Invariant subspace in $K\otimes V$ is generated by invariant vectors? I'm reading Popov and Vinberg's essay "Invariant Theory". (Link, apologies for paywall.) On p. 155, the following is stated without proof:

Lemma 2.4. Suppose $K$ is an extension of the field $k$ and $G$ is a group of its automorphisms. Suppose $V$ is a (not necessarily finite-dimensional) vector space over $k$ and $W$ is a subspace of the vector space $K\otimes V$ over $K$ that is invariant under the natural action of $G$. Then $W$ is generated by invariant vectors.

This is not at all obvious to me. What is the proof?
I think it's reasonable to assume in context that $G$ is a group of automorphisms of $K/k$, i.e. the elements of $k$ are all fixed points for the action of $G$ on $K$. The application of the lemma is to a situation where $G$ is an affine algebraic group defined over $k$ and $K$ is the field of rational functions of a variety over $k$; thus it cannot be assumed $G$ is finite or that $K$ is finite over $k$.
 A: In the comments, Alex Youcis links to these notes of J. Milne (see section c), which give the argument, in an ever-so-slightly-different notational regime. I'll reproduce and slightly reformulate the argument here, to prove exactly the statement made by Popov and Vinberg.
Recalling the setup, we have a field $k$, a $k$-vector space $V$, a field extension $K$ of $k$, and a group $G$ of $k$-automorphisms of $K$. We also have a $K$-subspace $W$ of the base-change $K\otimes_kV$ of $V$ to $K$. Note that in this setup, there is a natural action of $G$ on $K\otimes_k V$ via the action on the first tensor factor. The claim is that if $W$ is $G$-stable (as a set), then it is generated by vectors that are individually $G$-invariant.
As usual, $K^G$ is the invariant (i.e. pointwise-fixed) subfield of $K$ under the action of $G$, and likewise, if $T\subset K\otimes_k V$ is a $G$-stable $K$-subspace, then $T^G$ is the invariant (i.e. pointwise-fixed) subset. (NB: $T^G$ is always a $k$-linear space; in fact, it is always $K^G$-linear; but is usually not $K$-linear.) Note that, because $K\otimes_k V$ is a $K$-space, we can unambiguously write $cv$ if $c\in K$ and $v\in K\otimes_kV$. In particular, if $w\in W$, "$cw$" refers to a multiplication that utilizes the $K$-space structure of $K\otimes_k V$.
I'll use Milne's numbering for the lemmas, to ease matching what I'm saying to the source material, although I'm reformulating the argument into Popov and Vinberg's notation and ever-so-slightly-different setup. (Milne assumes $K^G=k$, which P and V do not, and he also starts from a $K$-vector space he calls $V(K)$ [well, actually, he uses $\Omega$ instead of $K$] and then $V\subset V(K)$ is a subspace subject to the assumption that $K\otimes_k V$ is isomorphic to $V(K)$. None of the authors have any finiteness constraint on $G$ or the dimensions of $V,W$.) As a consequence, the lemmas don't quite come out in their original order.
Throughout the below, let $\{e_i\}_{i\in I}$ be a $k$-basis of $V$, where $I$ is some indexing set. Then $\{1\otimes e_i\}_{i\in I}$ is a $K$-basis of $K\otimes_k V$.
Lemma 16.5: $(K\otimes_k V)^G = K^G\otimes_k V$.
Here the $=$ sign, as opposed to $\cong$, must be justified by embedding $K^G\otimes_k V$ in $K\otimes_kV$ via the embedding of $K^G$ in $K$. This can be done because $V$ is flat over $k$ since it is a field.
Proof: The functor of invariants commutes with flat base change. Alternatively, following Milne: for any $v=\sum c_i\otimes e_i\in K\otimes_k V$ to be $G$-invariant, the $K$-linear independence of the $1\otimes e_i$'s implies each $c_i$ must be individually $G$-invariant, i.e. in $K^G$.
Lemma 16.4: The following conditions on $W$ are equivalent:

*

*The $K^G$-subspace $W^G$ spans $W$ as a $K$-space.

*$W^G$ contains a $K$-basis for $W$.

*The map $K\otimes_{K^G}W^G \rightarrow W$ mapping $c\otimes v\mapsto cv$ is an isomorphism.

