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I am trying to find the minimum polynomial of $\alpha^5$ and using this primitive polynomial to define my operations. $p(X)=1+X^2+X^5$.

The conjugates of an element $\beta$ of $GF(2^m)$ are $\{\beta^{(2^i)},\;i\geq 0 \}$. The minimal polynomial is $$\phi(X)=\prod_{i=0}^{e-1} (X+\beta^{2^i}).$$ $$\phi(X) = (X+\beta)(X+\beta^2)(X+\beta^4)(X+\beta^8)(X+\beta^{16})$$ I have set $\beta = \alpha^5$ and this gives me, $$\phi(X)=X^5+X^4+(\alpha+\alpha^2+\alpha^4)X^3+(\alpha+\alpha^2+\alpha^3+\alpha^4)X^2+X+1$$ But the answer should be, $X^5+X^4+X^2+X+1$

Edit: I have figured it out, It was a calculation mistake and I missed some terms. Now I have got the correct answer, i.e. $$\phi(X)=X^5+X^4+X^2+X+1$$

Atif
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1 Answers1

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Now that the OP found the mistakes in their calculation I will post my trick answer.

The element $\alpha^2$ is a conjugate of $\alpha$, so it shares with it the minimal polynomial $p(X)=X^5+X^2+1$. We can use this by observing that $$\beta=\alpha^5=\alpha^5+p(\alpha)=1+\alpha^2.$$ Therefore $\beta$ must be a zero of $$ q(X)=p(1+X)=(1+X)^5+(1+X)^2+1=X^5+X^4+X^2+X+1. $$ After all, $$ q(\beta)=q(1+\alpha^2)=p(1+(1+\alpha^2))=p(\alpha^2)=0. $$

Jyrki Lahtonen
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  • This is amazing, and we don't have to go into the lengthy arithmetics.
    How can we use the same technique for $\alpha^7$, this can be decomposed into $\alpha^7 = \alpha^2+\alpha^4$     – Atif Nov 04 '21 at 09:52
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    This old answer of mine explains some tricks for finding more irreducible polynomials. Getting to $\alpha^7$ is a bit trickier. One way is the following. The minimal polynomial of $\alpha^{-1}$ is the reciprocal $X^5p(1/X)=X^5+X^3+1$. Given that $\alpha^3=\alpha^{-2}\alpha^5=1+\alpha^{-2}$, and $\alpha^{-2}$ is a conjugate of $\alpha^{-1}$ we then find the minimal polynomial of $\alpha^3$. Going reciprocal again then gives the minimal polynomial of $\alpha^{-3}$. But $\alpha^{-3}=\alpha^{28}$ is a conjugate of $\alpha^7$, so we are done. – Jyrki Lahtonen Nov 04 '21 at 10:06