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Let $p_1, p_2, t_1, t_2, a \in \mathbb{C}$ be constants. Consider the following plane complex curve in $\mathbb{C}^2$ ($c_1, c_2$ are indetermniates) \begin{align} & {p_1}^2 {p_2}^2 c_1 {t_1}^2 t_2 + {p_1}^2 p_2 {c_1}^2 c_2 {t_1}^2 t_2 + {p_1}^2 p_2 c_2 {t_1}^4 + {p_1}^2 c_1 {c_2}^2 {t_2}^3 \\ & + 2 p_1 p_2 {c_1}^3 c_2 {t_2}^3 + p_1 p_2 {c_1}^2 {c_2}^2 {t_1}^2 t_2 + p_2 {c_1}^3 {c_2}^2 {t_2}^3 = a {p_1}^2 p_2 c_1 c_2 {t_1}^2 t_2. \end{align} According to the article, the genus of the curve is the number of integer lattice points in the interior of the Newton polygon of the curve.

The Newton polygon of the curve is the convex hull of the points $(1,0)$, $(2,1)$, $(0,1)$, $(1,2)$, $(3,1)$, $(2,2)$, $(3,2)$, $(1,1)$. There are two integer lattice points in the Newton polygon: $(1,1)$, $(2,1)$. Therefore the genus of this curve is $2$. Is this correct? How to check that whether or not this curve is smooth? Is this curve an elliptic curve? Thank you very much.

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  • $\begingroup$ It is certainly not smooth, not even irreducible: it contains the line $c_2=0$. $\endgroup$
    – abx
    Commented Aug 2, 2018 at 7:59
  • $\begingroup$ I cannot follow the construction of the Newton polygon. Where does the term $(0,2)$ come from? Why is there no term $(1,1)$ (coming from the term on the right hand side)? The line $\{\,(\lambda,1)\mid \lambda\in\mathbb{R}\,\}$ seems to be a boundary component of the Newton polygon, so why does $(2,1)$ lie in the interior? Moreover, the answer depends on the choice of the constants since points in the Newton polygon vanish when some of the constants are zero. $\endgroup$
    – Philipp Lampe
    Commented Aug 2, 2018 at 8:08
  • $\begingroup$ @PhilippLampe, thank you very much. I will correct the post. $\endgroup$
    – Jianrong Li
    Commented Aug 2, 2018 at 8:21
  • $\begingroup$ @abx, thank you very much. I forgot some terms in the curve. I will add them. $\endgroup$
    – Jianrong Li
    Commented Aug 2, 2018 at 8:21
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    $\begingroup$ Macaulay2 or Sage can tell whether a curve is smooth and compute the genus. The answers depend on the chosen constants. For instance, the curve is not smooth for $p_1=b_1=0$ and $p_2=t_2=1$. Note that a curve of genus $2$ is not an elliptic curve since they have genus $1$. $\endgroup$
    – Philipp Lampe
    Commented Aug 2, 2018 at 10:38

1 Answer 1

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A computation with magma reveals that this curve is isomorphic, as a genus two curve over the function field $\mathbb{C}(p_1,p_2,t_1,t_2,a)$, to the hyperelliptic curve given by the Weierstrass equation $$y^2=f(x):=4t_2^6x^6 + 4p_1t_1^2t_2^4x^5 + (p_1^2t_1^4t_2^2 - 4p_1t_1^2t_2^4a - 4p_2t_1^2t_2^4)x^4 + (-2p_1^2t_1^4t_2^2a - 4p_1p_2t_1^4t_2^2 + 4p_1t_1^4t_2^3)x^3 + (2p_1^2t_1^6t_2 + p_1^2t_1^4t_2^2a^2 - 4p_1^2t_1^2t_2^4)x^2 - 2p_1^2t_1^6t_2ax + p_1^2t_1^8 $$ This defines an hyperelliptic curve for any value of the parameters such that the discriminant of the polynomial $f(x)$ is non-zero.

The discriminant is zero when $t_1=0$ or $t_2=0$ or $p_1=0$ (some cases one should consider apart) or a very large of huge degree irreducible polynomial in the parameters is zero.

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  • $\begingroup$ thank you very much. Since magma is not free, I do not have the software. I found that there is a webpage which could run magma: magma.maths.usyd.edu.au/calc. Could I have your codes which give the above result? I will try to run it on the webpage. I also have other curves which I need to compute the normal forms. Thank you very much. $\endgroup$
    – Jianrong Li
    Commented Aug 2, 2018 at 18:34
  • $\begingroup$ I used magma free page for the computations, which is fast. I am sending the code by email. $\endgroup$
    – Xarles
    Commented Aug 2, 2018 at 20:48

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