This post concerns Problem 3 from Griffin Macris’s list of open problems. It was originally stated as a question:

Question (Problem 3)

Say $\alpha$ is an algebraic number. Is there always a finite sequence of polynomials $f_1,f_2,\ldots,f_k$, all of which have integer coefficients and rational zeros, such that, when composed, $f_1(f_2(\ldots f_k(\alpha)\ldots)) = 0$?

It is known that for $\deg \alpha \leq 3$ such polynomials always exist. For example, if $\alpha=\frac{p}{q}$, then the polynomial $f_1(x)=qx-p$ works. For a quadratic irrational satisfying $A\alpha^2 - B\alpha + C = 0$, the sequence $f_1(x)=x+C$, $f_2(x)=x(Ax-B)$ works. The case for cubics was solved by Jordi Ribes:

Any cubic $x^3+ax^2+bx+c$ divides the polynomial $(x+a/3)^2((x+a/3)^2-(a^2-3b)/3)^2-((2a^3-9ab+27c)/27)^2$. This comes from computing the depressed cubic $t^3+dt+e$ (where $d$,$e$ depend on $a$,$b$,$c$) through the change $t=x+a/3$, and noting that $t^3+dx+e$ divides $t^2(t^2+d)^2-e^2$. So $f_1(x)=(x+a/3)^2$, $f_2(x)=x*(x-(a^2-3b)/3)^2$, $f_3(x)=x-((2a^3-9ab+27c)/27)^2$.

The open question is what happens in the case $\deg \alpha \geq 4$.

We first restate the problem in a more workable form, beginning with a name for the polynomials it allows.

Definition (Split)

A polynomial $f \in \Q[x]$ is split if $\deg f \geq 1$ and $f(x) = a\prod_{i=1}^n (x - r_i)$ for some $a \in \Q^\times$ and $r_1,\ldots,r_n \in \Q$.

Whether we demand integer or rational coefficients in the sequence of polynomials is immaterial here, since multiplying a polynomial in $\Q[x]$ by a common denominator of its coefficients will always put it in $\Z[x]$.

Lemma (Hitting $0$ is hitting $\Q$)

Let $\alpha$ be an algebraic number and $k \geq 1$. There exist split $f_1,\ldots,f_k$ with $f_1(f_2(\ldots f_k(\alpha)\ldots)) = 0$ if and only if there exist split $f_2,\ldots,f_k$ with $f_2(\ldots f_k(\alpha)\ldots) \in \Q$.

Proof

If $f_1$ is split and $f_1(\beta) = 0$, then $\beta$ is a zero of $f_1$, and all zeros of a split polynomial are rational, so $\beta \in \Q$. Conversely, if $\beta = p/q \in \Q$, then $f_1(x) = qx - p$ is split and $f_1(\beta) = 0$.

So Problem 3 is not about reaching $0$, but about whether $\Q$ can be reached. We define a grading scheme for the problem based on how many polynomials we are allowed to use.

Definition ($k$-collapsible)

Let $\alpha$ be an algebraic number over $\Q$ and $k \geq 1$. We say $\alpha$ is $k$-collapsible if there exist split polynomials $f_1,f_2,\ldots,f_k$ with \(f_1(f_2(\ldots f_k(\alpha)\ldots)) \in \Q.\) We say $\alpha$ is collapsible if it is $1$-collapsible, and eventually collapsible if it is $k$-collapsible for some $k \geq 1$.

Remark

By the lemma, Problem 3 asks precisely whether every algebraic number is eventually collapsible.

Theorem (Trinomials are $2$-collapsible)

Let $\alpha$ be an algebraic number satisfying \(\alpha^n + q\alpha^k + r = 0\) for some $q,r \in \Q$ and integers $1 \leq k < n$. Then $\alpha$ is $2$-collapsible.

Proof

Put $e = n-k$ and $t = e/\gcd(k,e)$. Since $\alpha^n = \alpha^k\alpha^e$, the relation factors as \(\alpha^k(\alpha^e + q) = -r,\) and raising it to the $t$-th power gives $\alpha^{kt}(\alpha^e+q)^t = (-r)^t$. By the choice of $t$ we have $kt/e = k/\gcd(k,e) \in \Z$, so the left-hand side is a polynomial in $\alpha^e$. Set \(f_2(y) = y^{k/\gcd(k,e)}(y+q)^t, \qquad f_3(z) = z^e.\) Both are split, their only zeros being $0$ and $-q$, and both have degree at least $1$ since $e \geq 1$. Now \(f_2(f_3(\alpha)) = (\alpha^e)^{kt/e}(\alpha^e+q)^t = \alpha^{kt}(\alpha^e+q)^t = (-r)^t \in \Q,\) so $\alpha$ is $2$-collapsible.

Note: This is a generalization of Jordi Ribes’s cubics ($n=3$, $k=1$). This result does not completely settle the degree $4$ case.

Collapsible reduction

The complexity of the problem arises from having to deal with composition of many functions. So restricting the problem to 1-collapsible makes an interesting case. There is no composition between functions, and this reveals certain structures.

