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    $\begingroup$ I think that the conclusion from this would in fact be that j(n.tau) satisfied a polynomial of low degree but with humungous coefficients. Computer calculation indicate that j(5.tau) is in fact not an integer for tau as above, although it is close to one. $\endgroup$
    – Kevin Buzzard
    Commented Nov 9, 2009 at 22:27
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    $\begingroup$ I don't think $j(n \tau)$ should be an integer. I think that would indicate that the order $Z[n \tau]$ has trivial class group, and I'm pretty sure that's not true. But I'm rusty on how the elliptic curve/class field dictionary works for nonmaximal orders. $\endgroup$
    – David E Speyer
    Commented Nov 9, 2009 at 22:30
  • $\begingroup$ I tried this on a computer. The conclusion is that j(5*tau) is a root of an irreducible degree 6 polynomial with coefficient of the order of 10^{100}. $\endgroup$
    – Kevin Buzzard
    Commented Nov 9, 2009 at 23:51
  • $\begingroup$ In fact the root we want is the one which is about -10^{87} and within 10^{-6} of an integer. The other roots are "random" complex numbers (not near integers, or even near the reals in general) of size about 10^5. $\endgroup$
    – Kevin Buzzard
    Commented Nov 9, 2009 at 23:54