# Is there an irreducible but solvable septic trinomial $x^7+ax^n+b = 0$?

The following irreducible trinomials are solvable:

$$x^5-5x^2-3 = 0$$

$$x^6+3x+3 = 0$$

$$x^8-5x-5=0$$

Their Galois groups are isomorphic to ${\rm D}_5$, ${\rm S}_3 \wr {\rm C}_2$ and $({\rm S}_4 \times {\rm S}_4) \rtimes {\rm C}_2$, respectively.

Question: Is there an irreducible septic trinomial $x^7+ax^n+b=0$ with solvable Galois group, where $n \in \{1, \dots, 6\}$, $a, b \in \mathbb{Z} \setminus \{0\}$?

P.S. For $n=1$, I did a search only for those with 1 real root and, if I did it correctly, there are none with integer $|a,b|<50$.

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Please add that you mean irreducible over ${\mathbf Q}$, presumably. Also, for completeness, it would be nice for the reader if you tell us what the Galois groups over ${\mathbf Q}$ are for the three polynomials in your question (even if a computer algebra package computes them). –  KConrad Nov 2 '13 at 20:54
Have you tried the search by the Maple code $$seq(seq(solve(x^7+a*x^2+b,x,explicit),a=-50..50),b=-50..50)$$ etc? –  user64494 Nov 2 '13 at 21:23
There is no such polynomial in the case $n=2,|a|\le 20,|b|\le20.$. –  user64494 Nov 2 '13 at 22:13
1. You mean $a\neq 0$, right? 2. Such a polynomial would have Galois group a solvable transitive subgroup of $S_7$. If I remember correctly, one can show by enumeration of cases that any such group must have a normal Sylow 7-subgroup, i.e., it must have Galois group a subgroup of the group of linear transformations of $\mathbb F_7$. So one might try to eliminate these cases. –  Will Sawin Nov 2 '13 at 22:46
@Will: that's true. More generally, for any prime $p$, Galois showed that every solvable transitive subgroup of $S_p$ is a group of linear transformations of $\mathbf{F}_p$. He deduced that, if an irreducible degree-$p$ polynomial is solvable, then any two roots generate its splitting field. –  Michael Zieve Nov 2 '13 at 23:10

If such polynomials exist, there will only be finitely many of them, up to composing on both sides with scalar polynomials $\alpha x$ with $\alpha\in\mathbf{Q}$. More generally, Guralnick and Shareshian proved that if $d=7$ or $d>8$ then there are only finitely many equivalence classes of irreducible degree-$d$ trinomials in $\mathbf{Q}[x]$ whose Galois group is not $A_d$ or $S_d$, where two trinomials $f(x)$ and $g(x)$ are called equivalent if $f(x)=\alpha\cdot g(\beta x)$ for some $\alpha,\beta\in\overline{\mathbf{Q}}$. They proved an analogous result over number fields, and also for reducible trinomials. See Theorem 1.4.3 of "Symmetric and Alternating Groups as Monodromy Groups of Riemann Surfaces I: Generic Covers and Covers with Many Branch Points", AMS Memoirs (2007), vol. 189, no. 886.

This explains why you were able to find the examples you found. The only pairs $(d,n)$ for which there are infinitely many equivalence classes of irreducible degree-$d$ trinomials in $K[x]$ with terms of degrees $0,n,d$ and Galois group not $A_d$ or $S_d$ (for some number field $K$) are those pairs with $0<n<d$ and $d\le 8$ and $d\ne 7$ and either $d=4$ or $\gcd(n,d)=1$. Also Guralnick-Shareshian listed the groups which occur for infinitely many equivalence classes over a number field.

In order to answer the question you asked, one can use the following approach by Elkies. Let $H=\text{AGL}(1,7)$ be the group of all linear maps on $\mathbf{F}_7$. For each $n\in\{1,2,\dots,6\}$, Guralnick-Shareshian say that Elkies told them that there is a bijection between the set of equivalence classes of solvable degree-$7$ trinomials in $\mathbf{Q}[x]$ with terms of degrees $0,n,7$ and the set of rational points on a certain Riemann surface $X$. The information Guralnick-Shareshian use about this Riemann surface is that it admits a degree-$120$ branched cover $X\to\mathbf{P}^1$ with monodromy group $S_7$ and three branch points, two of which have branch cycles $(1,n+1)$ and $(1,2,\dots,7)$. I haven't thought about how to prove this bijection. Anyway, assuming it's correct, then your question amounts to asking whether any of six specific Riemann surfaces have any rational points.

Added later: as Noam notes in his comment, there are really three Riemann surfaces, because the ones for $n$ and $7-n$ are the same. By my computations, all of these Riemann surfaces $X$ are defined over $\mathbf{Q}$, and the genus of $X$ is $17$, $10$, or $16$ according as $n=1$, $2$, or $3$. So the original question amounts to asking whether there are rational points on any of three specific curves over $\mathbf{Q}$, having genera $17$, $10$ and $16$.

