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A smooth algebraic curve over a field $k$ is an algebraic curve whose structural morphism to $\mathrm{Spec}k$ is smooth. Over a perfect field, this is equivalent to all its local rings being regular local rings. At each closed point, those rings have dimension one and are discrete valuation rings, so a local uniformizer measures orders of zeros and poles.

Over an imperfect field, regular local rings alone do not suffice. Let $k={\mathbb{F}}_p(s)$, let $K=k[u]/(u^p-s)$, and take $C=\mathrm{Spec}K[t]$. The polynomial ring $K[t]$ is regular. After the base change adjoining a $p$th root $v$ of $s$, the equation becomes $(u-v{)}^p=0$, giving a nonzero nilpotent element. The resulting curve is not geometrically reduced and cannot be smooth over $k$.

Thus smoothness tests regularity after extending the ground field, rather than only before that extension. The familiar nonsingular plane-curve Jacobian criterion over an algebraic closure is a geometric version of the same condition.^[ref]

The scheme $C^{(d)}$ parametrizes effective degree-$d$ Cartier divisors on a smooth algebraic curve. Near a Cartier divisor $\sum m_i{P}_i$, its completed local ring is a power-series ring in the elementary symmetric coordinates of $m_i$ local parameters at each $P_i$. It is therefore smooth of dimension $d$, even when the characteristic of a field divides $d!$. For projective $C$ it is projective.