point (proof: pick an open affine $\operatorname{Spec} R$, then $k\subset R\subset F$ exhibits $R$ as an integral domain trapped between a finite extension of fields and thus $R$ is also a field). Do you maybe mean something else? – KReiser Nov 3 '19 at 9:48
• Oh I guess I have typed something wrong. Actually I mean that if $X$ is just an integral scheme with function field $\mathcal{O}_{X,\eta}$ a finite extension of $k$, but we don't have $X$ can be covered by finitely many affine open sets, can we get the same result? – Great Dick Nov 3 '19 at 13:06
• I've just corrected my mistakes – Great Dick Nov 3 '19 at 13:08
• Yes, you can, and as I said in the previous comment, such a scheme is a point. Any open affine contains exactly one point by the previous comment, and therefore the scheme is a disjoint union of these points. But it's irreducible, so it must be a single point. – KReiser Nov 3 '19 at 17:51