Open nonempty subset of irreducible variety over $k$ has the same Krull dimension as the whole variety?

Question

Let $V$ be an irreducible variety of finite type over a field $k$, $V_{0}\subseteq V$ open and nonempty (or at least dense). Why is \begin{equation*}\operatorname{dim}_{\operatorname{Krull}}(V_{0})=\operatorname{dim}_{\operatorname{Krull}}(V)\text{?}\end{equation*}

My professor uses this result (without mentioning a proof) as if it were somehow self-evident. I have tried to prove it, but I am at a loss. Who can help?

The general statement is NOT true for arbitrary irreducible topological spaces, not even for varieties of finite type over an arbitrary integral domain, see this question for instance.

You can find the answer in "Algebraic Geometry", of Robin Hartshorne. — Somerandommathematician, Jan 13 '20 at 19:25
@Somerandommathematician can you be more specific, where exactly? I couldn't find it. — The Thin Whistler, Jan 13 '20 at 21:13
@GeorgesElencwajg and therefore "dense" is a weaker condition than "open and nonempty", hence the parenthetical remark makes sense. — The Thin Whistler, Jan 18 '20 at 15:27
Does this answer your question? Krull dimension of $\mathbb{C}[x_1, x_2, x_3, x_4]/\left< x_1x_3-x_2^2,x_2 x_4-x_3^2,x_1x_4-x_2 x_3\right>$ — amWhy, Nov 13 '20 at 12:30

score 1 · Accepted Answer · answered Jan 13 '20 at 21:21

1

This follows from the fact that the dimension of an irreducible variety is equal to the transcendence degree of its function field (over the base field $k$). Since $V_0$ is dense in $V$, it has the same function field.

answered Jan 13 '20 at 21:21

Eric Wofsey

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Actually we have this statements for integral schemes. How can you drop the reduced property? – T. Wildwolf Dec 17 '22 at 17:03
The statement is not true for arbitrary integral schemes. In any case reducedness is irrelevant since the statement is true for a scheme $X$ iff it is true for $X_red$. – Eric Wofsey Dec 17 '22 at 19:05
The statement you wrote is Proposition 2.5.19 of Liu's book for integral varieties. So it is correct. The problem is that you do not use integral but irreducible. How can you define the function field of an irreducible variety? Moreover, can you elaborate why your second sentence is true? I would like to understand why I can prove a property for Xred instead. – T. Wildwolf Dec 17 '22 at 19:48
It is not true for arbitrary integral schemes. It is true for integral schemes of finite type over a field, which I guess is what you mean by "integral varieties". $X$ is homeomorphic to $X_{red}$ and the statement only involves the topology. – Eric Wofsey Dec 17 '22 at 20:43
In any case "integral" and "irreducible" are equivalent for varieties, at least if you define varieties to be reduced (which most people do). – Eric Wofsey Dec 17 '22 at 20:48
Liu does not do it. – T. Wildwolf Dec 17 '22 at 21:57

Open nonempty subset of irreducible variety over $k$ has the same Krull dimension as the whole variety?

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