Doubt in a MO link involving different proof of Hilbert $90.$ I was trying to understand the following link in MO:(https://mathoverflow.net/q/21110/104333) where a few different proofs of Hilbert $90$ are discussed. I couldn't understand some part of Emerton's proof: for example, why the induced map from $L^n \to L^n$ is given by $(b_1,\ldots,b_n) \mapsto (a b_n, \sigma(a) b_1, \ldots, \sigma^{n-1}(a) b_{n-1}).$
What I tried: For $(b_1,\ldots,b_n) \in L^n$ there exist a polynomial $h(X) \in L[X]$ such that $h(X) \equiv b_i(\mod(X-\sigma^{i-1}(\alpha )))$ and if $h(X)=c_0+c_1X+\cdots +c_{n-1}X^{n-1}$ then the preimage of $(b_1,\ldots,b_n)$ in $L \otimes_K L$ is (lets say )$\beta = (c_0 \otimes \alpha)+(c_1\otimes \sigma(\alpha)+ \cdots (c_{n-1}\otimes \sigma^{n-1}(\alpha)))$ which is again followed by $T \otimes Id$ to get image of $\beta$ in $L\otimes_K L$ and finally I'm getting $(a \sigma(b_n),\ldots,a\sigma(b_1)$ in $L^n.$ I am not getting the expected one. 

Here $L/K$ is a cyclic extension of degree $n$ with $\sigma$
generating its Galois group
One is using a particular isomorphism $\Phi:L\otimes_KL\to L^n$
given (I think) by
$$\Phi(x\otimes y)=(xy,\sigma(x)y,\sigma(x^2)y,\ldots,\sigma^{n-1}(x)y).$$
For $a\in L$ there is a map $\mu_a:L\otimes_KL\to L\otimes_KL$ given
by $\mu_a(x\otimes y)=ax\otimes y$. The question is what does this induce on $L^n$, that is what is $\mu'_a=\Phi\circ\mu_a\circ\Phi^{-1}$?
Well,
$$\Phi\circ\mu_a(x\otimes y)=\Phi(ax\otimes y)
=(axy,\sigma(a)\sigma(x)y,\ldots,\sigma^{n-1}(a)\sigma^{n-1}(x)y).$$
If we define $\psi:L^n\to L^n$ by $\psi:(z_0,z_1\ldots,z_{n-1})
\mapsto(az_0,\sigma(a)z_1,\ldots,\sigma^{n-1}(a)z_{n-1})$
then $\Phi\circ\mu_a=\psi\circ\Phi$. This means that $\psi=\mu'_a$
which is what we wanted to prove.
