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Let $X$ be an irreducible quasiprojective variety. We define the dimension of $X$ to be the transcendence degree of $k(X)$ over $k.$

I'm trying to show that if $U \subset X$ is a nonempty open subset, then dim $U =$ dim $X.$

The results I've seen use the fact that $U$ is dense, and so its closure is the whole space: https://math.stackexchange.com/a/1790691/833271, or they just state the result without explaining it: pg. 33, Example 11.2 of http://math.uchicago.edu/~amathew/AGnotes.pdf

My question is how the fact that $U$ is dense relates back to transcendence degree, because I'm not seeing the immediate connection.

Smash
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  • The point is that if $U$ is dense then, as extensions of $k$, the fields $k(U)$ and $k(X)$ are isomorphic, by restriction of rational functions. If $U$ were not dense, there would be no well-defined restriction map. – Lazzaro Campeotti Oct 28 '20 at 15:58
  • Think about a reducible, non-equidimensional scheme (ignore this word if you don't know what it means, it isn't important): let $X$ be a line $L$ meeting a plane $P$ in a single point. If you take an open subset of $L$, its closure is just $L$, not all of $X$. And of course, $k(L)$ has transcendence degree $1$ while $k(P)$ has transcendence degree $2$, so if it made sense to talk about a function field of $X$ it would have transcendence degree $\ge 2$. – Tabes Bridges Oct 28 '20 at 16:09

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