Problem: Suppose $p \in \mathcal{P}(\mathbf{C})$ has degree $m$. Prove that $p$ has $m$ distinct roots if and only if $p$ and its derivative $p'$ have no roots in common.
My proof so far:
If $m=0$, then $p(z)=a_0\neq 0$ and $p$ has no roots. Then $p'(z)=0$ and has no roots. If $m=1$, then $p(z)=a_0+a_1z$ with $a_1 \neq 0$ so $p$ has
exactly one root, namely $-a_0/a_1$ and $p'(z)=a_1$ and has no roots. In both cases, $p$ and $p'$ have no roots in common.
Now suppose $m > 1$. We use induction on $m$, assuming that for every polynomial $r$ with $m-1$ distinct roots, $r$ and $r'$ have no roots in common. Let
$p$ be a polynomial of degree $m$ with distinct roots. There exists $q$ such that
\begin{align*}
p(z)=(z-\lambda)q(z)
\end{align*}
for all $z \in \mathbf{F}$. Since q has $m-1$ distinct roots, $q$ and $q'$ have no roots in common. By the chain rule,
\begin{align*}
p'(z)=(z-\lambda)q'(z)+q(z)
\end{align*}
We know that $\lambda$ is not a root of $p'$ since $p'(\lambda)=(\lambda-\lambda)q'(z)+q(\lambda)=0+q(\lambda)\neq 0$ as lambda is not a root of $q$. All
other roots of $p$ are roots of $q$. For these roots $\lambda_q$, $p'(\lambda_q)=(\lambda-\lambda_q)q'(\lambda_q)+q(\lambda_q)=(\lambda-\lambda_q)q'(\lambda_q) + 0 \neq 0$.
Now suppose $m > 1$. We use induction on $m$, this time assuming that for every polynomial $r$ of degree $m-1$ such that $r$ and $r'$ have no roots in common, $r$ has
$m-1$ distinct roots.
I'm stuck here, since I don't know how to how to manipulate the derivatives. I'm not sure if proof by induction is the best approach here as well and I'd appreciate your help!
Best Answer
Not sure that induction is really the way to go. Here is a sketch of a possible argument.
Firstly, if $p$ does not have $m$ distinct roots then it has a double (or higher order) root $a$ and so $$p(x)=(x-a)^2q(x)$$ for some polynomial $q$. You can now easily calculate $p'$ and show that it has a root in common with $p$.
Conversely, if $p$ and $p'$ have a common root $a$ we can write $$p(x)=(x-a)q(x)\quad\hbox{and}\quad p'(x)=(x-a)r(x)\ .$$ Differentiating the first equation and equating it with the second, $$(x-a)q'(x)+q(x)=(x-a)r(x)\ ,$$ which shows that $x-a$ is a factor of $q(x)$. So $a$ is a double (at least) root of $p$.