I. Four quintics?

The general quintic can be transformed in radicals to at least three one-parameter forms. For simplicity, assume this free parameter to be some generic "alpha". Hence,

$$x^5-10\alpha x^3+45\alpha^2x-\alpha^2=0\tag1$$ $$x^5-5\alpha x -\alpha = 0\tag2$$ $$x^5+5\sqrt{\alpha}\, x^2 -\sqrt{\alpha} = 0\tag3$$

which are the Brioschi, Bring-Jerrard, and Bring-Euler quintics, respectively. Naturally, these are $5T5$ with order $5!=120$. Their discriminants are,

\begin{align} d_1 &= 5^5\,(1-1728\alpha)^2\,\alpha^8\\ d_2 &= 5^5\,(1-256\alpha)\,\alpha^4\\ d_3 &= 5^5\,(1-108\alpha)\,\alpha^2 \end{align}

We can do a minor transformation to get their respective variants,

$$y^3(y^2+5y+40) = j_1\tag4$$ $$y(y-5)^4 = j_2\tag5$$ $$y^3(y-5)^2 = j_3\tag6$$

with discriminants,

\begin{align} D_1 &= 5^5\,(j_1-1728)^2\,{j_1}^2\\ D_2 &= 5^5\,(j_2-256)\,{j_2}^3\\ D_3 &= 5^5\,(j_3-108)\,{j_3}^3 \end{align}

However, it seems we are missing one quintic with discriminant $D_4$,

$$D_4 = 5^5(j_4-64)^a\,{j_4}^b\quad$$

which has level $p=6,7,8$ versions discussed in this MO post. (The octics in that post, after tedious manipulation, can be reduced to their deg-$7$ resolvents.)

II. Eta quotients and the Monster

Given Dedekind eta function $\eta(\tau)$, define the four eta quotients which in fact are the first four McKay-Thompson series 1A, 2A, 3A, 4A of the Monster,

\begin{align} \quad j_1 &=\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{8}+2^8 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{16}\right)^3 \\ \quad j_{2} &=\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{12}+2^6 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{12}\right)^2 \\ \quad j_{3} &=\left(\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{6}+3^3 \left(\frac{\eta(3\tau)}{\eta(\tau)}\right)^{6}\right)^2 \\ \quad j_{4} &=\left(\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{4} + 4^2 \left(\frac{\eta(4\tau)}{\eta(\tau)}\right)^{4}\right)^2 = \left(\frac{\eta^2(2\tau)}{\eta(\tau)\,\eta(4\tau)} \right)^{24} \end{align}

where $j_1$ is just the j-function. Let $\tau = \sqrt{-d}$ or $\tau =\frac12+ \sqrt{-d}$ such that the $j_i$ are radicals.

III. Question 1

Let $j_i(\tau)$ be the radicals defined as above. Then is it true that for the quintics,

$$y^3(y^2+5y+40) = j_1\tag4$$ $$y(y-5)^4 = j_2\tag5$$ $$y^3(y-5)^2 = j_3\tag6$$

the Galois group is now solvable, and the $y$ are solvable in radicals? For example, let,


which appears in Ramanujan's pi formula in the title. So, then a change of variable to $z$,

$$y(y-5)^4 = 396^4$$ $$z(z^4-5) = 396$$ $$z^5-5z-396 = 0$$

which is a solvable Bring-Jerrard quintic. (In fact, it factors).

IV. Question 2

However, there is still $j_4$. To recall, for level $p=7$ (in this MO post), the one-parameter formulas are complete for all four.

Q: So does this imply the general quintic can be reduced to a fourth one-parameter form (still unknown) and analogous to the three above?

V. Sextic version?

The "missing" quintic may have a sextic version (also with order $5! = 120$) and is given by,

$$j_4 =\frac{(x + 1)^5 (x + 5)}x\tag7$$

with expected discriminant,

$$\text{Discrim}_4 = 5^5\,(j_4-64)^2\,{j_4}^4$$

So what we're looking for might be its quintic subextension. But I am uncertain how to generate the correct quintic from this sextic. Ironically, the octics were easier. (Note: The correct quintic must have order $120$ for general $j_4$ but be solvable when $j_4 = j_4(\tau)$ as defined in Section II.)

