# automorphism of field

• Mar 26th 2009, 07:15 AM
knguyen2005
automorphism of field
I have been given this question:

Find Aut (F_2[x]/(x^3+x+1)) ?

I can do this question easily if the question asked: Find Aut (F_2[x]/(x^3+x+1))* where (F_2[x]/(x^3+x+1))* denotes the group

This is my work:
F_2 = {0,1}

Set x^3 + x + 1 = 0 then x^3 = -x -1 = x + 1

There are 8 elements in F_2[x]/(x^3+x+1): 0, 1, x, x+1, x^2, x^2 +x, x^2 +1, x^2 +x +1

Since x^3 + x + 1 is irreducible over F_2 , so (F_2[x]/(x^3+x+1)) is a field

If we exclude 0 then (F_2[x]/(x^3+x+1)) becomes a group, and denote this by (F_2[x]/(x^3+x+1))* = G

Since org (G) = |(F_2[x]/(x^3+x+1))*| = 7 then we conclude that
(F_2[x]/(x^3+x+1))* =~ C_7 where x is the generator of G

So: Aut (F_2[x]/(x^3+x+1))* =~ Aut(C_7) = C_6 (since 7 is prime)

BUT in this case the question is different. So I am wondering if Aut(R) and Aut(G) is the same? ( R = ring and G =group)

If I am wrong then, how do you solve this question.

• Mar 26th 2009, 07:43 AM
NonCommAlg
Quote:

Originally Posted by knguyen2005
I have been given this question:

Find Aut (F_2[x]/(x^3+x+1)) ?

I can do this question easily if the question asked: Find Aut (F_2[x]/(x^3+x+1))* where (F_2[x]/(x^3+x+1))* denotes the group

This is my work:
F_2 = {0,1}

Set x^3 + x + 1 = 0 then x^3 = -x -1 = x + 1

There are 8 elements in F_2[x]/(x^3+x+1): 0, 1, x, x+1, x^2, x^2 +x, x^2 +1, x^2 +x +1

Since x^3 + x + 1 is irreducible over F_2 , so (F_2[x]/(x^3+x+1)) is a field

If we exclude 0 then (F_2[x]/(x^3+x+1)) becomes a group, and denote this by (F_2[x]/(x^3+x+1))* = G

Since org (G) = |(F_2[x]/(x^3+x+1))*| = 7 then we conclude that
(F_2[x]/(x^3+x+1))* =~ C_7 where x is the generator of G

So: Aut (F_2[x]/(x^3+x+1))* =~ Aut(C_7) = C_6 (since 7 is prime)

BUT in this case the question is different. So I am wondering if Aut(R) and Aut(G) is the same? ( R = ring and G =group)

If I am wrong then, how do you solve this question.

$\displaystyle x^3 + x + 1$ is irreducible over $\displaystyle \mathbb{F}_2.$ thus $\displaystyle \frac{\mathbb{F}_2[x]}{<x^3 + x + 1>} \cong \mathbb{F}_8.$ so the automorphism group is $\displaystyle C_3,$ the cyclic group of order 3. the generator of this group is the Frobenius map.
• Mar 26th 2009, 07:59 AM
knguyen2005
Thanks alot 4 quick reply , but i dont understand whyhttp://www.mathhelpforum.com/math-he...c847abf0-1.gif so the automorphism group is http://www.mathhelpforum.com/math-he...db1455ba-1.gif How do you get that?

Sorry, I haven't learnt about Frobenius map yet. Can you explain to me please?

Thanks so much, I am really appreciated
• Mar 26th 2009, 08:07 AM
ThePerfectHacker
Quote:

Originally Posted by knguyen2005
Thanks alot 4 quick reply , but i dont understand whyhttp://www.mathhelpforum.com/math-he...c847abf0-1.gif so the automorphism group is http://www.mathhelpforum.com/math-he...db1455ba-1.gif How do you get that?

Sorry, I haven't learnt about Frobenius map yet. Can you explain to me please?

Thanks so much, I am really appreciated

The polynomial $\displaystyle x^3+x+1$ is irreducible over $\displaystyle \mathbb{F}_2$ therefore, $\displaystyle \mathbb{F}_2[x]/(x^3+x+1)$ is a field with $\displaystyle 2^3 = 8$ elements, that is why it is $\displaystyle \mathbb{F}_8$. Now, $\displaystyle \text{Aut}(\mathbb{F}_8) = \text{Gal}(\mathbb{F}_8/\mathbb{F}_2)$ where $\displaystyle \mathbb{F}_2$ is its prime subfield. However, this Galois group is cyclic of degree $\displaystyle 3$ and generated by $\displaystyle \sigma$ where $\displaystyle \sigma: \mathbb{F}_8 \to \mathbb{F}_8$ is defined by $\displaystyle \sigma(x) = x^2$.