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I am working through Algebraic Geometry: A Problem Solving Approach and am stuck on exercise 2.4.22.

The previous problem was to consider $y^2 = 4x^3 + b_2x^2 + 2b_4x + b_6$ and transform this with $x = x_1$ and $y= 2y_1$ into

$$y_1^2 = x_1^3 + \frac{b_2}{4}x_1^2 + \frac{b_4}{2}x_1 + \frac{b_6}{4}.$$

Then I'm told that we can factor the right side to get $$y_1^2 = (x_1 - e_1)(x_1 - e_2)(x_1 - e_3).$$

My problem is to show that $e_1, e_2,$ and $e_3$ are distinct. The hint is to recall that the cubic curve $$V((x_1 - e_1z)(x_1 - e_2z)(x_1 - e_3z) - y^2z)$$ is smooth in $\mathbb{P}^2$, so my idea was to take partial derivatives and maybe get a contradiction by finding a singular point when the roots are not distinct. However, I can't seem to get that to work out.

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  • $\begingroup$ Are we talking about real numbers here? Then your problem is not well defined. When $x$ is a very large negative number, $y^2$ is a negative number. Also, it will all depend on the $b_n$ numbers. What do you know about those? $\endgroup$ – Andrei Apr 9 at 17:51
  • $\begingroup$ @Andrei I am working over $\mathbb{C}$ so $x$ and $y$ as well as $b_i$ could be complex. $\endgroup$ – Smash Apr 9 at 17:56
  • $\begingroup$ A quadratic polynomial cannot be factored with three factors and have three roots. $y$ is not a polynomial. $\endgroup$ – Yves Daoust Apr 9 at 18:08
  • $\begingroup$ The distinctness of the roots depends on the discriminant of the cubic. The roots are distinct iff the discriminant is zero. $\endgroup$ – Somos Apr 9 at 19:30
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WLOG, assume that $e_1 = e_2$. Then the polynomial defining the curve is $$ f = y^2 - (x - e_1)^2 (x - e_3) $$ so $$ \frac{\partial f}{\partial y} = 2y \qquad \frac{\partial f}{\partial x} = 2(x - e_1)(x - e_3) + (x - e_1)^2 = (x - e_1)(2(x-e_3) + (x-e_1)) \, . $$ Then $\frac{\partial f}{\partial x}(e_1,0) = 0 = \frac{\partial f}{\partial y}(e_1,0)$, so the curve is not smooth at the point $(e_1, 0)$.

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