Thread: 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.

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.

is irreducible over thus so the automorphism group is the cyclic group of order 3. the generator of this group is the Frobenius map.

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

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 is irreducible over therefore, is a field with elements, that is why it is . Now, where is its prime subfield. However, this Galois group is cyclic of degree and generated by where is defined by .