A normal curve over a perfect field is nonsingular

Assume $X$ is normal and let $x\in X$. By [F1] its local ring is an integrally closed domain. At the generic point [F2] makes it a field, hence a regular local ring of dimension zero. At every other point [F2] makes it Noetherian of dimension one; it is not a field because its dimension is $1$, so [F3] makes it a discrete valuation ring and then a regular local ring. By [F4] the point $x$ is regular.

Conversely assume $X$ is regular and let $x\in X$. By [F4] the local ring ${\mathcal{O}}_{X,x}$ is a regular local ring, hence an integrally closed domain by [F5], so $x$ is a normal point by [F1].

Assume now that $k$ is perfect. The regularity criterion is pointwise on the given finite-type scheme, and [F6] identifies regularity with smoothness over $k$; hence $X$ is nonsingular, i.e. smooth over $k$, exactly when $X$ is regular. Combined with steps 1.1 and 1.2, for a curve over a perfect field the three conditions normal, regular and nonsingular agree.

Perfectness really is necessary for smoothness. Let $k={\mathbf{F}}_p(s)$ and $K=k(\alpha )$, where ${\alpha }^p=s$. The element $s$ is not a $p$th power in $k$ because its order at $s=0$ is one, whereas a $p$th power of a rational function has order divisible by $p$. The curve $X=\mathrm{Spec}K[t]$ is integral and finite type of dimension one over $k$. Polynomial division over the field $K$ makes every nonzero prime of $K[t]$ principal; its localization has dimension one and maximal ideal with one generator, while the generic localization is a field. Thus $X$ is regular by [F4], and normal by [F5]. After field extension to $K$, its affine ring is $K[t,u]\mathrm{/}(u^p)$, with $u=\alpha \otimes 1-1\otimes \alpha$. At the prime $(u)$ its local ring is $K(t)[u]\mathrm{/}(u^p)$, of dimension zero and embedding dimension one. This is not regular, so the geometric-regularity characterization in [F6] says $X$ is not smooth over $k$. Therefore normal and regular curves need not be smooth when perfectness is omitted.