Number Theory – Transforming the Kondo Quintic 5T2 into the Lehmer Quintic 5T1

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I. Kondo-Brumer quintic

The deceptively simple solvable quintic,

$$x^5 + (a – 3)x^4 + (-a + b + 3)x^3 + (a^2 – a – 1 – 2b)x^2 + b x + a=0$$

is quite important for imaginary quadratic fields. For example, let $a=1, b=0,$ and it becomes,

$$x^5-2x^4+2x^3-x^2+1=0$$

which is a Weber class polynomial and a solution is,

$$x= -e^{-2\pi i/48}\frac{\sqrt{2}\,\eta(2\tau)}{\quad\eta(\tau)}\approx -0.5764$$

with Dedekind eta function $\eta(\tau)$ and $\tau=\frac{1+\sqrt{-47}}2.$ Other $\sqrt{-d}$ with class number $h(d) = 5m$ are also possible.

II. Lehmer quintic

This is given by,

$$y^5 + n^2y^4 – (2n^3 + 6n^2 + 10n + 10)y^3 + (n^4 + 5n^3 + 11n^2 + 15n + 5)y^2 + (n^3 + 4n^2 + 10n + 10)y +1=0$$

and, on the other hand, is quite important for real fields. For example. let $n=-1$ then,

$$y^5+y^4-4y^3-3y^2+3y+1=0$$

and a solution is,

$$y = 2\cos(2\pi/11) \approx 1.6825$$

Of course, other $p=5m+1$ are also possible.

III. Subset of a, b

The two quintics seem to be radically different, the first generally a $5T2$ with order $10$, and the second a $5T1$ with order $5$. But it turns out (using a subset of $a,b$) we can transform the former into the latter. Their discriminants are,

$$D_1 = a^2\,\big( 4a – 91 a^2 + 40 a^3 + 4 a^4 – 4 a^5 – 2 (7 a + 17 a^2 – 12 a^3) b + (1 – 30 a + a^2) b^2 – 4 b^3 \big)^2$$

$$D_2 = (7 + 10 n + 5 n^2 + n^3)^2\,(25 + 25 n + 15 n^2 + 5 n^3 + n^4)^4$$

My first thought was to turn the cubic in $b$ into the square $c^2$ (so an elliptic curve). What I found to my surprise was it can also be solved as the $4$th power $c^4$ (a superelliptic curve),

$$\small{(4a – 91 a^2 + 40 a^3 + 4 a^4 – 4 a^5) – 2 (7 a + 17 a^2 – 12 a^3) b + (1 – 30 a + a^2) b^2 – 4 b^3 = \color{blue}{c^4}}$$

with solution,

$$\begin{align}
a &= -(7 + 10 n + 5 n^2 + n^3)\\
b &= -20 – 5 n + 10 n^2 + 12 n^3 + 5 n^4 + n^5\\
c &=\, 25 + 25 n + 15 n^2 + 5 n^3 + n^4
\end{align}$$

Incidentally, the $a,b,c$ have the nice linear relationship,

$$3a-b+nc = -1$$

So the new discriminant of the one-parameter Kondo quintic for these special $a,b$ neatly becomes,

$$D_1(\text{new}) = (7 + 10 n + 5 n^2 + n^3)^2\,(25 + 25 n + 15 n^2 + 5 n^3 + n^4)^{\color{blue}8}$$

and has an $8$th power.

IV. Tschirnhausen

To complete the transformation, there is a quartic Tschirnhausen between the one-parameter Kondo quintic (in $x$) and the Lehmer (in $y$) given by,

$$ax = y^4 – (4 + 3 n) y^3 – (8 + 13 n + 13 n^2 + 6 n^3 + n^4) y^2 + (30 + 60 n + 54 n^2 + 28 n^3 + 8 n^4 + n^5) y + (2 + n)$$

with $a$ as above. Incidentally, if we are to express the quartic Tschirnhausen in the form,

$$ax = y^4+py^3+qy^2+ry+s$$

then the coefficients $a,p,q,r,s$ also have a nice relationship,

$$3a – p + q s + r + s = – 1$$

V. Questions

Am I correct in assuming this one-parameter Kondo quintic is now a $5T1$ with order $5$?
More importantly, since I was trying to make the cubic in $b$ into a square, a mystery to me is why was it possible and easier to make it a 4th power and solve it as a superelliptic curve?

