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Let $F$ be a field with characteristic 2. Need to calculate the Galois group of $f$ defined as follows

(i)$f = x^3 + x + 1$ ,

(ii)$f = x^3 + x^2 + 1$ .

I know that if $\alpha$ is a root of (i), then so is $\alpha^2$. And hence the splitting field of $f$ over $F$ is $F(\alpha)$. Then how can I find $[F(\alpha):F]$? Is I thinking in the right way?

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    $\begingroup$ If you're doing Galois Theory then surely you know that the degree you seek is the degree of the minimal polynomial of $\alpha$? $\endgroup$ Apr 19 at 12:20
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    $\begingroup$ @ancientmathematician I know its the degree, but sadly I cannot figure out the minimal polynomial here.TAT .Please teach me how to do it $\endgroup$
    – s11r40ppp
    Apr 19 at 13:18

1 Answer 1

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(A) Let us first deal with the case when $F=\mathbb{F}_2$.

Your $\alpha$ is a root of $x^3+x+1$. This is a cubic polynomial. It is irreducible. (If it were reducible it would have a linear factor, and the only linear polynomials are $x$ and $x+1$ neither of which divides $x^3+x+1$.)

Hence the minimal polynomial of $\alpha$ is $x^3+x+1$, and so $|\mathbb{F}_2[\alpha]:\mathbb{F}_2|=3$.

In fact we now see that $x^3+x+1=(x-\alpha)(x-\alpha^2)(x-\alpha^4)$ in the splitting field.

I leave it to you to show that $x^3+x^2+1$ has the same splitting field and that its roots are $\alpha^{-1},\alpha^{-2},\alpha^{-4}$.

(B) In general note that $(x-1)(x^3+x+1)(x^3+x^2+1)=x^7-1$, so that every root of $x^3+x+1$ and every root of $x^3+x^2+1$ is a $7$-th root of unity and so lies in $\mathbb{F}_8$.

Hence

(a) either $\mathbb{F}_8\subseteq F$, and in this case $F$ splits both these polynomials;

(b) or $\mathbb{F}_8\not\subseteq F$, and in this case $|F(\alpha):F|=3$ and $F(\alpha)$ splits both polynomials.

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  • $\begingroup$ Thank a lot! It's helpful to me ! $\endgroup$
    – s11r40ppp
    Apr 19 at 13:49

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