A conservative vector field has zero curl wherever its scalar potential is sufficiently smooth. Under suitable conditions on the domain, the converse is also true. The important distinction is between a local differential condition and a global property of the region.

Gradients have zero curl

Suppose

$$\mathbf F=\nabla\phi$$

for a scalar potential $\phi$ with continuous second partial derivatives. In three dimensions,

$$\mathbf F=(\phi_x,\phi_y,\phi_z).$$

Its curl is

$$\nabla\times\mathbf F

(\phi_{zy}-\phi_{yz},;\phi_{xz}-\phi_{zx},;\phi_{yx}-\phi_{xy}).$$

Equality of mixed partial derivatives therefore gives

$$\boxed{\nabla\times\nabla\phi=\mathbf0}.$$

So zero curl is a necessary condition for a smooth vector field to be conservative.

When zero curl is enough

A region is simply connected when it has no holes that prevent closed loops from being continuously contracted to a point while remaining inside the region.

For a sufficiently smooth field on a suitable simply connected region, suppose

$$\nabla\times\mathbf F=\mathbf0.$$

For any closed curve $C$ that bounds a surface $S$ lying in the region, Stokes' theorem gives

$$\oint_C\mathbf F\cdot d\mathbf r

\iint_S(\nabla\times\mathbf F)\cdot\mathbf n,dS =0.$$

Thus closed-loop circulation vanishes. Line integrals are path-independent, so a scalar potential exists and $\mathbf F$ is conservative.

Under these conditions,

$$\boxed{ \nabla\times\mathbf F=\mathbf0 \quad\Longleftrightarrow\quad \mathbf F=\nabla\phi }.$$

The domain condition is essential.

A curl-free field that is not globally conservative

Consider

$$\mathbf F(x,y,z)= \left( -\frac{y}{x^2+y^2}, \frac{x}{x^2+y^2}, 0 \right),$$

which is defined everywhere except on the $z$-axis. Direct differentiation gives

$$\nabla\times\mathbf F=\mathbf0$$

at every point where the field is defined.

Now take the unit circle around the missing axis,

$$\mathbf r(t)=(\cos t,\sin t,0), \qquad 0\le t\le2\pi.$$

Along this circle,

$$\mathbf F(\mathbf r(t))=(-\sin t,\cos t,0)$$

and

$$\mathbf r'(t)=(-\sin t,\cos t,0).$$

Therefore

$$\oint_C\mathbf F\cdot d\mathbf r

\int_0^{2\pi}1,dt =2\pi.$$

The circulation is not zero, so the field cannot have one globally defined scalar potential on the whole domain. The missing axis creates a hole that the circle encloses.

Local versus global potentials

Even in a domain with holes, a smooth curl-free field has a scalar potential in sufficiently small hole-free neighborhoods. What can fail is the ability to combine those local potentials into one single-valued potential over the entire domain.

This is why the statement “curl-free means conservative” must always be read together with assumptions about the domain. Curl detects local circulation; global path independence can also depend on topology.