Let $k$ be a perfect field, $R$ a $k$-algebra and the $\bar{k}$ algebraic closure of $k$. define $S:= R \otimes_k \bar{k}$. we can let the Galois group $Gal(\bar{k}/k)$ act on $S$ in this way: $\sigma \in Gal(\bar{k}/k)$ acts on $S$ by $1_R \otimes \sigma: a \times b \mapsto a \otimes \sigma(b)$.
Q: prove that the sub-$\bar{k}$-algebra $T \subset S$ of invariant elements under $Gal(\bar{k}/k)$-action, i.e. $s \in S$ with $\sigma(s)=s$ for all $\sigma \in Gal(\bar{k}/k)$ is precisely $R$. more precisely the image of $R$ under canonical map $i: R \to S, r \mapsto r \otimes 1$.
I have worked out a "proof" and would like to know, if it works: let $s \in S_{\text{inv}}$ invariant. first, assume $s= a \otimes b$, with $a \in R, b \in \bar{k} \backslash k$. then under assumption on $s$ for all $\sigma$ we have $a \otimes b= a\otimes \sigma(b)$. because $b \in \bar{k} \backslash k$ and $k$ was assumed as perfect, $b$ is separable and it's minimal polynomial has different roots over $\bar{k}$: $f_b(x) = \prod^d _{i=1} (x -\sigma_i(b))$. since $s$ invariant, $a \otimes b= a\otimes \sigma_i(b)$, and thus $a^d \otimes b^d=(a \otimes b)^d= a^d \otimes \prod \sigma_i(b)= (\prod \sigma_i(b)) \cdot a^d \otimes 1$. last equality holds, because $\prod \sigma_i(b) \in k$.
now, does this imply that $b^d \in k$? is this obvious? I see two possible obstructions: what, if $a^d=0$? is thus the nilpotence of $R$ a nessesary condition for $i(R)=S_{\text{inv}}$?
the other problem is under assumption that $a^d \neq 0$, if $b^d \in k$ can be shown by a simple 'choice of basis' argument? concretly, let $e_1, e_2,...$ $k$-basis of $R$, $f_1, f_2,...$ $k$-basis of $\bar{k}$, then $e_j \otimes f_k$ form a basis of $S$ also in infinite dimensional case? the idea is that $i(R)$ can be identified with $k$-hull of the subbasis $e_i \otimes f_1$, if we choose $f_1 :=1_k$,i.e. the embedding $k \to \bar{k}$ is done by identifying $k$ with $k$-hull of $f_1$. does this argument work?
ps: the described problem popped up in Mumford's & Oda's Algebraic Geometry II on page 130
Take a $k$-basis for $R$, say $\{e_i\}$. Then every element in $S$ can be written uniquely as a finite sum $s=\sum_ie_i\otimes\lambda_i$, where $\lambda_i\in\bar k$. Then $\sigma(s)=\sum_ie_i\otimes\sigma(\lambda_i)$, so we have $\sigma(s)=s$ if and only if $\sigma(\lambda_i)=\lambda_i$ for all $i$.
It follows that the fixed points $T$ in $R\otimes_k\bar k$ under the action of the Galois group (which is a $k$-algebra, but not a $\bar k$-algebra as you have written) is precisely $R\otimes_kK$, where $K$ is the fixed field of $\bar k$. Since $k$ is perfect, we have $K=k$, and hence $T=R$.
Some good online notes on Galois descent theory can be found here:
https://kconrad.math.uconn.edu/blurbs/galoistheory/galoisdescent.pdf