# On discriminants of minimal Polynomials of the Ramanujan $t_{n}$ Class
Invariants

Sarth Chavan ID 1 Department of Mathematics, Massachusetts Institute of
Technology, Cambridge, MA, USA<EMAIL_ADDRESS>2 Euler Circle, Palo Alto,
California 94306, USA<EMAIL_ADDRESS>Dedicated to all my Rickoid
friends who turned into a family

###### Abstract.

We study the discriminants of the minimal polynomials $\mathcal{P}_{n}$ of the
Ramanujan $t_{n}$ class invariants, which are defined for positive $n\equiv
11\,(\textup{mod}\,24)$. The historical precedent for doing so comes from
Gross and Zagier, which is known for computing the prime factorizations of
certain resultants and discriminants of the Hilbert class polynomials $H_{n}$.
We show that $\Delta\left(\mathcal{P}_{n}\right)$ divides
$\Delta\left(H_{n}\right)$ with quotient a perfect square, and as a
consequence, we explicitly determine the sign of
$\Delta\left(\mathcal{P}_{n}\right)$ based on the class group structure of the
order of discriminant $-n$. We also show that the discriminant of number field
generated by $j\left(\frac{-1+\sqrt{-n}}{2}\right)$, where $j$ is the
$j$-invariant, divides $\Delta\left(\mathcal{P}_{n}\right)$. Moreover, we show
that 3 never divides $\Delta\left(\mathcal{P}_{n}\right)$ for all squarefree
positive integers $n\equiv 11\,(\textup{mod}\,24)$.

###### Key words and phrases:

Elliptic curves, Hilbert class polynomials, Class invariant, $j$-invariant,
Discriminant.

###### 1991 Mathematics Subject Classification:

Primary: 11R29, 11R09; Secondary: 11F03, 11R37.

The author was supported by The 2022 Spirit of Ramanujan Fellowship and The
Mehta Fellowship

##  1\. Introduction and Main Results

Let $E/\mathbb{C}$ be an elliptic curve $E$ over $\mathbb{C}$ that has
_complex multiplication_ (CM) by an imaginary quadratic order $\mathcal{O}$,
by which we mean that the endomorphism ring $\operatorname{End}(E)$ is
isomorphic to $\mathcal{O}$. Let $K$ denote the fraction field of
$\mathcal{O}$. The $j$-invariant of $E$ is an algebraic integer whose minimal
polynomial over $K$ is the _Hilbert class polynomial $H_{n}$_111Note that most
authors use the term _Hilbert class polynomial_ only when $\mathcal{O}$ is a
maximal order (they then use the term _ring class polynomial_ for the general
case); however, we will not make this distinction., where $n$ is the
discriminant of $\mathcal{O}$.

In particular, the Hilbert class polynomial $H_{n}$ is defined by

$H_{n}(x)\colonequals\prod_{j(E)\in\operatorname{Ell}_{\mathcal{O}}(\mathbb{C})}\left(x-j(E)\right),$

where
$\operatorname{Ell}_{\mathcal{O}}(\mathbb{C})\colonequals\\{j(E/\mathbb{C}):\operatorname{End}(E)\cong\mathcal{O}\\}$
is the set of $j$-invariants of elliptic curves $E/\mathbb{C}$ with complex
multiplication by the imaginary quadratic order $\mathcal{O}$ with
discriminant $n=\operatorname{disc}(\mathcal{O})$.

Moreover, $H_{n}\in\mathbb{Z}[x]$ and its splitting field over the imaginary
quadratic field $K$ is the _ring class field_ $K_{\mathcal{O}}$, which is an
abelian extension of $K$ whose Galois group
$\operatorname{Gal}\left(K_{\mathcal{O}}/K\right)$ is isomorphic to the class
group $\operatorname{Cl}\left(\mathcal{O}\right)$, via the Artin map. This is
indeed a remarkable result as it implies that of uncountably many isomorphism
classes of elliptic curves over $\mathbb{C}$, only countably many have CM.

Ramanujan, who made many beautiful and elegant discoveries in his short life
of 32 years, defined in his third notebook [15, Pages 392-393] the values

$t_{n}\colonequals\dfrac{f\left(\sqrt[3]{q_{n}}\right)f\left(q_{n}^{3}\right)}{f^{2}\left(q_{n}\right)}\sqrt{3}q_{n}^{1/18},$

where $q_{n}=\exp\left(-\pi\sqrt{n}\right)$, and
$f\left(-q\right)=\prod_{n\geqslant 1}\left(1-q^{n}\right).$

For all positive $n\equiv 11\,(\textup{mod}\,24)$, let $\mathcal{P}_{n}$ be
the minimal polynomial of $t_{n}$ over $\mathbb{Q}$. Without any further
explanation on how he found them, Ramanujan gave the following table of
polynomials $\mathcal{P}_{n}$ based on $t_{n}$ for first five values of
$n\equiv 11\,(\textup{mod}\,24)$:

$n$ | $\mathcal{P}_{n}(z)$
---|---
$11$ | $z-1$
$35$ | $z^{2}+z-1$
$59$ | $z^{3}+2z-1$
$83$ | $z^{3}+2z^{2}+2z-1$
$107$ | $z^{3}-2z^{2}+4z-1$
Table 1. $\mathcal{P}_{n}$ for $n=11,35,59,83,107.$

Berndt and Chan [3, Theorem 1.2] later verified his claims for
$n=11,35,59,83$, and $107$ using laborious computations involving Greenhill
polynomials and Weber class invariants, and proved that each $\mathcal{P}_{n}$
has $t_{n}$ as a root. However, due to computational complexity, their method
to contruct $\mathcal{P}_{n}$ could not be applied for higher values of $n$.
Thus, they asked for an efficient way of computing the polynomials
$\mathcal{P}_{n}$ for every $n\equiv 11\,(\textup{mod}\,24)$. Moreover, the
authors also proved the following:

######  Theorem 1.1 ([3, Theorem 4.1]).

Let $n\equiv 11\,(\textup{mod}\,24)$ be squarefree, and suppose that the class
number of $\mathbb{Q}(\sqrt{-n})$ is odd. Then $t_{n}$ is a real unit
generating the Hilbert class field of $\mathbb{Q}(\sqrt{-n})$.

