“Solvable Brioschi Quintics, other One-Parameter Forms, and the j-function”
Titus Piezas III
Abstract: By a series of quadratic Tschirnhausen transformations, the general quintic can be transformed into the solvable De Moivre quintic. The last step, naturally enough, involves solving a resolvent sextic which turns out to be: a) a polynomial identical to a formula for the j-function in terms of Dedekind eta quotients, and b) the Jacobi sextic in disguise. As a side effect, it gives the complete parametrization of solvable Brioschi quintics with rational coefficients. In addition, solutions of other one-parameter quintics will also be discussed.
Contents:
I. Introduction
II. Transforming the Brioschi and De Moivre Quintics
III. Connection to the j-function and Continued Fractions
IV. Jacobi Sextic
V. Bring-Jerrard Quintics
VI. Other One-Parameter Quintics
VII. Conclusion
In the paper “Deriving The Bring-Jerrard Quintic Using A Quadratic Transformation” [1] by this author, we continued the objective of looking for transformations of minimum degree to reduce the general quintic to simpler forms. It was found that in addition to the traditional quartic Tschirnhausen transformation, cubic or even quadratic ones could be used. For the latter, we used the sequence of quadratic transformations to obtain,
General → Principal → Brioschi → Bring-Jerrard
In this paper, it will be established that the following sequence,
General → Principal → Brioschi → De Moivre
is also possible with the difference that the last step involves solving a sextic which in general will not be solvable in radicals. However, this sextic surprisingly turns out to be of a special kind that is solvable in functions beyond the radicals, namely hypergeometric or elliptic ones.
The Brioschi quintic, named after Francesco Brioschi (1824-1897) is the one-parameter form,
w5 -10cw3 + 45c2w - c2 = 0
with polynomial discriminant D = 55c8(-1+1728c)2 and which is important to Klein’s solution of the general 5th degree equation in terms of hypergeometric functions. The explicit algorithm to transform the principal quintic and derive the parameter c in terms of its coefficients was given in [1]. On the other hand, we have the De Moivre quintic,
x5 + 5x3 + 5x = t
named after Abraham De Moivre (1667-1754) which can be generalized into a family of equations given by,
(x + Ö(x2 + 4a))k + (x - Ö(x2 + 4a))k = 2kt
which is solvable in radicals for all a,t and for all positive integer k. The explicit equation above is the case k = 5 where we can assume a = 1 without loss of generality. We wish to transform the Brioschi into the De Moivre using the transformation,
y = (mw+n)/(c-1w2-3)
with unknowns m,n and c as a fixed parameter. Using resultants to eliminate w, we get,
y5 + f3y2 + f4y + f5 = 0
where the fi explicitly are,
f3 = 5c(8n3 + mn2 + m3c + 72m2nc) /(-1+1728c)
f4 = 5c(-n4 + m3nc + 18m2n2c + 27m4c2) /(-1+1728c)
f5 = c(n5 - 10m2n3c + m5c2 + 45m4nc2) /(-1+1728c)
Next, we transform the De Moivre into principal form using the transformation y = x2+x+2 to get,
y5 - 5(2+t)y2 + 15y + (-t2+9t-22) = 0
By equating coefficients between the two we have a system of three equations in three unknowns m,n,t. By resolving this into a single equation in n, we end up with a 24th degree equation with the sextic factor,
-243(27+n) + (72-12n+n2)(396+42n+n2)2c = 0
We can reduce the size of the coefficients by letting n = 3u-27,
B6 := -u + (125-22u+u2)(-1-4u+u2)2c = 0
and we have the resolvent sextic B6 in the new variable u. This has discriminant,
D = 55(-1+1728c)4/c6
a form which is perhaps not surprising. If one can solve B6 then the Brioschi can be turned into the solvable De Moivre. It is easy to prove that for solvable quintics, B6 is also solvable. With appropriate roots wi of the Brioschi and yi of the De Moivre,
y1 = (mw1+n)/(c-1w12-3)
y2 = (mw2+n)/(c-1w22-3)
the unknowns m,n obviously can be solved in terms of the roots wi and yi by Gaussian elimination. If the roots are expressible in radicals, then so will be the unknowns. Furthermore, since irreducible but solvable quintics with rational coefficients do not need the cube roots of unity, then B6 must always be reducible for these cases, with either a linear (or perhaps quadratic) factor. If we then look at B6 another way, it is the complete formula or parametrization for c. And since c is conveniently only linear, it can be solved as c = u/k1 with k1 = (125-22u+u2)(-1-4u+u2)2. Or equivalently, in a more aesthetic form,
Theorem 1. The Brioschi quintic w5 -10cw3 + 45c2w - c2 = 0 with rational coefficiients and solvable in radicals have the complete parametrization,
c = 1/(1728+v)
where v = (u2-10u+5)3 /u.
