Brian Bi
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1. The automorphism $$\sigma$$ is an involution; the group it generates consists of only itself and the identity. Following Example 16.5.5, let $$K = \C(t + t^{-1})$$. Clearly $$K$$ is a subfield of the fixed field. Furthermore, $$t$$ is a root of the polynomial $$(x - t)(x - t^{-1}) = x^2 - (t + t^{-1})x + 1$$, which has coefficients in $$K$$, so the degree $$[\C(t) : K]$$ is at most 2. Since $$K$$ is a subfield of the fixed field and the fixed field has degree 2 (by the Fixed Field Theorem), $$K$$ itself is the fixed field.
2. The orbit of $$t$$ under $$\sigma$$ is $$t, it, -t, -it, t, \ldots$$, so $$\sigma$$ generates a cyclic group of order 4. Let $$K = \C(t^4)$$. Clearly the degree $$[\C(t) : K]$$ is at most 4, and $$K$$ is a subfield of the fixed field, over which $$\C(t)$$ has degree 4. Therefore $$K$$ is itself the fixed field.
3. Both $$\sigma$$ and $$\tau$$ are involutions, so the automorphism group is the Klein four-group. We claim that the fixed field is $$K = \C(t^2 + t^{-2})$$. This follows from the fact that $$K$$ is fixed by both $$\sigma$$ and $$\tau$$ and that $$t$$ is a root of the polynomial $$x^4 - (t^2 + t^{-2})x^2 + 1$$, so $$[\C(t) : K]$$ is at most 4.
4. $$\sigma$$ has order 3 and $$\tau$$ has order 2, with $$\sigma(\tau(t)) = \omega^{-1} t^{-1} = \omega^2 t^{-1} = \tau(\sigma^2(t))$$. Therefore the automorphism group is the symmetric group $$S_3$$ and the degree of $$\C(t)$$ over the fixed field is 6. By similar reasoning to part (c), we find that the fixed field is $$\C(t^3 + t^{-3})$$.
Exercise 16.5.3 Since $$\C$$ is algebraically closed, any element of an extension field of $$\C$$ that is not in $$\C$$ is automatically transcendental over $$\C$$. Explicitly, suppose $$u \in \C(t)$$ and $$u$$ is algebraic over $$\C$$. Then $$g(u) = 0$$ for some monic polynomial $$g \in \C[z]$$. By the fundamental theorem of calculus, we can write $$g(z) = (z - z_1) \ldots (z - z_n)$$ where $$n = \deg g$$ and $$z_i \in \C$$ for each $$i$$. Since $$g(u) = 0$$, this implies $$u = z_i$$ for some $$i$$, therefore $$u \in \C$$.