Singular point of an algebraic variety

Remarks

Jacobian criterion: let $X \subseteq k^n$ be an affine variety of dimension $d$ with $I(X) = (f_1, \dots, f_r)$. Then $p$ is nonsingular if and only if the Jacobian matrix $\left(\frac{\partial f_j}{\partial x_i}(p)\right)$ has rank $n - d$. Algebraic form: $p$ is nonsingular if and only if the local ring of $X$ at $p$ is a regular local ring. The singular points of $X$ form a closed subset different from $X$.

Example: for $X = V(f) \subseteq k^2$ with $f = y^2 - x^2(x + 1)$, a curve of dimension $1$, the partial derivatives are $\partial f / \partial x = -3x^2 - 2x$ and $\partial f / \partial y = 2y$. Assume $\operatorname{char} k \ne 2$. Both vanish together with $f$ only at $(0, 0)$, so the origin is the only singular point.

Non-example: on the circle $V(x^2 + y^2 - 1)$ with $\operatorname{char} k \ne 2$, the gradient $(2x, 2y)$ vanishes only at $(0, 0)$, which is not on the circle, so the circle has no singular point.