By Ash R.B.

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**Sample text**

Thus if bz = 0 (with b = 0 by choice of b), we have z = 0 and f is injective. But any linear transformation on a ﬁnite-dimensional vector space is injective iﬀ it is surjective. Therefore if b ∈ S and b = 0, there is an element c ∈ R[b] ⊆ S such that bc = 1. Consequently, S is a ﬁeld. 11 Preview Let S be integral over the subring R. We will analyze in great detail the relation between prime ideals of R and those of S. Suppose that Q is a prime ideal of S, and let P = Q ∩ R. ) Then P is a prime ideal of R, because it is the preimage of Q under the inclusion map from R into S.

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 . By hypothesis, ar ∈ R1 , so λr (Mn ) ⊆ Mn+1 . If Kn is the kernel, and Cn the cokernel, of λr : Mn → Mn+1 , we have the exact sequence G Kn G Mn λr G Mn+1 G Cn G 0. 0 Let K be the direct sum of the Kn and C the direct sum of the Cn , n ≥ 0.

A discrete valuation on K is a surjective map v : K → Z ∪ {∞}, such that for every x, y ∈ K, (a) v(x) = ∞ if and only if x = 0; (b) v(xy) = v(x) + v(y); (c) v(x + y) ≥ min(v(x), v(y)). A discrete valuation induces a nonarchimedean absolute value via |x| = cv(x) , where c is a constant with 0 < c < 1. 3. In Examples 1 and 2, we take v(pr m/n) = r. In Example 3, v(f /g) = deg g − deg f . In Example 4, ∞ v( i=r ai xi ) = r (if ar = 0). 3 Proposition If v is a discrete valuation on the ﬁeld K, then V = {a ∈ K : v(a) ≥ 0} is a valuation ring with maximal ideal M = {a ∈ K : v(a) ≥ 1}.