4
$\begingroup$

I am reading Joe Harris' book Algebraic Geometry: A first course. In the book he says:

The Grassmannian $G(k,n)$ contains, as a Zariski open subset, affine space $\mathbb{A}^{k(n-k)}$, and thus has dimension $k(n-k)$.

Is it true that a Zariski open subset of a variety has the same dimension as the variety? I am trying to find it in Harris' book, but I do not get it. Is it an inmediate consequence of any other result?

I would appreciate if you could explain this fact to me, or if you could tell me a book where it is shown.

$\endgroup$
4
$\begingroup$

Dimension of the variety $X$ is the transcendence degree of its function field $k(X)$ which is isomorphic to $k(U)$ for an open set $U \subset X$ so the dimension consequence.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.