Understanding the field trace equation $\text{Tr}_{L:K}(\alpha)=[L:K(\alpha)]\text{Tr}_{K(\alpha):K}(\alpha)$ According to this paper, given the finite field extension $L:K$ and $\alpha\in L$, we define the trace $\text{Tr}_{L:K}(\alpha)$ as the trace of the linear transformation $$m_\alpha:L\rightarrow L:x\mapsto \alpha x$$
The paper then argues that "if we build a $K$-basis of $L$ by first picking a basis of $K(\alpha )$ and then picking a basis of $L$ over $K(\alpha )$,
we get a ‘block’ matrix for $m_{\alpha}$ consisting of $[L : K(\alpha )]$ copies of the smaller square matrix for $m_{\alpha} : K(\alpha) \rightarrow K(\alpha)$
along the main diagonal, so
$\text{Tr}_{L:K}(\alpha)=[L:K(\alpha)]\text{Tr}_{K(\alpha):K}\big(\alpha\big)$".
I understand the idea of constructing a basis for $L$ over $K$ in the manner they describe, but I do not understand at all where this 'block' matrix comes from. In fact, it seems to me that for any $\vec{l}\in L$, regardless of our choice of basis, we have
$$\alpha \text{I}_n\vec{l}=\alpha\vec{l}$$
where $\text{I}_n$ is the identity matrix of dimension $n$. So that $\text{Tr}_{L:K}(\alpha)=n\alpha$.

What am I getting wrong?
 A: I think you're wrong.

*

*If the minimal polynomial of $\alpha$ over $K$ is $f(x)=x^s+a_1x^{s-1}+\ldots+a_s$,
then the basis $K(\alpha):K$ can be chosen as $1,\alpha,\ldots,\alpha^{s-1}$.


*Let $u_1,\ldots,u_t$ be a basis of extension $L:K(\alpha)$.


*Then the basis of $L:K$ is $\alpha_iu_j$, $0\leq i\leq s-1$, $1\leq j\leq t$.


*On each block $u_j,\alpha u_j,\ldots,\alpha^{s-1}u_j$ of the basis,
the linear transformation $m_\alpha$ acts as a shift
$$
m_\alpha(u_j)=\alpha u_j,\
m_\alpha(\alpha u_j)=\alpha^2 u_j,\
\ldots,
m_\alpha(\alpha^{s-1} u_j)=\alpha^s u_j=-(a_s+\ldots+\alpha^{s-1}a_1)u_j
$$
Therefore the matrix $m_\alpha$ in this basis has a block form.
On the diagonal there are $s\times s$-blocks of the form
$$
\begin{pmatrix}
0&0&\ldots & -a_s\\
1&0&\ldots &-a_{s-1}\\
0&1&\ldots &-a_{s-2}\\
\ldots&\ldots&\ldots&\ldots\\
0&0&\ldots&-a_1\end{pmatrix}
$$
and there are $t$ blocks of this kind.
It follows that $\operatorname{Tr}_{L:K}(m_\alpha)=-a_1t\in K$.
Addition.
Your formula is correct in the following form
$\operatorname{Tr}_{L:L}m_\alpha=\alpha$. But in this form it is of little interest.
To define a trace over $K$, we must define the matrix of multiplication by $\alpha$
in a basis $L$ over $K$ and then
the matrix coefficients necessarily lie in $K$, and hence the
trace lies in $K$.
Edit. Corrected the matrix and formula for the trace. Thanks @Leo.
