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Working through some problems in Rick Miranda's Algebraic Curves and Riemann Surfaces and wanna make sure I understand things correctly.

Problem V.1.A states: Let $X$ be the hyperelliptic curve defined by $y^2 = x^5 - x$. Note that $x$ and $y$ are meromorphic functions on $X$. Compute the principal divisors div$(x)$ and div$(y)$.

My reasoning: the function $x$, which is just projection onto the first coordinate, doesn't have any poles as $\infty$ is not included in our space at all. The only zero of $x$ is the point $(0,0)$ (as $x=0$ immediately gives $y=0$). I think that $(0,0)$ has order 2, because every other point $x_0$ has 2 points in its preimage (corresponding to the two roots of $x_0$) for $x_0\neq 0$. So div$(x) = 2\cdot (0,0)$.

Next I must address the function $y$, which is just projection onto the second coordinate.

Anyone care to comment/correct?

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    A small remark : I don't have this book with me right now but I think $\infty$ is included in your space, because Miranda might talk about the projective curve, given by the equation $y^2z^3 = x^5 - xz^4$ in $\mathbb P^2$. Else I would agree with you : looks like div$(x)$ = $2 [0:0:1] - 2 [0:1:0]$ and similar computations should give div$(y)$. – 2017-01-03
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    A principal divisor on a compact Riemann surface always has degree zero, which means that it has the same number of zeroes as poles, counted with multiplicity (cf. Lemma 1.5 of the same section). The equation $y^2 = x^5 - x$ only represents an affine piece of the surface: you need to figure out what happens at infinity, too. – 2017-01-03
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    Yeah but we're not talking about a projective curve here. The curve is defined in affine space... – 2017-01-03
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    Sorry. Meant "curve" not "surface" – 2017-01-03
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    As I said I didn't have the book with me, and since the purpose of the book is to study compact Riemann surface I though Miranda would take the projective closure of this curve. If you want to compute the divisor of the affine curve your computation is correct. – 2017-01-03
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    Thanks! Ha yeah it would make sense to take the projective closure, and he does consider projective curves (of course), but this problem is just about a vanilla affine plane curve ;P – 2017-01-03
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    As mentioned in other comments, usually when someone says "the hyperelliptic curve defined by" it is usually meant the projective curve. Also, it strikes me as odd that they would ask you to note that $x$ and $y$ are meromorphic functions on $X$ if they meant $X$ affine in which case they could've asked you to note that $x$ and $y$ are holomorphic functions on $X$. – 2017-01-03
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    I agree that it's odd...maybe it's a typo. Suppose I could skip this one for now and ask Dr. Miranda myself at some point haha. (I'm a grad. student at Colorado State where he works.) – 2017-01-03
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    Anyway if we let $X$ be in projective space, then we'd have the a zero of order 2 at $(0,0)$ (in projective space $[0:0:1]$ and a pole of order 2 at the point at infinity $[1:0:0]$ or $[0:1:0]$ or something like this? Does that sound about right? – 2017-01-03
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    You would need to be careful about how you are compactifying this curve. If you take the projective closure in $\bb{P}^2$ then you can calculate that the point at infinity will be a singularity of the projective curve. This is to be expected if you are aware of the fact that there can be no smooth hyperelliptic curves in $\bb{P}^2$. Instead you want the smooth normalization of this singular projective plane curve. That will have two points mapping to the singularity at infinity and those two points will be poles of the meromorphic function obtained by pulling back $x$ to the normalization. – 2017-01-03
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    Ah, you're right: from the other problems in the section, it looks like Miranda mentions explicitly when he is considering a projective curve (which he does not for this problem). – 2017-01-03

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