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My question is how to calculate the value of $j(2\sqrt{-163})$ and its minimal polynomial, where the $j$ is elliptic modular function (see https://mathworld.wolfram.com/j-Function.html). The class number of $\mathbb{Q}(\sqrt{-652})$ shows that $j(2\sqrt{-163})$ is degree 6, and its approximately equal to $e^{4\pi\sqrt{163}}\approx 4.75\times 10^{69}$. I tried ellj function in Pari/GP to compute minimal polynomial, but could only output approximation:

? minpoly(ellj(sqrt(-652)))

%2 = x - 4.750778730825177725463920957 E69

Thanks for your help.

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    $\begingroup$ Are you sure it's 3 and not 6? $\endgroup$
    – Will Sawin
    Commented Oct 4, 2023 at 17:20
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    $\begingroup$ Please use a high-level tag like "nt.number-theory". I added this tag now. $\endgroup$
    – GH from MO
    Commented Oct 4, 2023 at 18:27

1 Answer 1

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The minimal polynomial is

x^6 - 4750778730825177725463920948909726618214491718039471628856160047142000*x^5
+ 1247257156019654977752984724237035223986874495851369721538788760906203681342078134750000*x^4
- 3711837295929728841959711983585876317003061224025827146858434575359795370054804675000000000*x^3
+ 3160150517834696901784153329054103916683908447320628422609243420587305378116705388187500000000*x^2
- 700265800610377949731030701279743787907896705372015247301698864378963568331565476375000000000000*x
+ 327451677250026694198133278336402500637269636726169887842595798606445909367058576435277640625000000000000

It has degree $6$, not $3$, because $\sqrt{-652}$ generates the quadratic order of discriminant $-4^2 163$ which has index $4$ in the full ring of integers, and this order has class number $6$. The index-$2$ order is generated by $\sqrt{-163}$, and $j(\sqrt{-163})$ has cubic minimal polynomial

x^3 - 68925893036109279891085639286946000*x^2 
+ 102561728837719322645921325412908000000*x 
- 18095625621665522953693950872675200892692248000000000

gp's default precision is not nearly enough to find these polynomials. 200 digits suffice for the cubic, and 1200 for the sextic, e.g.

\p 1200
algdep(ellj(sqrt(-652)),6)

The cubic can also be obtained by plugging $j\bigl(\frac12(1+\sqrt{-163})\bigr) = -640320^3$ into the modular polynomial $\Phi_2(j,x)$; the sextic can be computed similarly with a bit more work.

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  • $\begingroup$ Thank you for your answer! In addition, for the imaginary quadratic field $K=\mathbb Q(\sqrt{D})(D<0)$, if $D\equiv 1(\text{mod} 4)$, then $\text{deg}(j(\sqrt{D}))=3h(K)$; If $D\equiv 2,3(\text{mod} 4)$, then $\text{deg}(j((1+\sqrt{D})/2))=2h(K)$? $\endgroup$
    – GuoJi
    Commented Oct 5, 2023 at 15:34
  • $\begingroup$ @Noam Can you kindly take a look at this parametric elliptic curve for $x^4+y^4+z^4=1$ in this post? $\endgroup$ Commented Feb 9 at 8:44

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