# Why the cofficients of the $j$-invariant polynomial are integers and how to compute them?

In the Page 333 of the book “Elliptic curves number theory and cryptogarphy” Lawrence C. Washington.

Given $\tau_i\in H$(up-half plane),and $j(\tau_i)$ is the j-invariant of the lattice $\mathbb Z_\tau+\mathbb Z$. Compute the values of $j(\tau_i)$ and then can form the polynomial $(x-j(\tau_1))(x-j(\tau_2))(x-j(\tau_3))(x-j(\tau_4))$.

My question is why $j(\tau_i)\in\mathbb C$, but the coefficiets are true integers? And, how to compute them?

• That's moonshine. If someone wants to completely fill that to an answer, that will take some time. But until then just look up moonshine. – AHusain May 8 '18 at 7:38
• This is not moonshine. It's just Galois theory with some complex multiplication. The very next sentence in the book tells you why the coefficients are integers - you have a Galois-invariant product of terms. That whole section is an explanation of how you compute the coefficients. – S. Carnahan May 8 '18 at 9:09
• @S.Carnahan The Galois invariance will imply the the coefficients are rational numbers. In order to prove that they are integers, one needs to know that CM $j$-invariants are algebraic integers. This is, of course, quite standard and proven in any book that covers CM elliptic curves. – Joe Silverman May 8 '18 at 11:13
• @ S. Carnahan and @ Joe Silverman: Thanks a lot! It is still hard for a non-mathematician to understand the computation process. Say just for a student who has background in computer science and knows how to write programs according explicitly description of the algorithms, where to find the consults for such algorithms? – Licheng Wang May 8 '18 at 11:18
• If you can compute the roots $j(\tau_i)$ then the polynomial coefficients are (up to sign) the elementary symmetric functions of the roots. – Somos May 8 '18 at 16:43