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How can we find minimal polynomials for $\alpha $ in $GF(2^{n})$. What is the general approach to find minimal polynomials. I know about minimal polynomials they are monic etc. In particular i want to know about primitive polynomials of $GF(32)$.

Infinity
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    For irreducible polynomials of degree five specifically you can take a look at this older question. Many ways to find them all, as you see. Because $2^5-1=31$ is a prime, all the irreducible quintics are primitive (see also Dietrich Burde's answer). – Jyrki Lahtonen Nov 17 '16 at 21:09

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$GF(32)^\times$ is a group of order $31$ and so is cyclic.

You need to factor $x^{31}-1 \bmod 2$ and pick a factor of degree $5$.

$x^5+x^2+1$ is one such factor.

lhf
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  • can u please describe in more detail , how do we factor $x^{31}-1 mod 2$? and get the $\alpha : m_{1}(x) = x^{5}+x^{2}+1$ – Infinity Nov 17 '16 at 19:21
  • @BreezyBoo, you don't need to factor $x^{31}-1$ completely, just find a factor of degree $5$, which can be done by inspection by trying simple candidates. – lhf Nov 17 '16 at 19:40
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There are $\frac{\phi(2^n-1)}{n}$ primitive polynomials of degree $n$ over $GF(2)$. For $n=5$ we have $30/5=6$ primitive polynomials, namely $$ x^5+x^2+1,\; x^5+x^3+1,\; x^5+x^3+x^2+x+1,\;x^5+x^4+x^2+x+1,\; $$

$$ x^5+x^4+x^3+x+1,\;x^5+x^4+x^3+x^2+1.\; $$ Indeed, the product of all of them, together with the factor $(x+1)$ gives $x^{31}-1$. Using a factorisation algorithm, e.g., the Berlekamp algorithm we obtain the factorization of $x^{31}-1$ over $GF(2)$.

Remark: All irreducible polynomials in $GF(2)[x]$ of degree $2, 3, 5$ are primitive.

References: See here.

Dietrich Burde
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  • i asked about minimal polynomial – Infinity Nov 17 '16 at 19:27
  • This is the same as irreducible, monic, and because of $n=5$, the same as primitive (you also asked about primitive). – Dietrich Burde Nov 17 '16 at 19:27
  • can you please tell me how do we get from $x^{5}+x^{2}+1$ to $x^{5}+x^{3}+1$, what factor we are multiplying. Actually i found somewhere written in form $\alpha : m_{1}(x)=x^{5}+x^{3}+1$ then in second row $\alpha ^{3} : m_{3}(x)=x^{5}+x^{4}+x^{3}+x^{2}+1$ – Infinity Nov 17 '16 at 19:52
  • "what factor we are multiplying?" -multiply all above factors $(x^5+x^2+1)(x^5+x^3+1)\cdots (x+1)$ together to obtain $x^{31}-1$. – Dietrich Burde Nov 17 '16 at 19:58