# A particular cubic can be written as $y_1^2 = (x_1 - e_1)(x_1 - e_2)(x_1 - e_3)$. Show that $e_1, e_2,$ and $e_3$ are distinct.

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.

• 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? – Andrei Apr 9 at 17:51
• @Andrei I am working over $\mathbb{C}$ so $x$ and $y$ as well as $b_i$ could be complex. – Smash Apr 9 at 17:56
• A quadratic polynomial cannot be factored with three factors and have three roots. $y$ is not a polynomial. – Yves Daoust Apr 9 at 18:08
• The distinctness of the roots depends on the discriminant of the cubic. The roots are distinct iff the discriminant is zero. – Somos Apr 9 at 19:30

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 $$(e_1, 0)$$.