Ten years later, Konstantinou and Kontogeorgis [9] generalized this result by
removing the constraint of the class number needing to be odd and also
provided an efficient method for constructing the minimal polynomials
$\mathcal{P}_{n}$ of $t_{n}$ over $\mathbb{Q}$ from the Ramanujan values
$t_{n}$ for $n\equiv 11\,(\textup{mod}\,24)$, using the Shimura reciprocity
law, and thus answered the demand made in [3] for a direct and an easily
applicable construction method. Moreover, the authors also proved that the
Ramanujan value $t_{n}$ is a class invariant for $n\equiv
11\,(\textup{mod}\,24)$ [9, Theorem 3.4]. Therefore, it follows that degree of
$\mathcal{P}_{n}$ equals the class number of the order of discriminant $-n$
for all positive $n\equiv 11\,(\textup{mod}\,24)$.

Let $\mathfrak{j}_{n}=j\left(\frac{-1+\sqrt{-n}}{2}\right)$, for all
$n\in\mathbb{N}$, where $j$ denotes the $j$-invariant as defined in Subsection
2.1. In a follow-up paper in 2010, Konstantinou and Kontogeorgis proved the
following:

######  Proposition 1.2 ([11, Lemma 3]).

Suppose $R_{\mathcal{P}}$ is a real root of a Ramanujan polynomial
$\mathcal{P}_{n}$. Then, the real number $R_{H}$ obtained from the equation

$R_{H}=\left(R_{\mathcal{P}}^{6}-27R_{\mathcal{P}}^{-6}-6\right)^{3},$

is a real root of the corresponding Hilbert class polynomial $H_{n}$.

In their proof [11, Pg. 12], the authors show that, letting
$R_{\mathcal{P}}=t_{n}$ gives
$\mathfrak{j}_{n}=\left(t_{n}^{6}-27t_{n}^{-6}-6\right)^{3}.$

It is interesting to point out that coefficients of the polynomial
$\mathcal{P}_{n}$ have remarkably smaller size compared to the coefficients of
the corresponding Hilbert class polynomial $H_{n}$, which is a clear
indication that their use in the CM method, which is used for the generation
of elliptic curves over prime fields, can be especially favoured. Proposition
1.2 also suggests that the polynomials $\mathcal{P}_{n}$ can be used in the CM
method because their roots can be transformed to the roots of $H_{n}$. For
more details on constructing elliptic curves with the CM method, see [1, 2,
12].

In this brief article we study the discriminant of $\mathcal{P}_{n}$; the
historical precedent for doing so comes from [8], which is known for computing
the prime factorization of certain resultants of Hilbert class polynomials.
Gross and Zagier [8] also computed the prime factorization of the discriminant
of the Hilbert class polynomial associated to the fundamental discriminant
$-p$, where $p\equiv 3\,(\textup{mod}\,4)$ is a prime. This result was later
generalized by Dorman [6], who extended the discriminant formula to the
Hilbert class polynomials associated with arbitrary fundamental discriminants.
Dorman’s result in turn was then extended by Ye [19], who computed the prime
factorization of Hilbert class polynomials associated to certain non-
fundamental discriminants. For more details, see [8, 6, 19].

###  1.1. Notations

Fix a positive integer $n\equiv 11\,(\textup{mod}\,24)$ and let
$K=\mathbb{Q}\left(\sqrt{-n}\right)$ be an imaginary quadratic number field.
Let $h_{n}$ and $\operatorname{Cl}\left(n\right)$ denote the class number and
ideal class group of the order of discriminant $-n$, respectively, and let
$\operatorname{Cl}\left(n\right)\left[2\right]$ be the subgroup of
$\operatorname{Cl}\left(n\right)$ consisting of elements of order at most $2$.
Note that $\mathbb{Z}\left[\mathfrak{j}_{n}\right]$ is contained in
$\mathbb{Z}\left[t_{n}\right]$, which follows from Section 3.

###  1.2. Main results

Our first main result, which relates $\Delta\left(H_{n}\right)$ to
$\Delta\left(\mathcal{P}_{n}\right)$, is as follows:

######  Theorem 1.3.

For all positive $n\equiv 11\,(\textup{mod}\,24)$, we have

$\Delta\left(H_{n}\right)=\Delta\left(\mathcal{P}_{n}\right)\left[\mathbb{Z}\left[t_{n}\right]:\mathbb{Z}\left[\mathfrak{j}_{n}\right]\right]^{2},$

where
$\left[\mathbb{Z}\left[t_{n}\right]:\mathbb{Z}\left[\mathfrak{j}_{n}\right]\right]$
is the index of $\mathbb{Z}\left[\mathfrak{j}_{n}\right]$ in
$\mathbb{Z}\left[t_{n}\right]$.

######  Remark 1.4.

Since the quotient
$\left[\mathbb{Z}\left[t_{n}\right]:\mathbb{Z}\left[\mathfrak{j}_{n}\right]\right]^{2}$
is a perfect square, we deduce that $\Delta\left(H_{n}\right)$ and
$\Delta\left(\mathcal{P}_{n}\right)$ have the same sign for all positive
integers $n\equiv 11\,(\textup{mod}\,24)$.

Let $\mathcal{D}\left(F\right)$ denote the discriminant of a number field $F$.
Our next main result is the following:

######  Theorem 1.5.

For all positive integers $n\equiv 11\,(\textup{mod}\,24)$, we have

$\Delta\left(\mathcal{P}_{n}\right)=\left[\mathcal{O}_{\mathbb{Q}\left(\mathfrak{j}_{n}\right)}:\mathbb{Z}[t_{n}]\right]^{2},$

where
$\left[\mathcal{O}_{\mathbb{Q}\left(\mathfrak{j}_{n}\right)}:\mathbb{Z}[t_{n}]\right]$
is the index of $\mathbb{Z}\left[t_{n}\right]$ inside
$\mathcal{O}_{\mathbb{Q}\left(\mathfrak{j}_{n}\right)}$.

