$K/k$ Galois with group $\mathfrak{g}$, $V_{K} \cong V_{k} \otimes K$. Then, $V_{k}$ consists of elements of $V_{K}$ invariant under $\mathfrak{g}$. I have been asked to read "Andre A. Weil. Algebras with involutions and the classical groups. J. Indian Math. Soc.(N.S.) 24 (1960), 589–623" as part of a project, and I am encountering some difficulties with some notations and notions used by Weil, in particular, the notion of a universal domain.
In the first page of the paper itself, Weil says

In Part I, all spaces, varieties, groups are.......allowed to have points in the universal domain, and not only in their field of definition. If $V$ is a variety defined over a field $k$, we shall denote by $V_{k}$ the set of points of $V$ with coordinates in $k$.

I think I understand what this means here, but I am confused when he carries the same thing for vector spaces for instance, later he says

If $V$ is a vector space of dimension $n$ (over the universal domain) defined over $k$, $V_{k}$ and $V_{K}$ are vector spaces of dimension $n$ over $k$ and over $K$, respectively. We have $V_{k} \subset V_{K}$, and we may identify $V_{K}$ with the tensor product $V_{k} \otimes K$ taken over $k$; $\mathfrak{g}$ operates in an obvious manner on $V_{K}$, and $V_{k}$ consists of the elements of $V_{K}$ which are invariant under $\mathfrak{g}$.

I am attaching here snaps of the paper for reference : Pertaining to the first quotation,
Pertaining to the second quotation
Note: We are under the following setting: We have fixed a universal domain of characteristic $0$, a groundfield $k$ inside it and a normal extension $K$ of $k$ of finite degree $d$, with Galois group $\mathfrak{g}$

*

*In the case of varieties, if we only think of affine varieties classically then we know they are embedded in some ambient space, so if $V$ is a variety over universal domain $U$, say, then we can think  of $V$ sitting inside $U^{m}$ for some $m$, and then it makes sense to say that $V_{k}$ is the set of points of $V$ with coordinates in $k$. But in the case of vector spaces, I don't understand what connection do $V, V_{k}$ and $V_{K}$ share? They are not just same as sets I guess. One way I could think of this was that I thought of $V$ as $U^{n}$ (as a vector space of dimension $n$ over $U$, the universal domain) and then $V_{k}, V_{K}$ are just the set of points in $U^{n}$ with coordinates in $k, K$ respectively.


*I am not able to see how exactly to identify $V_{K}$ with $V_{k} \otimes K$ if I go by my definition as in 1. My attempt was to construct the following linear function from $V_{k} \otimes K$ to $V_{K}$ sending $(x_{1}, x_{2}, ...., x_{n}) \otimes c \mapsto (cx_{1}, ..., cx_{n})$ and extending this linearly. Now both domain and codomain have the same dimension over $K$, so proving that this is one-one suffices to show the isomorphism. I got stuck here.


*I think the action of $\mathfrak{g}$ on $V_{K}$ is given by $\sigma(v \otimes c) = v \otimes \sigma(c)$. Using this definition of action, I couldn't prove that $V_{k}$ is the set of elements of $V_{K}$ invariant under $\mathfrak{g}$. One direction is trivial, in the other direction I had problems because of the fact that $m \otimes n = 0$ doesn't imply that $m= 0$ or $n= 0$.
I seek serious help with the above mentioned doubts, if anyone has any idea about any of it, please help. I would be grateful for the help. I am quite unsure what tags to use, feel free to add the relevant ones.
Update: (i) I got $3.$ figured out, since we are in a vector space, $m \otimes n = 0$ would indeed imply that $m = 0$ or $n = 0$.
(ii) In $2.$ instead of proving one-one, I proved it is onto like this : So, if we have a vector $(x_{1}, x_{2}, ..., x_{n})$ in the codomain, then consider the element $\sum_{i} e_{i} \otimes x_{i}$ in the domain of the function, its image is exactly $(x_{1}, x_{2}, ..., x_{n})$.
 A: Like think about this. If $X$ is a variety over $\mathbb{C}$ then the notation $X(\mathbb{Q})$ doesn't make sense as you yourself have noted in the first quote. But, if you say "Let $Y$ be a variety over $\mathbb{Q}$ and an isomorphism $i:Y_\mathbb{C}\to X$" then you can take $X(\mathbb{Q})$ to just mean $Y(\mathbb{Q})$ and, via the isomorphism $i$, you can think of $X(\mathbb{Q})$ as a subset of $Y(\mathbb{C})$.
Of course, what $Y$ you choose, and even what $i$ you choose, matters. For example, if $X=\mathbb{G}_{m,\mathbb{C}}$ (the multiplicative group over $\mathbb{C}$) then you could take $Y_1=\mathbb{G}_{m,\mathbb{Q}}$ or $Y_2=\mathrm{Spec}(\mathbb{Q}[x,y]/(x^2+y^2+1))$. Both of these are models for $X$ over $\mathbb{Q}$ and certainly have different $\mathbb{Q}$-points inside of $X(\mathbb{C})$.
Similarly, if you have a vector space $V$ over $\mathbb{C}$ it doesn't make sense to consider its '$\mathbb{Q}$-points' without choosing a model of $V$ over $\mathbb{Q}$. What is a model? Well, if $X=\text{Spec}(A)$ is an affine variety then a model over $\mathbb{Q}$ is an affine shcme $Y=\text{Spec}(B)$ over $\mathbb{Q}$ such that $Y_\mathbb{C}\cong X$. But, this just means that $B\otimes_\mathbb{Q}\mathbb{C}\cong A$. In other words, you need to choose a $\mathbb{Q}$-subalgebra $B$ of $A$ such that the inducd map $B\otimes_\mathbb{Q}\mathbb{C}\to A$ is an isomorphism.
So, it makes sense then that a model of $V$ over $\mathbb{Q}$ is a subspace $V_\mathbb{Q}$ of $V$ such that the induced map $V_\mathbb{Q}\otimes_\mathbb{Q}\mathbb{C}\to V$ is an isomorphism. Note though that if we choose a $\mathbb{Q}$-basis $\{e_\alpha\}$ of $V_\mathbb{Q}$ then the map $V_\mathbb{Q}\otimes_\mathbb{Q}\mathbb{C}\to V$ is just the map
$$v\otimes \alpha\mapsto \alpha v$$
which has image precisely $\text{Span}(\{e_\alpha\})$. From this it's not hard to see that the condition on $V_\mathbb{Q}$ can be phrased precisely as the condition that it's spanned over $\mathbb{Q}$ by a $\mathbb{C}$-basis of $V$.
I hope that this was enough to help clear the confusion.
I would suggest you look at this note by Milne which discusses these things at more depth. Specifically, your title question is then Lemma 16.5. You can also see Ben Blum-Smith's discussion when sorting through this note of Milne here.
