Exercise: Suppose $p \in P(\Bbb{C})$ has degree $m$. Prove that $p$ has $m$ distinct zeros if and only if $p$ and its derivative $p'$ have no zeros in common.
Proof: Suppose that $p\in P(\Bbb{C})$ with degree $m$. We will induct on the degree of $p$. Let $\deg p=1$ and $\lambda_1$ be a zero of $p$. One exists by the fundamental theorem of algebra. Then the derivative $p'=c$ for some constant $c$. For $c\ne 0$ we have that $p$ and $p'$ don't have any zeroes in common and $\lambda_1$ is distinct.
Now suppose inductively that for all polynomials $p\in P(C)$ with $\deg p<m$, the statement in question is satisfied. Consider the polynomial $p\in P(C)$ with $\deg p=m$. We have to show that $p$ has $m$ distinct zeroes if and only if $p$ and $p'$ have no zeroes in common. Because every polynomial of degree $m$ over $\Bbb{C}$ has $m$ zeroes, Suppose that $p$ has $m$ distinct zeroes. let $\lambda_1$ be a zero of $p$. Then there exists a polynomial $q\in P(C)$ such that $p(z)=(z-\lambda_1)q(z)$ with $\lambda_2,\dots,\lambda_m$ being the zeroes of $q(z)$. The derivative of $p$ is the polynomial $p'(z)=(z-\lambda_1)q'(z)+q(z)$. Because $\deg q=m-1$, by the induction hypothesis we know that $q$ and $q'$ don't have any zeroes in common. Thus, $p'(z)\ne 0$ for $\lambda_1,\dots,\lambda_m$ because $\lambda_1$ is not a zero of $q$ and $(z-\lambda_1)\ne 0$ when $z=\lambda_j$ for $j=2,\dots,m$ but $q(\lambda_j)=0$ for $j=2,\dots,m$. Which by the induction hypothesis implies that $q'(\lambda_j)\ne 0$ for $j=2,\dots,m$.
Conversely suppose that $p$ and $p'$ have no zeroes in common. And suppose that $\lambda_1,\dots,\lambda_m$ are not distinct zeroes. Let $\lambda_1$ be the zero that is repeated. Because $\lambda_1$ is a zero of $p$, there exists a polynomial $q\in P(\Bbb{C})$ such that $p(z)=(z-\lambda_1)q(z)$. Thus we have that $q(\lambda_1)$ also equals $0$. Now consider the derivative $p'(z)=(z-\lambda_1)q'(z)+q(z)$. Because $\lambda_1$ is a zero of $q(z)$, the above derivative implies that $p'(\lambda_1)=0$. Which contradicts the hypothesis that $p$ and $p'$ have no common zeroes. Thus, all the zeros $\lambda_1,\dots,\lambda_m$ of $p$ are distinct.
Is this proof correct? I deleted the original because it was completely wrong. I believe that this proof is somewhat correct.
Best Answer
The first part is not well written. You write about the “induction hypothesis”, without even saying what is it that you are proving by induction. I suppose that it is this: ff $p(x)$ has degree $m$ and $m$ distinct roots, then $p(x)$ and $p'(x)$ have no commun root. The idea of the proof is correct, yes.
What you should prove next is that if $p(x)$ has degree $m$ and less than $m$ roots, then $p(x)$ and $p'(x)$ have one common root. You cannot say “Let $\lambda_1$ be the zero that is repeated”; it might not be. It is better to say “We can assume without loss of generality that $\lambda_1$ is a multiple root”. What you wrote after is correct, but I think that it is easier to see (it's a matter of taste, really) that then $p(x)=(x-\lambda_1)^kq(x)$ for some $k\in\Bbb N\setminus\{1\}$ and for some polynomial $q(x)$. But then$$p'(x)=(x-\lambda_1)^kq'(x)+k(x-\lambda_1)^{k-1}q(x),$$and that therefore $\lambda_1$ is also a root of $p'(x)$.