I am trying the following problem which appeared on an exam for a grad course:

Determine the structure of finite field $F$ as a module over $F_p[x]$ when $xv=Lv$. Here $L(z)=z^p$ and $q=p^r=3^2$.

I am trying to find a presentation matrix but am not sure where to begin. Any help/hints will be greatly appreciated

  • $\begingroup$ If you use a normal basis, then $L$ will be presented by a permutation matrix given by an $r$-cycle. $\endgroup$ – Jyrki Lahtonen May 7 '14 at 19:21
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    $\begingroup$ do you know the structure theorem for finitely generated modules over PID ? $\endgroup$ – WLOG May 7 '14 at 21:12
  • $\begingroup$ @WLOG: Yes I have learnt the structure theorem $\endgroup$ – James Bond May 8 '14 at 16:47
  • $\begingroup$ @WLOG: Jyrki's answer is correct. Do you want an explanation of it? $\endgroup$ – Jack Schmidt May 21 '14 at 1:59
  • $\begingroup$ @JackSchmidt: yes an explanation, because this topic is interesting $\endgroup$ – WLOG May 21 '14 at 5:25

You should observe that $z^{p^r}-z=0$ for all $z\in\Bbb{F}_q$. Also this is the lowest degree polynomial vanishing at all the elements, because there are $p^r$ elements in all. You can reinterpret this as stating that $x^r-1$ is the minimal polynomial of $L$. As the characteristic polynomial of $L$ is of degree $r$, it must coincide with the minimal polynomial. Thus as a module over $\Bbb{F}_p[x]$ the extension field is cyclic, and we can conclude that $$ \Bbb{F}_{p^r}\cong \Bbb{F}_p[x]/\langle x^r-1\rangle. $$ Beware that this isomorphism is only as modules. The latter object also has natural structure as a ring, but this is not an isomorphism of rings.

Another point of view is the following. The extension $\Bbb{F}_q/\Bbb{F}_p$ has a so called normal basis. That is a basis $\{x_1,x_2,\ldots,x_r\}$ of the form $x_i=L^{i-1}(x_1)=x_1^{p^{i-1}}$ for all $i=2,3,\ldots,r$. With respect to such a basis the matrix of $L$ is a simple permutation matrix of an $r$-cycle. This leads to the same conclusion and allows us to describe the above isomorphism $\Phi: \Bbb{F}_p[x]/\langle x^r-1\rangle\to \Bbb{F}_{p^r}$ as $$ \Phi(f(x))=f(L)\cdot x_1. $$


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