By Ash R.B.

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2. COMPLETION OF A MODULE 5 2. Let R = A[x1 , . . , xn ] be a polynomial ring in n variables, and I the maximal ideal (x1 , . . , xn ). Let Mn = R/I n and θn (f + I n ) = f + I n−1 , n = 1, 2, . . An element of Mn is represented by a polynomial f of degree at most n − 1. ) The image of f in I n−1 is represented by the same polynomial with the terms of degree n − 1 deleted. Thus the inverse limit can be identified with the ring A[[x1 , . . , xn ]] of formal power series. Now let M be a filtered R-module with filtration {Mn }.

Since t ∈ M, we have t = ct for some c ∈ V . Thus 1 = v(t) = v(c) + v(t ) ≥ 0 + 1 = 1, which forces v(t ) = 1. 6 Proposition If t is a uniformizer, then every nonzero element a ∈ K can be expressed uniquely as a = utn where u is a unit of V and n ∈ Z. Also, K = Vt , that is, K = S −1 V where S = {1, t, t2 , . . }. Proof. Let n = v(a), so that v(at−n ) = 0 and therefore at−n is a unit u. To prove uniqueness, note that if a = utn , then v(a) = v(u) + nv(t) = 0 + n = n, so that n, and hence u, is determined by a.

DISCRETE VALUATION RINGS 7 Proof. 7), we have R ⊃ M ⊃ M2 ⊃ · · · ⊃ Mn = I, hence R/I ⊃ M/I ⊃ M2 /I ⊃ · · · ⊃ Mn /I = 0. By basic properties of composition length, we have, with l = lR , l(R/I) = l( R/I M/I ) + l(M/I) = l(R/M) + l( 2 ) + l(M2 /I). M/I M /I Continuing in this fashion, we get n−1 l(Mi /Mi+1 ). l(R/I) = i=0 Since M is generated by a uniformizer t, it follows that ti + Mi+1 generates Mi /Mi+1 . Since Mi /Mi+1 is annihilated by M, it is an R/M-module, that is, a vector space, over the field R/M.

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