Subscheme of projective space in general position Let $k$ be a field and let $\mathbb{P}^n(k)$ denote $n$-dimensional projective space over $k$. What is meant by a general linear space in $\mathbb{P}^n(k)$ of codimension $m$, in the language of schemes?
I believe linear space just means closed subscheme, but how does one encode the property of a closed subscheme being in "general position"?
My question is similar to What is a generic hyperplane of a projective space over $\mathbb{C}$?, but I am interested in a precise formulation in the language of schemes, not the geometric picture. I have read on Wikipedia that the notion of "generic point" is suppposed to formalize this, but I don't understand how.
[edit] Also is there any relation between being "in general position" and being a "generic hypersurface" (the latter in the sense described on page 123 of Eisenbud-Harris "GEOMETRY OF SCHEMES"?).
[edit2] i accidentally wrote "geometric point" instead of "generic point" before
 A: The linear subspaces $L$ of codimension $m$ in $\mathbb P^n$ are parametrized by a Grassmanniannan $\mathbb G=\mathbb G(n-m,\mathbb P^n)$ of dimension $(n-m+1)\cdot m$.
Thus to  every  $g\in  \mathbb G$ there corresponds a  codimension  $m$ plane $L_g\subset \mathbb P^n$.
A generic codimension $m$ plane is then a codimension $m$ plane corresponding to  some fixed subset $S\subset \mathbb G$ with nonempty interior.
So there is a subtlety here: the subset $S$ can vary according to the problem at hand.     
An example
Consider the lines  $L\subset \mathbb P^2$.
The corresponding grassmannian  $\mathbb G=\mathbb G(1,\mathbb P^2)$ is the dual  projective plane $\mathbb (\mathbb P^2)^* $ with coordinates $(a:b:c)$ and  to the point $g=(a:b:c)\in \mathbb G$ corresponds the line $L_g\subset \mathbb P^2$ of equation $ax+by+cz=0$.
Now what does it mean that the generic line in $\mathbb P^2$ does not go through the point $Q=(1:2:3)$?
It means that these lines correspond to  the open subset $S\subset \mathbb (\mathbb P^2)^*=\mathbb G$ given by the inequation  $a+2b+3c\neq 0$.
Of course the generic line does not go through $(1:-1:2)$ either and this means that they correspond to another open subset $S'\subset \mathbb (\mathbb P^2)^*=\mathbb G$, namely $a-b+2c\neq 0$.
