The question

Let $f$ be a nonconstant polynomial over $\mathbb{C}$. Let's say that a point $c \in \mathbb{C}$ is unusual for $f$ if every root $x$ of $f(x) - c$ is repeated. Can $f$ have more than one unusual point?

Short remarks

  • There can be exactly one unusual point, e.g. if $f(x) = x^2$. There can be none, e.g. if $f(x) = x^3 + 3x$.

  • There are at most $\deg(f) - 1$ unusual points, since every unusual point is the image under $f$ of a critical point.

  • The hypotheses "nonconstant" and "over $\mathbb{C}$" could be varied. I added them to rule out cases like the following: over a field $k$ of characteristic $p$, every point of $k$ is unusual for $x^p$ (since the derivative of $x^p$ is $0$). I'd be happy to change the hypotheses to "nonconstant polynomial over an algebraically closed field of characteristic 0". Maybe something like "polynomial whose derivative is nonzero, over an algebraically closed field" would also be sensible.

Weak reason to expect the answer to be "no"

Perhaps there's a very short answer to my question: I could be overlooking something elementary. But in case it's not so easy, I'll give a flimsy argument for why we might expect the answer to be "no" - that is, for why we might expect every polynomial to have at most one unusual point.

My question is equivalent to: is there a nonconstant polynomial $f$ over $\mathbb{C}$ for which $1$ and $-1$ are both unusual? For $1$ to be unusual means that every root of $f(x) - 1$ is also a root of $f'(x)$. Writing $d = \deg(f)$, this is equivalent to $$ (f(x) - 1) \mid f'(x)^d. $$ Similarly, for $-1$ to be unusual means that $(f(x) + 1) \mid f'(x)^d$. Both together are equivalent to $$ (f(x)^2 - 1) \mid f'(x)^d, $$ that is, $$ (f(x)^2 - 1)\cdot g(x) = f'(x)^d $$ for some $g(x) \in \mathbb{C}[x]$. This forces $\deg(g) = d(d - 3)$ (and so $d \geq 3$).

So, can we find $f$ and $g$ satisfying the last displayed equation? Comparing coefficients, what we have here is a system of $d(d - 1) = d^2 - d$ equations in $(\deg(f) + 1) + (\deg(g) + 1) = d^2 - 2d + 2$ unknowns. There are $d - 2$ more equations than unknowns, and $d \geq 3$, so a first guess is that it can't be done.

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    $\begingroup$ Such values are known as "totally ramified values" in complex function theory. For an entire function, it follows from Nevanlinna theory that there are at most two (finite) totally ramified values; examples are, of course, given by $\sin$ and $\cos$. For meromorphic functions, there can be at most four totally ramified values. As noted in the answers, for polynomials you can show that there is at most one using an elementary count of critical points. $\endgroup$ – Lasse Rempe-Gillen Nov 13 '12 at 21:21
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    $\begingroup$ Over the reals a polynomial can have arbitrarily many unusual points each with a single double root. Just take one whose derivative is a product of degree two polynomials each with a distinct double root e.g. if $p'(x) = (x+1)^2(x-1)^2$ then $p$ is increasing and has unusual points at $c = p(1)$ and $c = p(-1)$. $\endgroup$ – Pablo Lessa Jun 25 '17 at 15:17

This is impossible by the Mason-Stothers theorem (which holds over any algebraically closed field of characteristic zero).

We want to find $f, g, h$ such that $f + g = h$ where $g$ is a constant and $f, h$ have all of their roots repeated. If $g$ is nonzero, $f, h$ must be relatively prime. Letting $d = \deg f$, it follows that $fgh$ has at most $d$ roots, but by Mason-Stothers $fgh$ must have at least $d+1$ roots; contradiction.

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    $\begingroup$ Right. Unwinding the (elementary) proof of Mason-Strothers [a.k.a. polynomial ABC], this amounts to counting zeros of $f'$: if $c$ is unusual then the roots of $f=c$ contribute at least $\deg(f)/2$ to the roots of $f'$ counted with multiplicity, and unless $f'$ is identically zero the total number of roots with multiplicity is less than $\deg(f)$, QED. $\endgroup$ – Noam D. Elkies Nov 23 '11 at 19:09
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    $\begingroup$ Great! I'm particularly happy to see this answer because Wilson Stothers was a colleague of mine, and an excellent person to work with, until, sadly, he passed away a couple of years ago. I'd never learned the Mason-Stothers theorem before (though I'd heard of it), so I'm very pleased that unexpectedly I'm now using it. $\endgroup$ – Tom Leinster Nov 23 '11 at 19:11
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    $\begingroup$ Really nice answer, Qiaochu. $\endgroup$ – Todd Trimble Nov 24 '11 at 0:39

Encouraged by Pierre-Yves Gaillard here is a quick solution to the original problem. It is based on Tom Leinster's answer which in turn is an elaboration of Noam Elkies's comment to Qiaochu Yuan's answer.

