# Are these sufficient conditions for point on an abstract curve to be regular?

Let $p\in{X}$ where $X$ is a curve-- here the definition of a curve is an integral, seperated, 1 dimensional scheme of finite type over a field $k$ (not necessarily algebraically closed). Moreover, suppose $X$ is smooth at $p$, so that the sheaf of differentials $\Omega_{X/k}$ is free of rank 1 on some neighborhood of $p$. From this alone, can we conclude that $X$ is regular at $p$, i.e. is $\mathcal{O}_{X,p}$ is a regular local ring?

I realize that the answer is yes if $k$ is perfect, or if $k(p)$, the function field at p, is just $k$, or even if $\mathcal{O}_{X,p}$ contains a subfield isomorphic to $k(p)$, but I can find absolutely nothing in the literature without these restrictions. Can anyone set me straight here?

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## 1 Answer

Yes, if $X$ is a variety over an arbitrary field $k$ ( i.e. a scheme of finite type over $k$, not necessarily of dimension $1$), then every smooth closed point of $X$ is regular.
The converse is true if the residue field $\kappa (x)$ (which is finite over $k$) is separable over $k$.
So the converse is true if $x$ is rational over $k$, or if $k$ is perfect (for example: finite, of characteristic zero, algebraically closed,...) .

An excellent reference is Qing Liu's Algebraic Geometry and Arithmetic curves, Ch.4, Proposition 3.30 .

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