Example, let u = 1, then c = 1/1664 and so on for all u (other than that which causes division by zero) proving that, like the Bring-Jerrard and not counting scaled versions, there is an infinite number of this form with rational coefficients solvable in radicals.
III. Connection to the j-function and Continued Fractions
In addition to giving the parametrization for the Brioschi, the resolvent sextic B6 is yet another way to look at the connection of quintics to the j-function. Note that k1 - 1728u is a cube, or,
(125-22u+u2)(-1-4u+u2)2 -1728u = (u2-10u+5)3>
so B6 can also be expressed as,
(u2-10u+5)3c + (-1+1728c)u = 0
Defining the variable V = (-1+1728c)/c, then,
V = -(u2-10u+5)3/u
and which explains the alternative and more aesthetic formula for c (with v = -V). However, the above, after a minor change in sign of the ui is analogous to one of the formulas (of order 5) for the j-function j(t)!
Fm5: = j(t) = (x2+10x+5)3/x
where all six roots x are in terms of the elliptic Dedekind eta function h(t). Let argument t be the half-period ratio and defining the following six functions,
f55(t) = 51/2 h(5t)/h(t), f50+k(t) = exp(pik/6) h(t1)/h(t)
where t1 = (t+2k)/5 for k = 0 to 4, then the six roots are given by x = fi6 for i = 50 to 55. For example, let t = (1+Ö-1)/2, so j(t) = 123. Two of the roots of Fm5 are x1 = -2+Ö5 and x2 = -2-Ö5 which are identical to x1 = (f50(t))6 and x2 = (f55(t))6. Theoretically, by equating V with j(t), then the ui can also be expressed in terms of the Dedekind eta function, though for arbitrary V presumably it would be tricky to find the appropriate argument t to use. However, since B6 is just the Jacobi sextic in disguise (as we will see later), there should be a slicker way to express its roots in terms of elliptic functions.
Another connection between quintics and the j-function can be given by using a second formula (of order 25),
Fm25 := j(t) =
-(r20-228r15+494r
where one of the ri can be given by,
r -1- r = h(t/5)/h(5t) + 1
Fm25 is called the icosahedral equation (Duke, p.3). This can be derived from Fm5 using the simple substitution,
x = -125r5/(r10+11r5-1)
Note that the numerator and denominator (with a constant) of Fm25 can be combined to form a square,
1728r5(r10+11r5-1)5 + (r20-228r15+494r10+228r5+1)3 = (r30+522r25-10005r20-10005r10-522r5+1)2
These in fact are invariant polynomials of the icosahedral group and they appear in the hypergeometric formula for the Brioschi quintic w5 –10Zw3 + 45Z2w – Z2 = 0. This is given by,
y5(y10+11y5-1)5-(y30+522y25-10005y20-10005y10-522y5+1)2Z = 0
which sometimes is also called the icosahedral equation. The solution y of this 60th degree equation can then be given in terms of the hypergeometric function 2F1 as y = h1/h2 (Weisstein, “Quintic Equation”) where,
h1 = Z-1/60 2F1(-1/60, 29/60, 4/5, 1728Z)
h2 = Z11/60 2F1(11/60, 41/60, 6/5, 1728Z)
with 2F1 implemented in Mathematica as Hypergeometric2F1[n1, n2, n3, n4]. With y known, a root w of the Brioschi can then be solved by using a quadratic relation (in w) between the roots wi and yi,
w2y(y10+11y5-1) = (y6+2y5-5y4-5y2-2y+1)2Z
where the right hand side is a polynomial invariant of the regular octahedron (Dickson, p. 239). This relation can be derived by eliminating the parameter Z between the Brioschi and the icosahedral equation and after some tedious simplication.