Assume $a\neq b$ are unusual for $f(x)$. Then $f'(x)^2$ is divisible by $f(x)-a$ and $f(x)-b$, hence also by their product. This would show $\deg(f')\geq\deg(f)$, a contradiction.

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    $\begingroup$ This uses that $f' \ne 0$, thus revealing that the only obstruction to this result holding in positive characteristic is the one that @TomLeinster observed. (EDIT: Oops, I see that you already mentioned exactly this in your comment mathoverflow.net/questions/81732/… on which you are elaborating here.) $\endgroup$ – LSpice May 14 '16 at 19:56
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    $\begingroup$ unusal -> unusual? $\endgroup$ – Faheem Mitha Jun 25 '17 at 8:37
  • $\begingroup$ @FaheemMitha: Corrected, thank you! $\endgroup$ – GH from MO Jun 25 '17 at 13:13
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    $\begingroup$ I replaced the broken planetmath link by a wikipedia link. $\endgroup$ – Neil Strickland Jun 25 '17 at 13:29
  • $\begingroup$ @NeilStrickland: Thank you, but this comment should be below Qiaochu Yuan's response, not mine. $\endgroup$ – GH from MO Jun 25 '17 at 18:01

Extend $f$ to a regular morphism $\bar f:\mathbb P^1\to \mathbb P^1$ and write down the Hurwitz formula: $$K_{\mathbb P^1}\sim \bar f^*K_{\mathbb P^1} +R$$ where $R$ is the ramification divisor. Since $f$ is a polynomial, $\bar f$ is completely ramified at $\infty$, so $R$ contains that point with multiplicity $d-1$. Therefore, by the above equivalence the rest of $R$ has degree $d-1$. Now for a point on the target (your "$c$") for which all the points in the preimage are multiple, the degree of the ramification divisor above this point has to be at least $d-\frac d2=\frac d2$ (it's the degree of the map minus the number of points). If you had two such points their combined degree would be at least $d$ contradicting the previous observation that it should be at most $d-1$.

  • $\begingroup$ Thanks! So which fields does this work for? $\endgroup$ – Tom Leinster Nov 23 '11 at 19:04
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    $\begingroup$ Two complete answers came almost simultaneously, politeness dictates that I should accept one, and it's not possible to accept two... I'm going to accept Qiaochu's (partly for the sentimental reasons mentioned), but of course this means nothing. $\endgroup$ – Tom Leinster Nov 23 '11 at 19:19
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    $\begingroup$ It works in characteristic zero. Over an arbitrary field, it's true iff $f'$ is nonzero, so the only exceptions are $f(x^p)$ in characteristic $p$ where every value is taken to multiplicity at least $p$. See my comment to Q.Yuan's answer. $\endgroup$ – Noam D. Elkies Nov 23 '11 at 19:19
  • $\begingroup$ Got you. Thanks. After I saw your other (very useful) comment, I realized that nonzero f' was all we needed. $\endgroup$ – Tom Leinster Nov 23 '11 at 19:25

The question can be restated as follows: can there be a branched $n$-fold cover $S^2\to S^2$ with at least 3 critical values, one of index $n$ and all preimages of critical values being critical points?

If such a map existed, then in the preimage of any critical point other than the infinity we would have $\leq \frac{n}{2}$ points. The Riemann-Hurwitz formula for a branched $n$-fold cover $M\to N$ of 2-manifolds reads $$\chi(M)=n(\chi(N)-k)+a_1+\cdots +a_k$$ where $k$ is the number of the critical points and $a_i$ is the cardinality of the preimage of the $i$-th critical point. In our case this gives $$2=n(2-k)+1+a$$ with $a\leq\frac{n}{2}(k-1)$, so the right hand side is $\leq n(2-k+\frac{k-1}{2})+1=n(\frac{3}{2}-\frac{k}{2})+1$ and can't be 2 when $k\geq 3$.


