# How to prove that every polynomial in an infinite family is irreducible over Q?

Consider the bivariate polynomial $$p(X,Y) = X^5 - (2 Y + 1) X^3 - (Y^2 + 2) X^2 + Y (Y-1) X + Y^3.$$ For every integer $y \ge 4$, I conjecture that $p(X,y)$ is irreducible in $\mathbb{Q}[X]$. How can this be proved?

## Motivation

This infinite family of polynomials comes from a recent paper of mine. We wanted to show that the roots of every polynomial in this family satisfy what we call the lattice condition (Definition 6.3). For this family of polynomials, it is sufficient that they are each irreducible. However, we are unable to prove that they are in fact irreducible. Instead, with much effort, we find a different way to show that the roots of all these polynomials do indeed satisfy the lattice condition.

I am wondering if our proof could be simplified by proving the sufficient condition that these polynomials are irreducible.

## What I Know

1. $p(X,y)$ has three distinct real roots and two nonreal complex conjugate roots (since the discriminant is negative).
2. The algebraic curve defined by $p(X,Y)$ has genus 3 (computed by Maple).
3. From Theorem 1.2 in Hilbert’s irreducibility theorem for prime degree and general polynomials by Peter Müller, it follows that $p(X,y)$ is reducible in $\mathbb{Q}[X]$ for at most a finite number of $y \in \mathbb{Z}$. (Note that this result is not effective.)
4. The only values of $y \in \mathbb{Z}$ for which I know that $p(X,y)$ is reducible are $$p(X,y) = \begin{cases} (X - 1) (X^4 + X^3 + 2 X^2 - X + 1) & y = -1\\ X^2 (X^3 - X - 2) & y = 0\\ (X + 1) (X^4 - X^3 - 2 X^2 - X + 1) & y = 1\\ (X - 1) (X^2 - X - 4) (X^2 + 2 X + 2) & y = 2\\ (X - 3) (X^4 + 3 X^3 + 2 X^2 - 5 X - 9) & y = 3. \end{cases}$$ These five reducible cases also give five integer solutions to $p(X,Y) = 0$.
5. With help from Aaron Levin and Bjorn Poonen, we prove (Lemma 7.6) that the only integer solutions to $p(X,Y) = 0$ are the five solutions given by the five factorizations above. (The proof uses Puiseux series expansions.)
6. Using Mathematica, I checked that $p(X,y)$ is irreducible for $4 \le y \le 2.1 \times 10^9$.

## Conclusion

From item 5, if $p(X,y)$ is reducible, then it must be a product of an irreducible quadratic and an irreducible cubic. Then using item 1 and some Galois theory, we prove (Lemma 7.7) that the roots of $p(X,y)$ satisfy the lattice condition in this case.

I am wondering if we can avoid this line of reasoning and instead prove that $p(X,y)$ is in fact irreducible in $\mathbb{Q}[X]$ for every integer $y \ge 4$.

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Can you use the Newton polygon for the largest prime dividing $y$ to eliminate the possibility of a cubic times a quadratic? –  S. Carnahan Apr 16 '14 at 1:04
@S.Carnahan If $y$ were a fixed integer, then I understand how I might use the Newton polygon to help prove irreducibility. But since $y$ is not fixed, I don't know how to use the Newton polygon. Do you know how one might do this? –  Tyson Williams Apr 16 '14 at 1:08
I'm not sure I understand your question, but it looks like you are asking about the polynomials given by choosing integer values for $y$. For any choice of $y$, you can choose a valuation using the largest prime dividing $y$, and construct the corresponding Newton polygon. My wild guess is that it should be possible to reason about this polygon without much explicit knowledge about $y$, e.g., if $y$ is not a power of 2, then perhaps the polynomial can only split as a linear factor times a quartic. –  S. Carnahan Apr 16 '14 at 1:22
I assume that there is a typo in (3) andyou mean that it is "reducible" for at most a finite number of $y$, not "irreducible" for at most a finite number of $y$. –  Joe Silverman Apr 16 '14 at 3:15
What happens numerically for $y<-1$? –  Kevin Ventullo Apr 16 '14 at 4:15