######  Remark 1.6.

Note that Dorman explicitly computed
$\mathcal{D}\left(\mathbb{Q}\left(\mathfrak{j}_{n}\right)\right)$ in [6]. More
precisely, he proved that, for a squarefree positive integer $n\equiv
11\,(\textup{mod}\,24)$, we have [6, Proposition 5.1]

$\mathcal{D}\left(\mathbb{Q}\left(\mathfrak{j}_{n}\right)\right)=D_{0}^{\frac{h_{n}}{2}}\cdot
D_{1}^{\frac{h_{n}-2^{\mathfrak{t}-1}}{2}},$

where where $\mathfrak{t}$ is the number of distinct prime factors of $n$, and
$n=D_{0}D_{1}$ with

$D_{1}=\begin{cases}1&\textrm{if at least }2\textrm{ primes congruent to
}$3$\textrm{ mod }4\textrm{ divide }D,\\\ p&\textrm{if }p\textrm{ is the
unique prime congruent to }3\textrm{ mod }4\textrm{ dividing }D.\end{cases}$

Next, we explicitly determines the sign of
$\Delta\left(\mathcal{P}_{n}\right)$.

######  Theorem 1.7.

For all positive $n\equiv 11\,(\textup{mod}\,24)$,
$\Delta\left(\mathcal{P}_{n}\right)>0$ if and only if

$h_{n}\equiv\left|\operatorname{Cl}\left(n\right)\left[2\right]\right|\,(\textup{mod}\,4),$

where $\operatorname{Cl}\left(n\right)\left[2\right]$ is the subgroup of
$\operatorname{Cl}\left(n\right)$ consisting of elements of order at most $2$.

Note that using the ambiguous class number formula from genus theory, we find
that for all positive squarefree inetgers $n\equiv 11\,(\textup{mod}\,24)$, we
have

$\left|\operatorname{Cl}\left(n\right)\left[2\right]\right|=2^{\mathfrak{t}-1},$

where $\mathfrak{t}$ is the number of distinct prime factors of $n$.
Therefore, we have the following corollary:

######  Corollary 1.8.

For all positive squarefree $n\equiv 11\,(\textup{mod}\,24)$,
$\Delta\left(\mathcal{P}_{n}\right)>0$ if and only if

$h_{n}\equiv 2^{\mathfrak{t}-1}\,(\textup{mod}\,4),$

where, as usual, $\mathfrak{t}$ denotes the number of distinct prime factors
of $n$.

Moreover, when the ideal class group is a cyclic group, that is
$\operatorname{Cl}\left(n\right)\cong\left(\mathbb{Z}/h_{n}\mathbb{Z}\right)$,
which is true for all $n\equiv 11\,(\textup{mod}\,24)$, with $11\leqslant
n\leqslant 995$ as illustrated in Table 2, it is easy to see that

$\left|\operatorname{Cl}\left(n\right)[2]\right|=\begin{cases}1&\textrm{if
}h_{n}\textrm{ is odd}\\\ 2&\textrm{if }h_{n}\textrm{ is even.}\end{cases}$

Thus, we have the following corollary:

######  Corollary 1.9.

For all $n\equiv 11\,(\textup{mod}\,24)$, if
$\operatorname{Cl}\left(n\right)\cong\left(\mathbb{Z}/h_{n}\mathbb{Z}\right)$,
then $\Delta\left(\mathcal{P}_{n}\right)>0$ if and only if

$h_{n}\equiv 1,2\,(\textup{mod}\,4).$

######  Remark 1.10.

For example, the two positive values of $n\equiv 11\,(\textup{mod}\,24)$ for
which $\operatorname{Cl}\left(n\right)\not\cong\mathbb{Z}/h_{n}\mathbb{Z}$ are
1235 and 2555. In particular, we find that222The structure of the ideal class
group $\operatorname{Cl}\left(n\right)$ for positive $n\equiv
11\,(\textup{mod}\,24)$ was computed using Sage [18].
$\operatorname{Cl}\left(1235\right)\cong\operatorname{Cl}\left(2555\right)\cong\left(\mathbb{Z}/6\mathbb{Z}\right)\times\left(\mathbb{Z}/2\mathbb{Z}\right)$.

In Table 2, it can be observed that 3 seems to be the only prime that never
appears in the prime factorization of $\Delta\left(\mathcal{P}_{n}\right)$ for
all positive $n\equiv 11\,(\textup{mod}\,24)$. To prove this, we show that 3
never divides $\Delta\left(H_{n}\right)$, and thus
$\Delta\left(\mathcal{P}_{n}\right)$, using Ye’s explicit computation of
$\Delta\left(H_{n}\right)$ [19, Corollary 1.2] for squarefree $n\equiv
11\,(\textup{mod}\,24)$, for more details see Section 6.

Thus, we have the following important Theorem:

######  Theorem 1.11.

For all positive squarefree integers $n\equiv 11\,(\textup{mod}\,24)$, we have
$3\nmid\Delta\left(\mathcal{P}_{n}\right)$.

###  1.3. Organization of the paper

We start with a section on the necessary preliminaries and prove some
important results that will later play a very crucial role in the proofs of
our main results. Proofs of Theorem 1.3, 1.5, 1.7, and 1.11 are provided in
Section 3, 4, 5 and 6 respectively. In Table 2, we have computed the class
number and the ideal class group structure of $\mathbb{Q}(\sqrt{-n})$, the
prime factorization and sign of $\Delta\left(\mathcal{P}_{n}\right)$ for all
positive $n\equiv 11\,(\textup{mod}\,24)$ where $11\leqslant n\leqslant 995$
as an illustration of our main results. Finally, in Section 7, we show that
all our main results hold true for $n=227$, as an example. All the
computations were performed using Sage [18].

