Skip to main content
6 events
when toggle format what by license comment
Aug 26, 2010 at 15:55 comment added Charles Staats BCnrd: Your comment took me a bit to parse, so let me see if I understand it. Let's define a variety over an algebraically closed field to be a separated integral scheme of finite type over $k$. I have never actually seen a scheme-theoretic definition of variety over a non-algebraically closed field, but in my comment, I was assuming this still means "separated integral scheme of finite type over $k$." You point out that if one instead requires that the pullback to $\bar{k}$ be a variety over $\bar{k}$, then inkspot's original statement was correct.
Aug 26, 2010 at 1:22 comment added BCnrd Charles, this is a good illustration of why "$k$-variety" should never be used with clarifying the intended meaning: no one knows if (in addition to "finite type and separated") it means "irreducible and reduced", or "geometrically integral over $k$", etc. If one takes the definition in Hartshorne, namely to require geometric integrality (with any $k$) then what inkspot originally wrote is correct. Inkspot's new condition that $k$ be algebraically closed in the $k$-finite domain ${\rm{H}}^0(Y,O_Y)$ is the same as the more concrete statement ${\rm{H}}^0(Y,O_Y) = k$ (via the canonical map).
Aug 25, 2010 at 18:29 comment added inkspot Charles, you're quite right, thank you. I've added an appropriate hypothesis.
Aug 25, 2010 at 18:28 history edited inkspot CC BY-SA 2.5
Corrected error pointed out by Charles Siegel
Aug 25, 2010 at 18:22 comment added Charles Staats Your statement $H^0(S, \mathcal{O}_S) = H^0(S \times Y, \mathcal{O}_{S \times Y})$ for any proper $k$-variety $Y$ is not true if $k$ is not algebraically closed. For instance, there are proper (even projective) varieties over $\mathbb{R}$ having global sections $\mathbb{C}$. However, the statement should hold if $Y$ is, specifically, a projective space.
Aug 25, 2010 at 16:00 history answered inkspot CC BY-SA 2.5