Let \(K/F\) be a finite Galois extension and let \(n=[K:F]\). There is a theorem (often referred to as the “normal basis theorem”) which states that there exists an irreducible polynomial \(f(x)\in F[x]\) whose roots form a basis for \(K\) as a vector space over \(F\). You may assume that theorem in this problem.
Let \(G=\Gal(K/F)\). The action of \(G\) on \(K\) makes \(K\) into a finite-dimensional representation space for \(G\) over \(F\). Prove that \(K\) is isomorphic to the regular representation for \(G\) over \(F\).
The regular representation is defined by letting \(G\) act on the group algebra \(F[G]\) by multiplication on the left.
Suppose that the Galois group \(G\) is cyclic and that \(F\) contains a primitive \(n^{\text{th}}\) root of unity. Show that there exists an injective homomorphism \(\chi:G\rightarrow F^{\times}\).
Show that \(K\) contains a non-zero element \(a\) with the following property: \[\begin{align*} g(a)=\chi(g)\cdot a .\end{align*}\]
for all \(g\in G\).
If \(a\) has the property stated in (c), show that \(K=F(a)\) and that \(a^n\in F^{\times}\).
Solution
1. Part (a): \(K \cong F[G]\) as \(F[G]\)-modules (regular representation):
1.1. By the Normal Basis Theorem, there exists an element \(\alpha \in K\) such that \(\{g(\alpha) : g \in G\}\) is an \(F\)-basis for \(K\).
Normal Basis Theorem for finite Galois extensions.
1.2. Define the \(F\)-linear map \(\Phi: F[G] \to K\) by \(\Phi\left(\sum_{g \in G} c_g g\right) = \sum_{g \in G} c_g g(\alpha)\).
definition of \(\Phi\).
1.3. For any \(h \in G\):\[ \Phi\left(h \cdot \sum_{g \in G} c_g g\right) = \Phi\left(\sum_{g \in G} c_g (hg)\right) = \sum_{g \in G} c_g (hg)(\alpha) = h\left(\sum_{g \in G} c_g g(\alpha)\right) = h \cdot \Phi\left(\sum_{g \in G} c_g g\right). \]Thus \(\Phi\) is a homomorphism of \(F[G]\)-modules.
\(G\)-action on \(K\) is by field automorphisms.
1.4. Since \(\Phi\) sends the standard basis \(\{g : g \in G\}\) of \(F[G]\) to the basis \(\{g(\alpha) : g \in G\}\) of \(K\), \(\Phi\) is an isomorphism of \(F[G]\)-modules.
a linear map sending a basis to a basis is an isomorphism.
2. Part (b): Injective homomorphism \(\chi: G \to F^\times\):
2.1. Let \(G = \langle \sigma \rangle \cong \mathbb{Z}_n\), and let \(\zeta_n \in F\) be a primitive \(n\)-th root of unity.
setup.
2.2. Define \(\chi: G \to F^\times\) by \(\chi(\sigma^k) = \zeta_n^k\) for all \(k \in \mathbb{Z}\).
definition of \(\chi\).
2.3. Since \(\chi(\sigma^n) = \zeta_n^n = 1\), \(\chi\) is a well-defined group homomorphism.
universal property of cyclic groups.
2.4. \(\ker(\chi) = \{\sigma^k : \zeta_n^k = 1\} = \{\sigma^k : n \mid k\} = \{1\}\).Thus \(\chi\) is injective.
\(\zeta_n\) has multiplicative order \(n\).
3. Part (c): Existence of an eigenvector \(a \in K^\times\) for the \(G\)-action:
3.1. Consider the Lagrange resolvent map \(L: K \to K\) defined by:\[ L(x) = \sum_{j=0}^{n-1} \chi(\sigma^j)^{-1} \sigma^j(x) = \sum_{j=0}^{n-1} \zeta_n^{-j} \sigma^j(x). \]
definition of Lagrange resolvent.
3.2. By Dedekind’s Theorem on the linear independence of distinct field automorphisms, the linear combination \(\sum_{j=0}^{n-1} \zeta_n^{-j} \sigma^j\) is not identically zero on \(K\).
Dedekind’s Theorem.
3.3. Thus there exists some \(x \in K\) such that \(a = L(x) \neq 0\).
2.
3.4. Compute \(\sigma(a)\):\[ \sigma(a) = \sum_{j=0}^{n-1} \zeta_n^{-j} \sigma^{j+1}(x) = \zeta_n \sum_{j=0}^{n-1} \zeta_n^{-(j+1)} \sigma^{j+1}(x) = \zeta_n L(x) = \zeta_n a = \chi(\sigma) a. \]
re-indexing the sum modulo \(n\) using \(\zeta_n^{-n} = 1\) and \(\sigma^n = \operatorname{id}\).
3.5. By induction, \(\sigma^k(a) = \chi(\sigma^k) a\) for all \(k\), so \(g(a) = \chi(g) a\) for all \(g \in G\).
\(g = \sigma^k\).
4. Part (d): Show \(K = F(a)\) and \(a^n \in F^\times\):
4.1. Compute \(\sigma(a^n) = (\sigma(a))^n = (\zeta_n a)^n = \zeta_n^n a^n = a^n\).
field automorphism property \(\sigma(a^n) = (\sigma(a))^n\).
4.2. Since \(\sigma\) generates \(G\), \(g(a^n) = a^n\) for all \(g \in G\).By Galois theory, \(a^n \in K^G = F\). Since \(a \neq 0\), \(a^n \in F^\times\).
Fundamental Theorem of Galois Theory (\(K^G = F\)).
4.3. Compute the stabilizer \(\operatorname{Stab}_G(a) = \{g \in G : g(a) = a\}\):If \(g \in \operatorname{Stab}_G(a)\), then \(\chi(g) a = a \implies (\chi(g) - 1)a = 0 \implies \chi(g) = 1\) since \(a \neq 0\).By injectivity of \(\chi\) from Part (b), \(g = 1\).
2 and 3.
4.4. By the Galois correspondence, \([K : F(a)] = |\operatorname{Stab}_G(a)| = 1\), so \(K = F(a)\).
Galois correspondence.
5. Conclusion:\(K \cong F[G]\) as regular representation, \(\chi\) is an injective character, \(a\) exists via Lagrange resolvents, and \(K = F(a)\) with \(a^n \in F^\times\). Q.E.D.
1 through 4.