Let \(p\) be a prime number and let \(F\) be a field of characteristic \(p\).
Show that if \(a \in F\) is not a \(p\)-th power in \(F\), then the polynomial \(f(x) = x^p - a\) is irreducible in \(F[x]\).
Solution
Goal: Prove that \(x^p - a \in F[x]\) is irreducible in characteristic \(p\) by showing any non-trivial factor in a splitting field forces a \(p\)-th root of \(a\) to lie in \(F\) via Bézout’s identity.
1. Factorization in a splitting field:
1.1. Let \(K\) be a splitting field of \(f(x) = x^p - a\) over \(F\).
1.2. In \(K\), there exists a root \(\alpha \in K\) such that \(\alpha^p = a\).
1.3. Because \(\operatorname{char}(F) = p\), the Frobenius map \(t \mapsto t^p\) is an endomorphism, so\[(x - \alpha)^p = x^p - \alpha^p = x^p - a = f(x).\]
1.4. Thus in \(K[x]\), \(f(x)\) factors completely as \(f(x) = (x - \alpha)^p\).
2. Form of divisors of \(f(x)\) in \(F[x]\):
2.1. Suppose for contradiction that \(f(x)\) is reducible in \(F[x]\).
2.2. Then there exists a monic irreducible factor \(g(x) \in F[x]\) of \(f(x)\) with degree \(k = \deg g\) satisfying \(1 \le k < p\).
2.3. In \(K[x]\), \(g(x)\) divides \(f(x) = (x - \alpha)^p\).
2.4. Because \(K[x]\) is a unique factorization domain and \(x - \alpha\) is the only irreducible factor of \(f(x)\) in \(K[x]\), \(g(x)\) must be of the form:\[g(x) = (x - \alpha)^k.\]
3. Coefficient containment and Bézout’s identity:
3.1. Expanding \(g(x) = (x - \alpha)^k\), its constant term is \((-1)^k \alpha^k\).
3.2. Since \(g(x) \in F[x]\), all its coefficients lie in \(F\), so \((-1)^k \alpha^k \in F\).
3.3. Since \(-1 \in F\), \(\alpha^k \in F\).
3.4. Because \(p\) is prime and \(1 \le k < p\), \(\gcd(k, p) = 1\).
3.5. By Bézout’s identity, there exist integers \(u, v \in \mathbb{Z}\) such that\[u k + v p = 1.\]
3.6. Express \(\alpha \in K\) using this identity:\[\alpha = \alpha^1 = \alpha^{u k + v p} = (\alpha^k)^u \cdot (\alpha^p)^v = (\alpha^k)^u \cdot a^v.\]
3.7. Since \(\alpha^k \in F\) and \(a \in F\), and \(F\) is a field, the product \((\alpha^k)^u a^v\) lies in \(F\).
3.8. Thus \(\alpha \in F\).
4. Contradiction and conclusion:
4.1. Since \(\alpha \in F\), the element \(a = \alpha^p\) is a \(p\)-th power of an element of \(F\).
4.2. This contradicts the hypothesis that \(a\) is not a \(p\)-th power in \(F\).
4.3. Therefore \(f(x) = x^p - a\) has no factors of degree \(k\) with \(1 \le k < p\), so \(f(x)\) is irreducible in \(F[x]\).