1. Minimal Polynomials.
(a)
Let \(a(x)\) be a monic irreducible polynomial over a finite field \(F\text{.}\) Show that \(a(x)\) is the minimal polynomial of the element \(\xi\) in the extension field \(F[\xi]/a(\xi).\)
Solution.
(b)
Let \(\alpha\) be a nonzero element in an extension field \(E\) of \(F=\GF(q)\text{.}\) Show that the conjugacy classes of \(\alpha\) and \(\alpha^{-1}\) (with respect to \(F\)) are the same size.
Solution.
Let \(t=|C_{\alpha}|\text{,}\) so that \(\alpha^{q^t}=\alpha\) and no smaller positive power of \(q\) satisfies \(\alpha^{q^d}=\alpha\text{.}\) Then
\begin{equation*}
(\alpha^{-1})^{q^t}=(\alpha^{q^t})^{-1}=\alpha^{-1}
\end{equation*}
and by nearly the same computation if a smaller power of \(q\) satisfies
\begin{equation*}
(\alpha^{-1})^{q^d}=\alpha^{-1}
\end{equation*}
then it will also satisfy \(\alpha^{q^d}=\alpha\text{,}\) so \(t\) must be the smallest positive power possible and thus is the size of \(C_{\alpha^{-1}}\text{.}\)
(c)
Let \(M_{\alpha}(x)=\sum_{i=0}^m a_i x^i\) be the minimal polynomial of \(\alpha\) as in (b) (with respect to \(F\)). Show that the minimal polynomial of \(\alpha^{-1}\) is given by
\begin{equation*}
M_{\alpha^{-1}}(x)=\frac{1}{a_0} \sum_{i=0}^m a_{m-i}x^i = \frac{1}{a_0} M_{\alpha}(x)^{[-1]}\text{.}
\end{equation*}
That is, up to scaling by some nonzero element of \(F\text{,}\) the minimal polynomial of \(\alpha^{-1}\) is the reciprocal polynomial obtained by reversing the order of coefficients of the minimal polynomial of \(\alpha\text{.}\)
Solution.
From the previous part we know that \(m=\deg(M_{\alpha}(x))=\deg(M_{\alpha^{-1}}(x))\text{.}\) We will show that
\begin{equation*}
\frac{1}{a_0}\sum_{i=0}^m a_{m-i}x^i = \prod_{i=0}^m (x-(\alpha^{-1})^{q^i})
\end{equation*}
which proves the claim. To do this, first note that both polynomials are monic, so it suffices to show that they have the same roots. Since the conjugates of a root of a polynomial over \(F\) are also roots, it suffices to show that \(\alpha^{-1}\) is a root of the left-hand polynomial.
\begin{align*}
\frac{1}{a_0}\sum_{i=0}^m a_{m-i}(\alpha^{-1})^i \amp = \frac{1}{a_0}\sum_{i=0}^m a_{i}(\alpha^{-1})^{m-i}\\
\amp = \frac{1}{a_0}\sum_{i=0}^m a_{i}\alpha^{i-m}\\
\amp = \frac{1}{a_0\alpha^m}\sum_{i=0}^m a_{i}\alpha^{i}\\
\amp = \frac{1}{a_0\alpha^m}m_{\alpha}(\alpha)\\
\amp =0 \text{.}
\end{align*}
So the claim holds.
