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Definition: Let $\{a_1,\dots,a_n\}$ be any $n$ distinct points on the Riemann sphere $\mathbb{C}\cup\{\infty\}$ with coordinate $s$, and let $R$ be the ring of rational functions with poles allowed only in $\{a_1,\dots,a_n\}$. This ring is called $n$-point ring.

It is well known that

  • $\mathbb{C}[t^{\pm 1},u]/\langle u^2-t-a^2 \rangle \cong \mathbb{C}[t^{\pm 1},\frac{1}{t+1}] $ is a $3$-point ring, and

  • $\mathbb{C}[t^{\pm 1},u]/\langle u^2-(t^2-2bt+1) \rangle \cong \mathbb{C}[t^{\pm 1},\frac{1}{t+b+1},\frac{1}{t+b-1}] $ is a $4$-point ring.

It can be proven just using some isomorphisms.

Is $R=\mathbb{C}[x^{\pm 1},y]/\langle y^3-(x^2+ax+b) \rangle$ a $n$-point ring for some $n$?

Felipe
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1 Answers1

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No (assuming that $a^2 \neq 4b$, so the quadratic is not a square). In fact, the fields $\mathrm{Frac}(R)$ and $\mathbb{C}(t)$ are not isomorphic. From a sophisticated perspective, the point is that the former has genus one and the latter has genus $0$. In general, this is a hard notion to define in an elementary way; see the discussion here.

Let's do something elementary and check that $\mathbb{C}[x,y]/(y^2-x^3-1)$ is not an $n$-point ring. If it were, we could identify it with a subring of $\mathbb{C}(t)$, let $x$ correspond to $p(t)/q(t)$ and $y$ to $r(t)/s(t)$ in lowest terms. Comparing poles, we must have $q(t) = d(t)^3$ and $s(t) = d(t)^2$ for some polynomial $d$, so $(p(t)/d(t)^2)^3 = (r(t)/d(t)^3)^2+1$ or $p(t)^3=r(t)^2+s(t)^6$. This would violate the Mason-Stothers theorem, and you can mimic any proof of that theorem to see that this is impossible.

David E Speyer
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  • Thank you for your so detailed answer. Could you just clarify why any $n$-point ring needs to be a sub ring of $\mathbb{C}(t)$? – Felipe Mar 08 '18 at 17:13
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    The field of rational functions on $\mathbb{CP}^1$ is $\mathbb{C}(t)$. You defined $n$-point rings as subrings of this. – David E Speyer Mar 08 '18 at 18:32
  • Thank you for answer. It is helping me a lot have this discussion here since I'm not an expert in algebraic geometry. I'm trying to understand better that to work on representations of Lie algebras.

    You claimed that there is some polynomial $d$ such that $q(t)=d(t)^3$ and $s(t)=d(t)^2$. Is this a Theorem or you choose $q$ and $s$ in order to satisfy this condition? What do you main about “Compairing poles”?

     – Felipe Mar 09 '18 at 10:26
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    A pole of a rational function is an input which makes the rational function infinite. Factor $q(t) = \prod_{\alpha} (t-\alpha)^{m_{\alpha}}$ and $s(t) = \prod_{\alpha} (t-\alpha)^{n_{\alpha}}$. Then $y^2 - x^3=1$ forces $2m_{\alpha} = 3n_{\alpha}$. So $m_{\alpha}= 2 k_{\alpha}$ and $n_{\alpha}= 3 k_{\alpha}$ for some $k_{\alpha}$. Then $q(t) = \prod_{\alpha} (t-\alpha)^{3k_{\alpha}}$, $q(t) = \prod_{\alpha} (t-\alpha)^{3 k_{\alpha}}$ and we have $q=d^3$, $s=d^2$ with $d(t) = \prod_{\alpha} (t-\alpha)^{k_{\alpha}}$. – David E Speyer Mar 09 '18 at 12:19
  • Hello! I'm still thinking about this question.

    Q1- When I go back to my first question, in the case $y^3=(x^2+ax+b)$, without using the notion of genus, how could I see that it isn't $n$-point? It looks like that the Mason-Stothers doesn't help us here.

    Q2- Using the notion of genus, the fact that $Frac(R)$ and $\mathbb{C}(t)$ are not isomorphic ensure us that if I'm in the case $y^m=p(t)$ (a superelliptic curve with $m\geq 3$) it has genus $\neq 0$?

    Q3- Do you have recommendations about exactly this topic that could help a beginner like me?

     – Felipe Nov 06 '18 at 20:47