Let \(D\) be the differentiation operator on the space of polynomials over the field of complex numbers. Let \(f\) be a monic polynomial over the field of complex numbers. Prove that \[ f = (x - c_1) \cdots (x - c_k) \] where \(c_1, \ldots, c_k\) are distinct complex numbers if and only if \(f\) and \(Df\) are relatively prime. In other words, \(f\) has no repeated root if and only if \(f\) and \(Df\) have no common root. (Assume the Fundamental Theorem of Algebra.)
Solution
By the Fundamental Theorem of Algebra, write \[ f(x)=\prod_{j=1}^r (x-c_j)^{m_j} \] with distinct \(c_j\) and positive integers \(m_j\).
If some \(m_j\ge2\), then \((x-c_j)\) divides both \(f\) and \(Df\): differentiating the displayed factorization shows every term of \(Df\) still contains at least one factor \((x-c_j)\). Hence \(f\) and \(Df\) are not relatively prime.
Conversely, suppose all \(m_j=1\). Fix a root \(c_j\). Writing \[ f(x)=(x-c_j)q_j(x) \] with \(q_j(c_j)\ne0\), we have \[ f'(c_j)=q_j(c_j)\ne0. \] Thus no root of \(f\) is a root of \(Df\). Over \(\CC\), two nonconstant polynomials have a nonconstant common divisor exactly when they have a common root. Therefore \(f\) and \(Df\) are relatively prime.
Hence \(f\) has distinct roots if and only if \(\gcd(f,Df)=1\).