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Extra resources for An easier solution of a Diophantine problem about triangles, in which those lines from the vertices which bisect the opposite sides may be expressed rationally

Example text

We will often identify them with there images S+ = s+ (X) and S− = s− (X). 1. C(X, D) \ S± ∼ = C(X, D)∓ . Proof. Suppose that D is Cartier and let L = OX (D). Then s+ corresponds to the surjection S • (L ⊕ OX ) → S • OX defined by the projection L ⊕ OX → OX . The section s− corresponds to the surjection S • (L ⊕ OX ) → S • L defined by the projection L ⊕ OX → L. We have already explained in the beginning of the section that P(L ⊕ OX ) \ s± (X) = V(L∓1 ). 20), where rD is Cartier. It follows from the definition of the sections s± that the composition s± qr of X → C(X, D) → C(X, rD) is the zero section (resp.

The homomorphism of graded rings S • (L ⊕ OX ) ∼ = S • L[t] → k[C¯X ] = k[T0 , . . , Tn ][Tn+1 ] defines a morphism p¯ : P(L ⊕ OX ) → C¯X . Its restriction over CX coincides with the composition V(L) → Spec A → CX . It is a partial resolution of the vertex of C¯X . We will show in the next Lecture that the partial resolution morphisms are the blowing-up morphisms with center at the vertex. Next let X be a normal integral scheme over a field X and D be an ample Cartier Q-divisor on X. 4. CYLINDER CONSTRUCTIONS AX (D)+ = OX (iD), 49 AX (D)− = i≥0 OX (iD), i≤0 Let π± : C(X, D)± := Spec AX (D)± → X, π : C(X, D)∗ := Spec AX (D) → X be the corresponding affine schemes over X.

Dr ) be the least common multiple and hi = d/di . Let B be the graded subalgebra of A generated by xhi i , i = 1, . . , r, all of the same degree d. It is clear that B ⊂ A(d) . I claim that the corresponding B-module A(d) is generated by the monomials xi11 · · · xirr , ij < dj , j = 1, . . , r, d1 i1 + . . + dr ir ≡ 0 mod d. It suffices to show that any monomial xn1 1 · · · xnr r ∈ A(d) can be written as a linear combination of the monomials from above with coefficients in A0 . We write ni = di qi + mi , where 0 ≤ mi < di , to obtain mr 1 xn1 1 · · · xnr r = (xd11 )q1 · · · (xdr r )qr xm 1 · · · xr .

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