Let \(F\) be a field. Let \(f(x) \in F[x]\) be an irreducible polynomial of degree \(n\), and let \(g(x) \in F[x]\). Let \(p(x) \in F[x]\) be an irreducible factor of degree \(m = \deg p(x)\) of the composition \(f(g(x))\).
Prove that \(n\) divides \(m\).
Use this to prove that if \(r \in \mathbb{Z}\) is an integer that is not a perfect square, and \(n\) is a positive integer, then every irreducible factor of \(x^{2n} - r\) in \(\mathbb{Q}[x]\) has even degree.
Solution
Goal: Prove that the degree of any irreducible factor of \(f(g(x))\) is a multiple of \(\deg f\) using a tower of field extensions in (a), and apply this to \(x^{2n} - r\) in (b).
1. Part (a): \(n\) divides \(m\).
1.1. Let \(\overline{F}\) be an algebraic closure of \(F\).
1.2. Choose a root \(\alpha \in \overline{F}\) of the irreducible factor \(p(x)\), so \(p(\alpha) = 0\).
1.3. Since \(p(x) \in F[x]\) is irreducible, \(p(x)\) is a scalar multiple of the minimal polynomial \(\operatorname{irr}(\alpha, F)\), which implies\[[F(\alpha) : F] = \deg p(x) = m.\]
1.4. Since \(p(x)\) divides \(f(g(x))\), \(f(g(\alpha)) = 0\).
1.5. Define the element \(\beta = g(\alpha) \in F(\alpha)\).
1.6. Then \(f(\beta) = f(g(\alpha)) = 0\).
1.7. Because \(f(x) \in F[x]\) is irreducible, \(f(x)\) is a scalar multiple of the minimal polynomial \(\operatorname{irr}(\beta, F)\), which implies\[[F(\beta) : F] = \deg f(x) = n.\]
1.8. Since \(\beta = g(\alpha) \in F(\alpha)\), the subfield \(F(\beta)\) is contained in \(F(\alpha)\), giving the tower of finite extensions:\[F \subseteq F(\beta) \subseteq F(\alpha).\]
1.9. By the Tower Law for extension degrees:\[[F(\alpha) : F] = [F(\alpha) : F(\beta)] \cdot [F(\beta) : F].\]
1.10. Substituting the degrees gives:\[m = [F(\alpha) : F(\beta)] \cdot n.\]
1.11. Since \([F(\alpha) : F(\beta)]\) is a positive integer, \(n\) divides \(m\).
2. Part (b): Every irreducible factor of \(x^{2n} - r\) over \(\mathbb{Q}\) has even degree.
2.1. Define \(f(y) = y^2 - r \in \mathbb{Q}[y]\) and \(g(x) = x^n \in \mathbb{Q}[x]\).
2.2. The composition is\[f(g(x)) = (x^n)^2 - r = x^{2n} - r.\]
2.3. Irreducibility of \(f(y)\):- \(\deg f = 2\).- A quadratic polynomial in \(\mathbb{Q}[y]\) is reducible if and only if it has a root in \(\mathbb{Q}\).- If \(y^2 - r = 0\) has a rational root \(y = a/b \in \mathbb{Q}\) with \(\gcd(a, b) = 1\), then \(a^2 = r b^2\).- By the Rational Root Theorem (or unique prime factorization in \(\mathbb{Z}\)), \(b = 1\), so \(r = a^2\) is a perfect square in \(\mathbb{Z}\).- By hypothesis, \(r\) is not a perfect square, so \(f(y)\) has no rational roots.- Therefore \(f(y)\) is irreducible in \(\mathbb{Q}[y]\), with degree \(\deg f = 2\).
2.4. Application of Part (a):- Let \(p(x) \in \mathbb{Q}[x]\) be any irreducible factor of \(x^{2n} - r = f(g(x))\).- By Part (a), \(\deg f(y) = 2\) divides \(\deg p(x)\).- Therefore \(\deg p(x)\) is even.
3. Conclusion:
$[F(\alpha) : F(\beta)] \cdot [F(\beta) : F] = m \implies n \mid m$, and since $y^2 - r$ is irreducible over $\mathbb{Q}$, every irreducible factor of $x^{2n} - r$ has degree divisible by 2.