91Ó°ÊÓ

First, let's understand the given information: \(k\) is a field such that every finite extension is cyclic. This means that for any finite extension \(L\) of \(k\), the Galois group \(\text{Gal}(L/k)\) is cyclic. A cyclic group can be generated by a single element, say \(\sigma\). Therefore, there exists an automorphism \(\sigma \in \text{Gal}(L/k)\) such that \(\text{Gal}(L/k) = \langle \sigma \rangle\).

By the Fundamental Theorem of Galois Theory, there is a correspondence between subgroups of the Galois group and intermediate fields (subfields containing \(k\)) of the extension \(L/k\). If \(H\) is a subgroup of \(\text{Gal}(L/k)\), then the corresponding fixed field is $$ L^H = \{x \in L \mid \sigma(x) = x \text{ for all } \sigma \in H\}. $$ Since \(\sigma\) is a generator of \(\text{Gal}(L/k)\), the subgroup \(\{e, \sigma\}\) generated by the identity automorphism \(e\) and \(\sigma\) corresponds to a fixed field \(L^{\{e, \sigma\}}\) such that $$ [L^{\{e, \sigma\}}:k] = [L:k]/|\{e, \sigma\}| = 2, $$ where \([A:B]\) denotes the degree of the field extension \(A\) over \(B\).

Now, let \(k^a\) be the algebraic closure of \(k\). We want to construct an automorphism \(\sigma\) of \(k^a\) over \(k\) such that \(k\) is the fixed field of \(\sigma\). To do this, note that each element in \(k^a\) is algebraic over \(k\), meaning that they are all roots of some polynomial with coefficients in \(k\). Let \(x \in k^a\) be an arbitrary element. Then there exists a finite extension \(L\) of \(k\) containing \(x\). As we have seen in Step 1, there exists an automorphism \(\sigma \in \text{Gal}(L/k)\) such that \(\text{Gal}(L/k) = \langle \sigma \rangle\). We can then define an automorphism \(\sigma\) acting on \(k^a\) by defining it on such finite extensions \(L\) containing each element of \(k^a\). Since \(k^a\) is the union of all finite extensions of \(k\), this gives us a well-defined automorphism \(\sigma\) acting on \(k^a\).

Now, we want to show that the fixed field of \(\sigma\) is \(k\). First, note that elements in \(k\) are fixed by \(\sigma\) by definition. Therefore, \(k \subseteq k^{\sigma}\), where $$ k^{\sigma} = \{x \in k^a \mid \sigma(x) = x\}. $$ To show the reverse inclusion, let \(x \in k^{\sigma}\). We want to show that \(x \in k\). Since \(x \in k^a\), there exists a finite extension \(L\) of \(k\) containing \(x\). As we have seen in Step 2, the fixed field of the identity automorphism \(e\) and \(\sigma\) is \(L^{\{e, \sigma\}}\), and \([L^{\{e, \sigma\}}:k] = 2\). Since \(x\) is fixed by \(\sigma\), we have that \(x \in L^{\{e, \sigma\}}\). But since \(x\) is also fixed by the identity automorphism \(e\), we have that \(x \in k\). Hence, \(k^{\sigma} \subseteq k\). Therefore, we have shown that the fixed field of \(\sigma\), which is \(k^{\sigma}\), is equal to the field \(k\). This completes the proof.