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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?