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I'm trying to compute the Galois group of the quintic polynomial $f(x) = x^5 + x - 1$.

I first decomposed $f(x)$ into irreducible factors $f(x) = g(x)h(x)$ where $g(x) = x^2 - x + 1$ and $h(x) = x^3 + x^2 - 1$. I denoted by $K_g$, $K_h$ the splitting fields of $g(x), h(x)$ over $\mathbb{Q}$, respectively, and saw that $G_g = {\rm Gal}(K_g/\mathbb{Q}) \cong \mathbb{Z}_2$, $G_h = {\rm Gal}(K_h/\mathbb{Q}) \cong S_3$ where $S_3$ is the symmetric group on three letters. Then the splitting field of $f(x)$ over $\mathbb{Q}$ is $K_gK_h$ (compositum).

Guessing that the Galois group $G_f$ of $f(x)$ is $\mathbb{Z}_2 \times S_3$, I have tried to show that $K_g \cap K_h = \mathbb{Q}$. Actually, it suffices to show that $\sqrt{-3}$ does not belong to $K_h$ but I'm stuck on here. Could anyone give me an idea?

(If $K_g \cap K_h \neq \mathbb{Q}$, then $\mathbb{Q} \subsetneq K_g \subsetneq K_h$, so Galois theorem asserts ${\rm Gal}(K_h/K_g) \cong A_3$, but failed to reach a contradiction.)

Thank for your answer.

user112265
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1 Answers1

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if there were an element $\alpha \notin \mathbb{Q} $ in both $K_g$ and $K_h$ it would be a zero of an irreducible polynomial p(x) of degree 2 and an irreducible polynomial q(x) of degree 3 over $\mathbb{Q}$. then by division we would have:

$$ q(x) = (ax+b)p(x) + c $$ for $a,b,c \in \mathbb{Q} $

but since $p(\alpha)=0$ and $q(\alpha)=0$ this forces $c=0$ contradicting the irreducibility of $q(x)$

David Holden
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  • Thank you answer. Can I ask you why $\alpha$ must be a root of an irreducible polynomial of degree 3? (I see that is is a root of a degree 2 ploynomial) – user112265 Nov 29 '13 at 14:57
  • maybe i misunderstand? but if an element is in an extension of degree 3 over F, and is not in F, what is your take on it? – David Holden Nov 30 '13 at 00:03