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A long Weierstrass equation is an equation of the form $$y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6$$ Why are the coefficients named $a_1, a_2, a_3, a_4$ and $a_6$ in this manner, corresponding to $xy, x^2, y, x$ and $1$ respectively? Why is $a_5$ absent?

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If you make a standard change of variables $X=u^2x$ and $Y=u^3y$, the shape of the Weierstrass equation is preserved, and the coefficients of the new equation are $$Y^2+a_1uXY+a_3u^3Y =X^3+a_2u^2X^2+a_4u^4X+a_6u^6.$$ In other words, the new coefficients replacing the $a_i$ are of the form $a_iu^i$, for $i=1,2,3,4,6$. Since no coefficient changes by a power $u^5$, we don't call any of them by $a_5$.

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    $\begingroup$ Do you really think that is the original reason for the labels $a_i$? An explanation that has always seemed more compelling for me is that when you use coordinates $(z,w) = (-x/y,-1/y)$ with $z$ being a uniformizer at the identity, the formula for $w$ as a power series in $z$ has coefficients in ${\mathbf Z}[a_1,\dots,a_6]$ where the coefficient of $z^n$ is homogeneous of degree $n$ provided $a_i$ gets weight $i$. That's part of the first proposition in the chapter on formal groups in Silverman's Arithmetic of Elliptic Curves). $\endgroup$ – KCd Apr 26 '14 at 3:14
  • $\begingroup$ I don't claim to know that this is the original reason... it's just one reason, and an easy one to remember. Moreover, this reason appears in Chapter III of Silverman, while yours appears in Chapter IV! $\endgroup$ – Álvaro Lozano-Robledo Apr 28 '14 at 3:09
  • $\begingroup$ @KCd, isn't it from the weights of the Eisenstein series in the classical (not "long") Weierstrass equation? I mentioned it at math.stackexchange.com/questions/821187/… but probably you know the history much better. $\endgroup$ – zyx Jun 9 '14 at 20:34
  • $\begingroup$ @zyx: In my previous comment all I meant is that the explanation I like for the choice of notation $a_i$, omitting $i = 5$, is from formal groups, not that this is the original reason. I believe the notation for "long" Weierstrass equations (and minimal Weierstrass equations) was first set up systematically by Tate. See Stein's scans modular.math.washington.edu/Tables/antwerp/tate/tate.pdf and modular.math.washington.edu/Tables/antwerp/deligne from the Antwerp volumes and section 6.3 of Milne's survey paper jmilne.org/math/xnotes/Tate.pdf. $\endgroup$ – KCd Jun 9 '14 at 23:07
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This is because one thinks of each symbol having a weight. If $y$ has weight 3, $x$ has weight 2, and each $a_i$ has weight $i$, then the total weight of each term is exactly 6.

But where do these numbers come from? To answer that question, one has to look at the order of the pole at $\infty$. Consider the homogeneous form of the Weierstrass equation: $$v^2 w + a_1 u v w + a_3 v w^2 = u^3 + a_2 u^2 w + a_4 u w^2 + a_6 w^3$$ The point at $\infty$ has homogeneous coordinates $(u : v : w) = (0 : 1 : 0)$. Taking $x = u / w$ and $y = v / w$ recovers the original affine coordinates. On the other hand, if we take $\xi = u / v$ and $\eta = w / v$, we get: $$\eta + a_1 \xi \eta + a_3 \eta^2 = \xi^3 + a_2 \xi^2 \eta + a_4 \xi \eta^2 + a_6 \eta^3$$ One can check that $\xi$ is a local coordinate function near $(\xi, \eta) = (0, 0)$, i.e. near $\infty$. On the other hand, $$\eta = \frac{\xi^3}{1 + a_1 \xi + a_3 \eta - a_2 \xi^2 - a_4 \xi \eta - a_6 \eta^2}$$ and the denominator does not vanish at $\infty$, so we deduce $\eta$ has a zero of order 3 at $\infty$. Hence, $y = 1 / \eta$ has a pole of order 3 at $\infty$, and $x = \xi / \eta$ has a pole of order 2. That is why $y$ is considered to have weight 3 and $x$ is considered to have weight 2.

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