Assuming the Fundamental Theorem of Algebra, prove the following. If \(f\) and \(g\) are polynomials over the field of complex numbers, then g.c.d.\((f, g) = 1\) if and only if \(f\) and \(g\) have no common root.
Solution
Goal: Prove that for polynomials \(f, g \in \mathbb{C}[x]\), \(\gcd(f, g) = 1\) if and only if \(f\) and \(g\) have no common root in \(\mathbb{C}\).
1. Forward direction: \(\gcd(f, g) = 1 \implies f\) and \(g\) have no common root.Proof:
1.1. Assume \(\gcd(f, g) = 1\).
1.2. Suppose for contradiction that there exists \(\alpha \in \mathbb{C}\) such that \(f(\alpha) = 0\) and \(g(\alpha) = 0\).
1.3. By the Factor Theorem, the linear polynomial \((x - \alpha) \in \mathbb{C}[x]\) divides both \(f(x)\) and \(g(x)\).
1.4. Thus \((x - \alpha)\) is a common divisor of \(f\) and \(g\).
1.5. By definition of the greatest common divisor in a PID \(\mathbb{C}[x]\), \((x - \alpha)\) must divide \(\gcd(f, g) = 1\).
1.6. But \(\deg(x - \alpha) = 1 > 0 = \deg(1)\), so \((x - \alpha)\) cannot divide 1 in \(\mathbb{C}[x]\), a contradiction.
1.7. Therefore \(f\) and \(g\) share no common roots in \(\mathbb{C}\).
2. Backward direction: \(f\) and \(g\) have no common root \(\implies \gcd(f, g) = 1\).Proof:
2.1. Assume \(f\) and \(g\) have no common root in \(\mathbb{C}\).
2.2. Let \(d(x) = \gcd(f, g)\) denote the unique monic greatest common divisor of \(f\) and \(g\) in \(\mathbb{C}[x]\).
2.3. Suppose for contradiction that \(\deg d(x) \ge 1\).
2.4. By the Fundamental Theorem of Algebra, every non-constant polynomial in \(\mathbb{C}[x]\) has at least one root in \(\mathbb{C}\).
2.5. Thus there exists \(\alpha \in \mathbb{C}\) such that \(d(\alpha) = 0\).
2.6. Because \(d(x)\) divides \(f(x)\) and \(g(x)\), there exist polynomials \(q_1, q_2 \in \mathbb{C}[x]\) such that \(f(x) = d(x) q_1(x)\) and \(g(x) = d(x) q_2(x)\).
2.7. Evaluating at \(\alpha\) gives \(f(\alpha) = d(\alpha) q_1(\alpha) = 0\) and \(g(\alpha) = d(\alpha) q_2(\alpha) = 0\).
2.8. This means \(\alpha\) is a common root of \(f\) and \(g\), contradicting the hypothesis.
2.9. Hence \(\deg d(x) = 0\). Since \(d(x)\) is monic, \(d(x) = 1\).
3. Conclusion:Proof:By 1 and 2, \(\gcd(f, g) = 1\) if and only if \(f\) and \(g\) have no common root in \(\mathbb{C}\).