6
$\begingroup$

Question.

If $p$ is a polynomial of degree $n$ with $p(\alpha)=0$, what do we know of the polynomial $q$ (with degree $n-1$) such that the numbers $(q^k(\alpha))_{k=1}^n$ contain all of the zeroes of $p$?

Here I denote $q(q(\cdots q(\alpha)))=q^k(\alpha)$.


Notes.

We know for a fact such $q$ exists, since there always exists a polynomial of degree $n-1$ through $n$ given points. $q$ is not unique, however, since there are multiple permutations we can put the zeroes in.


Examples.

For linear $p$ (write $p(x)=a_1x+a_0$), this is obvious; $q(x)=\tfrac{-a_0}{a_1}=\alpha$ suffices. If $p(\alpha)=0$, then $q(\alpha)=\alpha$ indeed are all the zeroes of $p$.

If $p$ is quadratic, write $p(x)=a_2x^2+a_1x+a_0$, and have $p(\alpha)=0$ again; now $q(x)=\frac{-a_1}{a_2}-x$.

If $p$ is cubic, write $p(x)=a_3x^3+a_2x^2+a_1x+a_0$. This is where I get stuck, since the roots of cubic equations aren't expressions that are easy to work with.


Attempts.

First I see (denote the (not necessarily real) zeroes by $z_1,z_2,\cdots,z_n$) that $z_1+\cdots+z_n=\frac{-a_{n-1}}{a_n}$ and $z_1z_2\cdots z_n=\frac{(-1)^na_0}{a_n}$. We can produce similar expressions for the other coefficients, but I doubt this is useful; they're not even solvable for $n>4$. We also have (given $z_1$) $$-a_nz_1^n=a_{n-1}z_1^{n-1}+\cdots+a_1x+a_0$$ with which we can reduce every expression of degree $n$ or larger in $z_1$ to an expression of degree $n-1$ or smaller.

For $n=3$ (let's do some specific examples), we could write $q(x)=b_2x^2+b_1x+b_0$, and take for example $p(x)=x^3-x-1$. Then, if $\alpha$ is a zero of $p$, then $\alpha^3=\alpha+1$, and so $q(\alpha)^3=q(\alpha)+1$, which is

$$(b_2\alpha^2+b_1\alpha+b_0)^3=b_2\alpha^2+b_1\alpha+b_0+1$$

working out the constant terms gives $b_2^3+b_1^3+b_0^3+6b_0b_1b_2=b_0+1$ which isn't very useful either.

Please, enlighten me. Has there been done work on this subject, am I missing something obvious, or perhaps you see something that I missed?

$\endgroup$
  • 1
    $\begingroup$ For $n=3$ and $p$ with all roots real, see math.stackexchange.com/questions/1767252/…. $\endgroup$ – lhf Mar 14 '17 at 10:03
  • $\begingroup$ Your justification for the existence of $q$ does not convince me. I think you need $p$ irreducible and some Galois theory. $\endgroup$ – lhf Mar 14 '17 at 10:06
  • $\begingroup$ @lhf $q$ always exist but the field its coefficients are in depends on the galois group of $p$ $\endgroup$ – mercio Mar 14 '17 at 10:08
  • $\begingroup$ $p$ doesn't need to be irreducible as I found a corresponding $q$ for quadratic $p$, and quadratic polynomials are never irreducible $\endgroup$ – vrugtehagel Mar 14 '17 at 10:09
  • 1
    $\begingroup$ @lhf, that's interesting. Because it has a root with multiplicity $2$, of course $q(0)=0$ and $q(q(0))=1$ isn't possible. We can however simply pick $q(x)=-\tfrac12x^2+-\tfrac12x+1$ to have $q(0)=1$ and $q(q(1))=0$ $\endgroup$ – vrugtehagel Mar 14 '17 at 10:47
3
$\begingroup$

For any sequence of $n$ distinct (complex) numbers $\alpha_1 \ldots \alpha_n$, there is a unique interpolating polynomial $Q$ of degree $n-1$ with $P(\alpha_i) = \alpha_{i+1}$.

In the generic case, the coefficients of $Q$ are in $\Bbb Q[\alpha_i, \delta^{-1}]$, where $\delta = \prod_{i<j} (\alpha_i - \alpha_j)$ (to be more precise, $\delta Q \in \Bbb Z[\alpha_i]$)

In the case where the $\alpha_i$ are roots of a polynomial $P$ with no repeated roots, then for any $n$-cycle $\sigma$, $Q_\sigma$ is defined over the splitting field of $P$, but sometimes it is defined over a smaller field.

If $G$ is the Galois group of $P$, the action of $G$ on the $Q_\sigma$ corresponds to its action on the set of $n$-cycles by conjugation.

If $K = \Bbb Q(a_i)$ is the field of definition of $P$ and $L_\sigma$ is the field of definition of $Q_\sigma$ for an $n$-cycle $\sigma$, then $[ L_\sigma : K ]$ is exactly the size of the orbit of $\sigma$ under this conjugation action. Since the normalizer of an $n$-cycle in $S_n$ is the subgroup of $S_n$ generated by the $n$-cycle, we have that the subgroup of $G$ fixing the $n$-cycle is $G_\sigma = G \cap \langle \sigma \rangle$, and then $L_\sigma = \Bbb Q(\alpha_i)^{G_\sigma}$, and finally $[L_\sigma : K] = |G| / |G_\sigma|$

For example, if $n=2$, there is only one $2$-cycle, so there is only one $Q$, and $L= K$ which means that there are formulas for the coefficients of $Q$ in terms of the $a_i$.

If $n=3$, there are two $3$-cycles. If $G \not \subset A_3$ then $G$ will switch them so you will get an orbit of size $2$, so the coefficients of $Q$ will be in a degree $2$ extension of $K$. If $G \subset A_3$ then $G$ won't switch them, which tells you that in this case $L_\sigma = K$. In fact you have formulas for the two $Q$ in terms of the $a_i$ and $\delta = \pm \sqrt \Delta$ (and you switch between the two $Q$ by switching the sign of $\delta$)

If $n=4$, there are $3! = 6$ $4$-cycles, living in $6/\phi(4) = 3$ different cyclic subgroups of order $4$, so lots of stuff can happen. In the worst case, $G = S_4$ and they are all defined in $3$ different (non-normal) extensions of order $6$ of $K$. $S_4$ is still solvable so if you really want to you can get formulas for $Q$ with square roots and cube roots.

If $n \ge 5$ well then $G$ may not even be solvable anymore so you won't have those formulas anymore.

The only way you can have $L_\sigma = K$ is when $|G_\sigma| = |G|$, which means $G \subset \langle \sigma \rangle$, and for this to happen $G$ has to be cyclic too (not necessarily of order $n$ if $P$ is not irreducible, though).

$\endgroup$
0
$\begingroup$

Let $\alpha=\alpha_1, \alpha_2, \ldots, \alpha_m$ be the distinct roots of $p$.

Choose a permutation $\sigma$ of $1,2,\dots,m$ without fixed points. For instance, an $m$-cycle such as $(12\cdots m)$.

Let $q$ be the unique polynomial such that $q(\alpha_i)=\alpha_{\sigma(i)}$. That will work, but won't have degree $n-1$.

$\endgroup$
  • 1
    $\begingroup$ This is quite trivial though, I my question is more about what we know about $q$ besides the property we defined it with $\endgroup$ – vrugtehagel Mar 14 '17 at 11:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.