\[ \DeclareMathOperator{\ker}{ker} \DeclareMathOperator{\im}{im} \DeclareMathOperator{\diag}{diag} \DeclareMathOperator{\char}{char} \DeclareMathOperator{\lcm}{lcm} \newcommand\divides{\mathbin |} \newcommand\ndivides{\mathbin \nmid} \newcommand\d{\mathrm{d}} \newcommand\p{\partial} \newcommand\C{\mathbb{C}} \newcommand\N{\mathbb{N}} \newcommand\Q{\mathbb{Q}} \newcommand\R{\mathbb{R}} \newcommand\Z{\mathbb{Z}} \newcommand\pref[1]{(\ref{#1})} \DeclareMathOperator{\Aut}{Aut} \DeclareMathOperator{\Gal}{Gal} \]

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Section 16.5. Fixed Fields

Exercise 16.5.1

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 \(u \in \C\).