# Galois group of $x^4-1$ over $\mathbb{Q}$.

Let $p$ be the polynomial $x^4-1$. What is $\text{Gal}(p/\mathbb{Q})$?

The splitting field is $\;\mathbb{Q}[i]\;$. The order of the Galois group is equal to the degree of the splitting field of $\;p\;$ over $\;\mathbb{Q}\;$, so it is $\;[\mathbb{Q}[i]:\mathbb{Q}]\,$.

$x^2+1$ is an irreducible polynomial in $\mathbb{Q}[i]$ which has degree 2, so $[\mathbb{Q}[i]:\mathbb{Q}]=|\text{Gal}(p/\mathbb{Q})|=2$. Furthermore, $\text{Gal}(p/\mathbb{Q})\subset S_4$, since $f$ has 4 roots. The only subgroup of $S_4$ with order 2 is $S_2$, so $\text{Gal}(p/\mathbb{Q})\cong S_2$.

Am I correct? Is this general method for finding the Galois group correct?

• Yes this is correct – happymath Apr 23 '15 at 10:22
• This is indeed correct. But you should keep in mind, that a galois extension of degree $2$ is such an easy case, that it would be weird to deduce a general method from this. – MooS Apr 23 '15 at 10:22
• Thank you for your answers. Do you recommend a particular website or book with practice exercises for determining the Galois group of polynomials on this kind of level? – Heinz Doofenschmirtz Apr 23 '15 at 10:27
• If you look at the list of Related questions on this page, you will see many "find the Galois group" problems, and you may be able to learn quite a bit from the answers. – Gerry Myerson Apr 23 '15 at 11:10
• It's easy to describe $\text{Gal}(p / \mathbb{Q})$ explicitly too: The roots of the irreducible polynomial $x^2 + 1$ are $\pm i$, and so the nonidentity field isomorphism of $\mathbb{Q}[i]$, induced by transposing those roots, is just $a + bi \mapsto a - bi$, that is, complex conjugation. – Travis Apr 23 '15 at 11:42