Find the minimal polynomial of $\alpha^i$ for $i$, $0 \le \le 48$, over $GF(7)$.

Question

Let $\alpha$ be a root of the polynomial $x^2 + x + 3$ in $GF(49)$.

Question: Find the minimal polynomial of $\alpha^i$ for $i$, $0 \le \le 48$, over $GF(7)$.

I know that if $\alpha \in GF(P^n)$ then it's minimal polynomial can be factored as $(x-\alpha)(x-\alpha^p)\cdots(x-\alpha^{p^{t-1}})$, where $t$ is the degree of the minimal polynomial of $\alpha$ over $GF(P)$. If we take $\alpha$ to be the element we adjoin to $GF(P)$ in order to obtain $GF(P^n)$ we have that $t=n$ and everything else is the same.

I also know that this shouldn't be too difficult because we only need methods from linear algebra. I guess from here I'll need to choose some $\alpha$ to be zero of some irreducible polynomial, and then I'll need to find a minimal polynomial for i such that $\beta = a^i$. But I'm not quite sure how it's done and how to continue from here. Any help is appreciated.

Answer 1

The Frobenius automorphism of $GF(7^2)$ is the map $z \mapsto z^7$. It is of order $2$ and generates the Galois group (which is the cyclic group of order $2$). So, in general, the factorization of the minimal polynomial of $\alpha^i$ is of degree $2$ and contains the factor $x-\alpha^i$. The other factor $x-\beta$ must be such that the coefficients $\alpha^i+\beta$ and $\alpha^i\beta$ are invariant under this automorphism, so the only choice for $\beta$ is $\alpha^{7i}$. I said "in general" because for $i \in \{8,16,24,30,40\}$ we have that $\alpha^i$ is already invariant, so for these elements the minimal polynomial is of the first degree.

Answer 2

I'll use $a$ instead of $\alpha$ for easier typing. Since $GF(49)$ is two dimensional over $GF(7)$ I'll use the vector basis $a,1$ . Then $a$ acts as a linear transformation on $GF(49)$ as follows $1 \mapsto a$ and $a \mapsto -a-3$ so $a$ can be represented as the matrix $M = \begin{pmatrix}0 & -3 \\ 1 & -1\end{pmatrix}$. This avoids lengthy calculations with polynomials as in $a^3 = a(-a - 3) = -a^2 - 3a = -(-a-3)-3a = -2a+3$, etc., whereas one can directly calculate $M^3 = \begin{pmatrix}3 & 6 \\ -2 & 5\end{pmatrix}$ and its characteristic polynomial $x^2-8x+27$, which modulo $7$ gives $x^2-x-1$. It's interesting to note that M satisfies its own characteristic polynomial: $M^2+M + 3I =0I $. Introducing the Frobenius automorphism $a \mapsto a^7$ we know that everything in $GF(49)$ and is fixed by it automatically lives in $GF(7)$ (and is represented by a scalar matrix), for example $a+a^7 \mapsto a^7 + a$ and indeed : $M + M^7 = \begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix}$ also $MM^7\mapsto M^7M$ giving $M^8 = \begin{pmatrix}-39 & -105 \\ 35 & -74\end{pmatrix} \stackrel{\bmod 7}{\rightarrow}\begin{pmatrix}3 & 0 \\ 0 & 3\end{pmatrix}$ from which $a^8 = 3$. The introduction of a matrix operator does not only restrict itself to this case but can be used for calculations in more general (finitely generated) rings like algebraic extentions etc.