Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers 2

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?

share|improve this answer
    
Dear Mariano, Since the intersection of even an infinite number of closed sets remains closed, you don't need the finite generatedness assumption (i.e. you could just look at the set of $U$ such that $a(U) \subseteq U$ for all $a \in A$). (Sorry for such a minor nitpick!) Regards, –  Matt E Mar 15 '11 at 18:13
1  
@Matt: yeah, I noticed that after posting (I am used to think about the variety of modules, and in that situation one does need finite generation so as to get a something which is finite dimensional at first sight---I got the two things mixed up) –  Mariano Suárez-Alvarez Mar 15 '11 at 18:17
add comment

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.

share|improve this answer
    
please use comments to make comments. My guess is, you did not register so you do not have enough repution to make comments, though... –  Mariano Suárez-Alvarez Mar 16 '11 at 1:19
    
Did you try writing out an example explicitly? This might make the general case clearer. –  Matt E Mar 16 '11 at 3:37
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.