# An addition property of Weierstrass $\wp$

I want to show

$$\left( \begin{array}{ccccc} &1 &\wp(v) &\wp'(v) \\ &1 &\wp(w) &\wp'(w) \\ &1 &\wp(v+w) &-\wp'(v+w) \end{array} \right)=0$$

where $\wp$ denotes the Weierstrass elliptic function.

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You might find this thread useful. – t.b. Mar 18 '12 at 10:22
In particular, look at Bruno's answer. Your determinant vanishing is equivalent to the fact that $(P(v), P'(v))$, $(P(w), P'(w))$ and $(-P(v+w), -P'(v+w))$ are colinear, which is shown in the last paragraph of Bruno's answer. – David Speyer Mar 18 '12 at 15:41

As already mentioned in David's comment and Bruno's answer, this determinant is merely saying that the triangle formed by the three points $(\wp(v),\wp^\prime(v))$, $(\wp(w),\wp^\prime(w))$, and $(-\wp(v+w),-\wp^\prime(v+w))$ has area zero (i.e. they are collinear). (I presume the determinantal formula for the area of a triangle was taught to you in your geometry classes, no?)

In particular, you'll want to use the relation $\wp^\prime(z)^2=4\wp(z)^3-g_2 \wp(z)-g_3$. Your determinant is now equivalent to saying that a line in general position must intersect the elliptic curve $y^2=4x^3-g_2 x-g_3$ (or parametrically, $x=\wp(u)\quad y=\wp^\prime(u)$) in three points. I presume you already know how to show that a line and a cubic will have three intersection points at most. If you do the computations, you can obtain an expression for the third intersection point of a line going through the points $(\wp(v),\wp^\prime(v))$ and $(\wp(w),\wp^\prime(w))$ with the elliptic curve in terms of the coordinates of those two given points.

To get the equivalent expression in terms of $\wp(v+w)$ and $\wp^\prime(v+w)$, as Bruno mentions in his answer, one assembles the linear combination $\wp(u)+a\wp^\prime(u)+b$ and note that it has a triple pole at the origin (since $\wp^\prime(u)$ itself has a triple pole from the $-\frac2{u^3}$ term in its lattice series expansion). One can then find appropriate values of $a$ and $b$ such that $v$ and $w$ are zeroes of $\wp(u)+a\wp^\prime(u)+b$, and then find that $u=-v-w$ is a zero as well due to the properties of the Weierstrass functions within their fundamental period parallelogram. Make use of the symmetry of the two Weierstrass functions ($\wp$ is even; $\wp^\prime$ is odd), and you're golden.

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