What does it mean to say a polynomial has an isolated singularity In algebraic geometry, what does it mean when people say a polynomial $f$ has an isolated singularity at the origin?
 A: If we're talking about a polynomial defined over the field $k = \mathbb R$ or $\mathbb C$, it means that the zero locus $V(f) := \{x\in k^n: f(x_1,\ldots,x_n)=0\}$ is not smooth at the origin, but that there is some ball around the origin in which the zero locus is smooth, other than at the origin.
More formally, let $f\in k[x_1,\ldots,x_n]$ be a polynomial over a field $k$. Just as in the theory of vector calculus, we can then form its Jacobian matrix $J(f)$, the $1\times n$ matrix whose $i$th entry is the polynomial $\frac{\partial f}{\partial x_i}$. For $a\in k^n$, denote by $J(f)|_a$ the matrix obtained by evaluating the entries of $J(f)$ at $a$. The singular locus of $f$ is defined to be
$$\operatorname{Sing}(f) = \{a\in V(f): J(f)|_a = 0\}.$$ This is a subvariety of $V(f)$, and as expected by the name, in the case where $k=\mathbb R$ or $\mathbb C$ this is exactly the set of non-smooth points of $V(f)$. In fact, this defines the notion of "smooth point" for an affine variety over an arbitrary field.
More generally, if $F = (f_1,\ldots,f_m):k^n\to k^m$ is a polynomial function, its Jacobian matrix is $m\times n$ with entries $\frac{\partial f_j}{\partial x_i}$, and the condition for a point $a$ to be singular is for the rank of $J(f)|_a$ to be strictly less than $m$.
Now what it means for a polynomial $f$ to have an isolated singularity at a point $a$ is that $\{a\}$ is a (connected) component of the variety $\operatorname{Sing}(f)$.
