# Splitting Fields and Galois Groups

• Mar 6th 2012, 09:43 AM
leshields
Splitting Fields and Galois Groups
Hi,

I've been set a question and think I've managed about half of it but stuck on some other bits..

Show that $f := x^3 - 3x + 1$ is irreducible over $\mathbb{Q}$. Let $a\in\mathbb{C}$ be given as a root of $f$. Show that $(1-a)^{-1}$ is also a root of $f$. Hence, find a splitting field $L$ for $f$ over $\mathbb{Q}$. Identity the Galois group $\tau(L : \mathbb{Q})$. (You must justify any claim that a particular map is an isomorphism. Be sure to state carefully
any facts that you assume in doing this.)

first part is easy just reducing by $\mathbb{Z}_{2}$ to get $\bar{f} := x^3 +x + 1\in\mathbb{Z}$\ $\mathbb{Z}_{2}(x)$ which has no zeros in $\mathbb{Z}_{2}$ and thus irreducible over $\mathbb{Z}$ and hence over $\mathbb{Q}$ also.

I think I can identify the Galois group using Prop 22.4 which gives $\Delta= -27(-1)^2 - 4(-3)^3 = 81$ , a perfect square thus the Galois Group is $\mathbb{A}_{3}\cong C_{3}$

I can't think how to show that $(1-a)^{-1}$ is also a root of $f$. For the splitting field all I can think to do is say that as $a\in\mathbb{C}$ is a root of $f$then $f$ splits in $\mathbb{C}(a)$ and so $\mathbb{C}(a)$ is the splitting field for $f$ over $\mathbb{Q}$.

Any help for this last bit or if you can point out any errors I've already made would be greatly appreciated!
• Mar 6th 2012, 09:48 AM
leshields
Re: Splitting Fields and Galois Groups
As I posted just thought that to show $(1-a)^{-1}$ is also a root of $f$ you would say something along the lines of $f$ factors into $f:= (x-a)(x^2+bx-\frac{1}{a})$ for some $b\in\mathbb{C}$ and somehow this ends up with $(1-a)^{-1}$ as a root... I know this is obviously not correct but am I along the right lines??
• Mar 6th 2012, 02:21 PM
Deveno
Re: Splitting Fields and Galois Groups
one can verify $\frac{1}{1-a}$ is a root directly:

$f(\frac{1}{1-a}) = \left(\frac{1}{1-a}\right)^3 - \frac{3}{1-a} + 1$

$= \frac{1}{(1-a)^3} - \frac{3(1-a)^2}{(1-a)^3} + \frac{(1-a)^3}{(1-a^3)}$

$= \frac{1 - 3 + 6a - 3a^2 + 1 - 3a + 3a^2 - a^3}{(1-a)^3}$

$= \frac{-(1 - 3a + a^3)}{(1-a)^3} = 0$
• Mar 6th 2012, 03:39 PM
leshields
Re: Splitting Fields and Galois Groups
Haha, cheers; I'm a fool. Other than that does the rest seem OK?