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Let $E$ be an elliptic curve over an algebraically closed field, and let $R$ be the coordinate ring of $E \setminus \{\infty\}$. I have read somewhere that $R$ has no principal maximal ideal. But I cannot find it again. Does someone know a reference? Or perhaps a simple proof?

Edit: Already two proofs are given below. I am still interested in a reference to the literature, because I would like to cite this.

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I don't have a precise reference, but (as perhaps you know) what you want is more or less equivalent to the calculation of $\mathrm{Pic}(E)$. You can find that in many textbooks, although I don't know offhand what level of detail they provide... –  Asal Beag Dubh Apr 27 '13 at 16:50
    
How is this connected to the Picard group? –  Martin Brandenburg Apr 27 '13 at 18:57
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Suppose there was a point $p \in E \setminus \{\infty\}$ such that $m_p=(f)$, a principal ideal. Thinking of $f$ as a rational function on $E$, the divisor $div(f)$ would then have to be $p-\{\infty\}$ (because a principal divisor on a curve has degree zero). That is, the points $p$ and $\{\infty\}$ would be linearly equivalent as divisors on $E$, which is never the case for distinct points on an elliptic curve. –  Asal Beag Dubh Apr 27 '13 at 20:53

2 Answers 2

OK, let me make my comment an answer as requested.

Suppose there was a point $p \in E \setminus \{\infty\}$ such that $m_p=(f)$, a principal ideal. Thinking of f as a rational function on E, the divisor $div(f)$ would then have to be $p−\infty$ (because a principal divisor on a curve has degree zero). That is, the points $p$ and $\infty$ would be linearly equivalent as divisors on $E$, which is never the case for distinct points on an elliptic curve, as is proved (hopefully without invoking the fact we want to prove!) in standard texts on algebraic geometry.

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Thank you. That distinct points are not linearly equivalent follows from IV.1.3.7 in Hartshorne's book (the bijection between the closed points and the divisor classes of degree $0$). Do you know a reference for the statement that principal maximal ideals don't exist? I would like to use this fact in some paper, and don't want to add the proof there since it seems to be standard. A reference would be great. –  Martin Brandenburg Apr 27 '13 at 23:42
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@MartinBrandenburg: Dear Martin, This is indeed very standard, but I'm not sure where it's written down. One quick way to justify it in a paper is to cite the result of Hartshorne you mentioned, and then to also cite the result from Hartshorne in Ch. II, section 6 or 7, where he gives the relationship between Picard/Class groups for a variety and the open subset obtained by deleting a closed subvariety. (If you look you will find an exact sequence, and you can apply taking the original variety to be the complete curve $E$, and the closed subvariety to the be the point at infinitey.) Regards, –  Matt E Apr 27 '13 at 23:48
    
Dear Martin, I'm afraid I don't know a direct reference. –  Asal Beag Dubh Apr 27 '13 at 23:52

Here is a direct proof, using a presentation of the elliptic curve as the solutions to a Weierstrass cubic, and Bezout's theorem.


Present our elliptic curve in terms of a Weierstrass cubic in the affine $x,y$-plane. Let $X,Y,Z$ be the corresponding homogenous coordinates in the projective plane. Then the point at infinity $O$ is $[0,1,0]$, and we can take the affine coords. there to be $x = X/Y, z = Z/Y.$ The key fact is that the tangent line to the curve at $O$ is the line at infinity $z = 0$, and it meets the curve at $O$ with order $3$. This means that we can take $x$ to be a uniformizer at $O$, and when we expand $z$ as a power series in $x$ in the complete local ring at $O$, it has leading term $x^3$.

Now suppose that some maximal ideal in the coordinate ring of $E\setminus \{O\}$ is principal, say generated by $f(x,y)$. By the Nullstellensatz, this maximal ideal corresponds to a point $P \neq O$ of $E$, and to say that $f(x,y)$ generates the maximal ideal associated to $P$ is to say that $f(x,y)$ vanishes to exact order $1$ at $P$, and vanishes at no other point of $E \setminus \{O\}$.

Let $C$ be the projective closure (i.e. Zariski closure in $\mathbb P^2$) of $f(x,y) = 0$. If $d$ is the maximal degree of a monomial in $f(x,y)$, then $C$ has degree $d$, and by Bezout it meets $E$ in $3d$ points, counted with multiplicity. From what we've said, it must meet $E$ at $O$ with mult. $3 d -1$.

Now let $g(x,z)$ be the equation for the intersection of $C$ with the $(x,z)$-plane. To compute the mult. with which $C$ meets $E$ at $O$, we have to regard $g(x,z)$ as an elt. of the complete local ring at $O$, and see what order zero it has.

Now $g(x,z)$ is a linear combination of monomials $x^i z^j$ with $i + j \leq d$. Recalling that $z = x^3 + $ higher order terms, we find that if $g(x,z)$ vanishes to order $\geq 3d - 1$ at $O$, then in fact the only non-zero monomial it contains is $z^d$, and then it actually vanishes to order $3d$ at $O$.

In short, it's not possible to find a curve of degree $d$ meeting $O$ with multiplicity precisely equal to $3d - 1$, and so the maximal ideal of $P$ cannot be principal after all. QED


This argument is related to the intersection theory arguments that give a direct proof that the chord-tangent law makes the points of $E$ an algebraic group. Thus, while it is perhaps not obvious, this argument is related to the relationship between $E$ and its Picard group described in the other answer.

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Dear Matt, thanks for giving this direct argument, which happily avoids the overkill of my answer. –  Asal Beag Dubh Apr 27 '13 at 23:57
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@AsalBeagDubh: Dear Asal, No problem, and I don't think of your answer as overkill. Indeed, it's definitely the conceptually correct explanation! I just wanted to give the direct argument in order to illustrate that the statement from your answer isn't as hard as people might think. (Since we can always find a Weierstrass model which puts any given point at infinity, the above gives a proof that no two points on an elliptic curve are linearly equivalent just using the basics of plane curves.) Regards, –  Matt E Apr 28 '13 at 0:08
    
@AsalBeagDubh: P.S. Probably we should avoid using RR to get the Weierstrass models, since if you admit RR then it is pretty easy to just prove that no two points are linearly equivalent directly. But if you have a point on a plane cubic, it's not hard to prove directly that you can find an isomorphism between that plane cubic and a Weierstrass cubic in which the chosen point gets taken to the point at infinity. –  Matt E Apr 28 '13 at 0:09

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