# How to prove that shifting does not change the discrimintant?

I have an elliptic curve given of the form:

$$E \hspace{2mm} : \hspace{2mm} y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6.$$

The discriminant of $E$ is given by $$\Delta(E) = -b_2^2b_8-8b_4^3-27b_6^2+9b_2b_4b_6,$$where $b_2=a_1^2+4a_4$, $b_4=2a_4+a_1a_3$, $b_6=a_3^2+4a_6$ and $b_8=a_1^2a_6+4a_2a_6-a_1a_3a_4+a_2a_3^2-a_4^2$. I now want to show that the discriminant is $\neq0$ if and only if the function $E$ is non-singular.

In order to show one direction I assume that there is a singular point in E, namely $(x_0,y_0)$ . Now a proof in a book says within one sentence: "Since the discriminant doesn't change by transformation $x=x+x_0$ and $y=y+y_0$ w.l.o.g. we can assume that the singular point is $(0,0)$."

But is there a simple way to see that the discriminant doesn't change within such a modification?

Thanks a lot for help

• You can directly compute. It's "simple," as in "elementary." Though "complex," as in "tedious." – Lorenzo Najt Apr 23 '17 at 0:05
• I know that I can directly compute it but I was just hoping there might be a more simple and elegant way to prove it - do you have an idea? :-) – user299124 Apr 23 '17 at 0:08
• The discriminant of a cubic in $x$ can be written as a product of the pairwise differences of the three roots. But these pairwise differences don't change if each root is shifted by the same amount. (Not sure how including $y$ changes things, so this is not a full answer.) – Semiclassical Apr 23 '17 at 0:12
• You could define an isometric change of variables. – Chickenmancer Apr 23 '17 at 0:12
• There are other ways to define the discriminant , but if you want to relate them to the given equation you are going to need to do computation at some point. Something that makes elliptic curves very tractable is that much can be done explicitly, using equations. A general theory will only be able to say things about general curves. If you are satisfied with a general argument that there is a polynomial which checks if a cubic is singular, I can give you one, but it won't be as powerful as having these formula. – Lorenzo Najt Apr 23 '17 at 0:13