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 Sep 28 awarded Yearling Sep 23 awarded Nice Question Jun 24 awarded Citizen Patrol Jun 24 comment Properties of schemes determined by field valued points I have edited the question to address the comments thus far. Is the phrasing still problematic? Can it be considered for re-opening? Jun 16 comment Properties of schemes determined by field valued points @WillSawin I have made edits to state the question more precisely. Mostly I want to consider $X(K)$ as just a topological space. Is there other structure of $X(K)$ from a functorial point of view that come to mind? Jun 16 revised Properties of schemes determined by field valued points added 858 characters in body Jun 15 asked Properties of schemes determined by field valued points Feb 9 awarded Nice Question Jan 9 awarded Notable Question Oct 15 awarded Notable Question Sep 28 awarded Yearling Sep 26 comment Conceptual algebraic proof that Grassmannian is closed in Plucker embedding Both of the previous claims can be boiled down to a Lie algebra calculation using that $Gr_{k,n} = GL_n/P$. Factor $P = GL_k\times GL_{n-k}\times U$. Let $U^-$ be the transpose of $U$. Then tangent space of $GL_n/P$ is naturally $Lie(U^-)$ and can check tangent map is injective because $Lie(U^-)$ acts non trivially on highest wt. vector. Sep 26 comment Conceptual algebraic proof that Grassmannian is closed in Plucker embedding Closed immersion? The map $GL_n\to \mathbb{P}(\wedge^k \mathbb{C}^n)$ is not even injective so cannot be an immersion. It is a submersion; since $GL_n$ acts transitively on the image so its enough to check map on tangent spaces is surjective at identity. Similarly, the map from $Gr_{k,n} \to \mathbb{P}(\wedge^k \mathbb{C})$ is a closed embedding again because map on tangent spaces is injective in this case. Aug 16 awarded Nice Question Aug 14 comment vector bundles on $\mathbb{C}[x,y,z]/(x y - z^k)$ The idea is that if $P$ is a nontrivial $G$ bundle on $U$ and $W$ is any etale neighborhood of $(u^k,v^k,u v)$ then $P$ restricted to $W \cap U$ remains nontrivial. Aug 14 comment vector bundles on $\mathbb{C}[x,y,z]/(x y - z^k)$ I believe this argument can be adapted to show that on $Spec\ A$ every principal $G$ bundle is trivial whenever $G$ is a split reductive group. Aug 13 accepted vector bundles on $\mathbb{C}[x,y,z]/(x y - z^k)$ Aug 13 comment vector bundles on $\mathbb{C}[x,y,z]/(x y - z^k)$ If I understand correctly then the conclusion should be the same for higher rank: any vector bundle on $V$ is trivial so $Spec\ A$ you will see $B_{r_1} \oplus ... \oplus B_{r_n}$ and the fiber at $(u^k,v^k, u v)$ will be bigger than $n$ unless all $r_i = 0$. So $Spec\ A$ has only trivial vector bundles. Aug 13 comment vector bundles on $\mathbb{C}[x,y,z]/(x y - z^k)$ Great! The point I was missing is if a locally free on $U$ is to extend to a locally free on $Spec A$ it must be the reflexive extension. Aug 13 asked vector bundles on $\mathbb{C}[x,y,z]/(x y - z^k)$