I need to find the minimum splitting field of the polynomial over GF(2): $$x^5+x^4+1.$$ I find that $x^5+x^4+1 = (x + \alpha)^2(x + \alpha + 1)(x + \alpha^2)(x + \alpha^2 + \alpha)$ over GF(8). But I don't know if it's right, because $x =\alpha $ is a multiple root. Am I right? And what is right? Help me please!

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    $\begingroup$ The derivative of that polynomial is $x^4$. A multiple root would need to be also a root of the derivative. But the only root of $x^4$ is $x=0$, which is not a root of the original polynomial. $\endgroup$ – user551819 May 10 '18 at 22:07

$x^5+x^4+1=(x^2 + x + 1) (x^3 + x + 1)$ in $GF(2)$.

These factors are irreducible because they have no root in $GF(2)$.

The splitting field of $x^2 + x + 1$ has degree $2$ over $GF(2)$.

The splitting field of $x^3 + x + 1$ has degree $3$ over $GF(2)$.

Therefore, the splitting field of $x^5+x^4+1$ has degree $6$ over $GF(2)$.

  • $\begingroup$ Thanks very much! But what the theorem did you use? $\endgroup$ – alexhak May 10 '18 at 22:17
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    $\begingroup$ @alexhak: if $f$ is an irreducible polynomial over a finite field $F$ of degree $n$, then adding one root of $f$ adds all roots, and so the splitting field of $f$ has degree $n$. $\endgroup$ – lhf May 10 '18 at 22:24
  • $\begingroup$ Roots of $x^2+x+1$ are $\alpha$ and $\alpha + 1$ over GF(4). And roots of $x^3+x+1$ are $\alpha^2$ and $\alpha^2+\alpha$ over GF(8). But $\alpha$ can't be a multiple root (take the derivative), so we have $GF(2) \subset GF(2)(x^2+x+1) = GF(4) \subset GF(4)(x^3+x+1) = GF(64)$. Am I right now? $\endgroup$ – alexhak May 10 '18 at 22:44
  • $\begingroup$ Do you know how to solve this problem click here? $\endgroup$ – alexhak May 13 '18 at 18:09

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