Galois descent for dimension of vector spaces

Question

Let $L/K$ be a Galois extension (I am interested in $\overline{\mathbb{Q}}/\mathbb{Q}$ so I do not assume it to be finite).

Let $V\subset L^n$ be a $L$-subvector space, of dimension $d$, such that $g(V)=V$ for each $g\in \mathrm{Gal}(L/K)$. Do you have a reference for the fact that the dimension of the $K$-vector space $V\cap K^n$ is equal to $d$?

It seems to me that it corresponds to a Galois descent and I've looked into books for this notion but I did not find something with this precise statement.

Answer 1

What you need is the corollary on the bottom of page 60 of Bourbaki, Algèbre, chapitre V : Corps commutatifs ; section 10 (Extensions galoisiennes), subsection 4 (Descente galoisienne).

The statement given there is more precise: given an $L$-subspace $W$ of $L^n$ which is stable under Galois, the $K$-subspace $V=W\cap K^n$ of $K^n$ satisfies $W=V\otimes_K L$ and is the unique one with this property. Since dimension is preserved by base change, one has $\dim_L(W)=\dim_K(V)$, as you requested.

Answer 2

Just to elaborate on Adel BETINA's comment, $V \cap K^n$ is exactly $V^G$, so the part "$f$ is one-to-one" of the proof of Theorem 2.14 in Conrad's text does the job, because from $L\otimes_K V^G\cong V$ follows $\dim_LV=\dim_KV^G$.