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!


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 $p$.


Try writting $p(x)=(x-\lambda)^kq(x)$ where $q(\lambda)\neq 0$ then $$p^{\prime}(x)=k(x-\lambda)^{k-1}q(x) +(x-\lambda)^kq^{\prime}(x)$$ so $$(x-\lambda)|(p(x),p^{\prime}(x))$$ of and only if $k>1$. So the root $\lambda $ is a multiple root if and only if $p$ and $p^{\prime}$ have $\lambda$ as common root. Thus if $p$ and $p^{\prime}$ have no common roots, all roots of $p$ are simple and there must be $m$ of them.

  • $\begingroup$ Thanks! This is a very simple and elegant proof! $\endgroup$ Jun 4 '14 at 7:13
  • $\begingroup$ So would it be fine to combine this argument with the induction argument for the converse (I'm sure the induction is not efficient but it seems to work) to have a valid proof? $\endgroup$ Jun 4 '14 at 7:17

If $p(z)$ has $m$ distinct roots $\rho_j$, $1 \le j \le m$, then we may write

$p(z) = a \prod_1^m (z - \rho_j), \tag{1}$

where $0 \ne a \in \Bbb C$ is a complex constant. Then the derivative of $p(z)$ is

$p'(z) = a \sum_{k = 1}^m \prod_{j = 1, j \ne k}^m (z - \rho_j), \tag{2}$

and successively evaluating $p'(z)$ at $\rho_1$, $\rho_2$, $\ldots$, $\rho_m$ we see that

$p'(\rho_l) = a \prod_{j = 1, j \ne l}^m (\rho_l - \rho_j) \ne 0 \tag{3}$

since the $\rho_j$ are distinct; thus $p(z)$ and $p'(z)$ have no common zero.

If on the other hand we assume that $p(z)$has a multiple zero $\rho$, then

$p(z) = (z - \rho)^kq(z) \tag{4}$

for some $k \ge 2$ and $q(z) \in \Bbb C[z]$. We thus have

$p'(z) = k(z - \rho)^{k - 1}q(z) + (z - \rho)^kq'(z), \tag{5}$

which, since $k \ge 2$, shows that $(z - \rho) \mid p'(z)$; by contraposition, this in turn implies that if no $\rho$ satisfying $p(\rho) = p'(\rho) = 0$ exists, that is, there is no $\rho$ such that $(z - \rho) \mid p(z)$ and $(z - \rho) \mid p'(z)$, the roots of $p(z)$ must be distinct from one another.


Note Added in Response to Comment by Srinivas Vasudevan, Wednesday 4 June 2014, 4:43 PM PST: The transition 'twixt (1) and (2) follows the ordinary rules of differentiation applied to polynomials. In a nutshell, we have the Leibniz Product Rule for the differentiation operation: for $f(z), g(z) \in \Bbb C[z]$:

$(f(z)g(z))' = f'(z)g(z) + f(z)g'(z). \tag{6}$

(6) in fact applies to any differentiable functions, and may be validated by a standard $\epsilon$-$\delta$ proof built around the ordinary definition of derivative as

$f'(z) = \lim_{\Delta z \to 0} \dfrac{f(z + \Delta z) - f(z)}{\Delta z}, \tag{7}$

or it may be had for polynomials in $\Bbb C[z]$ in a purely algebraic manner wherein the derivative $Df(z) = f'(z)$ is defined as a linear operator on $\Bbb C[z]$ which in addition satisfies $Dc = 0$ for all constant $c \in \Bbb C$ and $Dz^n = nz^{n - 1}$ for all integer $n \ge 1$; the ordinary formula for the derivative of a polynomial follows easily, algebraically, from these defining properties, as does (6). Granting that (6) holds for any two polynomials $f(z), g(z) \in \Bbb C[z]$, we see that for any three polynomials $f(z)$, $g(z)$, $h(z)$,

$(f(z)g(z)h(z))' = (f(z)(g(z)h(z)))' = f'(z)g(z)h(z) + f(z)(g(z)h(z))'$ $= f'(z)g(z)h(z) + f(z)(g'(z)h(z) + g(z)h'(z))$ $= f'(z)g(z)h(z) + f(z)g'(z)h(z) + f(z)g(z)h'(z). \tag{8}$

It is easy, given (6) and (7), to show inductively that for any integer $m \ge 2$ and any collection $f_j(z) \in \Bbb C[z]$, $1 \le j \le m$ of polynomials, that

$(\prod_1^m f_j(z))' = \sum_{k = 1}^m (f'_k(z) \prod_{j = 1, j \ne k}^m f_j(z)); \tag{9}$

I'll leave the details to the reader. If we set $f_j(z) = z - \rho_j$ in (9), we have $f'_j(z) = 1$ and the implication $(1) \Rightarrow (2)$ is immediately had. End of Note.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

  • $\begingroup$ Could you share what theorem/algebraic manipulation you used to go from step (1) to step (2)? $\endgroup$ Jun 4 '14 at 16:13
  • 1
    $\begingroup$ @Srinivas Vasudevan: Of course, I'll be glad to; but it may take me a few hours to respond fully since I have business this morning. I'll message you via comment when I've added the remarks you request. Cheers! $\endgroup$ Jun 4 '14 at 16:33
  • $\begingroup$ @SrinivasVasudevan: OK, I've added a note to my post addressing your concerns. Please feel free to contact me via comment if you have any further questions! Cheers! $\endgroup$ Jun 4 '14 at 23:46
  • $\begingroup$ Thank you Robert! This makes sense now. This is a very interesting approach. $\endgroup$ Jun 5 '14 at 2:09
  • $\begingroup$ @SrinivasVasudevan: my pleasure, sir! Indeed, direct calculation of $p'(z)$ from (1), (2) is illuminating. The rest of my argument is pretty standard and similar to some of the other answers . . . $\endgroup$ Jun 5 '14 at 2:16

Any polynomial $p(x)$ can be written as $\prod_{i=1}^mx-r_i$ if $r_i$ is the $i$th root of $p(x)$

The constant polynomial has no roots, so it cannot have any roots in common with its derivative.

A polynomial $x-r_1$ has the derivative $1$, so it has no roots in common with its derivative.

A polynomial $(x-r_1)(x-r_2)$ has derivative $2x-r_1-r_2$, which has a single root at $\frac{r_1+r_2}{2}$, which only has a root shared with $(x-r_1)(x-r_2)$ if $r_1=r_2$. This means that if and only if $(x-r_1)(x-r_2)$ has a multiple root, the derivative will also have that root.

Iff $p(x)$ has a multiple root then the derivative shares that root, that can be shown for all $p(x)$ inductively, in which they either are a polynomial of degree 2, in which case it is shown above, or they're of a higher degree. And it is shown below.


This means that if $p(x)=0$ and $p'(x)=0$, then the equation above is also $0$. If only $p(x)=0$ then the other term will evaluate to a nonzero value unless $x=r$, in which case $p(x)(x-r)$ will have a multiple root. If $p'(x)$ but not $p(x)(x-r)$ evaluates to zero, then the above equation will not be zero and they do not share a root.

This is not completely properly formulated but I hope you get how I'm trying to show this.


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