If \(\mathbb{F}_q\) is a field with \(q\) elements, show that \(\mathbb{F}_q^\times\) is a cyclic group.
Show that for each integer \(n \geq 1\), there exists an irreducible polynomial over \(\mathbb{F}_q\) of degree \(n\).
Consider the map \(\phi: \mathbb{F}_{q^n} \to \mathbb{F}_{q^n}\) given by \(\phi(x) = x^q\). Note that \(\phi\) is an \(\mathbb{F}_q\)-linear endomorphism of the \(\mathbb{F}_q\)-vector space \(\mathbb{F}_{q^n}\). Find the characteristic and minimal polynomials of \(\phi\).
Solution
1. Part (a): \(\mathbb{F}_q^\times\) is cyclic:
1.1. Let \(G = \mathbb{F}_q^\times\) be the multiplicative group of the finite field \(\mathbb{F}_q\), which is an abelian group of order \(|G| = q - 1\).Let \(m = \exp(G)\) denote the exponent of \(G\), which is the least common multiple of the orders of all elements in \(G\).By Lagrange’s Theorem, \(m \le q - 1\).
1.2. By definition of exponent, every element \(x \in \mathbb{F}_q^\times\) satisfies \(x^m - 1 = 0\).The polynomial \(x^m - 1 \in \mathbb{F}_q[x]\) has degree \(m\) and can have at most \(m\) roots in the field \(\mathbb{F}_q\).Because all \(q - 1\) elements of \(\mathbb{F}_q^\times\) are roots:\[ q - 1 \le m. \]
1.3. Combining \(m \le q - 1\) and \(q - 1 \le m\) gives \(m = q - 1\).For any finite abelian group, there exists an element whose order equals the exponent \(m\).Thus there exists \(\alpha \in \mathbb{F}_q^\times\) with \(o(\alpha) = q - 1 = |\mathbb{F}_q^\times|\), so \(\mathbb{F}_q^\times = \langle \alpha \rangle\) is cyclic.
2. Part (b): Existence of an irreducible polynomial of degree \(n\) over \(\mathbb{F}_q\):
2.1. Consider the finite field extension \(\mathbb{F}_{q^n}\), which is an \(n\)-dimensional vector space over \(\mathbb{F}_q\), so \([\mathbb{F}_{q^n} : \mathbb{F}_q] = n\).
2.2. By Part (a), the multiplicative group \(\mathbb{F}_{q^n}^\times\) is cyclic.Let \(\gamma\) be a generator of \(\mathbb{F}_{q^n}^\times\).Then \(\mathbb{F}_{q^n} = \mathbb{F}_q(\gamma)\).
2.3. Let \(m_\gamma(x) \in \mathbb{F}_q[x]\) be the minimal polynomial of \(\gamma\) over \(\mathbb{F}_q\).By field theory, \(m_\gamma(x)\) is irreducible over \(\mathbb{F}_q\), and its degree is:\[ \deg(m_\gamma) = [\mathbb{F}_q(\gamma) : \mathbb{F}_q] = [\mathbb{F}_{q^n} : \mathbb{F}_q] = n. \]Thus \(m_\gamma(x)\) is an irreducible polynomial of degree \(n\) over \(\mathbb{F}_q\).
3. Part (c): Characteristic and minimal polynomials of the Frobenius endomorphism \(\phi\):
3.1. The map \(\phi: \mathbb{F}_{q^n} \to \mathbb{F}_{q^n}\) given by \(\phi(x) = x^q\) is the Frobenius \(\mathbb{F}_q\)-automorphism.For any \(x \in \mathbb{F}_{q^n}\), \(\phi^n(x) = x^{q^n} = x\), so \(\phi^n = \operatorname{id}_{\mathbb{F}_{q^n}}\).Thus \(\phi\) satisfies the polynomial \(T^n - 1 \in \mathbb{F}_q[T]\).
3.2. By Dedekind’s Theorem on the linear independence of distinct group characters/automorphisms, the automorphisms \(\{\operatorname{id}, \phi, \phi^2, \ldots, \phi^{n-1}\}\) are linearly independent over \(\mathbb{F}_{q^n}\) (and hence over \(\mathbb{F}_q\)).
3.3. Therefore, no non-zero polynomial in \(\mathbb{F}_q[T]\) of degree strictly less than \(n\) can annihilate \(\phi\).Since \(T^n - 1\) is a monic polynomial of degree \(n\) annihilating \(\phi\), the minimal polynomial is:\[ m_\phi(T) = T^n - 1. \]
3.4. The characteristic polynomial \(\chi_\phi(T)\) is a monic polynomial of degree \(\dim_{\mathbb{F}_q}(\mathbb{F}_{q^n}) = n\) which is divisible by \(m_\phi(T)\).Since \(\deg(m_\phi) = n = \deg(\chi_\phi)\) and both are monic:\[ \chi_\phi(T) = T^n - 1. \]
4. Conclusion:\(\mathbb{F}_q^\times\) is cyclic, an irreducible polynomial of degree \(n\) exists for every \(n \ge 1\), and both the minimal and characteristic polynomials of \(\phi\) are \(T^n - 1\). Q.E.D.