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Ar belonging to R1 . If M is a finitely generated graded R-module, define h(M, n) = lR0 (Mn ), n ∈ N. Then h, as a function of n with M fixed, is polynomial-like of degree at most r − 1. Using slightly loose language, we call h the Hilbert polynomial of M . Proof. We argue by induction on r. If r = 0, then R = R0 . Choose a finite set of homogeneous generators for M over R. If d is the maximum of the degrees of the generators, then Mn = 0 for n > d, and therefore h(M, n) = 0 for n >> 0. Now assume r > 0, and let λr be the endomorphism of M given by multiplication by ar .

Sometimes, Rn is referred to as the nth graded piece and elements of Rn are said to be homogeneous of degree n. The prototype is a polynomial ring in several variables, with Rd consisting of all homogeneous polynomials of degree d (along with the zero polynomial). A graded module over a graded ring R is a module M expressible as ⊕n≥0 Mn , where Rm Mn ⊆ Mm+n . Note that the identity element of a graded ring R must belong to R0 . For if 1 has a component a of maximum degree n > 0, then 1a = a forces the degree of a to exceed n, a contradiction.

If a/b ∈ V , then a + b = b(1 + a/b) ∈ M (because if b(1 + a/b) were a unit, then b would be a unit as well). Similarly, if b/a ∈ V , then a + b ∈ M. If r ∈ V and a ∈ M, then ra ∈ M, else a would be a unit. Thus M is an ideal. 4. V is integrally closed. Let α be a nonzero element of K, with α integral over V . Then there is an equation of the form αn + cn−1 αn−1 + · · · + c1 α + c0 = 0 with the ci in V . We must show that α ∈ V . If not, then α−1 ∈ V , and if we multiply the above equation of integral dependence by α−(n−1) , we get α = −cn−1 − cn−2 α−1 − · · · − c1 α−(n−2) − c0 α−(n−1) ∈ V.

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A course in commutative algebra by Ash R.B.

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