Question (Collapsible)

Which algebraic numbers $\alpha$ are $1$-collapsible? That is, for which $\alpha$ does there exist a polynomial $f\in \Q[x]$ that splits in $\Q$ such that $f(\alpha)\in \Q$?

The cases $\deg \alpha\leq 2$ are already covered by the previously known results stated above. But the case $\deg \alpha = 3$ is currently open.

When the minimal polynomial has a convenient shape, collapsibility can be exhibited by hand. Suppose $\alpha^3 - s^2\alpha + C = 0$ for some $s \in \Q$, and take the split polynomial with roots $0, s, -s$:

\[f(x) = x(x-s)(x+s) = x^3 - s^2 x, \qquad f(\alpha) = -C \in \Q .\]

Or suppose $\alpha^3 - 3\alpha + C = 0$; then the roots $1,1,-2$ do the same job:

\[f(x) = (x-1)^2(x+2) = x^3 - 3x + 2, \qquad f(\alpha) = 2 - C \in \Q .\]

Neither computation used anything about $\alpha$ beyond its minimal polynomial $m$, and in both cases $f$ was rigged so that $f - m$ is a rational constant. That is forced rather than lucky: $m$ generates the ideal of polynomials vanishing at $\alpha$, so $f(\alpha) = c$ holds exactly when $m$ divides $f - c$. Collapsibility of $\alpha$ is therefore a statement about $m$ alone,

\[\alpha \text{ is collapsible} \iff mh + c \text{ is split for some } h \in \Q[x]\setminus\{0\},\ c \in \Q,\]

and the search for $f$ is a search for a rational root set whose elementary symmetric functions match the coefficients of such an $mh + c$.

The cubic case

Before specialising into the cubic case, we show that collapsibility only sees $\alpha$ up to an affine change of variable.

Lemma (Affine invariance)

Let $\alpha$ be an algebraic number, $\lambda \in \Q^\times$ and $b \in \Q$. Then $\alpha$ is collapsible if and only if $\lambda\alpha + b$ is.

Proof

Let $f(x) = a\prod_{i=1}^n(x-r_i)$ be split with $f(\alpha) \in \Q$, and set $g(x) = f\big((x-b)/\lambda\big)$. Then

\[g(x) = a\prod_{i=1}^n\Big(\tfrac{x-b}{\lambda} - r_i\Big) = a\lambda^{-n}\prod_{i=1}^n\big(x - (\lambda r_i + b)\big),\]

so $g$ is split of the same degree $n \geq 1$, with rational roots $\lambda r_i + b$ and leading coefficient $a\lambda^{-n} \in \Q^\times$, and $g(\lambda\alpha + b) = f(\alpha) \in \Q$. The converse is the same argument applied to $\lambda^{-1}$ and $-b/\lambda$, which invert $x \mapsto \lambda x + b$.

Writing $m(x) = x^3+ax^2+bx+c$, the lemma lets us replace $\alpha$ by $\alpha + a/3$: the roots of the new minimal polynomial sum to $-a + 3\cdot\tfrac{a}{3} = 0$, so it is depressed,

\[m(x) = x^3 + dx + e .\]

Lemma (Cubic criterion)

Let $\alpha$ have minimal polynomial $m(x) = x^3+dx+e$. There is a split $f$ of degree $3$ with $f(\alpha)\in\Q$ if and only if

\[-d = \rho_1^2 + \rho_1\rho_2 + \rho_2^2 \qquad\text{for some } \rho_1,\rho_2 \in \Q.\]

In that case $f(x) = (x-\rho_1)(x-\rho_2)(x+\rho_1+\rho_2)$ and $f(\alpha) = \rho_1\rho_2(\rho_1+\rho_2) - e$.

Proof

Suppose $f$ is split of degree $3$ with $f(\alpha) \in \Q$. Dividing by its leading coefficient we may take $f$ monic, and we may write $f(\alpha) = c$. Then $f - c$ is monic of degree $3$ and vanishes at $\alpha$, so it is divisible by $m$; comparing degrees, $f = m + c$. Writing $f = \prod_{i=1}^3 (x-\rho_i)$ with $\rho_i \in \Q$ and comparing coefficients with $m + c = x^3 + dx + (e+c)$ gives

\[\rho_1+\rho_2+\rho_3 = 0, \qquad \rho_1\rho_2+\rho_1\rho_3+\rho_2\rho_3 = d.\]

Substituting $\rho_3 = -(\rho_1+\rho_2)$ into the second equation turns it into $-(\rho_1^2+\rho_1\rho_2+\rho_2^2) = d$.

Conversely, given such $\rho_1,\rho_2$, set $\rho_3 = -(\rho_1+\rho_2)$ and $f = \prod_{i=1}^3(x-\rho_i)$. Then $f$ is split, its $x^2$-coefficient vanishes and its $x$-coefficient is $d$, so $f - m$ is the constant $-\rho_1\rho_2\rho_3 - e$, whence $f(\alpha) = f(\alpha)-m(\alpha) = \rho_1\rho_2(\rho_1+\rho_2)-e \in \Q$.