Let \(\overline{\mathbb{F}}_p\) denote an algebraic closure of the finite field \(\mathbb{F}_p\). Show that the absolute Galois group \(G = \operatorname{Gal}(\overline{\mathbb{F}}_p/\mathbb{F}_p) \cong \widehat{\mathbb{Z}}\) has no non-trivial finite subgroups (i.e. is torsion-free).
Solution
Goal: Prove that the profinite group \(\operatorname{Gal}(\overline{\mathbb{F}}_p/\mathbb{F}_p) \cong \widehat{\mathbb{Z}}\) is torsion-free, so it has no non-trivial finite subgroups.
1. Structure of the Galois Group \(\operatorname{Gal}(\overline{\mathbb{F}}_p/\mathbb{F}_p)\):Proof:
1.1. The finite subfields of \(\overline{\mathbb{F}}_p\) are the unique fields \(\mathbb{F}_{p^n}\) of size \(p^n\) for each integer \(n \ge 1\).
1.2. The extension \(\overline{\mathbb{F}}_p/\mathbb{F}_p\) is the direct limit (union) of the finite Galois extensions \(\mathbb{F}_{p^n}/\mathbb{F}_p\):\[\overline{\mathbb{F}}_p = \bigcup_{n=1}^\infty \mathbb{F}_{p^n}.\]
1.3. For each \(n \ge 1\), the finite Galois group \(\operatorname{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p) \cong \mathbb{Z}/n\mathbb{Z}\), generated by the Frobenius automorphism \(\operatorname{Frob}_p(x) = x^p\).
1.4. By the Fundamental Theorem of Infinite Galois Theory, the absolute Galois group is the inverse limit (profinite completion):\[G = \operatorname{Gal}(\overline{\mathbb{F}}_p/\mathbb{F}_p) \cong \varprojlim_{n} \operatorname{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p) \cong \varprojlim_{n} \mathbb{Z}/n\mathbb{Z} = \widehat{\mathbb{Z}}.\]
1.5. By the Chinese Remainder Theorem for profinite rings:\[\widehat{\mathbb{Z}} \cong \prod_{q \text{ prime}} \mathbb{Z}_q\]where \(\mathbb{Z}_q\) is the ring of \(q\)-adic integers (under addition, \(\mathbb{Z}_q \cong \varprojlim_k \mathbb{Z}/q^k\mathbb{Z}\)).
2. Torsion-Freeness of \(\widehat{\mathbb{Z}}\):Proof:
2.1. Let \(\sigma = (\sigma_q)_{q} \in \prod_q \mathbb{Z}_q \cong \widehat{\mathbb{Z}}\) be an element of finite order \(m \ge 1\).
2.2. Then \(m \cdot \sigma = 0 \in \widehat{\mathbb{Z}}\), which means for every prime \(q\):\[m \cdot \sigma_q = 0 \in \mathbb{Z}_q.\]
2.3. The additive group of the \(q\)-adic integers \((\mathbb{Z}_q, +)\) is an integral domain (char 0), so it is a torsion-free abelian group:\[\text{If } m \in \mathbb{Z}_{> 0} \text{ and } m \cdot x = 0 \text{ in } \mathbb{Z}_q, \text{ then } x = 0.\]
2.4. Therefore, \(\sigma_q = 0\) for every prime \(q\).
2.5. This implies \(\sigma = 0 = \operatorname{id}\).
2.6. Thus the only element of finite order in \(\operatorname{Gal}(\overline{\mathbb{F}}_p/\mathbb{F}_p)\) is the identity \(\operatorname{id}\).
3. Absence of Non-Trivial Finite Subgroups:Proof:
3.1. Let \(H \le G\) be a finite subgroup of order \(|H| = k \ge 1\).
3.2. By Lagrange’s Theorem, every element \(h \in H\) satisfies \(h^k = 1\).
3.3. By Step 2, since \(G\) is torsion-free, the only element satisfying \(h^k = 1\) is \(h = 1\).
3.4. Thus \(H = \{1\}\).
4. Alternative Direct Field-Theoretic Proof:Proof:
4.1. Suppose \(H \le \operatorname{Gal}(\overline{\mathbb{F}}_p/\mathbb{F}_p)\) is a finite subgroup with \(|H| = k > 1\).
4.2. Let \(K = (\overline{\mathbb{F}}_p)^H\) be the fixed field of \(H\).
4.3. By Artin’s Theorem on Galois extensions, \([\overline{\mathbb{F}}_p : K] = |H| = k < \infty\).
4.4. Thus \(\overline{\mathbb{F}}_p\) would be a finite-degree extension of the field \(K\).
4.5. Since \(K\) is of characteristic \(p\), \(K\) cannot have an algebraically closed finite extension of degree \(> 1\) other than the Artin-Schreier theorem case where the closure is algebraically closed (which requires char 0 and degree 2, \(\mathbb{R} \subset \mathbb{C}\)).
4.6. Specifically, by the Artin-Schreier Theorem on formally real fields, if an algebraically closed field is a finite non-trivial extension of a subfield \(K\), then the characteristic must be 0 and the extension degree must be 2.
4.7. Since \(\operatorname{char}(\overline{\mathbb{F}}_p) = p > 0\), this is impossible.
4.8. Thus no such finite subgroup \(H\) can exist.
5. Conclusion:\(\operatorname{Gal}(\overline{\mathbb{F}}_p/\mathbb{F}_p) \cong \prod_q \mathbb{Z}_q\) is torsion-free, hence has no non-trivial finite subgroups. Q.E.D.