Let $A = k[x]$. Then, any ideal of $A$ is a principal ideal $(f) \subset A$, for some polynomial$$f = x^n + c_{n - 1}x^{n - 1} + \ldots + c_0 \in k[x], \quad f \neq 0.$$Let $M = k[x]/(f)$, an $A$-module. Then we can choose $1$, $x$, $\ldots$, $x^{n - 1}$ as a $k$-basis of $M$. Multiplication by $x$ sends$$1 \to x \to x^2 \to \ldots \to x^{n - 1} \to x^n = -(c_{n - 1}x^{n - 1} + \ldots + c_0)$$(because $f = 0$ in $M$). Thus, we have an isomorphism $k[x]/(f) \cong k^n$, as $k$-vector spaces, and the multiplication by $x$ operator has matrix$$\begin{pmatrix} 0 & 1 & & & & 0 \\ & 0 & 1 & & & 0 \\ & & \ddots & \ddots & & 0 \\ & & & & 1 & 0 \\ & & & & 0 & 0 \\ -c_0 & -c_1 & & \cdots & & -c_{n - 1} \end{pmatrix} \quad \text{("Frobenius block")}.$$
My question is, what is the intuition behind the definition of Frobenius block here? Can anyone explain a bit more in detail what's going on here?
Of course, $k[x]/(f)$ is a $k$-module, which is to say a $k$-vector space.
Moreover, the map $T(g(x)) = xg(x)$ is a linear map. As such, we can characterize this map with respect to a certain basis via a matrix.
In particular, we note that $\{1,x,\dots,x^{n-1}\}$. We see that $T(1) = x$, so (if $n \geq 2$) our corresponding matrix $M$ should satisfy $$ \pmatrix{1&0&0 & \cdots & 0} M = \pmatrix{0 & 1 & 0 & \cdots &0} $$ Similarly, since $T(x) = x^2$, we should have (if $n\geq 3$) $$ \pmatrix{0&1&0&0 & \cdots & 0} M = \pmatrix{0&0 & 1 & 0 & \cdots &0} $$ Of course, $T(x^{n-1}) = x^n = -c_0 - c_1 x - \cdots - c_{n-1}x^{n-1}$. Correspondingly, we have $$ \pmatrix{0 & \cdots & 0 & 1} M = \pmatrix{-c_0 & -c_1 & \cdots & -c_{n-1}} $$ Of course, this transformation will have a minimal polynomial dividing $f$ since for any $g$, we have $$ f(T)g = f(x)g = 0g = 0 $$ On the other hand, there is no polynomial of lower degree such that $p(T)$ is the zero transformation. Thus, this Frobenius block has characteristic and minimal polynomial $f(x)$.