# identify galois group

• Mar 22nd 2009, 08:19 PM
dopi
identify galois group
f= t^3 -6t +6 in Q[t]

basically my first part of the question was to find the zeros of f, i have done that part already but the part i got stuck on was that i want to identify the galois group of f, but not sure how to, so can anyone explain why it is that galois group thanks
• Mar 22nd 2009, 08:45 PM
ThePerfectHacker
Quote:

Originally Posted by dopi
f= t^3 -6t +6 in Q[t]

basically my first part of the question was to find the zeros of f, i have done that part already but the part i got stuck on was that i want to identify the galois group of f, but not sure how to, so can anyone explain why it is that galois group thanks

This irreducible polynomial has 1 real root and 2 complex roots that are complex conjugates. If $\displaystyle \tau$ is complex conjugation then we see that $\displaystyle \tau$ is a transposition (2-cycle) if viewed as an element of $\displaystyle S_3$ (remember that the Galois group is a subgroup of $\displaystyle S_3$). Now if $\displaystyle \alpha$ is a root of $\displaystyle f$ then $\displaystyle \mathbb{Q}(\alpha)/\mathbb{Q}$ is a degree $\displaystyle 3$ extension and so $\displaystyle 3$ divides the degree of the splitting field over $\displaystyle \mathbb{Q}$. Thus, $\displaystyle 3$ divides the order of the Galois group, hence there is an element of order 3 in the Galois group i.e. a 3-cycle. But any subgroup of $\displaystyle S_3$ generated by a 2-cycle and a 3-cycle is $\displaystyle S_3$ itself, so the Galois group is $\displaystyle S_3$.
• Mar 22nd 2009, 09:48 PM
dopi
Quote:

Originally Posted by ThePerfectHacker
This irreducible polynomial has 1 real root and 2 complex roots that are complex conjugates. If $\displaystyle \tau$ is complex conjugation then we see that $\displaystyle \tau$ is a transposition (2-cycle) if viewed as an element of $\displaystyle S_3$ (remember that the Galois group is a subgroup of $\displaystyle S_3$). Now if $\displaystyle \alpha$ is a root of $\displaystyle f$ then $\displaystyle \mathbb{Q}(\alpha)/\mathbb{Q}$ is a degree $\displaystyle 3$ extension and so $\displaystyle 3$ divides the degree of the splitting field over $\displaystyle \mathbb{Q}$. Thus, $\displaystyle 3$ divides the order of the Galois group, hence there is an element of order 3 in the Galois group i.e. a 3-cycle. But any subgroup of $\displaystyle S_3$ generated by a 2-cycle and a 3-cycle is $\displaystyle S_3$ itself, so the Galois group is $\displaystyle S_3$.

awsome thank you
• Mar 23rd 2009, 01:15 AM
ZetaX
You could also find the discriminant of the polynomail, which is -108 and since it is not a square in Q(rationals) and the polynomail is irreducible 6-Eisenstein, you have that the Galois group is S_3.