# Galois group of reducible polynomial

I want to find Gaolois group of $(x^3-x+1)(x^2+1)$ over $\mathbb Q$. The polynomial of degree third is irreducible and has discriminant $-23$ so it's Galois group is $S_3$. Galois group of the other polynomial is clearly $C_2$. I thought that the galois group of the product was $S_3$ since both the polynomials share conjugation as a permutation of the roots. Too bad the answer is $S_3 \times C_2$. What am I missing?

The nontrivial element of the Galois group of $x^2+1$ over $\Bbb Q$ is the automorphism of $\Bbb Q(i)$ that fixes $\Bbb Q$ (of course) and exchanges the two roots $i$ and $-i$ of $x^2+1$. It is true that this is the restriction of complex conjugation on $\Bbb C$ to $\Bbb Q(i)$; however, intrinsically it is completely disconnected from complex conjugation.
In particular, there is a well-defined automorphism of the splitting field of $(x^3-x+1)(x^2+1)$ over $\Bbb Q$ that exchanges $i$ and $-i$, yet does not exchange the two complex roots of $x^3-x+1$. This is one example of the "missing" elements of your Galois group.
• Perhaps a simpler example illustrating the same point: there is an automorphism of ${\bf Q}(i,i\sqrt2)$ that takes $i$ to $-i$ but fixes $i\sqrt2$. – Gerry Myerson Oct 5 '14 at 1:42
• @GerryMyerson but takes $\sqrt(2)$ to $-\sqrt(2)$. But how? – user1118686 Oct 5 '14 at 2:15
• @GerryMyerson but takes $\sqrt(2)$ to $-\sqrt(2)$. Forgive me stubbornness, but how? Autmorphism means "renaming" with something inside the set. When you rename the $i$ as $-i$, how come there's no renaming in $i\sqrt(2)$? Is it because of the renaming of $\sqrt(2)$? – user1118686 Oct 5 '14 at 2:25
• Yes. If $f(i)=-i$, and $f(\sqrt2)=-\sqrt2$, then $f(i\sqrt2)=i\sqrt2$. – Gerry Myerson Oct 5 '14 at 4:37