# What is the Galois group of the modular equation?

Let $$j(\tau)$$ be the modular invariant. Let $$\mathcal M_n^*$$ be the set of all $$2$$-by-$$2$$ matrices with relatively prime integer entries and with determinant $$n$$. The modular equation is defined by $$\Phi_n(X,j(\tau))=\prod_{M\in \operatorname{SL}_2(\mathbf Z)\backslash\mathcal M_n^*}(X-j(M\tau)).$$ What is the Galois group of the polynomial $$\Phi_n(X,X)$$?

For more details on the modular equation, see

Zagier: Elliptic Modular Forms and Their Applications, p. 68

Andrew Sutherland: The modular equation

• @Bullet51, we have $j\circ\gamma = j$ whenever $\gamma\in \operatorname {SL}_2(\mathbf Z)$. The product in question is extended over some set of right coset representatives of $\mathcal M_n ^*$ modulo $\operatorname {SL}_2(\mathbf Z)$. – Shimrod Sep 6 '19 at 14:30
• Your definition of ${\mathcal M}_n$ does not seem to depend on $n$ – Venkataramana Sep 6 '19 at 15:04
• Galois group over what field: ${\bf Q}(j(\tau))$ or ${\bf C}(j(\tau))$? It does matter for all but a few $n$. – Noam D. Elkies Sep 6 '19 at 15:37
• @NoamD.Elkies, The polynomial $\Phi_n(X,X)$ has integer coefficients, so the question is what is its Galois group over $\mathbf Q$. I would be also interested in the Galois group of $\Phi_n(X,Y)$ over $\mathbf Q(Y)$ and what happens when we specialize the $j$-invariant at a particular CM point. I would really appreciate if you could comment on these questions. – Shimrod Sep 6 '19 at 15:54
• You may check the book of Clemens Adelmann "The decomposition of primes in torsion point fields" ; Section 5.2 contains an interesting answer of your question in a special case. – Zakariae.B Sep 7 '19 at 14:51

## 1 Answer

A long comment

It is easier to look at $$A_n = \{ M\in M_n(\Bbb{Z}), \det(M)=n\}$$ $$\psi_n(X,z) = \prod_{M \in SL_2(\Bbb{Z})\setminus A_n} (X-j(Mz))$$ The coefficients of the polynomial are meromorphic $$SL_2(\Bbb{Z})$$-invariant with poles only at $$i\infty$$ where they have rational $$q$$-expansion so that $$\psi_n(X,z)= \Psi_n(X,j(z)), \qquad \Psi_n(X,Y)\in \Bbb{Q}[X,Y]$$ $$j$$ is an isomorphism $$SL_2(\Bbb{Z})\setminus\Bbb{H} \to \Bbb{C}$$ thus the zeros of $$\Psi_n(j(z),j(z)) =\prod_{M\in SL_2(\Bbb{Z})\setminus A_n}(j(z)-j(Mz))$$ are the $$j(z),\Im(z)>0$$ such that $$SL_2(\Bbb{Z}) z \in SL_2(\Bbb{Z}) A_n z\implies z \in A_n z$$ $$\implies z= \frac{az+b}{cz+d}, \qquad ad-bc=n$$ It means $$cz^2+(a-d)z-b=0$$ so $$cz+d$$ is the algebraic integer $$cz+d = t/2\pm \sqrt{(t/2)^2-n}\in O_K,\qquad K=\Bbb{Q}(\sqrt{(t/2)^2-n}), t=a+d,(t/2)^2-n<0$$

$$\Bbb{Z}+z\Bbb{Z}=\Bbb{Z}+\frac{az+b}{cz+d}\Bbb{Z}$$ means $$(cz+d)(\Bbb{Z}+z\Bbb{Z})=(cz+d)\Bbb{Z}+(az+b)\Bbb{Z}$$ so the lattice $$\Bbb{Z}+z\Bbb{Z}$$ has CM by $$\Bbb{Z}[cz+d]$$.

Conversely if a lattice $$\Bbb{Z}+w\Bbb{Z}$$ has CM by $$\Bbb{Z}[cz+d]$$ then $$cz+d = Cw+D \in \Bbb{Z}+w\Bbb{Z}$$ and $$(Cw+D) w \in \Bbb{Z}+w\Bbb{Z}$$ so that $$w =\frac{Aw+B}{Cw+D}$$ which is the same equation as above with $$N=AD-BC$$ instead of $$n$$, and since $$t,n$$ can be recovered just from $$cz+d$$ it means $$N=n$$ and $$\Psi_n(j(z),j(z))=0$$.

Thus for a lattice $$L$$ then $$\Psi_n(j(L),j(L))=0$$ iff for some $$t^2<4n$$, $$(t/2\pm \sqrt{(t/2)^2-n})L \subset L$$.

To enumerate all such lattices we could enumerate the $$t^2<4n$$, the rings $$O_K\supset R\supset \Bbb{Z}[ t/2\pm \sqrt{(t/2)^2-n}]$$ and the classgroup $$Cl(R^\land)$$ of the ring with CM exactly by $$R$$, then $$\Psi_n(j(L),j(L))=0$$ iff $$L = r \mathfrak{c}$$ for some $$\mathfrak{c} \in Cl(R^\land)$$.

The Galois group of $$\Psi_n(X,X)\in \Bbb{Q}[X]$$ will permute and exchange $$Cl(R^\land),Cl(\overline{R}^\land)$$ where $$\overline{R}$$ is the complex conjugate ring.