##  2\. Preliminaries

###  2.1. The $j$-invariant of a lattice

A _lattice_ is defined to be an additive subgroup $L$ of $\mathbb{C}$ which is
generated by two complex numbers $\omega_{1}$ and $\omega_{2}$ that are
linearly independent over $\mathbb{R}$. We express this by writing
$L=[\omega_{1},\omega_{2}].$ The $j$-invariant $j(L)$ of a lattice $L$ is
defined to be the complex number

$j\left(L\right)=1728\dfrac{g_{2}\left(L\right)^{3}}{g_{2}\left(L\right)^{3}-27g_{3}\left(L\right)^{2}},$

where

$g_{2}\left(L\right)=60\sum_{\omega\in
L\backslash\\{0\\}}\dfrac{1}{\omega^{4}}=60\sum_{\begin{subarray}{c}m,n=-\infty\\\
\left(m,n\right)\neq\left(0,0\right)\end{subarray}}^{\infty}\dfrac{1}{\left(m\tau+n\right)^{4}},$

and

$g_{3}\left(L\right)=140\sum_{\omega\in
L\backslash\\{0\\}}\dfrac{1}{\omega^{6}}=140\sum_{\begin{subarray}{c}m,n=-\infty\\\
\left(m,n\right)\neq\left(0,0\right)\end{subarray}}^{\infty}\dfrac{1}{\left(m\tau+n\right)^{6}},$

where $L=[1,\tau]$ with $\tau\in\mathbb{H}$, the upper-half plane.

######  Proposition 2.1.

Let $L$ be a lattice, and let $\overline{L}$ denote the lattice obtained by
complex conjugation. Then
$g_{2}(\overline{L})=\overline{g_{2}(L)},g_{3}(\overline{L})=\overline{g_{3}(L)}$
and $j(\overline{L})=\overline{j(L)}$.

###### Proof.

From the definition of $g_{2}(L)$, we have

$\displaystyle g_{2}(\overline{L})$
$\displaystyle=60\sum_{\omega\in\overline{L}\backslash\\{0\\}}\dfrac{1}{\omega^{4}}=60\sum_{\omega\in
L\backslash\\{0\\}}\dfrac{1}{\overline{\omega^{4}}}=\overline{60\sum_{\omega\in
L\backslash\\{0\\}}\dfrac{1}{\omega^{4}}}=\overline{g_{2}(L)}.$

A similar argument can be implemented to show that
$g_{3}(\overline{L})=\overline{g_{3}(L)}$.

Finally, putting all things together produces

$j\left(\overline{L}\right)=1728\dfrac{g_{2}\left(\overline{L}\right)^{3}}{g_{2}\left(\overline{L}\right)^{3}-27g_{3}\left(\overline{L}\right)^{2}}=1728\dfrac{\overline{g_{2}\left(L\right)^{3}}}{\overline{g_{2}\left(L\right)^{3}}-27\overline{g_{3}\left(L\right)^{2}}}=\overline{j(L)},$

which is the desired result. ∎

We say that two lattices $L$ and $L^{\prime}$ are _homothetic_ if there is a
nonzero complex number $\lambda$ such that $L^{\prime}=\lambda L$. Note that
homothetic lattices have the same $j$-invariant.

######  Lemma 2.2 ([5, Theorem 10.9]).

If $L$ and $L^{\prime}$ are lattices in $\mathbb{C}$, then
$j(L)=j(L^{\prime})$ if and only if both the lattices $L$ and $L^{\prime}$ are
homothetic.

Next, we prove an important result that will play a crucial role in the proof
of Theorem 1.7.

######  Proposition 2.3.

Let $\mathfrak{a}$ be a proper fractional $\mathcal{O}$-ideal, where
$\mathcal{O}$ is an order in an imaginary quadratic number field. Then
$j(\mathfrak{a})$ is a real number if and only if the class of $\mathfrak{a}$
has order at most 2 in the ideal class group
$\operatorname{Cl}\left(\mathcal{O}\right)$.

###### Proof.

From Proposition 2.1, we know that $j(\mathfrak{a})$ is a real number if and
only if $j(\mathfrak{a})=j(\overline{\mathfrak{a}})$. Now Lemma 2.2 tells us
that this is only possible when $\mathfrak{a}$ and $\overline{\mathfrak{a}}$
are homothetic. Or equivalently, when they represent the same ideal in the
ideal class group $\operatorname{Cl}\left(\mathcal{O}\right)$. By computing
$\mathfrak{a}\overline{\mathfrak{a}}$ and comparing the $j$-invariants by a
similar argument that we just sketched above, it is easy to see that

$\mathfrak{a}=\mathfrak{a}^{-1}=\overline{\mathfrak{a}}.$

Note that this is not true as a statement about ideals, here we are explicitly
referring to ideal classes, that is, elements of
$\operatorname{Cl}\left(\mathcal{O}\right)$. Putting all things together gives
us the desired result. ∎

As a consequence of the above Proposition, we have the following important
corollary:

######  Corollary 2.4.

The $j$-invariant $j(\mathcal{O})$ is a real number for any order
$\mathcal{O}$.

###  2.2. Discriminant of an algebraic number field

Let $\mathcal{D}\left(K\right)$ denote the discriminant of an algebraic number
field $K$. The discriminant of a nonzero finitely generated
$\mathbb{Z}$-submodule $\mathfrak{a}$ of a number field $K$ is defined as

$\mathcal{D}\left(\mathfrak{a}\right)=\mathcal{D}\left(\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right),$

where
$\mathfrak{a}=\mathbb{Z}\alpha_{1}+\mathbb{Z}\alpha_{2}+\cdots+\mathbb{Z}\alpha_{n}$
and spans $K$ as a $\mathbb{Q}$-vector space.

######  Proposition 2.5 ([14, Prop. 2.12]).

If $\mathfrak{a}\subset\mathfrak{a}^{\prime}$ are two nonzero finitely
generated $\mathbb{Z}$-submodules of a number field $K$ that span $K$ as a
$\mathbb{Q}$-vector space, then the index
$[\mathfrak{a}^{\prime}:\mathfrak{a}]$ is finite and

$\mathcal{D}\left(\mathfrak{a}\right)=\left[\mathfrak{a}^{\prime}:\mathfrak{a}\right]^{2}\mathcal{D}\left(\mathfrak{a}^{\prime}\right).$

######  Remark 2.6.

Note that although Neukirch [14, Prop. 2.12] only states that Proposition 2.5
is true for finitely generated $\mathcal{O}_{K}$-submodules, the same result
and proof also works in the more general case, that is, for finitely generated
$\mathbb{Z}$-submodules.

###  2.3. Number fields $\mathbb{Q}\left(\mathfrak{j}_{n}\right)$ and
$\mathbb{Q}\left(t_{n}\right)$

The main goal of this subsection is to show that
$\mathbb{Q}\left(\mathfrak{j}_{n}\right)$ equals
$\mathbb{Q}\left(t_{n}\right)$. This important result indeed proves Theorem
1.5 in a very simple and efficient way as explained later in Section 4.

