Let K be an algebraically closed field of arbitrary characteristic. Then there is a Grassmannian of subspaces of fixed dimension say m of n-dimentional space and it is a projective variety. If you have an algebra A over field K you can vew say left modules over this algebra and for a given module of dimension n you can vew a Grassmannian of m-dimentional submodules. I want to know why the Grassmannian of submodules is closed in the Grassmannian of subspaces.
I have a proof for characteristic 0. Take a subspace U, U is a submodule iff for any $a \in A$ $a(U) \subseteq U$, for any $a \in A$ there is $c \in K$ (that is true only if charK=0) such that multipaing the whole module by a-c1 is an isomorphism. But $a(U) \subseteq U$ iff for any c $(a-c1)(U) \subseteq U$ or also there exists c such that $(a-c1)(U) \subseteq U$. So you can chose c such that (a-c1) induces an isomorphism and have $(a-c1)(U) = U$. If you have an iso induced by multiplication by b on the whole module than you have a morphism on the Grassmannian of submodules say b'. So the Grassmannian of submodules is {subspaces U such that b'(U)=U for any iso b} and it is a set of equasions and so it is closed and the Grassmannian is projective.
But if charK=p than it does't work. And I don't know what to do.