Incidentally, the formula Fm25 also has a connection to the famous Rogers-Ramanujan continued fraction R(q). For an illustration and details, see http://mathworld.wolfram.com/Rogers-RamanujanContinuedFraction.html. In his first letter to Hardy, Ramanujan gave a beautiful example of R(q) with q = e-2p which was the q-analog of the continued fraction for the golden ratio. Amazingly, it had the exact value,
R(e-2p) = - f + Ö(f+2)
where f is the golden ratio. It turns out that for the broad class q = e2pit = exp(2pit), the value of R(q) can be evaluated by Fm25 and can be given simply as R(q) = r. (Duke) In other words, if the value of j-function j(t) is known, then by solving for r one of its values will be R(q)! For example, R(e-2p) used the argument t = Ö-1 = i, so j(t) = 123. Substituting this into Fm25, one of its factors will be a quartic, with one root given by,
r1 = - f + Ö(f+2)
precisely the value Ramanujan found.
The Jacobi sextic is given by,
J6 - 10zJ3 + 12z2J + 5z2 = 0
with parameter z, discriminant 2123655(-1+z)2z10 and which Kronecker found could be solved by means of elliptic functions (Rutgers, p.107-108). The equation derived in the previous section,
V = -(u2-10u+5)3/u
is just this sextic in disguise. By a change of variables V = 123z and u = J3/z, the resulting 18th degree equation factors as,
(J6 - 10zJ3 + 12z2J + 5z2)*P(J) = 0
with the first factor obviously the Jacobi sextic and P(J) as a 12th degree polynomial.
In Weisstein’s entry on the “Quintic Equation”, we find the statement that in 1885, Runge, Cadenhad, and Young showed that all irreducible but solvable Bring-Jerrard quintics x5 + ax + b = 0 with rational coefficients have the form,
x5 + 5u4(4v+3)/(v2+1)x + 4u5(2v+1)(4v+3)/(v2+1) = 0
and that in 1994, Spearman and Williams also showed that they have the form,
x5 + 5u4(3-4aw)/(w2+1)x – 4u5(11a+2w)/(w2+1) = 0
for a = +/-1. (Minor changes have been made in the variables.) So which is it? Note that by letting a = -1 the two x terms are essentially identical. It turns out the parametrizations are just two sides of the same coin. Using the latter, choose a = -1 without loss of generality and equating coefficients with the generic Bring-Jerrard,
5u4(3+4w)/(w2+1) = a
- 4u5(-11+2w)/(w2+1) = b
Eliminating w between the two using resultants, and letting u = x/2, we get a simple sextic resolvent,
ax6 + 5bx5 - 4a2x2 + 4abx – 5b2 = 0
which has discriminant D = b2(256a5+3125b4)3/a10. Since the degrees of a,b are conveniently small, we can solve for b,
10b = 4ax+5x5 +/- xÖ((5x4-4a)(5x4+16a))
and we have a 2-parameter solution for b. Note that by scaling the Bring-Jerrard, we can effectively set u = 1, thus x is simply x = 2 and we can have the simplified solution,
5b/4 = a+20+/- 2Ö((20-a)(5+a))
If we wish to have b rational, then obviously the objective is to set y2 = (20-a)(5+a) and we can come up with a second theorem.
Theorem 2. The Bring-Jerrard quintic z5 + au4z + bu5 = 0 with rational coefficients and solvable in radicals must satisfy the quadratic curve,
y2 = (20-a)(5+a)
for rational a,y and where b is a function of the two, given by b =(4/5)(a+20+/-2y).