######  Proposition 2.7.

For all positive integers $n\equiv 11\,(\textup{mod}\,24)$, we have
$\mathbb{Q}\left(\mathfrak{j}_{n}\right)=\mathbb{Q}\left(t_{n}\right)$.

###### Proof.

To start with, consider the following field diagram:

${H}$${K=\mathbb{Q}\left(\sqrt{-n}\right)}$${\mathbb{Q}\left(\mathfrak{j}_{n}\right)}$${\mathbb{Q}\left(t_{n}\right)}$${\mathbb{Q}}$$\scriptstyle{h_{n}}$$\scriptstyle{2}$
Figure 1.

We freely use some standard facts about the ring class fields of $K$
throughout the proof.

From class field theory, recall that if $H$ is the ring class field associated
to the discriminant of order $-n$, then $H$ is a finite Galois extension of
$K$ and $[H:K]=h_{n}$, where $h_{n}$ denotes the class number of the order of
discriminant $-n$. Therefore, we have $[H:K]=[H:\mathbb{Q}(\sqrt{-n})]=h_{n}$.
Next, it is easy to see that $[K:\mathbb{Q}]=2$ since $\\{1,\imath\sqrt{n}\\}$
forms a basis. Moreover, since
$H=\left(\mathbb{Q}\left(\mathfrak{j}_{n}\right)\right)(\sqrt{-n}),$ we have
$\left[H:\mathbb{Q}\left(\mathfrak{j}_{n}\right)\right]\leqslant 2$.
Theorefore, using the tower law, we deduce that either
$\left[\mathbb{Q}\left(\mathfrak{j}_{n}\right):\mathbb{Q}\right]=h_{n}$ or
$\left[\mathbb{Q}\left(\mathfrak{j}_{n}\right):\mathbb{Q}\right]=2h_{n}$. Now
let $\mathfrak{f}$ be the minimal polynomial of $\mathfrak{j}_{n}\in H$ over
$K$. Note that $\mathfrak{f}\in K[x]$ has degree $h_{n}$. From Corollary 2.4,
we know that $\mathfrak{j}_{n}\in\mathbb{R}$, that is
$\overline{\mathfrak{j}_{n}}=\mathfrak{j}_{n}$, thus we have
$\overline{\mathfrak{f}}=\mathfrak{f}$, which implies
$\mathfrak{f}\in\mathbb{Q}[x]$, and therefore, we deduce that
$[\mathbb{Q}\left(\mathfrak{j}_{n}\right):\mathbb{Q}]\leqslant h_{n}$.

Putting all things together finally produces
$\left[\mathbb{Q}\left(\mathfrak{j}_{n}\right):\mathbb{Q}\right]=h_{n}$ and
$\left[H:\mathbb{Q}\left(\mathfrak{j}_{n}\right)\right]=2.$ Moreover, since
$t_{n}\in\mathbb{R}$, which easily follows from the Ramanujan’s definition of
$t_{n}$, a similar argument can be applied to show that
$\left[\mathbb{Q}\left(t_{n}\right):\mathbb{Q}\right]=h_{n}$ and
$\left[H:\mathbb{Q}\left(t_{n}\right)\right]=2.$

All this discussion can be summarised by the following field diagram:

${H}$${K=\mathbb{Q}\left(\sqrt{-n}\right)}$${\mathbb{Q}\left(\mathfrak{j}_{n}\right)}$${\mathbb{Q}\left(t_{n}\right)}$${\mathbb{Q}}$$\scriptstyle{h_{n}}$$\scriptstyle{2}$$\scriptstyle{2}$$\scriptstyle{2}$$\scriptstyle{h_{n}}$$\scriptstyle{h_{n}}$
Figure 2.

Now observe that we have the tower of fields
$\mathbb{Q}\subseteq\mathbb{Q}\left(\mathfrak{j}_{n}\right)\subseteq\mathbb{Q}\left(t_{n}\right)\subseteq
H$, where
$\mathbb{Q}\left(\mathfrak{j}_{n}\right)\subseteq\mathbb{Q}\left(t_{n}\right)$
follows from Proposition 1.2. Therefore, by the tower law, we have

$\left[\mathbb{Q}\left(t_{n}\right):\mathbb{Q}\left(\mathfrak{j}_{n}\right)\right]\left[H:\mathbb{Q}\left(t_{n}\right)\right]=\left[H:\mathbb{Q}\left(\mathfrak{j}_{n}\right)\right]\implies\left[\mathbb{Q}\left(t_{n}\right):\mathbb{Q}\left(\mathfrak{j}_{n}\right)\right]=1,$

which is only possible when
$\mathbb{Q}\left(\mathfrak{j}_{n}\right)=\mathbb{Q}\left(t_{n}\right)$. This
completes the proof. ∎

##  3\. Proof of Theorem 1.3

Since $t_{n}\in\mathbb{Z}\left[t_{n}\right]$ and the constant term of
$\mathcal{P}_{n}$ is always $\pm 1$, which follows from [11, Section 3] where
the authors state that $t_{n}$ is a unit, it follows that
$t_{n}^{-1}\in\mathbb{Z}\left[t_{n}\right]$. From Proposition 1.2, we have

$\mathfrak{j}_{n}=\left(t_{n}^{6}-27t_{n}^{-6}-6\right)^{3}.$

Therefore, we deduce that $\mathfrak{j}_{n}\in\mathbb{Z}\left[t_{n}\right]$
and thus
$\mathbb{Z}\left[\mathfrak{j}_{n}\right]\subseteq\mathbb{Z}\left[t_{n}\right]$.
Now plugging in $\mathfrak{a}=\mathbb{Z}\left[\mathfrak{j}_{n}\right]$ and
$\mathfrak{a}^{\prime}=\mathbb{Z}\left[t_{n}\right]$ in Proposition 2.5
produces

