Is the curl of every non-conservative vector field nonzero at some point? Counterexamples? Intuitively, why?
Thanks for any answers.
As a side note, in what math class are gradient, divergence and curl taught typically?
 A: The question to ask is:

If there is a smooth vector field $\mathbf{v}$ such that 
  $$\nabla \times \mathbf{v} = 0 \quad \text{ in } \Omega,$$
  is there a smooth function $\phi$ such that 
  $$\mathbf{v} = \nabla\phi?$$



*

*If the domain is simply-connected (for example, a sphere, a box, but not cup with a handle), then this field must be a conservative field, i.e., $\mathbf{v} = \nabla \phi$.

*In general, we can have non-conservative field having zero curl. Let 
$$Z = \mathrm{ker}(\nabla\times) := \{\mathbf{w} \text{ is smooth}: \nabla \times \mathbf{w} = 0\},$$ 
which is the space of the fields having zero curl and 
$$B = \mathrm{im}(\nabla) = \{\nabla \psi: \psi \text{ is smooth}\},$$
which is the range of the gradient operator (i.e., all gradient fields). Then 
$$
\mathrm{dim}\big(Z/B\big) = \beta_1\tag{1}
$$
where $\beta_1$ is the first Betti number of the domain of interest $\Omega$, and 
$$\beta_1 = (\# \text{ of holes in the domain})$$
roughly speaking. (1) essentially means:
$$
\text{Field whose curl is zero} = \text{Conservative vector field } + \text{"Something"}. \tag{2}
$$
This "something" here is a $\beta_1$-dimensional space. 

*(1) is the by the coincidence of the dimension of the de Rham cohomology group and homology group. For simply connected domain $\beta_1 =0$. If the domain has $k$ holes, the difference is a $k$-dimensional space. 

*Intuition: To be honest I don't know the intuition here either, if someone knows please enlighten me as well. 

*How to find this "something": Let's use that example in the wiki's entry of conservative fields AWertheim mentioned in the first comment, but modify it a little bit. For an infinite cylinder $C_{x=y=0,z\in (-\infty,\infty)}(1)$ with radius 1 in $\mathbb{R}^3$ that contains the origin, we dig a hole of radius $\epsilon$ along $z$-axis: 
$$\Omega = C_{x=y=0,z\in (-\infty,\infty)}(1)\backslash  C_{x=y=0,z\in (-\infty,\infty)}(\epsilon)$$
Then by above, that "something" is a dimension 1 space. Let $\mathbf{A}$ be a member of this something, we can pose a boundary value problem which is the technique used in Helmholtz decomposition of vector fields:
$$\left\{
\begin{aligned}
\nabla \times \mathbf{A} &= 0\quad \text{ in }\Omega
\\
\nabla \cdot \mathbf{A} &= 0 \quad \text{ in }\Omega
\\
\frac{1}{2\pi}\oint_{\gamma_C} \mathbf{A} \cdot d\mathbf{s} &= 1 
\\
\mathbf{A} \cdot \mathbf{n} &= 0 \quad \text{ on }\Gamma,
\end{aligned}
\right.$$
where $\gamma_C $ is a counterclockwise closed curve lived on that interior cylinder surface with winding number 1, and $\Gamma$ is the boundary of the exterior cylinder. We can find that this $\mathbf{A}$ is what wikipedia's entry has:
$$
\mathbf{A} = \left( \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2}, 0 \right).
$$
That "something" in (2) is just a multiple of $\mathbf{A}$. And now we can say that if $\nabla \times \mathbf{v} = 0$ in $\Omega$ above, then
$$
 \mathbf{v} = \nabla \phi + c\mathbf{A},
$$
where $c$ is a constant.

*Relevant question: What is the solution to Nash's problem presented in "A Beautiful Mind"?

*Your last question: You can learn div, curl, and grad in Calculus III I believe. Like Green theorem, Divergence theorem, and Stokes theorem.