The same Puiseux-series tactic works (effectively) to list all $y$ for which $p(X,y)$ has a quadratic factor. For example, the linear coefficient of such a factor is $a = -(r_i+r_j)$ for some distinct $i,j$ where $r_1,\ldots,r_5$ are the roots of $p(X,y)$. But for large $y$ there are three real and two imaginary roots, and thus four choices of $\{i,j\}$ that make $r_i+r_j$ real; and in each case you can expand $a$ in a Laurent series in $y^{-1/6}$ with rational coefficients. You can then concoct a polynomial in $a$ and $y$ with integer coefficients that decays as a positive power of $1/y$ but is not identically zero, and thus cannot be an integer once $y$ is large enough. Surely your computation up to $10^9$ will then be way more than enough to establish the desired result.

[Added later] Here are explicit power series and polynomials that can be used for this purpose. Let $z = y^{-1/6}$. Then for large $y>0$ (i.e. small $z>0$) the roots of $p(X,y)$ are $$r_1, r_2 = \pm z^{-3} - z^6 \mp z^9 \pm \frac32 z^{15} + 5 z^{18} + O(z^{21}),$$ $$r_3 = z^{-4} + \frac13 z^{-2} - \frac1{81} z^2 + \frac{82}{243} z^4 + \frac23 z^6 + \frac{5585}{6561} z^8 + O(z^{10}),$$ and complex conjugate roots $r_4,r_5$ obtained from $r_3$ by multiplying $z$ by a cube root of unity. Let's warm up by showing that for large $y$ there are no integral roots, i.e. none of $r_1,r_2,r_3$ can be an integer (we needn't worry about $r_4,r_5$ because they're not even real). For $x=r_1,r_2$ it is enough to observe that $$x^2 - y = \mp 2 z^3 - 2 z^6 + 4 z^{12} \pm 12 z^{15} + 12 z^{18} + O(z^{21}),$$ which for large $y$ is a small nonzero real number and thus cannot be an integer. For $x=r_3$ we use the $11$ monomials in $x,y$ that have a pole of order at most $9$ in $z^2$, and find the polynomial $$y^3 + (3-2x) y^2 + (-x^3-3x^2+4x-4) y + 3x^4-3x^3-x^2-7x+4$$ with power series $$4 z^4 - \frac43 z^6 - \frac{104}{9} z^8 - \frac{980}{81} z^{10} - \frac{32}{3} z^{12} + O(z^{14}),$$ which again is nonzero but small for large $y$.

For a quadratic factor the candidates are $(X-r_1)(X-r_2)$, $(X-r_1)(X-r_3)$, $(X-r_2)(X-r_3)$, and $(X-r_4)(X-r_5)$. The first of these has $r_1+r_2 = -2z^6 + O(z^{18})$ which is already small and nonzero. For the last, let \begin{align*} a &= r_4 + r_5\\ &= -(r_1+r_2+r_3)\\ &= -z^{-4} - \frac13 z^{-2} + \frac1{81} z^2 - \frac{82}{243} z^4 + \frac43 z^6 - \frac{5585}{6561} z^8 + O(z^{10}) \end{align*} [sorry, I switched from the earlier $a = -(r_4+r_5)$], and proceed as we did before for $r_3$, finding $$3y^3 + (7a-5)y^2 + (3a^3-10a^2+2a+52)y + 10a^4-5a^3-22a^2-a-20$$ with a somewhat more complicated power series $$82 z^6 - 34z^8 - \frac{58}{3} z^{10} - \frac{400}{3} z^{12} + O(z^{14}),$$ but still enough for our purposes once $|y|$ exceeds $100$ or so.

