Singular points query

Question

I have learnt that a point is singular if all partial derivatives vanish there. However I have recently come across with worked examples where they also require that the function vanishes at the point. Can anyone clarify? Thanks.

Answer 1

Answered in the comments by Brenin:

If $f$ does not vanish at $P$, then $P$ is not a point of the variety defined by $f$, so it does not make sense to ask whether it is singular there.

Answer 2

I shall suppose that you are talking about a hypersurface $H=V(f)$ given by the vanishing of a single non zero polynomial $f$ and that the base field is of characteristic zero .
For a point $p$ to be a singular point of $H$ it is certainly necessary that all partial derivatives of $f$ vanish at $p$, namely $\frac { \partial f }{\partial x_i}(p)=0$ .
But are these conditions sufficient? It depends on what you are talking about!

I) If you are talking about a hypersurface $H\subset \mathbb A^n$ the vanishing of the partial derivatives of $f(x_1,\cdots,x_n)$ at $p$ is not enough: you must add the condition $f(p)=0$ .
For example, take $n=2$ and $f(x_1,x_2)=x_1^2+x_2^2-1$ .
The polynomials $\frac { \partial f }{\partial x_1}=2x_1$ and $\frac { \partial f }{\partial x_2}=2x_2 $ certainly vanish at $p=(0,0)$.
But is $p$ a singular point of the hypersurface $H=V(f)=\{f=0\}$?
Of course not! The point $p$ is the center of the circle $H$ and certainly not on the circle : $p\notin H$.
The circle has no singularity, a result the man in the street in Greece knew 2300 years ago...

II) What about a hypersurface $H=V(f(x_0,x_1,\cdots,x_n))\subset \mathbb P^n$ given by a homogeneous polynomial $f$ of degree $d$ ?
Is the vanishing $\frac { \partial f }{\partial x_i}(p)=0$ of the partial deivatives at $p$ sufficient for $p$ to be a singular point of $H$ ?
Yes!
But doesn't $p$ have to lie on $H$ in order to be a singular point of $H$ ?
Yes, it has to !
The explanation of this apparent paradox is that in the projective case the vanishing of all the partial derivatives $\frac { \partial f }{\partial x_i}$ at $p$ implies that also $f(p)=0$.
This follows from Euler's identity for a homogeneous polynomial $f$ of degree $d$ : $$ \sum_{i=0} ^nx_i\cdot \frac { \partial f }{\partial x_i}=d\cdot f(x_0,x_1,\cdots,x_n)$$