$\mathcal{D}\left(\mathbb{Z}\left[\mathfrak{j}_{n}\right]\right)=\mathcal{D}\left(\mathbb{Z}\left[t_{n}\right]\right)\left[\mathbb{Z}\left[t_{n}\right]:\mathbb{Z}\left[\mathfrak{j}_{n}\right]\right]^{2},$

or equivalently,

$\Delta\left(H_{n}\right)=\Delta\left(\mathcal{P}_{n}\right)\left[\mathbb{Z}\left[t_{n}\right]:\mathbb{Z}\left[\mathfrak{j}_{n}\right]\right]^{2},$

which is the desired result. $\blacksquare$

##  4\. Proof of Theorem 1.5

Simply notice that
$\mathbb{Z}\left[t_{n}\right]\subseteq\mathcal{O}_{\mathbb{Q}\left(t_{n}\right)}=\mathcal{O}_{\mathbb{Q}\left(\mathfrak{j}_{n}\right)}\subseteq\mathbb{Q}\left(\mathfrak{j}_{n}\right)$,
where
$\mathcal{O}_{\mathbb{Q}\left(t_{n}\right)}=\mathcal{O}_{\mathbb{Q}\left(\mathfrak{j}_{n}\right)}$
follows from the fact that
$\mathbb{Q}\left(\mathfrak{j}_{n}\right)=\mathbb{Q}\left(t_{n}\right)$, as
shown in Proposition 2.7. Now, as we did earlier, plugging in
$\mathfrak{a}=\mathbb{Z}\left[t_{n}\right]$ and
$\mathfrak{a}^{\prime}=\mathcal{O}_{\mathbb{Q}\left(\mathfrak{j}_{n}\right)}$
in Proposition 2.5 gives us the desired result. $\blacksquare$

##  5\. Proof of Theorem 1.7

From the definition of Hilbert class polynomial, recall that its roots are
given by

$\\{j(\mathfrak{a})\mid\mathfrak{a}\in\operatorname{Cl}\left(\mathcal{O}\right)\\}.$

From Proposition 2.3, we deduce that the real roots of Hilbert class
polynomial $H_{n}$ are given by

$\\{j(\mathfrak{a})\mid\mathfrak{a}\in\operatorname{Cl}\left(\mathcal{O}\right),\textrm{order
of }\mathfrak{a}\textrm{ at most }2\\}=\operatorname{Cl}\left(n\right)[2].$

Therefore, the number of non-real roots of $H_{n}$ is given by
$\left|\operatorname{Cl}\left(n\right)\right|-\left|\operatorname{Cl}\left(n\right)\left[2\right]\right|=h_{n}-\left|\operatorname{Cl}\left(n\right)\left[2\right]\right|$.
Recall that the discriminant of a polynomial in $\mathbb{Q}$ is positive if
and only if the number of non-real roots of the polynomial is divisible by 4.
Putting all things together gives us the desired result. $\blacksquare$

##  6\. Proof of Theorem 1.11

Assume that $n$ is a positive squarefree integer. Ye [19, Corollary 1.2] very
recently explicitly computed the discriminant of Hilbert class polynomials
$H_{n}$ as

$\displaystyle\log\left|\Delta\left(H_{n}\right)\right|$
$\displaystyle=-\dfrac{h_{K}}{4}\sum_{\begin{subarray}{c}[\mathfrak{a}]\in\operatorname{Cl}_{K}\\\
[\mathfrak{a}]\neq[\mathcal{O}_{K}]\end{subarray}}\sum_{\ell=0}^{n-1}\sum_{X,Y=-\infty}^{\infty}\kappa\left(1-\dfrac{n\left(2AX+BY\right)^{2}+\left(nY-2A\ell\right)^{2}}{4An},\dfrac{B\ell}{n}f_{1}^{(\mathfrak{a})}-\dfrac{2A\ell}{n}f_{2}^{(\mathfrak{a})}+L_{-}^{(\mathfrak{a})}\right)$

where
$\mathfrak{a}=\left[A,\dfrac{B+\sqrt{-n}}{2}\right],f_{1}^{(\mathfrak{a})}=\begin{pmatrix}-1&B\\\
0&A\end{pmatrix},f_{2}^{(\mathfrak{a})}=\begin{pmatrix}0&C\\\
1&0\end{pmatrix},h_{K}$ is the class number of
$K=\mathbb{Q}\left(\sqrt{-n}\right)$, $L_{-}^{(\mathfrak{a})}$ denotes the
lattice $\mathbb{Z}f_{1}^{(\mathfrak{a})}+\mathbb{Z}f_{2}^{(\mathfrak{a})}$,
and

$\displaystyle\kappa\left(m,\mu\right)$
$\displaystyle=-\dfrac{1}{h_{K}}\left(\sum_{q\,\textrm{inert}}\xi_{q}\left(m,\mu\right)\left(\operatorname{ord}_{q}(m)+1\right)\rho_{K}\left(mn/q\right)\log
q+\rho_{K}\left(mn\right)\sum_{q\mid
n}\xi_{q}\left(m,\mu\right)\operatorname{ord}_{q}(mn)\log q\right)$

where $\xi_{q}$ and $\rho_{K}$ are as defined in [19, Corollary 1.2].

To prove Theorem 1.11, we will show that $\log 3$ never appears in
$\kappa\left(m,\mu\right)$ for any choice of parameters $m$ and $\mu$. From
the definition of $\kappa\left(m,\mu\right)$, it suffices to prove that 3
always splits in $\mathbb{Q}\left(\sqrt{-n}\right)$ and does not divide $n$
for all positive squarefree $n\equiv 11\,(\textup{mod}\,24)$.

