Let \(X\) be an elliptic curve over \(k\), with \(\characteristic k \neq 2\), let \(P \in X\) be a point, and let \(R\) be the graded ring \(R=\bigoplus_{n \geq 0} H^0(X, \OO_X(nP))\). Show that for suitable choice of \(t, x, y\) \[ R \cong k[t, x, y] /\left(y^2-x\left(x-t^2\right)\left(x-\lambda t^2\right)\right), \] as a graded ring, where \(k[t, x, y]\) is graded by setting \(\deg t=1\), \(\deg x=2\), \(\deg y=3\).
Solution
We use Hartshorne’s standing convention that \(k\) is algebraically closed. Write \[ R_n=H^0(X,\mco_X(nP)). \]
1. For every \(n\ge1\),\[ \dim_k R_n=n. \]
Since \(X\) has genus one, its canonical divisor is linearly equivalent to zero. Riemann–Roch gives \[ \ell(nP)-\ell(-nP)=n. \] For \(n>0\) the divisor \(-nP\) has negative degree, so \[ \ell(-nP)=0. \] Hence \[ \ell(nP)=n. \]
2. There are homogeneous elements\[ t\in R_1, \qquad x\in R_2, \qquad y\in R_3 \]such that, after dehomogenizing by \(t\),\[ u=\frac{x}{t^2}, \qquad v=\frac{y}{t^3} \]have poles at \(P\) of exact orders \(2\) and \(3\), respectively.
Use the rational-function description \[ H^0(X,\mco_X(nP)) = \{f\in k(X):(f)+nP\ge0\}\cup\{0\}. \] Let \(t\) denote the constant function \(1\), regarded as a homogeneous element of degree \(1\). Multiplication by \(t\) realizes the natural inclusions \[ R_{n-1}\hookrightarrow R_n. \]
By step 1, \[ \dim R_1=1, \qquad \dim R_2=2, \qquad \dim R_3=3. \] Choose \[ x\in R_2\setminus kt^2. \] Its dehomogenization \(u=x/t^2\) lies in \(L(2P)\) but not in \(L(P)\), so its only pole is \(P\) and that pole has exact order \(2\).
Next choose \[ y\in R_3\setminus tR_2. \] Then \(v=y/t^3\) lies in \(L(3P)\) but not in \(L(2P)\), hence has exact pole order \(3\) at \(P\) and no other poles.
3. The elements \(t,x,y\) generate the graded ring \(R\).
For every \(n\ge1\), consider the degree-\(n\) elements \[ t^n, \] together with \[ x^a t^{n-2a} \qquad (1\le a,\ 2a\le n) \] and \[ x^a y t^{n-(2a+3)} \qquad (0\le a,\ 2a+3\le n). \] After division by \(t^n\), these become \[ 1, \qquad u^a, \qquad u^a v. \] Their pole orders at \(P\) are, respectively, \[ 0, \qquad 2a, \qquad 2a+3. \] Thus their pole orders are exactly \[ 0,2,3,4,\ldots,n, \] with no repetition. They are therefore linearly independent. There are exactly \(n\) of them, and step 1 says \(\dim_kR_n=n\). Hence they form a basis of \(R_n\).
Every basis element is a monomial in \(t,x,y\), so these three elements generate all graded pieces of \(R\).
4. After replacing \(y\) by another degree-\(3\) generator, there arecoefficients \(a_3,a_2,a_1,a_0\in k\), with \(a_3\ne0\), such that\[ y^2 = a_3x^3+a_2x^2t^2+a_1xt^4+a_0t^6. \]
Step 3 gives the following basis of \(R_6\): \[ t^6, \quad xt^4, \quad x^2t^2, \quad x^3, \quad yt^3, \quad xyt. \] Therefore \(y^2\in R_6\) has a unique expression \[ y^2 = a_3x^3+a_2x^2t^2+a_1xt^4+a_0t^6 +b_1xyt+b_0yt^3. \] The coefficient \(a_3\) is nonzero: after dehomogenizing, \(v^2\) has pole order \(6\), while every displayed basis term except \(u^3\) has pole order at most \(5\).