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Three surfaces, actually: $n$ and $7-n$ are equivalent. –  Noam D. Elkies Nov 3 '13 at 0:24
Noam, is your bijection proved in the literature or on your website? (The bijection in my post -- I can prove the bijection in your comment.) –  Michael Zieve Nov 3 '13 at 0:27
Nils Bruin and Noam D. Elkies: Trinomials $ax^7+bx+c$ and $ax^8+bx+c$ with Galois Groups of Order 168 and 8*168, Lecture Notes in Computer Science 2369 (proceedings of ANTS-5, 2002; C.Fieker and D.R.Kohel, eds.), 172-188. –  Noam D. Elkies Nov 3 '13 at 0:33
Igor Rivin has extended the search radius for $x^7+ax+b=0$ irreducible over $\Bbb Q(x)$ for $|a,b|<1000$ and found no solvable case. –  Tito Piezas III Nov 3 '13 at 18:18
By the way, I found by accident that there is an irreducible quadrinomial that is solvable, namely $x^7-7x^4-14x^3-7=0$. –  Tito Piezas III Nov 3 '13 at 18:51

The good news is that in this book:

Generic Polynomials: Constructive Aspects of the Inverse Galois Problem By Christian U. Jensen, Arne Ledet, Noriko Yui (pp. 52 and up).

There is a complete criterion, as follows:

$x^7 + a x + b$ is solvable (we assume it's irreducible) if and only if the polynomial (which I am intentionally leaving in Mathematica input form, so you can type it in yourself).

P35[a_, b_, x_] := x^35 + 40 a x^29 + 302 b x^28 - 1614 a^2 x^23 + 2706 a b x^22 + 3828 b^2 x^21 - 5072 a^3 x^17 + 2778 a^2 b x^16 - 18084 a b^2 x^15 + 36250 b^3 x^14 - 5147 a^4 x^11 - 1354 a^3 b x^10 - 21192 a^2 b^2 x^9 - 26326 a b^3 x^8 - 7309 b^4 x^7 - 1728 a^5 x^5 - 1728 a^4 b x^4 + 720 a^3 b^2 x^3 + 928 a^2 b^3 x^2 - 64 a b^4 x - 128 b^5

has factorization pattern [degrees of irreducible factors] one of (14, 21), (7, 7, 7, 21), (7, 7, 7, 14), (7, 7, 7, 7, 7)

The bad news is that I have run this for all pairs (a, b) where both coordinates are between -1000 and 1000, and there is not a single success.

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Thank you! Playing with rational $a,b$, I found that $P35$ factors into (15,20) when $$x^7+nx+(n+1)=0$$ Unfortunately, that septic is reducible, having a linear and sextic factor. The sextic is not generally solvable, but some $n$ are, like $n=7$ and $n=-203/320$ yielding $$x^7+7x+8=0$$ $$x^7-\frac{203}{320}x+\frac{117}{320}=0$$ and the sextic factors into cubics over the extension $\sqrt{-7}$ and $\sqrt{-35}$, respectively. –  Tito Piezas III Nov 3 '13 at 18:01
@Tito: these latest reducible examples you found are of a different flavor than your earlier ones, which might shed light on the problem. Their Galois group is the wreath product of $S_3$ by $S_2$. There are only finitely many septic trinomials of the form $x^7+ax+b$ having this Galois group (up to composing with scalar multiples on both sides). This is because the curve parametrizing such examples has genus $2$. I think the curve must be special in some way in order to have two such "small" points as you found. –  Michael Zieve Nov 3 '13 at 23:21
@Michael: I used the (15,20) factorization of $P35$ to find those two. There are other $a,b$, but the sextic factor of the septic is reducible over $\Bbb{Q(x)}$. –  Tito Piezas III Nov 3 '13 at 23:31
There is another one: $x^7-14x-13=0$. The sextic factor is still solvable though does not factor over a square root extension. It does over a cubic subfield, so I assume this has a different Galois group from those two above. –  Tito Piezas III Nov 3 '13 at 23:45
@Tito: yes, that one has Galois group of order $24$. In this case the curve parametrizing polynomials $x^7+ax+b$ with this Galois group has genus $12$, so again I wouldn't expect to see such small points. It would be really interesting if you could find lots of such examples. The reason is that this seems to be a novel way to produce curves over $\mathbf{Q}$ of genus $g>1$ which have "many" rational points. Any such curve has only finitely many points, and it is conjectured that there is a uniform bound $C(g)$ on the number of points. Your examples could shed light on $C(g)$. –  Michael Zieve Nov 4 '13 at 0:09