Proof: Any spanning set can be pruned to a basis (see here) and any basis is spanning, so 1$\Leftrightarrow$2. Surjectivity of the map in 3 is equivalent to 1. Injectivity always holds (regardless of conditions 1,2); we see this as follows. Consider the embedding $\iota:W^G\hookrightarrow (K\otimes_kV)^G$. Since $K$ is flat over $K^G$ since the latter is a field, $\text{id.}\otimes \iota:K\otimes_{K^G} W^G\rightarrow K\otimes_{K^G}(K\otimes_kV)^G$ is also injective. By Lemma 16.5, $(K\otimes_kV)^G$ is $K^G\otimes_kV$, so we recognize the image of this map as $K\otimes_{K^G}(K^G\otimes_kV)=K\otimes_kV$, and thus the map itself as precisely the map from 3.
Remark: the device used in this proof to conclude injectivity in 3 allows us to conclude in general that the base change of a $K^G$-subspace of $(K\otimes_kV)^G$ to $K$ can be seen as a $K$-subspace of $K\otimes_kV$. We will use this fact several times in the below, without further comment.
The following is the heart of the matter.
Lemma 16.6: Assume $W$ is $G$-stable and nontrivial. Then $W^G$ is nontrivial.
Proof: An arbitrary nontrivial element of $K\otimes_kV$ can be written (uniquely) as a finite linear combination of some $1\otimes e_i$'s with nonzero coefficients $c_i\in K$. Since $W$ is nontrivial, we can choose a nontrivial element of $W$ whose representation in this form has a minimal number of terms; call it $w$. By scaling by an appropriate element of $K$, we can guarantee that one of $w$'s terms is $1\otimes e_{i^\star}$ for some $i^\star$:
$$w = 1\otimes e_{i^\star} + \sum_{i\in J} c_i\otimes e_i$$
where $J$ is a finite set of indices, $i^\star\notin J$, all the $c_i$'s are in $K$ and nonzero, and the representation of any nonzero element of $W$ on the basis $\{1\otimes e_i\}_{i\in I}$ has at least $1+|J|$ terms. Then, since $W$ is $G$-stable, $gw$ is also in $W$ for any $g\in G$, and thus
$$w - gw = \sum_{i\in J} (c_i-gc_i)\otimes e_i$$
is also in $W$. But the sum on the right side has fewer than $1+|J|$ terms, so it must be zero, regardless of $g\in G$. In other words, $gw=w,\;\forall g\in G$, i.e. $w\in W^G$. Thus $W^G$ is nontrivial.
Popov and Vinberg's Lemma 2.4: Assume $W\subset K\otimes_k V$ is $G$-stable. Then $W$ is generated as a $K$-space by $W^G$.
Remark: Milne's Lemma 16.7 is actually equivalent to this statement and its converse. Note that the conclusion of this statement is exactly condition 1 of Lemma 16.4, thus the real story is that the equivalent conditions of Lemma 16.4 are further equivalent to the $G$-stability of $W$! I'll phrase the proof toward Popov and Vinberg's statement, but it's not hard to massage it into this stronger statement.
Proof: Let $W'$ be the $K$-space spanned by $W^G$. Obviously $W'\subset W$, and our goal is to show equality. Let $L$ be any complement of $W^G$ in the $K^G$-vector space $(K\otimes_kV)^G = K^G\otimes_k V$, so that
$$(K\otimes_kV)^G = W^G\oplus L.$$
Base changing from $K^G$ to $K$, the left side of this equation becomes $K\otimes_kV$, as in the proof of Lemma 16.4; thus we have
$$K\otimes_kV = \left(K\otimes_{K^G}W^G\right) \oplus \left(K\otimes_{K^G}L\right).$$
Note that $K\otimes_{K^G}W^G$ is exactly $W'$. Now $K\otimes_{K^G}L$ is a $G$-stable $K$-subspace of $K\otimes_kV$. Call it $KL$ for short. Thus we can write the above as
$$K\otimes_kV = W' \oplus KL.\tag{1}$$
We get $(KL)^G = L$ by applying Lemma 16.5 with $L$ in the place of $V$ and $K^G$ in the place of $k$. So $W\cap KL$ is also $G$-stable, and we have
$$(W\cap KL)^G = W^G\cap(KL)^G = W^G\cap L = 0,$$
with the final equality because $L$ and $W^G$ are complements in $(K\otimes_kV)^G$. Applying Lemma 16.6 with $W\cap KL$ in the place of $W$, we get $W\cap KL = 0$, since if it were nonzero, Lemma 16.6 would assert that $(W\cap KL)^G$ was also nonzero.
Now evidently $W$ is a $K$-subspace of $K\otimes_kV$ containing $W'$. Since we now know it only intersects $KL$ trivially, we can conclude from the direct sum decomposition (1) that actually $W=W'$, as desired.
