# Galois group of a polynomial

• Jan 3rd 2010, 02:59 AM
AlexanderW
Galois group of a polynomial
Hello.

I have to find a splitting field $L$ of the polynomial $f=x^4-4x^2+9 \in \mathbb Q[x]$ over $\mathbb Q$ and the galois group and all intermediate fields of the algebraic extension $L/\mathbb Q$.

I found out, that $f$ is irreducible, the roots of $f$ are:
$a=\sqrt{2-i\sqrt{5}}, \quad b=-\sqrt{2+i\sqrt{5}}, \quad c=\sqrt{2+i\sqrt{5}}=-b, \quad d=-\sqrt{2-i\sqrt{5}}=-a$ and that $\mathbb Q(a)$ is a splitting field of $f$.

Furthermore I find out, that the three homomorphisms
$a \mapsto \pm b, \ a \mapsto -a$ have order 2, so the galois group is isomorph to $\mathbb Z/2\mathbb Z \times \mathbb Z/2\mathbb Z$.

But now I don't know how to find the fixed fields, which belongs to the homomorphisms.

For example: Which fixed field belongs to the homomorphism $\sigma_1: a \mapsto b=\frac{-3}{a}$ ?

The definition says:
$\mathbb Q(a)^{\left \{ id, \sigma_1 \right \}}=\left \{ z \in \mathbb Q(a)\ |\ \sigma_1(z)=z \right \}$

...and this don't help me, because I need a description of the fixed field like this: $\mathbb Q(\beta)$ (with suitable $\beta$).

I hope, you can help me.

Bye,
Alex
• Jan 3rd 2010, 04:55 AM
NonCommAlg
Quote:

Originally Posted by AlexanderW
Hello.

I have to find a splitting field $L$ of the polynomial $f=x^4-4x^2+9 \in \mathbb Q[x]$ over $\mathbb Q$ and the galois group and all intermediate fields of the algebraic extension $L/\mathbb Q$.

I found out, that $f$ is irreducible, the roots of $f$ are:
$a=\sqrt{2-i\sqrt{5}}, \quad b=-\sqrt{2+i\sqrt{5}}, \quad c=\sqrt{2+i\sqrt{5}}=-b, \quad d=-\sqrt{2-i\sqrt{5}}=-a$ and that $\mathbb Q(a)$ is a splitting field of $f$.

Furthermore I find out, that the three homomorphisms
$a \mapsto \pm b, \ a \mapsto -a$ have order 2, so the galois group is isomorph to $\mathbb Z/2\mathbb Z \times \mathbb Z/2\mathbb Z$.

But now I don't know how to find the fixed fields, which belongs to the homomorphisms.

For example: Which fixed field belongs to the homomorphism $\sigma_1: a \mapsto b=\frac{-3}{a}$ ?

The definition says:
$\mathbb Q(a)^{\left \{ id, \sigma_1 \right \}}=\left \{ z \in \mathbb Q(a)\ |\ \sigma_1(z)=z \right \}$

...and this don't help me, because I need a description of the fixed field like this: $\mathbb Q(\beta)$ (with suitable $\beta$).

I hope, you can help me.

Bye,
Alex

well, it's very straightforward: $\{1,a,a^2,a^3 \}$ is a $\mathbb{Q}$-basis for $\mathbb{Q}(a).$ so for any $z \in \mathbb{Q}(a)$ there exist unique $q_i \in \mathbb{Q}$ such that $z=q_0 + q_1a+q_2a^2+q_3a^3.$

now $\sigma_1(z)=z$ is equivalent to $\sum_{k=0}^3q_ka^k=\sum_{k=0}^3 q_k b^k=\sum_{k=0}^3 (-3)^k q_k a^{-k}.$ multiplying both sides by $a^3$ gives us $\sum_{k=0}^3q_ka^{k+3}=\sum_{k=0}^3 (-3)^kq_ka^{3-k}.$ we also have

$a^4=4a^2-9, \ a^5=4a^3-9a,$ and $a^6=7a^2-36.$ then you'll get $q_2=0, \ q_3=-q_1.$ therefore the fixed field of the automorphism $\sigma_1$ is:

$\{q_0 + q_1(a-a^3) : \ q_0, q_1 \in \mathbb{Q} \}=\mathbb{Q}(\sqrt{2}i).$ (note that $a-a^3$ is a root of the irreducible polynomial $x^2 + 18$).
• Jan 3rd 2010, 06:10 AM
AlexanderW