  • $\begingroup$ In fact, years ago, I found a fourth one-parameter form the general quintic can be reduced to in this MO post. But it doesn't seem to have the correct discriminant $D_4$, nor is it solvable in radicals if its parameter is equated to some multiple of $j_4$. $\endgroup$ Commented May 9, 2023 at 5:47
  • $\begingroup$ I think you'll find the "missing" quintic is just a quadratic pullback of the Bring-Jerrard one, so in terms of transformation in radicals, there's nothing new to be seen here. $\endgroup$ Commented May 9, 2023 at 15:44
  • $\begingroup$ @JoachimKönig A quintic I derived from the sextic $(7)$ is $${j_4}^2 + (5 x^2 + 15x + 8) j_4 - x^5 = 0$$ with discriminant $$\text{Discrim} = 5^5(j_4 - 1)^2(j_4 - 64)^2 {j_4}^4$$ It has group $120$ but is solvable when $j_4 = j_4(\tau)$. So almost what I was looking for, other than the extra square factor. Still checking out other avenues. $\endgroup$ Commented May 9, 2023 at 16:19
  • $\begingroup$ This quintic is given by $f(\frac{-\alpha}{(\alpha-64)^2},X)$, where $f(\alpha,X)$ is the Bring-Jerrard form above. My point was that this shape (without the concrete coefficients) could be seen a priori just by comparing the combinatorial data (=ramification types) of the two. Of course the quadratic transform can be inverted in radicals, so this is not a "new" form (or if you'd count it as one, there would be infinitely many others). That the polynomial discriminant happens to have an extra divisor is irrelevant for the main question, since the quintic field is given uniquely by the sextic. $\endgroup$ Commented May 10, 2023 at 7:44
  • $\begingroup$ @JoachimKönig: Anyway, my other objective was to find a simple transformation (if any) that can reduce the general quintic to the one-parameter $j_4$ quintic. I found that by passing through the Brioschi (instead of the Bring-Jerrard), then the transformation was simpler than I thought. $\endgroup$ Commented May 10, 2023 at 9:11

1 Answer 1


(This addresses Question 2.)

I. In general, it asks if we can reduce the general quintic to a one-parameter form with a specified discriminant $D_i$ that is different from the other well-known forms. These $D_i$ involve the integers $\color{blue}{1728, 256, 108, 64}$, numbers which appear per level in Ramanujan's theory of elliptic functions to alternative bases, and his pi formulas with radii of convergence $\color{blue}{1/1728, 1/256, 1/108, 1/64}.$ ("Rational analogues of Ramanujan's series for 1/Pi", Cooper and Chan, p. 23).

II. As a further constraint, we require its single parameter $j$, if equated to the appropriate eta quotient in Section II, must render the quintic solvable in radicals.

III. For the 4th form, it turns out one way is to start with the Brioschi quintic,

$$w^5 - 10 c w^3 + 45c^2 w - c^2 = 0\tag1$$

then using a rational Tschirnhausen transformation $x = P(w)/Q(w)$,

$$\left(w^2-3c\right)x - \left(\frac{j\,w}{j+512}-24c\right)=0\tag2$$

$$c= \frac{j^2}{(j-64)(j+512)^2}\tag3$$

Eliminating $w$ between $(1)$ and $(2)$ using resultants, then factoring, one gets the new form,

$$x^5 - 5j\, x^2 - 15j\, x - (8j + j^2) = 0\tag4$$

which, for general $j$, has order $5! = 120$. However, it has the desired property that if $j = j_4(\tau)$, then it becomes solvable in radicals. The two quintics have the discriminants,

$$d_1 = 5^5\,(1-\color{blue}{1728}c)^2\,c^8$$ $$D_4 = 5^5\,(j−\color{blue}{64})^2(j-1)^2j^4$$

respectively. Note that $(3)$ is a just a cubic in $j$, so the general quintic has been transformed to the one-parameter form $(4)$ in radicals.

IV. As an afterword, just like the Bring quintic, the parameter of the Brioschi has infinitely many rational $c$ such that it is solvable in radicals, namely,

$$ c = \frac{n}{(n^2 - 10n + 5)^3 + \color{blue}{1728}n}$$

for rational $n$. And since $c$ in $(3)$ is a cubic in $j$, then the new quintic $(4)$ is solvable for infinitely many cubic radicals $j$, in addition to when $j = j_4(\tau)$ are eta quotients.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.