Noam Elkies left a very useful comment underneath Qiaochu Yuan's answer, providing a completely elementary and self-contained solution to my question. For the benefit of anyone interested, I'll spell it out here. (And I'll make this answer community wiki as this is Noam's solution, not mine.)

Let $k$ be an algebraically closed field. For $f \in k[x]$ and $\alpha \in k$, write $\mu(f, \alpha)$ for the multiplicity of $\alpha$ as a root of $f$; that is, $\mu(f,\alpha) = \sup \{n \in \mathbb{N} : (x - \alpha)^n|f(x)\}$. Some very basic facts:

  • $\mu(f,\alpha) > 0$ iff $f(\alpha) = 0$

  • $\sum_{\alpha \in k} \mu(f,\alpha) = \sum_{\alpha \in f^{-1}(0)} \mu(f,\alpha) = \deg(f)$ as long as $f \neq 0$

  • $\mu(f',\alpha) \geq \mu(f,\alpha) - 1$.

A point $a \in k$ is unusual for $f$ iff $\mu(f-a,\alpha) \geq 2$ for all $\alpha \in f^{-1}(a)$.

For nonconstant $f$, it's easy to see that if $a$ is unusual then $|f^{-1}(a)| \leq \deg(f)/2$. Indeed, $$ \deg(f) = \deg(f - a) = \sum_{\alpha \in f^{-1}(a)} \mu(f - a,\alpha) \geq 2|f^{-1}(a)|. $$

Theorem  Let $f \in k[x]$. If $f' = 0$ then every point of $k$ is unusual for $f$. If $f' \neq 0$ then at most one point of $k$ is unusual for $f$.

Proof  The first statement is clear. For the second, suppose that $a$ and $b$ are unusual, with $a \neq b$. We have $$ \begin{aligned} \sum_{\alpha \in f^{-1}(a)} \mu(f', \alpha)& = \sum_{\alpha \in f^{-1}(a)} \mu((f - a)', \alpha)\\\ &\geq \sum_{\alpha \in f^{-1}(a)} \bigl[\mu((f - a), \alpha) - 1\bigr]\\\ & = \deg(f-a) - |f^{-1}(a)|\\\ & \geq \frac{1}{2}\deg(f). \end{aligned} $$ The same goes for $b$. But $f^{-1}(a) \cap f^{-1}(b) = \emptyset$ and $f' \neq 0$, so $$ \begin{aligned} \deg(f') & = \sum_{\gamma \in k} \mu(f', \gamma)\\\ & \geq \sum_{\alpha\in f^{-1}(a)} \mu(f', \alpha) + \sum_{\beta\in f^{-1}(b)} \mu(f',\beta)\\\ & \geq \frac{1}{2}\deg(f) + \frac{1}{2}\deg(f) = \deg(f), \end{aligned} $$ giving $\deg(f') \geq \deg(f)$, a contradiction.

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    $\begingroup$ I would tell this as follows. Assume $f(x)-a=\prod_i (x-\alpha_i)^{m_i}$ and $f(x)-b=\prod_j (x-\beta_j)^{n_j}$ with $a\neq b$ and $m_i,n_j\geq 2$. Then $f'(x)$ is divisible by $\prod_i (x-\alpha_i)^{m_i-1}\prod_j (x-\beta_j)^{n_j-1}$, hence either $f'(x)=0$ or we have $\deg(f')\geq \sum_i m_i/2+\sum_j n_j/2=\deg(f)$, a contradiction. $\endgroup$ – GH from MO Nov 28 '11 at 20:09
  • $\begingroup$ I don't understand: If @GH's argument is incomplete, what's missing? Otherwise, why is it a comment, not an answer? $\endgroup$ – Pierre-Yves Gaillard Nov 29 '11 at 8:40
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    $\begingroup$ @Pierre: I have now posted a variant as an answer. I hope you like it. $\endgroup$ – GH from MO Nov 29 '11 at 12:00
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    $\begingroup$ @GH: I don't like it, I love it!!! You compactified the compactification. I was the first one (and hopefully not the last one) to vote for your answer. Arguments like this are very helpful for people like me who already have a hard time understanding simple things. Thank you very much! $\endgroup$ – Pierre-Yves Gaillard Nov 29 '11 at 12:15
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    $\begingroup$ @Pierre-YvesGaillard: This was almost 6 years ago, but thank you for your kind words! $\endgroup$ – GH from MO Jun 26 '17 at 3:17

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