# Finite fields....

• May 17th 2009, 07:36 AM
AAM
Finite fields....
Hi everyone,

I'm trouble solving the following question. (note - calculator not allowed)

Prove that this only 1 distinct intermediate subgroup between K = finite field of order 2 adjoined with Theta a root of X^4+X^3+1 & the finite field of order 2.

I attempted to show that the elements fixed by sigma, sigma^2 & sigma^3 are the same (where sigma is the Frobenius map A |---> A^2 ), but I just get lost in calculation! :-s

Is there an easier method?

Many thanks. x
• May 17th 2009, 11:40 AM
ThePerfectHacker
Quote:

Originally Posted by AAM
Hi everyone,

I'm trouble solving the following question. (note - calculator not allowed)

Prove that this only 1 distinct intermediate subgroup between K = finite field of order 2 adjoined with Theta a root of X^4+X^3+1 & the finite field of order 2.

I attempted to show that the elements fixed by sigma, sigma^2 & sigma^3 are the same (where sigma is the Frobenius map A |---> A^2 ), but I just get lost in calculation! :-s

Is there an easier method?

Many thanks. x

The polynomial $\displaystyle x^4+x^3+1$ is irreducible over $\displaystyle \mathbb{F}_2$. Let $\displaystyle \theta$ be a root in some larger extension field. Construct $\displaystyle K = \mathbb{F}_2(\theta)$. Now $\displaystyle [K:F] = 4$ and $\displaystyle \text{Gal}(K/F) = \left< \sigma \right>$ where $\displaystyle \sigma$ is the Frobenius automorphism i.e. $\displaystyle \sigma: K\to K$ by $\displaystyle \sigma (x) = x^2$. Therefore, the Galois group is cyclic. The intermediate subfields of degree 2 over $\displaystyle \mathbb{F}_2$ correspond to subgroup of the Galois group of index 2. However, $\displaystyle \text{Gal}(K/F) \simeq \mathbb{Z}_4$ and so there is only one subgroup of index 2.