The remaining candidates, $(X-r_1)(X-r_3)$ and $(X-r_2)(X-r_3)$, are tougher, though related by $z \leftrightarrow -z$ so we can treat them together. This time $a = r_1+r_3$ is $z^{-4} + z^{-3} + \frac13 z^{-2} - \frac1{81} z^2 + \ldots$ involves powers of both $z^2$ and $z^3$, and it takes a while before we can find a polynomial with no singular part. It's easier to use also $b = r_1 r_3 = z^{-7} + \frac13 z^{-5} - \frac1{81} z^{-1} + \frac{82}{243} z - z^2 + \ldots$, which must also be an integer if $(X-r_1) (X-r_3)$ is to be an integral factor of $q(X,y)$. Even then we need to use the monomials up to degree $22$, $${}^{1,\; a,\; y,\; b,\; a^2,\; ay,\; ab,\; a^3,\; y^2,\; by,\; b^2,\; a^2b,\; a^4,\; ay^2,\; aby,\; a^3y,\; y^3,\; ab^2,\; a^3b,\; by^2,\; a^5,\; a^2y^2,\; b^2y,\; a^2by,\; b^3,\; a^4y,\; ay^3,\; a^2b^2}$$ because the first few combinations we find, such as $y^3 + by^2 + 2aby + 2ab^2 + (-a^3+a+2)b$, are identically zero. But the combination with coefficients $${}^{120,\; 44,\; -83,\; -159,\; 103,\; 82,\; 144,\; -42,\; -5,\; -36,\; -9,\; -27,\; 9,\; 24,\; 86,\; -19,\; 24,\; 13,\; 42,\; 0,\; -14,\; 1,\; -12,\; -10,\; -8,\; 9,\; 0,\; -9}$$ succeeds, giving $$72z + 42z^2 - 214z^3 + 370z^4 - 134z^5 + \frac{142}{3}z^6 - \frac{1702}{9}z^7 + \frac{2674}{27} z^8 + \frac{1064}{3} z^9 + O(z^{10}).$$ The resulting upper bound on $y = z^{-6}$ still seems impractically large: certainly nowhere near the astronomical $10^{20000}$ that the OP obtained from the literature, but still a bit beyond the OP's computation to $10^9$ (it's about $72^6$, which exceeds $10^{11}$). However, we don't need to try all $y$ up to that bound, only those which make our polynomial nearly an integer (positive or negative, because replacing $r_1$ by $r_2$ yields the same power series with $-z$ in place of $z$); and there are only a few hundred such to compute. For each of them, check whether the nearest integer $y$ yields a reducible $p(X,y)$, and you're finally done.

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The biggest mystery to me (in both the proof of our Lemma 7.6 and your answer) is how to concoct this function (it is probably not a polynomial, right?) in $a$ and $y$ that decays for large $y$ but doesn't vanish. Can you elaborate on this part? –  Tyson Williams Apr 16 '14 at 8:42
Yes, they are polynomials; there are more monomials to use than power-series terms to kill (presumably this always happens unless the curve actually does have infinitely many integral points). See the new paragraphs. –  Noam D. Elkies Apr 16 '14 at 23:45
@Noam D. Elkies: I did not obtain the bound $10^{20000}$ from Runge, rather I doubted that carrying out Runge's algorithm would be that bad. Indeed, arxiv.org/abs/math/0512418 might be interesting in this regard. –  Peter Mueller Apr 17 '14 at 12:17
@Peter Mueller: sorry for the misattribution; I see now that the $10^{20000}$ estimate did not come from you. Will fix in the next edit. Tyson Williams: you're welcome. I computed the series expansions for the roots using Newton's method, starting from the approximations $Y^{2/3}$ and $\pm Y^{1/2}$. These approximations, in turn, come in effect from the two sides of the Newton polygon of $p(X,Y)$ at infinity: the dominant terms are $X^5 - X^2 Y^2$ for the $Y^{2/3}$ root, and $-X^2 Y^2 + Y^3$ for the $\pm Y^{1/2}$ root. –  Noam D. Elkies Apr 17 '14 at 21:13
In view of Noam Elkies' perfect answer, I removed my answer. Also, his construction of the auxiliary polynomial is similar, but more efficient than Runge's original argument. –  Peter Mueller Apr 17 '14 at 21:44