Singular varieties Let $y^2=x^3+ax+b$ and V be its affine variety. V is singular iff $y^2-x^3-ax-b$, 2y, and $3x^2+a$ have a common zero iff $x^3+ax+b$ and $3x^2+a$ have a common zero iff $x^3+ax+b$ has a multiple root.
Is it correct that the first iff occurs because elements of the variety are solutions to the polynomial equation, and 2y and $3x^2+a$ are the partial derivatives in terms of y and x respectively (which means the point is singular). 
Also, could someone explain the next two 'iffs'
 A: Let $f = y^2 - x^3 - ax - b$.
The first iff is the Jacobian test.  A point of $V = V(f)$ is singular if and only if the matrix of partial derivatives
$$\begin{bmatrix} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y}\end{bmatrix}$$
does not have rank $1$.  The only way this happens is if it's the zero matrix, so if $2y$ and $3x^2 + b$ are both zero.  The reason $f$ is also in this list is to ensure that you're talking about a point in $V$.
For the second iff note $2y = 0$ means $y = 0$ so we can plug this into the other polynomials and get $-(x^3 + ax + b)$ and $3x^2 + a$.
Finally observe that $3x^2 + a$ is the derivative of $x^3 + ax + b$.  It's well known that a polynomial has a solution in common with it's derivative if and only if that polynomial has a repeated root.  To prove this write $p(x) = (x - c)^n\cdot g(x)$ and assume $c$ is not a root of $g(x)$.  Now just take the derivative of $p(x)$ and check whether $a$ is also a root.  You'll find that it is when $n > 1$ but it is not when $n = 1$.
