By Alexandru Buium

ISBN-10: 270566226X

ISBN-13: 9782705662264

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We have: 0 ↓ 3 F: 0→ R − di ψ2 → i=1 φ2 ↓ ξ2 G : 0 → R(−d1 − d2 ) → R(−d1 − d3 ) ψ1 ψ0 ⊕ → R(−d3 ) → R(−d3 )/I → 0 R(−d2 − d3 ) φ1 ↓ R(−d1 ) ⊕ R(−d2 ) φ0 ↓ ξ1 → R · f3 ↓ ξ0 → R/I →0 π↓ R/ I, f 3 → 0 ↓ 0 How do we get a free resolution for R/ I, f 3 ? You might think that we could just take the cokernels of the vertical maps, but a moment of thought shows that it is not this simple. Notice we have a map from G 0 R onto R/ I, f 3 via π ◦ ξ0 ; the kernel of this map is generated by the images of φ0 and ξ1 .

For R = k[x, y, z] and I = x 2 − x z, y 3 − yz 2 , ask Macaulay 2 to compute the Hilbert polynomial of R/I . Draw a picture of the variety in P2 (work on the patch where z = 1), and verify that Bezout’s theorem holds. ✸ 32 Projective Space and Graded Objects The most naive possible generalization of Bezout’s theorem is false: suppose { f 1 , . . , f n } ⊆ k[x0 , . . , xn ] are polynomials which have no pairwise common factor; say degree f i = di . As the next example illustrates, it is not in general true that V ( f 1 , .

For emphasis, we say it again: when R is a polynomial ring over a ﬁeld, then studying graded maps between graded R-modules is nothing more than linear algebra! Now we prove our earlier observation that the Hilbert function becomes a polynomial, for i 0. 3. If M is a ﬁnitely generated, graded module, then there exists a polynomial f (x) ∈ Q[x] such that for i 0, H F(M, i) = f (i). The polynomial f (i) is called the Hilbert polynomial of M, written H P(M, i). Proof. Induct on the number of variables in the ring over which M is deﬁned, the base case being trivial.

### Differential algebra and diophantine geometry by Alexandru Buium

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