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Let S be the set of all homogeneous algebraic functions of order 3 with 3 variables. Putting f(x)=0 gives 10 terms with 9 degrees of freedom because all functions are projective. An elliptic function E belongs to S. a,b and c are 3 points in E. f1

f1 contains 3 lines with points (a,b,-(a+b)), (c,0,-c),(-(b+c),-a) f2

f2 contains 3 lines with points (b,c,-(b+c)),(a,0,-a),(-(a+b),-c) Functions

Functions f1 and f2 also belong to S. Define x as the point where the lines (-(b+c),-a) and (-(a+b),-c) meet. If x lies on E, therethen associativity is proved as x is then another name for (a+b)+c and a+(b+c). 

If the 8 points of E are all distinct, then functions which pass through all 8 points in S have only one degree of freedom remaining. Hence E must be a linear combination of functions f1 and f2. Since f1 and f2 are both zero at the point x, then E must also be zero at the point x.

This proof is a variation of the proof which uses Lame's Theorem (I couldnt find it on the web), but uses only basic algebra.

Let S be the set of all homogeneous algebraic functions of order 3 with 3 variables. Putting f(x)=0 gives 10 terms with 9 degrees of freedom because all functions are projective. An elliptic function E belongs to S. a,b and c are 3 points in E. f1 contains 3 lines with points (a,b,-(a+b)), (c,0,-c),(-(b+c),-a) f2 contains 3 lines with points (b,c,-(b+c)),(a,0,-a),(-(a+b),-c) Functions f1 and f2 also belong to S. Define x as the point where the lines (-(b+c),-a) and (-(a+b),-c) meet. If x lies on E, there associativity is proved as x is then another name for (a+b)+c and a+(b+c). If the 8 points of E are all distinct, then functions which pass through all 8 points in S have only one degree of freedom remaining. Hence E must be a linear combination of functions f1 and f2. Since f1 and f2 are both zero at the point x, then E must be zero at the point x.

This proof is a variation of the proof which uses Lame's Theorem (I couldnt find it on the web), but uses only basic algebra.

Let S be the set of all homogeneous algebraic functions of order 3 with 3 variables. Putting f(x)=0 gives 10 terms with 9 degrees of freedom because all functions are projective. An elliptic function E belongs to S. a,b and c are 3 points in E.

f1 contains 3 lines with points (a,b,-(a+b)), (c,0,-c),(-(b+c),-a)

f2 contains 3 lines with points (b,c,-(b+c)),(a,0,-a),(-(a+b),-c)

Functions f1 and f2 also belong to S. Define x as the point where the lines (-(b+c),-a) and (-(a+b),-c) meet. If x lies on E, then associativity is proved as x is then another name for (a+b)+c and a+(b+c). 

If the 8 points of E are all distinct, then functions which pass through all 8 points in S have only one degree of freedom remaining. Hence E must be a linear combination of functions f1 and f2. Since f1 and f2 are both zero at the point x, then E must also be zero at x.

This proof is a variation of the proof which uses Lame's Theorem (I couldnt find it on the web), but uses only basic algebra.

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Let S be the set of all homogeneous algebraic functions of order 3 with 3 variables. Putting f(x)=0 gives 10 terms with 9 degrees of freedom because all functions are projective. An elliptic function E belongs to S. a,b and c are 3 points in E. f1 contains 3 lines with points (a,b,-(a+b)), (c,0,-c),(-(b+c),-a) f2 contains 3 lines with points (b,c,-(b+c)),(a,0,-a),(-(a+b),-c) Functions f1 and f2 also belong to S. Define x as the point where the lines (-(b+c),-a) and (-(a+b),-c) meet. If x lies on E, there associativity is proved as x is then another name for (a+b)+c and a+(b+c). If the 8 points of E are all distinct, then functions which pass through all 8 points in S have only one degree of freedom remaining. Hence E must be a linear combination of functions f1 and f2. Since f1 and f2 are both zero at the point x, then E must be zero at the point x.

This proof is a variation of the proof which uses Lame's Theorem (I couldnt find it on the web), but uses only basic algebra.