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.
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.
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.
