MHB Automorphisms of an extension field

[Bingk1]

Hello, I found this question, and I was able to do the easier parts, but I'm really not comfortable with automorphisms in fields.

Let f(x)=x^2 + 1 = x^2 - 2 \in Z_3[x].
Let u= \sqrt{2} be a root of f in some extension field of Z_3.
Let F=Z_3(\sqrt{2}).

d)List the automorphisms of F which leave Z_3 fixed.

What I did is as follows:
a=1+\sqrt{2} generates the nonzero elements of F, which is a finite field.
The automorphisms of the multiplicative group of F is isomorphic to the group U(9)=\{1,2,4,5,7,8\}.
Since [F:Z_3]=2, we know that there will be 2 automorphisms which fix Z_3, and that these 2 automorphisms form a group, so the non-identity automorphism should have order 2.
This gives us the two automorphisms: \sigma_1: a \mapsto a and \sigma_2: a \mapsto a^8.

Is what I've done okay? Any comments/suggestions?
Also, I just wanted to make sure that I've remembered correctly. Technically, Z_3(\sqrt{2}) is supposed to consist of elements of the form \frac{f(u)}{g(u)} (taken modulo 3) where f,g \in Z_3[x] and g \neq 0, but because u is algebraic over Z_3, we can say that Z_3(\sqrt{2}) consists of elements of the form f(u). Is this right?

 

[Deveno]

there is really no need to mention Aut(U(9)).

as it turns out, you have the wrong group there, anyway: there are but 8 non-zero elements of Z₃(u), so the field automorphism group must be (isomorphic to) a subgroup of U(8) = {1,3,5,7}.

let's look at the powers of 1+u:

(1+u)² = 1 + 2u + u² = 2u

(1+u)³ = (1+u)(2u) = 2u + 2u² = 2u + 2(2) = 1 + 2u

(1+u)⁴ = (2u)² = u² = 2 <--this shows that 1+u has order 8.

(1+u)⁵ = (2)(1+u) = 2 + 2u

(1+u)⁶ = (1+u)(2+2u) = 2 + 2(2u) + 2u² = (2+1) + u = u

(1+u)⁷ = (u)(1+u) = u + u² = 2 + u

(1+u)⁸ = (2+u)(1+u) = 2 + (1+2)u + u² = 2 + u² = 2 + 2 = 1.

now a field automorphism σ must also be an automorphism of the multiplicative group, hence we have:

σ(1+u) = 1+u, (1+u)³, (1+u)⁵ or (1+u)⁷.

but a field automorphism must also be an additive homomorphism, so:

0 = σ(0) = σ(u² + 1) = σ(u²) + σ(1) = σ(u)² + 1, so σ(u) must also be a root of x² + 1 in Z₃[x]. clearly this "other root" is 2u (= -u):

(x - u)(x - 2u) = x² -(u + 2u)x + 2u² = x² + 2u² = x² + (2)(2) = x^​2 + 1.

so σ(1+u) = σ(1) + σ(u) = 1 + σ(u) = 1 + 2u = (1+u)³.

it would be nice to express σ solely in terms of what it does to u:

σ(u) = 2u = (u²)(u) = u³.

of course, this makes sense, in any ring of characteristic p, a→a^p, is a ring homomorphism.

we have not "yet" shown that σ so defined is actually an additive isomorphism, that is:

σ(a+bu) = 0 iff a = b = 0 (where a,b are in Z₃).

however, σ(a+bu) = σ(a) + σ(bu) = σ(a) + σ(b)σ(u)

= a + bσ(u) = a + bu³ = a + 2bu,

so if σ(a+bu) = 0, a = 0, 2b = 0, whence b = (2)(2b) = (2)(0) = 0.

(note that we used the fact that in Z₃, a³ = a for all a:

0³ = 0
1³ = 1
2³ = 2(2²) = 2((2)(2)) = 2(1) = 2).

it should be clear that {id,σ} forms a group of order 2:

σ²(a+bu) = σ(σ(a+bu)) = σ(a+2bu) = σ(a) + 2bσ(u) = a + 2b(2u) = a + bu,

so σ² = id.

it might be instructive for you to examine WHY:

σ:1+u → (1+u)⁵
σ:1+u → (1+u)⁷

do not lead to additive homomorphisms.