Suppose $X$ is a smooth, geometrically irreducible projective algebraic curve over the finite field $\mathbb{F}_q$ and fix a closed point $\infty \in X$. Denote by $A = \Gamma(X  \{\infty\}, \mathcal{O}_X)$ the ring of functions on $X$ regular away from $\infty$. My question is the following (which is probably an easy one): why do we have $Spec(A) = X  \{\infty\}$? Thanks
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5$\begingroup$ Because $X  \{\infty\}$ is an affine curve, as follows for instance from (Riemann's part of) RiemannRoch. $\endgroup$– Vesselin DimitrovAug 12, 2015 at 14:42
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