The Frobenius map is defined as a map from the finite field \(F_{p^n}\) to itself. It is denoted as \(\Phi\) and is defined by \(\Phi(\alpha) = \alpha^p\), where \(\alpha\) belongs to \(F_{p^n}\).
To show that \(\Phi\) is an automorphism of \(F_{p^n}\), we need to demonstrate that it satisfies the properties of an automorphism.
1. Injectivity: To prove that \(\Phi\) is injective, we need to show that for any two distinct elements \(\alpha\) and \(\beta\) in \(F_{p^n}\), \(\Phi(\alpha) \\eq \Phi(\beta)\). Since \(\Phi(\alpha) = \alpha^p\) and \(\Phi(\beta) = \beta^p\), if \(\alpha \\eq \beta\), then \(\alpha^p \\eq \beta^p\), ensuring injectivity.
2. Surjectivity: For surjectivity, we need to show that for any element \(\gamma\) in \(F_{p^n}\), there exists an element \(\alpha\) in \(F_{p^n}\) such that \(\Phi(\alpha) = \gamma\). By the definition of \(\Phi\), we can choose \(\alpha = \gamma^{p^{n-1}}\), and we will have \(\Phi(\alpha) = \alpha^p = (\gamma^{p^{n-1}})^p = \gamma^{p^n} = \gamma\).
3. Homomorphism: The Frobenius map is a homomorphism because it preserves addition and multiplication. Specifically, for any elements \(\alpha\) and \(\beta\) in \(F_{p^n}\), we have \(\Phi(\alpha + \beta) = (\alpha + \beta)^p = \alpha^p + \beta^p = \Phi(\alpha) + \Phi(\beta)\), and \(\Phi(\alpha \cdot \beta) = (\alpha \cdot \beta)^p = \alpha^p \cdot \beta^p = \Phi(\alpha) \cdot \Phi(\beta)\).
Since the Frobenius map satisfies the properties of injectivity, surjectivity, and homomorphism, it is an automorphism of \(F_{p^n}\).
In summary, the Frobenius map \(\Phi\) is an automorphism of \(F_{p^n}\) because it is injective, surjective, and a homomorphism.