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I took an exam a few hours ago and there was a question I couldn't do.

For those who speak French, here is the document http://www.concours-centrale-supelec.fr/CentraleSupelec/2014/MP/sujets/2012-006.pdf (precisely III.B.3).

Let $\alpha \in \mathbb C$.

Let $P_\alpha = X^2+\alpha$

Prove that for any $n \in \mathbb N$, there exists at most one polynomial $Q$ with degree $n$ such that $P_\alpha \circ Q=Q \circ P_\alpha $

If such $Q$ exists it must have $1$ as leading coefficient.

I tried rewriting $P_\alpha \circ Q=Q \circ P_\alpha$ in terms of coefficients of $Q$ and find a recurrence relation on these coefficients (this would grant the "at most") but computations got awful.

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Gabriel Romon
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2 Answers2

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If you let $Q_ = \sum_{0 \le k \le n} b_k X^k$ and write out the coefficients of $P_\alpha \circ Q$ and $Q \circ P_\alpha$, you should notice that, in front of $X^{2n-k}$, in $P_\alpha \circ Q$ you obtain a linear combination of $b_{n-i}b_{n-j}$ for $i+j=k$, and in particular, you get $2b_nb_{n-k} = 2b_{n-k}$. In $Q \circ P_\alpha$, you obtain an expression involving $b_n \ldots b_{n-\lfloor k/2 \rfloor}$ (because $2k$ is the greatest power of $X$ appearing with $b_k$).

Hence by taking $k=1, \ldots, n$ you obtain equations relating each $b_{n-k}$ to some expression in $b_{n-l}$ with $0\le l<k$. Since $b_n$ has a unique possible value, so does each coefficient. So by looking at the $n+1$ most dominant coefficients of the equality, you obtain $n+1$ equations that determine $Q$ uniquely.

Also note that this doesn't work for degree $0$ polynomials.

This still works if you change the degree of $P$ to $d>1$, the $\lfloor k/2 \rfloor$ turns into $\lfloor k/d \rfloor$, which is still strictly less than $k$ for $k > 0$. Only the first coefficient of $Q$ can vary a bit, bust the rest is determined by that coefficient.

mercio
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I feel very dumb now...

Suppose for contradiction sake that $Q$ and $R$ are distinct degree $n\geq 2$ polynomials that commute with $P_\alpha$.

Then Let $Z=Q-R$.

Since $Q$ and $R$ are monic, $\deg Z < n$

Furthermore, $Z \circ P_\alpha =P_\alpha\circ Q -P_\alpha\circ R=Q^2-R^2=Z(Q+R) $

Note that $\deg Z \circ P_\alpha=2\deg Z$ and $\deg Z(Q+R)=n+\deg Z$

Hence $\deg Z =n$

Contradiction.

Gabriel Romon
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