Finitely Generated General Linear Group

Question

Is there a $F$ over which $GL_n(F)$, the general linear group, is finitely generated? Obviously one generator won't work, since $GL_n(F)$ is never abelian for any field. This is possibly a silly question, which I can't really trace to any single source (just sort of came to me one day). By the way, $n$ neither has to be a fixed number nor arbitrary, so interpret the question as you wish (just so long as your interpretation doesn't render the question trivial).

EDIT:

As Morgan Rodgers points out, the question becomes relatively trivial by taking $F$ to be finite. So, perhaps I should stipulate that $F$ is infinite. What's the answer in that case?

Answer 1

If $\text{GL}_m(F)$ is finitely generated, then $F^*$ is finitely generated (take determinants of a finite set of generators). Note that any subgroup of a finitely generated Abelian group is finitely generated. If $F$ has characteristic zero, $F^*$ contains a copy of $\Bbb Q^*$ which is not finitely generated.

If $F$ has characteristic $p$, and infinite, then either it is (i) algebraic over $\Bbb F_p$ or (ii) transcendental. In (i) the group $F^*$ is torsion, so cannot be finitely generated (lest it be finite). In (ii) $F^*$ contains a copy of the $\Bbb F_p(X)^*$ which is not finitely generated.

Answer 2

Here is a proof that $GL_m(K)$ is not finitely generated, for any infinite field $K$ and $m\ge 2$, also applying, unlike the determinant one, to $PGL_m(K)$ and $SL_m(K)$.

Lemma 1: $A\subset B$ be subrings. If $A\neq B$ then all the inclusions $SL_m(A)\subset SL_m(B)$, $PGL_m(A)\subset PGL_m(B)$, $GL_m(A)\subset GL_m(B)$ are proper.

Proof: take the elementary matrix $e_{12}(b)$ for any $b\in B-A$. Then it does not belong to $GL_2(A)$ (and does not represent an element of $PGL_2(A)$).

Corollary 2: if $B$ is a commutative ring and is not finitely generated, and $m\ge 2$ then each of $GL_m(B)$, $PGL_m(B)$, $SL_m(B)$ is not finitely generated.

Proof: write $G$ for $GL_m$, $PGL_m$ or $SL_m$. Then $G(B)$ is the filtering union of $G(A)$ for $A$ ranging over finitely generated subrings of $B$, and the inclusion of $G(A)$ in $G(B)$ is proper for each such $A$, by the lemma. Whence the conclusion.

Lemma 3: a field which is a finitely generated ring is finite.

This is classical.

Corollary 4: for any infinite field $K$ and $m\ge 2$, none of $GL_m(K)$, $PGL_m(K)$, $SL_m(K)$ is finitely generated.

For $SL_m$ and $PGL_m$ of course we need $m\ge 2$ since for $m=1$ this is the trivial group. On the other hand $GL_1(K)=K^*$ is also not finitely generated, by the answer by "Lord Shark...".