Letting a = 5(4v+3)/(v2+1) and substituting into the simplified solution for b, using the negative case of the square root, we get,
b = 4(2v+1)(4v+3)/(v2+1)
which is the Runge-Cadenhad-Young parametrization. But by using the positive case, then,
b = - 4(-11+2v)/(v2+1)
which is the Spearman-Williams with a = -1, proving, as was said, that they were just two sides of the same coin.
In contrast to the first two forms, the Euler-Jerrard form R3: = x5 + E1x2 + E2 = 0 that is solvable and with rational coefficients only has a finite number of examples. In a 1996 paper, Spearman and Williams (and independently in 1999 by Elkies) established that there are exactly 5 of these quintics. To quote Elkies, “…these are parametrized by the elliptic curve,
C20 : y2 + xy +y = x3 + x2 + 35x - 28
which, over Q, has rank zero so there are only finitely many equivalence classes of the Euler-Jerrard trinomials with solvable Galois group.” These are,
x5 - 2x2 -1/5 = 0, x5 - 100x2 - 1000 = 0, x5 - 25x2 - 300 = 0,
x5 - 5x2 - 3 = 0, x5 - 5x2 + 15 = 0,
with the last two generating the same field. And finally, the last of the four known one-parameter reduced forms,
R4: = w5 – 5gw3 + 10g2w – g2 = 0
It is not known if R4 is solvable for an infinite number of rational g. However, heuristic evidence suggests it may not be the case. For one, while one kind of quadratic Tschirnhausen transformation changes it into the Bring-Jerrard form (discussed in [1]), another kind, given by,
y = (8aw+b)/(g-1w2-2)
with a change of variable g = (1-n2)/32 and b = 2a(-1+n), transforms it into the Euler-Jerrard form,
y5 – 5(n+1)(n-1)2a3 y2 + 2(n+1)(3n-1)(n-1)3a5 = 0
As was pointed out, there are only five solvable quintics of this form and by equating coefficients, one has to find n such that n2 is rational. Curiously, one can do this for the two R3 which generate the same field, namely,
x5 - 5x2 - 3 = 0, x5 - 5x2 + 15 = 0
and which are given by n1 = 7/9, a1 = 9/4 and n2 = -7/9, a2 = 9/8, respectively. This yields the same R4 with parameter g = (1-n2)/32 = 1/81. Looking for other rational g using a general sextic resolvent which has a linear factor whenever a quintic is solvable, only three g were found below a certain bound, namely, g = 1/16, 1/27, 1/81 (which obviously are all powers) with discriminants D1 = 2-34, D2 = 3-28, D3 = 3-3872. Whether there are more remains to be seen.
In this paper, an overview of some results regarding one-parameter reduced quintic forms was given as well as various parametrizations. We saw that they had interesting connections to elliptic and hypergeometric functions, continued fractions, to each other, and most especially to the j-function. As an end note to this paper, it should be pointed out that this curious function, which has a profound connection to the Monster group, has more formulas.
We discussed two of them here, but at least eight could be given. Surprisingly, these can be given explicitly for order n such that n-1 divides 24, hence for n = 2,3,4,5,7,9,13,25. In fact, going beyond the quintic, the one for order 7,
Fm7 := j(t) = (x2+5x+1)3(x2+13x+49)/x
with one root given as x = (f77(t))4 where f77(t) = 71/2 h(7t)/h(t), behaves in the same manner as Fm25, namely one can combine the numerator and denominator to form a perfect square,
1728x-(x2+5x+1)3(x2+13x+49) = (x4+14x3+63x2+70x-7)2
These are polynomial invariants associated with the heptagon and were used by Klein in his approach to solve the general 7th degree equation. Then there’s the one for n = 13…
Seems there’s much more to the theory of equations than meets the eye.
--End--
© Titus Piezas III
April 22, 2006
titus_piezasIII@yahoo.com (Pls. remove “III”)
www.oocities.org/titus_piezas/
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