How to show that a Grassmannian of submodules is a projective variety? 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.
 A: I will suppose in the following that you have a finitely generated $k$-algebra $A$ over a field $k$ and an $A$-module $M$ which is finite dimensional over $k$. If this is indeed the situation you have in mind, please do edit the question to make it explicit.
Let $m\geq0$ We want to show that the set $G^A_m(M)$ of $A$-submodules of $M$ is a closed subset of the grassmanian $G^k_m(M)$ of $m$-dimensional vector subspaces.
If $a_1$, $\dots$, $a_n$ generate $A$ over $k$ as an algebra, then an element $U\in G^k_m(M)$ is in $G^A_m(M)$ iff $a_i(U)\subseteq U$ for all $i\in\{1,\dots,n\}$. Since the intersection of closed sets is closed, it follows immediately that it is enough to show the following:

If $M$ is a $k$-vector space and $f:M\to M$ is a linear map, then the  $$\{U\in G^k_m(M):f(U)\subseteq U\}$$ is a closed subset of $G^k_m(M)$.

Can you see how to show this last statement?
A: Well, this was exactly my question: how to get a set of equations, if I have a condition that $f(U) \subseteq U,$ the only idea I can think of is that the matrix of f must have some zero blocks, but this is not an equation in terms of points of Grassmannian. So my question was exactly in the last statement.
