By N. S. Gopalakrishnan
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Additional info for Commutative algebra
Then the tangent to F at P is given by ∂f ∂f ∂f · X0 + · X1 + · X2 = 0. ∂x0 ∂x1 ∂x2 Proof. 3(b), since the trans∂f lation that maps (x, y) to (0, 0) sends ∂f ∂x · (X − x) + ∂y · (Y − y) to the leading form of the polynomial describing the aﬃne curve. (b) We may suppose without loss of generality that x0 = 0 and let f (X, Y ) := F (1, X, Y ). In aﬃne coordinates the tangent is given by ∂f ∂f · (X − x) + · (Y − y) = 0 ∂x ∂y with x := x1 x2 , y := , x0 x0 56 6 Regular and Singular Points of Algebraic Curves.
Let F be irreducible. Then (a) To each R ∈ X(F ) there is exactly one P ∈ V+ (F ) such that O F,P ⊂ R and mF,P = m ∩ O F,P , where m is the maximal ideal of R. There is therefore a natural mapping π : X(F ) → V+ (F ) (R → P ). (b) The mapping π is surjective. For P ∈ Reg(F ), the set π −1 (P ) consists of only one “point” R, namely R = O F,P . For P ∈ Sing(F ), π −1 (P ) is ﬁnite. Proof. (a) We write R(F ) = K(x, y), where K[x, y] is the aﬃne coordinate ring of F with respect to the line at inﬁnity X0 = 0.
It is clear that these fractions form a subﬁeld of K(X0 , X1 , X2 ). For a point P = x0 , x1 , x2 ∈ P2 (K) with ψ(P ) = 0, φ(x0 , x1 , x2 ) ψ(x0 , x1 , x2 ) is independent of the choice of homogeneous coordinates for P . Thus in fact a function r : P2 (K) \ V+ (ψ) → K P → φ ψ gives φ(P ) ψ(P ) that vanishes on V+ (φ). We call r a rational function on the projective plane, whose domain of deﬁnition, Def(r), is P2 (K)\V+ (ψ). We call ψ the pole divisor and φ the zero divisor of r. The diﬀerence φ − ψ in the divisor group D of P2 (K) is called the principal divisor belonging to r.