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The reference is http://www.math.columbia.edu/~masdeu/files/notes/FallSeminar.pdf, page 9:

Let now $C/R$ be a curve over a noetherian ring $R$; this means that $C$ is smooth, connected, integral, proper and of relative dimension $1$ over $R$. For curves it is always possible to find a covering by two open affines, say $\{U,V\}$.

To me this is not so clear. How would one find $U$ and $V$? If $C$ was a projective curve over a field $K$, given by $f(X,Y,Z) = 0$, I think we can find $U$ and $V$ by dehomogenizing with respect to $Z$ or $Y$, respectively. If $R$ is not a field, I'm not sure it makes sense to dehomogenize - basically 'dividing by $Z$' could be a problem. Can we still do this?

user84471
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I think this is not so easy, even when $C$ is projective over $R$.

The answer is yes if $R$ is a local ring. In general, even when $R$ is a Dedekind ring, I don't know whether this is true. There are some related discussions at https://mathoverflow.net/questions/112085/.

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  • If we do have a projective curve over $R$, maybe $C \rightarrow \mathbb{P}_R^n$, then $n+1$ is definitely enough for the cover, though? Because we would take the standard open cover of $\mathbb{P}_R^n$ and intersect it with $C$? – user84471 Jul 04 '13 at 11:16
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    Sure. But you don't have a good controle on $n$ (though it can be bounded with the genus of the fibers of $C$). In other direction, it is possible to prove that $C$ can be covered by $\dim R+2$ affine open subsets if $R$ has finite dimension. Finally, if $R$ is the ring of integers of a number field, then $C$ can actually be covered by two affine open subsets. All of this are non-trivial. –  Jul 05 '13 at 07:04