# Find Galois group of polynomials when char F=2

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?

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

(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.

• Thank a lot! It's helpful to me ! Apr 19 at 13:49