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I have a 4-torsion point on elliptic curve

$$E : y^2=x^3-3267x+45630$$ given by $(x,y)=(15-36\sqrt{-2},27\sqrt{256\sqrt{-2} - 160})$ which I have checked that it satisfies my elliptic curve. Let $\beta=\sqrt{-2}$. So we have $$(x,y)=(15-36\beta,27\sqrt{256\beta - 160})$$

But than, I was told that the coordinate can be written as $$(x,y)=(15-36\beta,27\alpha(\alpha^2-4\beta-5))$$ where $\alpha^4 - 5\alpha^2 - 32=0$ and $\beta=\sqrt{-2}$.

My question is how is the y-coordinate $27\sqrt{256\beta - 160}$ equals to $27\alpha(\alpha^2-4\beta-5)$ ? How can I obtain that?

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  • $\begingroup$ you have 4 possible values for $y$ : $\pm\sqrt{\pm256\beta - 160}$, and those should correspond to $\alpha_k(\alpha_k^2-4\beta-5)$ where $\alpha_1,\ldots,\alpha_4$ are the $4$ roots of $X^4-5X^2-32$ $\endgroup$ – reuns Nov 6 '16 at 23:03
  • $\begingroup$ @user1952009 But how did you know that? I mean it is not obvious that $y=\pm\sqrt{\pm256\beta-160}$ = $\alpha_k(\alpha_k^2-4\beta-5)$. Is there like a specific calculation I should do to obtain that? Because now I have another coordinate $$(x,y)=\bigg[-\dfrac{27}{2}\beta-\dfrac{1}{2}\sqrt{2430\beta+24768}-\dfrac{15}{2}, \sqrt{-264627\beta-\dfrac{1}{2}\sqrt{1180628067960\beta+5672869080264}-557685}\bigg]$$ where $\beta =\sqrt{17}$. How am I to write this in the form of $\alpha, \beta$ like above? $\endgroup$ – shahrina ismail Nov 7 '16 at 4:58
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This is a late answer to the question. For an easy typing, i will use the letters $b$ for a root of the polynomial $X^2+2\in\Bbb Q[X]$, and $a$ for a root of the polynomial $X^4 -5X^2-32\in\Bbb Q[X]$.

Note than as a short intermezzo the relation, used in the sequel in blue: $$ (a^2-5)^2=a^4-10a^2+25 =(a^4-5a^2-32)-5(a^2-5)+32=-5(a^2-5)+32\ . $$ Now we are ready to show / compute:

$$ \begin{aligned} a(a^2-4b-5) &=\pm \sqrt{-256b-160} \\ &\qquad\text{ in the equivalent, purely algebraic form} \\ \Big(\ a(a^2-4b-5) \ \Big)^2 &=-256b-160 \ . \end{aligned} $$

Computation:

$$ \begin{aligned} \Big(\ a(a^2-4b-5) \ \Big)^2 &= a^2(\ (a^2-5)-4b\ )^2 \\ &= a^2(\ \color{blue}{(a^2-5)^2}-8b(a^2-5)+4b^2\ ) \\ &= a^2(\ \color{blue}{-5(a^2-5)+32}-8b(a^2-5)-32\ ) \\ &= a^2(\ -5(a^2-5)-8b(a^2-5)\ ) \\ &= a^2(a^2-5)(\ -5-8b\ ) \\ &= 32(\ -5-8b\ ) \\ &=-256b-160 \ . \end{aligned} $$ This means that we have the points on the given elliptic curve: $$ \begin{aligned} P&=(15-36b,\ 27\sqrt{256b - 160}) \\ &=(15-36b,\ \pm a(a^2+4b-5))\ , \\ Q&=(15+36b,\ 27\sqrt{-256b - 160}) \\ &=(15+36b,\ \pm a(a^2-4b-5))\ . \end{aligned} $$ But they differ. So the information

  • I was told that the coordinate can be written as...

may be possibly traced back to only

  • I was told that the $y$-coordinate of some point in $E(K)$, for $K=\Bbb Q(\alpha,\beta)$, can be written as $27\alpha(\alpha^2−4\beta−5))$...

Note: Such computations are easily covered by using computer algebra systems. My choice of the wappon is sage, and we can type to get the above:

sage: R.<x>  = QQ[]
sage: R.<x>  = PolynomialRing(QQ)
sage: K.<a,b> = NumberField( [x^4 - 5*x^2 - 32, x^2+2] )
sage: ( a*(a^2-4*b-5) )^2
-256*b - 160

For the second case, things are "slightly more complicated". I would start by using a new letter for $\sqrt{17}$, maybe $c=\sqrt{17}$, although we could write it in terms of $a,b$ in the same field $K=\Bbb Q(a,b)$,

sage: sqrt(K(17))
-2/3*a^2 + 5/3

and working economically in $\Bbb Q(c)$ we factorize first as much as possible in $\Bbb Q$ and $K$:

sage: L.<c> = QuadraticField(17)
sage: gcd( 1180628067960, 5672869080264 ).factor()
2^3 * 3^15 * 17
sage: w = ( 1180628067960*c + 5672869080264 ) / (2^3 * 3^15 * 17)
sage: w
605*c + 2907
sage: factor(w)
(-34840*c + 143649) * (-c) * (-1/2*c - 3/2)^16 * (-1/2*c + 3/2)
sage: L.units()
(c - 4,)
sage: (c-4)^6
-34840*c + 143649
sage: w / ( (c-4)^3 * (c+3)^8 / 2^8 )^2
-3/2*c + 17/2
sage: _.norm().factor()
2 * 17

This gives a hint on how to go on...

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