Singular Points on Algebraic Curves

Question

What does it mean for a point on an algebraic curve (over any field) to be singular? Over $\mathbb{R}$ or $\mathbb{C}$ is means that all derivatives vanish, but how does this definition generalise to general fields?

Answer 1

Zariski introduced in 1947 the definition of regularity used today, namely:
On a curve $C$ defined over a completely arbitrary base field $k$ a point $P\in C$ is said to be regular (or non-singular) if and only if the local ring $\mathcal O_{C,P}$ is a discrete valution ring.

By contrast the curve $C$ is said to be smooth at the point $P\in C$ if all points $\bar P\in C_{\bar k }$ mapped onto $P$ after the base change map $ C_{\bar k }\to C$ ($\bar k$=algebraic closure of $k$) are regular on $C_{\bar k }$.
Smoothness (but not regularity!) can be checked by the usual Jacobian criterion involving partial derivatives and, most importantly, smoothness implies regularity.
The converse ot this implication however is false: here is

An example of a regular but not smooth point $Q$

Let $k$ be a non perfect field of odd characteristic $p\geq 3$ and $a\in k$ be an element without a $p$-th root in $k$.
The plane curve $C\subset \mathbb A^2_k$ defined by $y^2+x^p-a=0$ is regular at all its points but not smooth at the point $Q\in C$ defined by the maximal ideal ideal $\langle y,x^p-a \rangle\subset \mathcal O(C)=k[x,y]=k[X,Y]/\langle Y,X^p-a \rangle$

Answer 2

Roughly speaking (meaning I will voluntarily not being precise) being nonsingular is not being smooth. Over $\mathbf{R}$ or $\mathbf{C}$ you can define being smooth be the fact that the jacobian is invertible, as in the implict function theorem, and "being invertible" for the jacobian is an algebraic statement (meaning for instance that some ideal is equal to some whole ring) that you can translate to any field, and in fact any ring. (Modulo some additive assumptions/definitions.) I don't like pointing to wikipedia, but you could start with this and come back if you need more EGA 4 references.