By Ingold L.

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**Example text**

Moreover, the degree of pI,M is less than or equal to rank(I). Proof. Set I k /I k+1 , gr I = [I k M ]/[I k+1 M ]. 7, we see that gr M is a finitely generated graded module on Noether graded ring gr I. Let rank(I) = l and let {p1 , p2 · · · , pl } be a generating set of I. Then gr I, as a C/I-algebra, is generated by the images p˜i of pi in I/I 2 for i = 1, 2, · · · , l. Since deg(p˜i ) = 1 for i = 1, 2, · · · , l, Hilbert’s original results (cf. [AM, Chapter 11]) imply that for large integer k, dim [I k M ]/[I k+1 M ] is a polynomial of k with rational coefficients.

Since each function f in [I2 ] satisfies that p(D)f |λ = 0, p ∈ I2λ , λ ∈ Z(I2 )\Z(I1 ), this ensures that [I2 ] ∩ R = {0}. ˙ contains I1 , and [I2 ]+R ˙ is closed, we obtain that By the fact that [I2 ]+R ˙ = [I1 ]. [I2 ]+R Therefore, it holds that I1 /I2 ∼ = R. 5 is complete. 6 Let M be a finite codimensional submodule of X. Then we have the following. (1) Z(M ) = σp (Mz1 , Mz2 , · · · , Mzn ) ⊂ Ω, (2) M = λ∈Z(M ) Mλe , (3) codim M = card Z(M ) = ˆ dim Mλ . 6 shows that the codimension codim M of M in X equals the cardinality of zeros of M by counting multiplicities.

12 is true for n = 1. Fix (z1 , z2 , · · · , zn ) ∈ Ω. Let U = {z : (z, z2 , · · · , zn ) ∈ Ω}. Then U is a disk with the center 0 and p(z, z2 , · · · , zn ) has no zero in U . Thus |p(z1 , z2 , · · · , zn )/p(rz1 , z2 , · · · , zn )| ≤ 2d1 , where we denote by d1 the degree of p in the variable z1 . Applying the induction hypothesis to p(z1 , · · · ) we obtain that |p(rz1 , z2 , · · · , zn )/p(rz1 , rz2 , · · · , rzn )| ≤ 2d2 +···+dn , that is, |p(z1 , z2 , · · · , zn )/p(rz1 , rz2 , · · · , rzn )| ≤ 2d(p) .