Elementary explanation for why $j\left(\frac{1+\sqrt{-163}}2\right)\in\mathbb{Z}$

Question

I'm looking into the math behind Ramanujan's Constant $e^{\pi\sqrt{163}}$. The idea is to prove that $j\left(\frac{1+\sqrt{-163}}2\right)\in\mathbb{Z}$ and then use the $q$-expansion. The first part is the hardest.

I'm reading Cox's Primes of the form $p=x^2+ny^2$, and in Chapter 10, he gives an elementary proof that for an order $\mathcal{O}$ in an imaginary quadratic field and for a fractional ideal $\mathfrak{a}$, $j(\mathfrak{a})$ is algebraic with degree at most the class number of $\mathcal{O}$. This shows that $j\left(\frac{1+\sqrt{-163}}2\right)\in\mathbb{Q}$. In the next chapter, he proceeds to prove that it's an integer, but the proof makes use of class field theory, which is too hard for me to learn.

My question: is there a relatively elementary way to prove that $j\left(\frac{1+\sqrt{-163}}2\right)\in\mathbb{Z}$, without using class field theory?

Answer 1

There is no elementary proof?

Paraphrasing Silverman's Advanced Topics in the Arithmetic of Elliptic Curves as well as Cox and Peter Wu. For $N\ge 1$ not a square and $\Im(\tau)>0$ let $$F_N(\tau) = \prod_{ad=N,b\bmod d} (j(\tau)-j(\frac{a\tau+b}{d}))$$

• This is a non-zero modular function for the full modular group.

• It is holomorphic on the upper half-plane.

• Its Fourier expansion $F_N(\tau)= \sum_{n\ge -M_N} c_{N,n} e^{2i\pi n \tau}$ has integer coefficients.

• The negative-most coefficient $c_{N,-M_N}$ is a root of unity (because this is the case for each $j(\tau)-j(\frac{a\tau+b}{d})$ factor)

All this together implies that $F_N(\tau) =G_N(j(\tau))$ with $G_N(X) \in \Bbb{Z}[X]_{monic}$.

Finally $j(\frac{1+i\sqrt{163}}2)$ is a root of $G_{41}$ (where $41=|\frac{1+i\sqrt{163}}2|^2$)

So $j(\frac{1+i\sqrt{163}}2)$ is an algebraic integer.

Its distinct Galois conjugates give distinct ideal classes with CM by $\Bbb{Z}[\frac{1+i\sqrt{163}}2]$. As there is no such distinct class, $j(\frac{1+i\sqrt{163}}2)$ has no distinct Galois conjugate, so it is an integer.

Answer 2

Here's a sketchy proof, essentially rewriting those in Cox book: your elliptic curve has a endomorphism of degree $41$, which is prime. Assume you've already know there is a polynomial $j_N(x,y)$, s.t. $j_N(j(\tau),j(N\tau))=0$. We will show that the $j_N(x,x)$ is monimic, when $N$ is prime.

The order-$N$ sublattice of $[1,\tau]$ is $[1/N,\tau],\,[1,\tau/N],\dots,[1,(\tau+N-1)/N]$. Which indicates that $j_N(j(\tau),y)=(y-j(N\tau))(y-j(\tau/N))\cdots(y-j((\tau+N-1)/N))$. Thus the coefficients of $y^{N+1}$ is 1, and coefficients of $y^{N}$ is $-j(N\tau)-j(\tau/N)-\cdots-j((\tau+N-1)/N)$.

Nextly $j(N\tau)+j(\tau/N)+\cdots+j((\tau+N-1)/N)$ is a monomic polynomial of $j(\tau)$ with degree $N$, this can be done by comparing $q$-expansions, lowest term is $q^{-N}$ contributed by $j(N\tau)$ so the coefficients of $j(\tau)^N$ is 1.

Finally look at $j_N(x,x)$, the highest term is contributed by $x^{N}y^{N}$ in $j_N(x,y)$, which has coefficients $\pm 1$, which proves the integrality.