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The discriminant $\Delta$ of a Weierstrass equation over a field $K$ is an element of $K$ defined in terms of the Weierstrass coefficients. If the Weierstrass equation is $y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6,$ then $\Delta$ is given by a polynomial expression in $a_1,\dots,a_6$, namely, $\Delta=-b_2^2b_8 - 8 b_4^3 -27 b_6 ^2 + 9 b_2 b_4 b_6$ where \begin{aligned} b_2 &= a_1^2 + 4 a_2\\ b_4 &= 2a_4 + a_1 a_3\\ b_6 &= a_3^2 + 4 a_6 \\ b_8 &= a_1^2 a_6 + 4 a_2 a_6 - a_1 a_3 a_4 + a_2 a_3^2 - a_4^2. \end{aligned}

Then $\Delta\neq 0$ if and only if the equation defines a smooth curve, in which case its projective closure gives an elliptic curve.

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• Last edited by John Cremona on 2021-01-07 09:43:10
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