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We have an elliptic curve in its normal form:

$$y^2 = f(x) = x^3 +a x^2 + bx + c,$$ where $a,b,c$ are rational numbers. The discriminant here is said to be

$$-16(4b^3 + 27c^2) \quad \text{ or } \quad 4b^3 + 27c^2.$$

However, in Rational Points on Elliptic Curves (Silverman & Tate) it is defined as

$$-4a^3c + a^2b^2 + 18abc - 4b^3 - 27c^2.$$

It looks like the first is of the same form only when $a = 0$, so the second form is more general. My question is: where on earth is this derived from? There is no such derivation given in the book.

Clyde Kertzer
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  • Try computing the Jacobian and seeing which values of $(a,b,c)$ causes the curve to be singular. – David Lui Feb 10 '23 at 17:34
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    The linked question is talking about a different form though? It was talking about $y^2 = x^3 + \color{red}{ax + b}$. – VTand Feb 10 '23 at 17:35
  • Hmm, sorry. I don't know what the Jacobian of an elliptic curve is. – Clyde Kertzer Feb 10 '23 at 17:36
  • @VTand, well yes. But the point is the $x^2$ term has coefficient $0$ in the first case, regardless of how you notate it. – Clyde Kertzer Feb 10 '23 at 17:36
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    You can always reduce the form to kill off the quadratic term, so the discriminant is usually given in that form. If one applies that transformation (and its inverse), I imagine it moves between the statements of the discriminant. In any case, the discriminant you gave is just the general discriminant for a cubic (with constant term one). Just look up discriminant on Wikipedia and see how to calculate it... It's just a bit of linear algebra. – Brevan Ellefsen Feb 10 '23 at 17:40
  • Well sometimes isn’t quite difficult to remove the quadratic term? I was hoping someone could show how to derive it in normal form and not Weirestrass form – Clyde Kertzer Feb 10 '23 at 17:57

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Given the cubic $$ x^3 + ax^2 + bx + c, $$ replace $x$ with $X-\frac13 a$ to get $$ X^3 + BX + C = X^3 + (b-a^2/3)X + (c-ab/3+2a^3/27).$$ Now compute $4B^3 + 27C^2$ to get $$ 4b^3 + 27c^2 + 4a^3c - a^2b^2 - 18abc. $$

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