# Thread:

1. ## 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.

Thank you for reading my thread.

2. 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.

Thank you for reading my thread.
$x^3 + x + 1$ is irreducible over $\mathbb{F}_2.$ thus $\frac{\mathbb{F}_2[x]}{<x^3 + x + 1>} \cong \mathbb{F}_8.$ so the automorphism group is $C_3,$ the cyclic group of order 3. the generator of this group is the Frobenius map.

3. Thanks alot 4 quick reply , but i dont understand why so the automorphism group is 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

4. Originally Posted by knguyen2005
Thanks alot 4 quick reply , but i dont understand why so the automorphism group is 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 $x^3+x+1$ is irreducible over $\mathbb{F}_2$ therefore, $\mathbb{F}_2[x]/(x^3+x+1)$ is a field with $2^3 = 8$ elements, that is why it is $\mathbb{F}_8$. Now, $\text{Aut}(\mathbb{F}_8) = \text{Gal}(\mathbb{F}_8/\mathbb{F}_2)$ where $\mathbb{F}_2$ is its prime subfield. However, this Galois group is cyclic of degree $3$ and generated by $\sigma$ where $\sigma: \mathbb{F}_8 \to \mathbb{F}_8$ is defined by $\sigma(x) = x^2$.
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