Recall that a prime number $\mathfrak{p}$ splits in the imaginary quadratic
field $\mathbb{Q}\left(\sqrt{-n}\right)$ if and only if $-n$ is a nonzero
quadratic residue mod $\mathfrak{p}$. The quadratic residues mod 3 are 0 and
1, and it is easy to see that when $n\equiv 11\,(\textup{mod}\,24)$, we have
$-n\equiv 1\,(\textup{mod}\,3)$. Thus $-n$ is a nonzero quadratic residue mod
3 for all positive squarefree integers $n\equiv 11\,(\textup{mod}\,24)$. This
completes the proof. $\blacksquare$

##  7\. An Exmaple of Main Results

######  Example 7.1.

For $n=227$, we have [9, Table 1]

$\mathcal{P}_{227}(z)=z^{5}-5z^{4}+9z^{3}-9z^{2}+9z-1.$

Now since 227 is squarefree, it follows that the degree of $\mathcal{P}_{227}$
is equal to the class number of $\mathbb{Q}\left(\sqrt{-227}\right)$, and
thus, $h_{227}=5$. The class group structure of
$\mathbb{Q}\left(\sqrt{-227}\right)$ is computed using Sage as
$\mathbb{Z}/5\mathbb{Z}$, see Table 2. Now using Corollary 1.9, we have
$\Delta\left(\mathcal{P}_{227}\right)>0$, since

$h_{227}=5\equiv 1\,(\textup{mod}\,4).$

From Table 2, we have $\Delta\left(\mathcal{P}_{227}\right)=2^{4}\cdot
227^{2}$. Moreover, using the Sage database for Hilbert class polynomials and
their discriminants [18], it can be found that

$\displaystyle H_{227}(z)=$ $\displaystyle\,z^{5}+360082897644683264000z^{4}$
$\displaystyle-2562327002832961536000000000z^{3}$
$\displaystyle+18227340807938993794580480000000000z^{2}$
$\displaystyle-2111118203460821622718464000000000000z$
$\displaystyle+5085472193216544027705344000000000000000,$

and

$\Delta\left(H_{227}\right)=2^{316}\cdot 5^{60}\cdot 13^{20}\cdot 17^{20}\cdot
31^{4}\cdot 37^{6}\cdot 41^{4}\cdot 61^{4}\cdot 83^{2}\cdot 151^{2}\cdot
179^{2}\cdot 191^{2}\cdot 199^{2}\cdot 227^{2}.$

Therefore, we have

$\displaystyle\dfrac{\Delta\left(H_{227}\right)}{\Delta\left(\mathcal{P}_{227}\right)}$
$\displaystyle=\left(2^{156}\cdot 5^{30}\cdot 13^{10}\cdot 17^{10}\cdot
31^{2}\cdot 37^{3}\cdot 41^{2}\cdot 61^{2}\cdot 83\cdot 151\cdot 179\cdot
191\cdot 199\right)^{2},$

which verifies Theorem 1.3, since the quotient is also a perfect square.

Now using Remark 1.6, we have $D_{0}=1$, $D_{1}=227$, $\mathfrak{t}=1$, and
thus

$\mathcal{D}\left(\mathbb{Q}\left(\mathfrak{j}_{227}\right)\right)=227^{2}\mid
2^{4}\cdot 227^{2}=\Delta\left(\mathcal{P}_{227}\right).$

Moreover, notice that 3 does not appear in the prime factorization of
$\Delta\left(\mathcal{P}_{227}\right)$ and $\Delta\left(H_{227}\right)$.