Best Answer

Regarding your first question, the answer is yes. In fact, your quartic Tschirnhausen formula shows that. In particular, this shows that the one-parameter Kondo quintic has a root in the splitting field of the Lehmer quintic, which is a degree $5$ Galois extension of $\mathbb{Q}(n)$. Because it's Galois, it's normal, and since it's normal, the one-parameter Kondo quintic must split in this field. This implies that the Galois group of the one-parameter Kondo quintic is $5T1$. (If you want, you can also compute the Galois group of a polynomial over a one-variable function field with Magma and independently check this.)

Your second question is a bit trickier - the why questions in mathematics can be hard to answer. If you look at the elliptic curve that comes from setting the discriminant to be a square, namely $$ y^{2} = x^{3} + (1-30a+a^{2})x^{2} - 8(7a + 17a^{2} - 12a^{3})x + 16(4a - 91a^{2} + 40a^{3} + 4a^{4} - 4a^{5}), $$ then this elliptic curve has $(n^{4} + 5n^{3} + 15n^{2} + 25n + 25)^{5}$ as a factor of its discriminant, which is an unusually high power. Moreover, at this place of bad reduction (which corresponds to choosing $n$ in $\mathbb{Q}(\zeta_{5})$), the reduction is split multiplicative of type $I_{5}$. In particular, the component group $E/E_{0}$ is cyclic of order $5$. The solution $a$, $b$, $c$ you specified gives an integral point $P$ on this elliptic curve (that is, the coordinates of $P$ are polynomials in $n$, rather than rational functions). It is somewhat remarkable in that $P$, $2P$ and $3P$ are all integral points as well. One consequence of the reduction properties at this place is that $n^{4} + 5n^{4} + 15n^{3} + 25n + 25$ divides the $y$-coordinate of $mP$ to the power $2$ whenever $m \equiv 1, 4 \pmod{5}$. This general phenomena among other things forces $(n^{4} + 5n^{4} + 15n^{3} + 25n + 25)^{2}$ to divide the $y$-coordinate of $P$. These things are sensitive to the particular set up. I'm not sure if you first tried to find a parametric family of $5T1$ specializations of the Brumer-Kondo quintic by setting $a = 7 + 10n + 5n^{2} + n^{3}$ (the negative of what you chose). You may have quickly figured out that this doesn't work: in this situation the Mordell-Weil rank of the elliptic curve over $\mathbb{Q}(n)$ is zero, so you do not get a parametric family of $5T1$ specializations.

Related Solutions

Solvable Octic Equation – Analysis of x^8-x^7+29x^2+29=0

To answer your second question, there are soluble octics with the same Galois group involving other $p$th roots of unity. Take $p\equiv 1\mod 7$, and $K={\mathbb Q}(\alpha)$ the unique degree 7 extension of ${\mathbb Q}$ in ${\mathbb Q}(\zeta_p)$. E.g. take $\alpha=\sum_i \zeta_p^i$ where $i$ ranges over all seventh powers in ${\mathbb F_p}$.

If $f(x)$ is the minimal polynomial of $\alpha$ (degree 7), then $f(x^2)$ is the minimal polynomial of $\sqrt\alpha$, which defines, generally, a 'random' quadratic extension of $K$. That is, its Galois group over ${\mathbb Q}$ is $$ G=C_2\wr C_7\cong C_2^7:C_7. $$ Viewing $C_2^7$ as a 7-dimensional representation of $C_7$ over ${\mathbb F}_2$, it decomposes as a 1-dimensional (trivial) representation plus two distinct 3-dimensional ones. (The reason for this is that $2^3\equiv 1\mod 7$.) Factor out $C_2^4\triangleleft G$, which is one of those plus the trivial one. This gives a Galois group $C_2^3:C_7$ that you are after, and a subgroup $C_7$ in it cuts out the required octic field.

Here is a Magma code that can be used in the Magma online calculator to get such an octic:

p:=43;        // or some other p = 1 mod 7

K<z>:=CyclotomicField(p);
alpha:=&+[z^i: i in [1..p] | IsPower(GF(p)!i,7)];    

R<x>:=PolynomialRing(Rationals());
f:=Evaluate(MinimalPolynomial(alpha),x^2);
K:=NumberField(f);
assert exists(a){a: a in ArtinRepresentations(K) | #Kernel(Character(a)) eq 16};
F:=Field(Minimize(a));
DefiningPolynomial(F);

You can also stick in a Tschirnhaus transformation, say,

alpha:=alpha^3+alpha+1;

in the 5th line to vary the generator of $K$ - in this way you get all possible $C_2^3:C_7$-extensions involving $p$th roots of unity.

