If Fq\mathbb{F}_q is the field of characteristic qq then Fnq\mathbb{F}_q^n is a vector space over Fnq\mathbb{F}_q^n of dimension nn.

It is apparently true that if V⊂FnqV\subset \mathbb{F}_q^n is a subspace of dimension mm then VV has exactly qmq^m elements but I can’t find a proof of this. It is easy to check if m=1m=1 or m=nm=n (: but even the case m=2m=2 is eluding me.

=================

=================

2 Answers

2

=================

If V⊆FnqV\subseteq\mathbb{F}_q^n is a subspace, it is a vector space over Fq\mathbb{F}_q in its own right. If dimFq(V)=m\dim_{\mathbb{F}_q}(V)=m, then V≃FmqV\simeq\mathbb{F}_q^m, since vector spaces over a given field are determined by their dimension up to isomorphism. Hence #V=#Fmq=qm\#V=\#\mathbb{F}_q^m=q^m.

If VV is an mm-dimensional subspace of an nn-dimensional vector space WW over Fq\mathbb{F}_q, then we can choose a basis {β1,…,βn}\{\beta_1,\dots,\beta_n\} for WW such that {β1,…,βm}\{\beta_1,\dots,\beta_m\} is a basis for VV.

Using coordinates with respect to this basis, we can identify VV with the subspace

{(c1,…,cm,0,…,0):c1,…,cm∈Fq} \{(c_1,\dots,c_m,0,\dots,0):c_1,\dots,c_m\in\mathbb{F}_q\}

of Fnq\mathbb{F}_q^n, and this subspace has qmq^m elements because there are qq choices for each of c1,…,cmc_1,\dots,c_m.