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thanks in advance.

I am trying to prove that the Weierstrass equation y^2+a1xy+a3y=x^3+a2x^2+a4x+a6 can be written in the form of: y2=x3+ax+b ,curves are over some field K where char K≠2,3.

I found out that this could be done by completing the square and a substitution that will eliminates some coefficients. But couldn't find any step by step explanation. and I tried to do it for hours, but not succeeded. If I do just the RHS (x^3+a2x^2+a4x+a6) and substitue x->x-a/3 then the coefficient of x^2 becomes 0, and I keep try to manipulate the LHS (y^2+a1xy+a3y) , no idea what to do now Can anyone please help me please.

Student X
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1 Answers1

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Write the left as $y^2+(a_1x+a_3)y$ then complete the square

$$y^2+(a_1x+a_3)y=\left(y+\frac{1}{2}(a_1x+a_3)\right)^2-\left(\frac{1}{2}(a_1x+a_3)\right)^2$$

So you have $$\left(y+\frac{1}{2}(a_1x+a_3)\right)^2=x^3+ a_2x^2+a_4x+a_6+ \left(\frac{1}{2}(a_1x+a_3)\right)^2$$

Now use the trick you described on the RHS.

Rene Schipperus
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