For your questions in the comments, the roots might be real or complex, and the constant term may or may not be a square - this depends on whether $\alpha$ is chosen to be totally real, and on the way 'Minimize' works; you can always use an additional Tschirnhaus transformation to modify the final output or Pari's 'polredabs' function to try and get the coefficients smaller.

I do not know the reason why for $p=29$ there is such an elegant octic, this is very curious. It is a bit like the Trinks polynomial $x^7-7x+3$ with Galois group PSL(2,7), and I wonder whether simple polynomials having interesting Galois group is such a statistical blip, or there is a reason behind it.

Rogers-Ramanujan Continued Fraction – Parameterizing Fermat Quintic

(Courtesy of a comment by Nemo who suggested Huber's paper.)

Part I. $x^5+y^5 = 1$

In "A Theory of Theta Functions to the Quintic base", Tim Huber defines four theta functions which can be ultimately expressed in terms of the Rogers-Ramanujan identities. Define $q=e^{2\pi i z}$ and, $$P(z):=q^{11/60}H(q)=q^{11/60}\prod_{n=1}^\infty \frac1{(1-q^{5n-2})(1-q^{5n-3})}$$ $$Q(z):=q^{-1/60}G(q)=q^{-1/60}\prod_{n=1}^\infty \frac1{(1-q^{5n-1})(1-q^{5n-4})}$$ and the Rogers-Ramanujan continued fraction, $$R(z)=q^{1/5}\frac{H(q)}{G(q)}=\cfrac{q^{1/5}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\ddots}}}}$$

then Huber's four functions in simplified form are,

$$\begin{aligned} a(\tau) &=\eta^{2/5}(\tau)\;P(\tau)\\[2mm] b(\tau) &=\eta^{2/5}(\tau)\;Q(\tau)\\[2mm] c(\tau) &= 5^{1/4}\phi^{1/2}\,\eta^{2/5}(5\tau)\;P\big(\tfrac{-1}{5\tau}\big)\\[2mm] d(\tau) &= \frac{5^{1/4}}{\phi^{1/2}}\,\eta^{2/5}(5\tau)\;Q\big(\tfrac{-1}{5\tau}\big)\end{aligned}$$ with golden ratio $\phi$. These obey, $$\Big(\frac{a\,\phi}{b}\Big)^5+\Big(\frac{c}{b}\Big)^5 = 1\tag1$$ $$-\Big(\frac{a}{b\,\phi}\Big)^5+\Big(\frac{d}{b}\Big)^5 = 1\tag2$$ It then follows that the ratio of $a,b$ as well as $c,d$ are Rogers-Ramanujan cfracs, $$\frac{a(\tau)}{b(\tau)} = R(\tau)\tag3$$ $$\frac{c(\tau)}{d(\tau)} = \phi\, R\big(\tfrac{-1}{5\tau}\big)\tag4$$ so $R(\tau)$ in a way can parameterize the Fermat quintic. Some manipulation will also show that, $$\frac{c(\tau)}{d(\tau)} = \phi\,\frac{1-\phi\, R(5\tau)}{\phi+R(5\tau)}\tag5$$

Part II. $x^5+y^5+z^5 = 1$

It turns out that just like, $$x^3+y^3=1$$ can be solved by the cubic continued fraction $C(q)$ and $C(q^3)$, its quintic analogue $$x^5+y^5+z^5 = 1$$ has a beautiful solution using the Rogers-Ramanujan cfrac $R(q)$ and $R(q^5)$. This is given by,

$$\alpha^5\phi^5+\beta^5\phi^5+\alpha^5\beta^5 = 1\tag6$$

where, $$\alpha = R(q),\quad\beta = \frac{1-\phi R(q^5)}{\phi+R(q^5)}$$ and which can be derived by combining $(1),(2)$.

Best Answer

Related Solutions

Solvable Octic Equation – Analysis of x^8-x^7+29x^2+29=0

Rogers-Ramanujan Continued Fraction – Parameterizing Fermat Quintic

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