Principal maximal ideals in coordinate ring of an elliptic curve 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.
 A: 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.
A: 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.
