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I am attempting to understand equations of the form $z^2 = x^3 + Ax + B + Cy + y^3$, which is a generalisation of an elliptic curve $y^2 = x^3 + Ax + B$. I know that to make sure an elliptic curve has nice properties, we require that $f(x) = x^3 + Ax + B$ has three distinct roots. This leads to $4A^3 + 27B^2 \neq 0$. I wondered if the generalised equation has a similar inequality in A, B and C which forces the surface to have particularly "nice" properties, such as a group law.

I guessed that if $g(x,y) = x^3 + Ax + B + Cy + y^3$ could be factored into distinct linear factors then it would be sufficiently "nice". Trying to find the required restriction on the coefficients, I first set $g(x,y) = (a_1 x + b_1 y + c_1)(a_2 x + b_2 y + c_2)(a_3 x + b_3 y + c_3)$ and expanded the RHS. This results in the following system of equations, $$a_1 a_2 a_3 = 1$$ $$b_1 b_2 b_3 = 1$$ $$a_3 b_2 c_1 + a_2 b_3 c_1 + a_3 b_1 c_2 + a_1 b_3 c_2 + a_2 b_1 c_3 + a_1 b_2 c_3 = 0$$ $$a_2 a_3 b_1 + a_1 a_3 b_2 + a_1 a_2 b_3 = 0$$ $$a_3 b_1 b_2 + a_2 b_1 b_3 + a_1 b_2 b_3 = 0$$ $$a_2 a_3 c_1 + a_1 a_3 c_2 + a_1 a_2 c_3 = 0$$ $$b_2 b_3 c_1 + b_1 b_3 c_2 + b_1 b_2 c_3 = 0$$ $$A = a_3 c_1 c_2 + a_2 c_1 c_3 + a_1 c_2 c_3$$ $$B = c_1 c_2 c_3$$ $$C = b_3 c_1 c_2 + b_2 c_1 c_3 + b_1 c_2 c_3$$ But I don't see how I can use these to deduce a restriction on the coefficients. Hopefully someone can find a simpler approach to this problem!

By "nice" I mean that every line meeting the surface must meet it in exactly three points (counting multiplicities and involving points at infinity if necessary). My ultimate goal is to find the weakest, necessary condition for such a surface to have this property.

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  • $\begingroup$ What of $4A^3+27(B+Cy+y^3)^2\ne 0$? $\endgroup$
    – Allawonder
    Feb 25 at 21:51
  • $\begingroup$ @Allawonder (assuming this a condition which must be satisfied for all y - i.e. the LHS has no real roots) I'm not sure whether that is sufficient for the property I want the surface to have $\endgroup$ Feb 25 at 22:15
  • $\begingroup$ would the fact that, for every choice of y, there is an elliptic curve in the xz-plane mean that the the equation describes an elliptic surface? (en.wikipedia.org/wiki/Elliptic_surface) $\endgroup$ Feb 25 at 22:26
  • $\begingroup$ original reference jstor.org/stable/1968156?seq=1 and written up, in English, in the 2000 book by Schinzel, Polynomials with special regard to Reducibility, cambridge.org/core/books/… $\endgroup$
    – Will Jagy
    Feb 25 at 23:10
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Turns out the reference for part reduction was a book by Schinzel, he of the Hypothesis.

enter image description here

there is a nice answer to this sort of thing. Throw in a variable $z$ to homogenize, $$ f = x^3 + A x z^2 + B z^3 + C y z^2 + y^3 $$

Next, find the second partial derivatives of $f$ by $x,y,z$ and write out the Hessian matrix. The entries in $H$ are linear in $x,y,z,$ as they begin degree three.

Finally, calculate the determinant of $H.$ This is a homogeneous degree three. The form $f$ factors into linear terms over $\mathbb C$ if and only if the determinant is a constant multiple of $f.$ The reference is by Brookfield, I believe I saved part of it, let me look.

Gary Brookfield (2016) Factoring Forms, The American Mathematical Monthly, 123:4, 347-362

enter image description here

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  • $\begingroup$ thanks! I will look into the Hessian matrix $\endgroup$ Feb 25 at 21:31
  • $\begingroup$ @ThomasKing finished it, the determinant is not a constant multiple of the original form, as the determinant form has no $x^3$ or $y^3$ or $z^3$ from my letters. So, your original does not factor even if we allow complex coefficients; there rationals are also ruled out. It is still permissible to have it factor as a linear times a quadratic form, when written including my letter $z.$ However, that does not seem to hold either. $\endgroup$
    – Will Jagy
    Feb 25 at 21:50
  • $\begingroup$ factoring as a linear times a quadratic may be sufficient, the cubic $x^3 - x - y + y^3$ factors as $(x+y)(x^2 + y^2 - xy - 1)$ so I think the level curve at $z=0$ in $z^2 = x^3 - x - y + y^3$ has the stated property but I am not sure about the rest of the surface $\endgroup$ Feb 25 at 22:20
  • $\begingroup$ @ThomasKing I think it is impossible for most of your triples $A,B,C$ There is a more complicated condition for deciding this less strict case, linear times quadratic. Don't immediately remember where. $\endgroup$
    – Will Jagy
    Feb 25 at 22:30

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