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Automorphisms of an extension field

Bingk

New member
Jan 26, 2012
16
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 [TEX]f(x)=x^2 + 1 = x^2 - 2 \in Z_3[x][/TEX].
Let [TEX]u= \sqrt{2}[/TEX] be a root of [TEX]f[/TEX] in some extension field of [TEX]Z_3[/TEX].
Let [TEX]F=Z_3(\sqrt{2})[/TEX].

d)List the automorphisms of [TEX]F[/TEX] which leave [TEX]Z_3[/TEX] fixed.

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

Is what I've done okay? Any comments/suggestions?



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

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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→ap, 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.