Since \(\characteristic k\ne2\), set \[ y_1 = y-\frac12\left(b_1xt+b_0t^3\right). \] This differs from \(y\) by an element of \(tR_2\), so its dehomogenization still has exact pole order \(3\) and \(t,x,y_1\) still generate \(R\). Completing the square gives \[ y_1^2 = a_3x^3+a_2'x^2t^2+a_1'xt^4+a_0't^6 \] for suitable \(a_2',a_1',a_0'\in k\). Rename \(y_1\) as \(y\) and the primed coefficients as \(a_2,a_1,a_0\).
5. The relation in step 4 is the only relation among \(t,x,y\).
Let \[ A = \frac{k[t,x,y]} {\left(y^2-a_3x^3-a_2x^2t^2-a_1xt^4-a_0t^6\right)}, \] with weights \[ \deg t=1, \qquad \deg x=2, \qquad \deg y=3. \] Step 4 gives a surjective graded homomorphism \[ A\longrightarrow R. \]
3. Their images in \(R_n\) are the basis constructed there,so they are linearly independent already in \(A_n\). Thus\[ A_n\xrightarrow{\sim}R_n \]for every \(n\), and hence\[ A\cong R. \]
6. The cubic\[ q(u)=a_3u^3+a_2u^2+a_1u+a_0 \]has three distinct roots in \(k\).
Localizing the section ring at \(t\) and taking degree zero gives \[ R[t^{-1}]_0 = \bigcup_{n\ge0}L(nP) = \Gamma(X\setminus\{P\},\mco_X). \] The nonconstant function \(u\) has pole divisor \(2P\), so it defines a finite morphism \[ u:X\longrightarrow\PP^1 \] with \(u^{-1}(\infty)=\{P\}\) set-theoretically. Hence \(X\setminus\{P\}=u^{-1}(\AA^1)\) is affine, and the displayed ring is its coordinate ring.
By step 5 this ring is \[ k[u,v]/\left(v^2-q(u)\right). \] Thus \(X\setminus\{P\}\) is the affine curve \[ v^2=q(u). \]
The coefficient \(a_3\) is nonzero by step 4, so \(q\) has degree \(3\). Because \(k\) is algebraically closed, it splits completely. If \(\alpha\) were a repeated root, then the point \[ (u,v)=(\alpha,0) \] would satisfy both partial derivative equations \[ 2v=0, \qquad q'(u)=0. \] It would therefore be a singular point of \(X\setminus\{P\}\), contrary to the nonsingularity of the elliptic curve \(X\). Hence \[ q(u)=a_3(u-\alpha)(u-\beta)(u-\gamma) \] with \(\alpha,\beta,\gamma\) pairwise distinct.
7. After homogeneous changes of the degree-\(2\) and degree-\(3\)generators, the relation becomes\[ y^2=x(x-t^2)(x-\lambda t^2) \]with \(\lambda\in k\setminus\{0,1\}\).
Set \[ \lambda = \frac{\gamma-\alpha}{\beta-\alpha} \] and replace \(x\) by \[ x' = \frac{x-\alpha t^2}{\beta-\alpha}. \] Then the relation in step 6 becomes \[ y^2 = c\,x'(x'-t^2)(x'-\lambda t^2), \qquad c=a_3(\beta-\alpha)^3\ne0. \] Since \(k\) is algebraically closed, choose \(\mu\in k^\times\) with \[ \mu^2=c. \] Replacing \(y\) by \(y'=y/\mu\) gives \[ y'^2 = x'(x'-t^2)(x'-\lambda t^2). \] The roots are distinct, so \(\lambda\ne0,1\). Renaming \(x',y'\) as \(x,y\) proves the asserted presentation.
8. Q.E.D.
Steps <1>1–<1>3 construct homogeneous generators of degrees \(1,2,3\). Steps <1>4–<1>5 determine the unique degree-\(6\) relation, and steps <1>6–<1>7 put its cubic factor into the required Legendre form. Therefore \[ \boxed{ R \cong k[t,x,y]/\left(y^2-x(x-t^2)(x-\lambda t^2)\right) } \] as graded rings.