I'm looking for an elementary proof of the fact that $\ell(nP) = \dim L(nP) = n$, where $L(nP)$ is the linear (Riemann-Roch) space of certain rational functions associated to the divisor $nP$, where $n > 0$ is an integer and $P$ is a point on a curve of genus 1.

This fact is used to prove that every elliptic curve is isomorphic to a curve given by a Weierstrass equation. See for instance Silverman, The Arithmetic of Elliptic Curves, p. 59, Proposition 3.1.(a).

I know that it's a corollary to the Riemann-Roch theorem (it follows from the corollary that says if $\deg D > 2g - 2$ then $\ell(D) = \deg D + 1 - g$). The proof of the full Riemann-Roch theorem still looks a little intimidating to me, so I'm hoping someone can provide a more elementary proof of this special case, perhaps by making the relevant simplifications in the proof of the Riemann-Roch theorem.

Thanks in advance.

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    In the definition of an elliptic curve you need to include that it has a point in the field of definition. For example, the curve defined in the projective plane by $3x^3 + 4y^3 + 5z^3 = 0$ has genus 1 but no solution $[x,y,z]$ where the coordinates are all in $\mathbf Q$, so it would not be considered an elliptic curve over $\mathbf Q$. – KCd Jun 1 '13 at 0:13
up vote 4 down vote accepted

Let $X$ be an elliptic curve over $k$, and $P \in X$. We have to prove that for $D=nP$ we have $$ l(D)=\deg(D)=n $$ If $D$ satisfies the above equation, and $E$ is a divisor such that $E \geq D$, then we have $l(E)=\deg(E)$ (cf. Fulton, Corollary 8.3.1). Hence it is enough to show that $l(P)=1$.

Clearly $l(P) > 0$, since $k \subset L(P)$. On the other hand, $l(P) > 1$ would imply that there exists $x \in k(X)$ such that $x$ has a simple pole at $P$, and no other poles. But this would imply that the map $$ x: X \rightarrow \Bbb{P}^1 $$ is an isomorphism, which is a contradiction. Thus $l(P)=1$.

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    Of course, it remains to be shown that elliptic curves are not rational curves. This is usually done using a genus argument, but perhaps the OP does not want to use that... – Zhen Lin Jun 1 '13 at 14:49
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    @ZhenLin: True. You have to start somewhere, and it is not quite clear where the OP draws the line. Still, when you define the genus as is e.g. done in Fulton, then it is tautological (using resolution of singularities for curves) that birational curves have the same genus, hence elliptic curves cannot be rational (if you know that $\Bbb{P}^1$ has genus zero, of course). – Nils Matthes Jun 1 '13 at 15:18

Here is an elementary solution in a very special case. It assumes that the curve is already embedded in the plane using a Weierstrass-like equation, so you can't use it to later prove that every elliptic curve can be given by a Weierstrass equation, but perhaps it will still be instructive.

We assume the base field is of characteristic $0$ and algebraically closed. Suppose $E$ is the projective curve in $\mathbb{P}^2$ defined by the following equation in $\mathbb{A}^2$, $$y^2 = (x - \lambda_1)(x - \lambda_2)(x - \lambda_3)$$ where $\lambda_1, \lambda_2, \lambda_3$ are non-zero and distinct. It can be shown that $E$ is a smooth curve with a unique point at infinity, namely $P = (0 : 0 : 1)$. As Nils Matthes has indicated, to show that $\ell (n P) = n$ for all $n \ge 1$, it suffices to prove the case $n = 1$.

Suppose, for a contradiction, that $\ell (P) > 1$. We know that for all divisors $D$, $\ell (D) \le \deg D + 1$, so that means $\ell (P) = 2$, and so there is some non-constant rational function $t$ such that $L (P)$ is spanned by $1$ and $t$. Moreover, $t$ has a simple pole at $P$ (and nowhere else), so $t^n$ has a pole of order $n$ at $P$ (and nowhere else). As such, $L (n P)$ is spanned by $1, t, \ldots, t^n$. However, we know that $y$ is in $L (2 P)$ and $x$ is in $L (3 P)$, so that means, for some constants, \begin{align} x & = a_0 + a_1 t + a_2 t^2 \\ y & = b_0 + b_1 t + b_2 t^2 + b_3 t^3 \end{align} and by rescaling $t$ if necessary, we may assume that $a_2 = b_3 = 1$. (Note that $a_2^3 = b_3^2$.) We may also complete the square to assume that $a_1 = 0$. But then substituting into the original equation, we get $$(b_0 + b_1 t + b_2 t^2 + t^3)^2 = (t^2 + a_0 - \lambda_1) (t^2 + a_0 - \lambda_2) (t^2 + a_0 - \lambda_3)$$ which is impossible, since the RHS has six distinct zeros.

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