Number of embeddings $F_q^n \rightarrow F_q^{n+m}$

Question

Let $F=F_q$ be a finite field where, $q$ is a power of prime.

I wish to compute the number of extnesions $$ 0 \rightarrow F^m \rightarrow F^{n+m} \rightarrow F^m \rightarrow 0 $$ This is in page 8, of this notes.

What I don't understand is that it is claimed the number of such extensions i given by the following sequence of identities

$$|\{V \subseteq F^{n+m}| V\cong F^n \}|= \frac{|F^n \hookrightarrow F^{n+m}|}{|GL_n(F)|}=\frac{|(q^{n+m}-1) \cdots (q^{n+m}-q^{m-1})}{|(q^m-1)\cdots (q-1)|}$$

It would be great if some explaination is given for each of the equality. Especially the counting part.

Answer 1

Counting embeddings $\Bbb F^m\hookrightarrow\Bbb F^{m+n}$ amounts to counting subsets of $m$ linearly independent vectors in $\Bbb F^{m+n}$. Any such subset $$ \{v_1,...,v_m\} $$ can be constructed as follows:

• $v_1$ is any non zero vector,
• $v_2$ is any vector not in the span of $v_1$,
• $v_3$ is any vector not in the span of $v_1$ and $v_2$, and so on.

Since the span of $k$ linearly independent vectors consists of $q^k$ vectors all the above choices can be made in $$ (q^{m+n}-1)(q^{m+n}-q)\cdots(q^{m+n}-q^{m-1}) $$ ways.

Then if you want to count subspaces you need to mod out by the natural ${\rm GL}_m$ action and you obtain the formula observing that $$ |{\rm GL}_m(F)|=(q^m-1)(q^m-q)\cdots(q^m-q^{m-1}) $$ basically by the same principle (to give an invertible $m\times m$ matrix amounts to give $m$ linearly independent vectors).

(The formula you wrote seems to have a wrong factor)