Galois group of reducible polynomial

Question

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?

Answer 1

This is actually a really good question for anyone learning Galois theory.

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.