TITLE: Automorphisms of finite-dimensional vector spaces without the Axiom of Choice
QUESTION [1 upvotes]: Let $k$ be an infinite field (for example, $\mathbb{C}$), and let $V$ be a finite-dimensional vector space over $k$ in a ZF-model without the Axiom of Choice (AOC). (If I am not mistaken, a finitely generated vector space over $k$ has a well-defined dimension without AOC, so we could start from $k^n$, $n$ finite.)
My question: is the automorphism group of $V$ the same as in the case where we assume AOC ? (My guess is yes, due to the well-defined finite dimension.)
Second question (which probably is equivalent to my first question): what about the automorphism group of the projective space $P(V)$? (Which, if we assume AOC, would be as follows: each automorphism is of the form
$$x \mapsto Ax^\sigma, $$
where x is a point written in homogeneous coordinates (column vector), $A$ is a nonsingular $(n \times n)$-matrix over $k$, and $\sigma$ is a field automorphism of $k$.)
Thanks !

REPLY [2 votes]: Yes. The same automorphism group.
To see why, note that any automorphism is uniquely determined by its action on the basis vectors of any fixed basis, e.g. the standard one. Since the basis is finite, everything is fully determined the same way it would in $\sf ZFC$.
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