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I found the x-coordinate of my 4-torsion point on E: $y^2=x^3-3267x+45630$ given by $x=15\pm 36B$ with $B^2=-2$. My question here is how am I to find the y-coordinate of this 4-torsion point?

shahrina ismail
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    $y= \pm \sqrt{x^3-3267x+45630}$ – reuns Nov 03 '16 at 05:47
  • @user1952009. i tried doing that with no success.. by trying an error so many times I manage to get $y$-coordinate as $27a(a^2-4B-5)$ where $a^4-5a^2-32=0$. But how to get it straight away without trying an error. – shahrina ismail Nov 03 '16 at 06:53
  • Have you done the reduction modulo $p$ ? – reuns Nov 03 '16 at 07:15
  • What do you mean by finding the $y$-coordinate of the point? Do you expect it to be defined over $\mathbb{Q}(\sqrt{-2})$? What kind of expression are you looking for? – Álvaro Lozano-Robledo Nov 03 '16 at 15:50
  • @ÁlvaroLozano-Robledo Yes I want it to be defined over $\mathbb{Q}(\sqrt{-2})$ I actually try an error so many times until I get the answer y-coordinate = $27a(a^2−4B−5)$ where $a^4−5a^2−32=0$ using Sage. I did such a way that I equate LHS expression of my elliptic curve to RHS and play around to make them equal to get y-coordinate. But is there a way I can get it straight away without having to try an error. – shahrina ismail Nov 03 '16 at 23:44
  • I am confused, because the $y$ coordinate is not defined over $\mathbb{Q}(\sqrt{-2})$ but over a quartic extension of $\mathbb{Q}$. In fact, the poynomial defining $a$ in your comment here is a quartic, and then you are writing $y$ defined over this quartic extension. – Álvaro Lozano-Robledo Nov 04 '16 at 00:40
  • @ÁlvaroLozano-Robledo sorry my mistake I have been confused. Since I have the $x$-coordinate in $\mathbb{Q}\sqrt{-2}$, so I assumed that $y$-coordinate should be over $\mathbb{Q}\sqrt{-2}$ as well. y-coordinate is actually defined over the quartic extension. And the thing is that I got the $y$-coordinate by trying an error. Is there a way I can know the y-coordinate. I sub the x-coordinate into my elliptic curve expression and got this $y^2=186624*B - 116640$. – shahrina ismail Nov 04 '16 at 02:08
  • I think that's the best you can do. The field of definition of the $y$ coordinate is a quartic extension of $\mathbb{Q}$ that is not Galois. The Galois closure of $\mathbb{Q}(y)$ is of degree 8 over $\mathbb{Q}$. – Álvaro Lozano-Robledo Nov 04 '16 at 15:56
  • Thank you @ÁlvaroLozano-Robledo.. Your feedback has been so helpful.. – shahrina ismail Nov 05 '16 at 12:02

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I was looking at the [elliptic-curves] tagged questions these days, found this post, which is already answered in the comments. Maybe i should say some words, that would not fit in a comment. Here is what i am doing in such situations. We want to compute $$ \sqrt{x^3-3267x+45630}\ ,\qquad\text { for the special value }x=15+36B\ ,\ B=\sqrt{-2}\ . $$ By chance, the number under the square root may be a square in the field $K=\Bbb Q(\sqrt{-2})=\Bbb Q(B)$. Let us see if this is the case.

sage: K.<B> = QuadraticField(-2)
sage: def f(x):    return x^3 - 3267*x + 45630

sage: f(15+36*B)
-186624*B - 116640

sage: f(15+36*B).factor()
(-B - 1)^8 * (B - 1)^6 * B^10 * (-2*B + 3)

sage: (3-2*B).norm()
17

sage: (3-2*B).is_square()
False

We need to adjoin a root of it. So the factor $3-2B=3-2\sqrt{-2}$ of norm $17$ is the problem in this case. (All other factors come at even powers.)

dan_fulea
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