$n$ | $h_{n}$ | $\operatorname{sgn}(\Delta)$ | Prime factorisation of $\left|\Delta\left(\mathcal{P}_{n}\right)\right|$ | $\operatorname{Cl}\left(n\right)$
---|---|---|---|---
$11$ | $1$ | $+$ | $1$ | $\mathbb{Z}/1\mathbb{Z}$
$35$ | $2$ | $+$ | $5$ | $\mathbb{Z}/2\mathbb{Z}$
$59$ | $3$ | $-$ | $59$ | $\mathbb{Z}/3\mathbb{Z}$
$83$ | $3$ | $-$ | $83$ | $\mathbb{Z}/3\mathbb{Z}$
$107$ | $3$ | $-$ | $107$ | $\mathbb{Z}/3\mathbb{Z}$
$131$ | $5$ | $+$ | $2^{4}\times 131^{2}$ | $\mathbb{Z}/5\mathbb{Z}$
$155$ | $4$ | $-$ | $2^{4}\times 5^{2}\times 31$ | $\mathbb{Z}/4\mathbb{Z}$
$179$ | $5$ | $+$ | $2^{6}\times 179^{2}$ | $\mathbb{Z}/5\mathbb{Z}$
$203$ | $4$ | $-$ | $2^{2}\times 7\times 29^{2}$ | $\mathbb{Z}/4\mathbb{Z}$
$227$ | $5$ | $+$ | $2^{4}\times 227^{2}$ | $\mathbb{Z}/5\mathbb{Z}$
$251$ | $7$ | $-$ | $2^{10}\times 251^{3}$ | $\mathbb{Z}/7\mathbb{Z}$
${275}$ | $4$ | $-$ | $2^{2}\times 5^{3}\times 7^{2}\times 11$ | $\mathbb{Z}/4\mathbb{Z}$
$299$ | $8$ | $-$ | $2^{14}\times 13^{4}\times 23^{3}\times 47^{2}$ | $\mathbb{Z}/8\mathbb{Z}$
$323$ | $4$ | $-$ | $2^{2}\times 7^{2}\times 17^{2}\times 19$ | $\mathbb{Z}/4\mathbb{Z}$
$347$ | $5$ | $+$ | $2^{4}\times 347^{2}$ | $\mathbb{Z}/5\mathbb{Z}$
$371$ | $8$ | $-$ | $2^{20}\times 7^{3}\times 17^{2}\times 53^{4}$ | $\mathbb{Z}/8\mathbb{Z}$
$395$ | $8$ | $-$ | $2^{18}\times 5^{4}\times 13^{4}\times 79^{3}$ | $\mathbb{Z}/8\mathbb{Z}$
$419$ | $9$ | $+$ | $2^{22}\times 167^{2}\times 419^{4}$ | $\mathbb{Z}/9\mathbb{Z}$
$443$ | $5$ | $+$ | $2^{6}\times 7^{2}\times 443^{2}$ | $\mathbb{Z}/5\mathbb{Z}$
$467$ | $7$ | $-$ | $2^{12}\times 5^{4}\times 467^{3}$ | $\mathbb{Z}/7\mathbb{Z}$
$491$ | $9$ | $+$ | $2^{28}\times 7^{2}\times 23^{2}\times 491^{4}$ | $\mathbb{Z}/9\mathbb{Z}$
$515$ | $6$ | $+$ | $2^{6}\times 5^{3}\times 7^{2}\times 13^{2}\times 103^{2}$ | $\mathbb{Z}/6\mathbb{Z}$
${539}$ | $8$ | $-$ | $2^{28}\times 7^{7}\times 11^{4}\times 13^{2}$ | $\mathbb{Z}/8\mathbb{Z}$
$563$ | $9$ | $+$ | $2^{18}\times 5^{4}\times 311^{2}\times 563^{4}$ | $\mathbb{Z}/9\mathbb{Z}$
$587$ | $7$ | $-$ | $2^{10}\times 5^{4}\times 13^{4}\times 587^{3}$ | $\mathbb{Z}/7\mathbb{Z}$
$611$ | $10$ | $+$ | $2^{30}\times 7^{2}\times 13^{9}\times 47^{4}$ | $\mathbb{Z}/{10}\mathbb{Z}$
$635$ | $10$ | $+$ | $2^{28}\times 5^{5}\times 13^{2}\times 127^{4}\times 383^{2}$ | $\mathbb{Z}/{10}\mathbb{Z}$
$659$ | $11$ | $-$ | $2^{40}\times 7^{4}\times 191^{2}\times 659^{5}$ | $\mathbb{Z}/{11}\mathbb{Z}$
$683$ | $5$ | $+$ | $2^{6}\times 7^{2}\times 683^{2}$ | $\mathbb{Z}/5\mathbb{Z}$
$707$ | $6$ | $+$ | $2^{8}\times 7^{2}\times 13^{2}\times 19^{2}\times 101^{3}$ | $\mathbb{Z}/6\mathbb{Z}$
$731$ | $12$ | $-$ | $2^{52}\times 17^{6}\times 19^{2}\times 43^{5}\times 263^{2}\times 479^{2}$ | $\mathbb{Z}/{12}\mathbb{Z}$
$755$ | $12$ | $-$ | $2^{38}\times 5^{6}\times 41^{4}\times 71^{2}\times 151^{5}\times 503^{2}$ | $\mathbb{Z}/{12}$
$779$ | $10$ | $+$ | $2^{40}\times 19^{8}\times 41^{5}\times 311^{2}$ | $\mathbb{Z}/{10}\mathbb{Z}$
$803$ | $10$ | $+$ | $2^{26}\times 5^{4}\times 11^{4}\times 19^{8}\times 73^{5}$ | $\mathbb{Z}/{10}\mathbb{Z}$
$827$ | $7$ | $-$ | $2^{12}\times 7^{2}\times 13^{6}\times 827^{3}$ | $\mathbb{Z}/7\mathbb{Z}$
$851$ | $10$ | $+$ | $2^{44}\times 7^{2}\times 13^{2}\times 23^{4}\times 37^{5}\times 167^{2}$ | $\mathbb{Z}/{10}\mathbb{Z}$
${875}$ | $10$ | $+$ | $2^{32}\times 5^{13}\times 7^{4}\times 19^{2}\times 37^{4}\times 89^{2}$ | $\mathbb{Z}/{10}\mathbb{Z}$
$899$ | $14$ | $+$ | $2^{72}\times 13^{2}\times 19^{4}\times 29^{7}\times 31^{6}\times 647^{2}$ | $\mathbb{Z}/{14}\mathbb{Z}$
$923$ | $10$ | $+$ | $2^{30}\times 5^{4}\times 13^{5}\times 19^{2}\times 61^{4}\times 71^{4}$ | $\mathbb{Z}/{10}\mathbb{Z}$
$947$ | $5$ | $+$ | $2^{4}\times 13^{2}\times 19^{2}\times 947^{2}$ | $\mathbb{Z}/5\mathbb{Z}$
$971$ | $15$ | $-$ | $2^{78}\times 41^{4}\times 71^{2}\times 503^{2}\times 719^{2}\times 971^{7}$ | $\mathbb{Z}/{15}\mathbb{Z}$
$995$ | $8$ | $-$ | $2^{22}\times 5^{4}\times 7^{2}\times 13^{4}\times 19^{2}\times 23^{2}\times 199^{3}$ | $\mathbb{Z}/8\mathbb{Z}$
Table 2. Computation of the class number $h_{n}$, the sign
$\operatorname{sgn}\left(\Delta\right)$ of
$\Delta\left(\mathcal{P}_{n}\right)$, the structure of the class group
$\operatorname{Cl}\left(n\right)$, and the prime factorization of
$\left|\Delta\left(\mathcal{P}_{n}\right)\right|$ for all $n\equiv
11\,(\textup{mod}\,24)$ with $11\leqslant n\leqslant 995$.

##  8\. Concluding Remarks and Further Research

We can now replace the Hilbert class polynomials $H_{n}$ and their
discriminants $\Delta\left(H_{n}\right)$ with the Ramanujan polynomials
$\mathcal{P}_{n}$ and their discriminants $\Delta\left(\mathcal{P}_{n}\right)$
to simplify computations wherever needed. Ramanujan polynomials
$\mathcal{P}_{n}$ can be used in the generation of special curves, such as MNT
curves [13, 16], and in the generation of elliptic curves that do not
necessarily have prime order [1].

Moreover, problems such as primality testing and proving [1], the generation
of elliptic curve parameters [11] and the representability of primes by
quadratic forms [5] could be considerably improved once we know more about
Ramanujan polynomials $\mathcal{P}_{n}$ and the discriminant and traces.

##  9\. Acknowledgements

The majority of this research was done during the Research Science Institute
(RSI) at MIT in the summer of 2022. First I would like to thank my mentor Alan
Peng for his much-valued mentorship, guidance and support, which included
having daily meetings, sharing ideas, and clarifying any confusion, throughout
the completion of this work. I would like to thank Prof. Andrew Sutherland
from the MIT Math Department for suggesting this research project and Dr.
Tanya Khovanova, Prof. Ankur Moitra, Prof. David Jerison and Dr. John Rickert
for their helpful comments and suggestions on a rough draft of this
manuscript, and for making this research project possible. I would also like
to thank Prof. Aristides Kontogeorgis for giving us the pari-gp code to
compute $\mathcal{P}_{n}$ for higher values of $n$. Thank you to the MIT and
CEE for giving me the opportunity to attend the RSI and work on this project.
Thank you to my sponsors, Mr. Nicholas Nash and Ms. Phalgun Raju, for making
my participation in RSI possible. Finally, I would like to thank all my
Rickoid friends who turned into a family and made my summer a memorable one.

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