By Dan Laksov
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Example text
For every prime ideal p of A the height of p in A is the codimension of the irreducible subset V (a) of Spec(A). That is, the height of p is the supremum of the lengths n of the chains p = p0 ⊃ p1 ⊃ · · · ⊃ pn of prime ideals pi of A. We denote the height of p by ht(p) = htA (p). When there exists arbitrary long chains we say that A has infinite dimension. 16) Example. We have that dim(Z) = 1. 17) Example. Let p be a prime number and Z(p) the ring of rational numbers of the form m/n such that p does not divide n.
Since p is maximal in I the ideals f A + p and f A + p of A both intersect S. Hence there are elements g, g in A, h, h in p, and s, s in S such that f g + h = s and f g + h = s . Since ss ∈ S the product (f g + h)(f g + h ) is not in p. Since h and h both are in p, we therefore have that f f can not be in p, as we wanted to prove. 17) Remark. Let A be a ring and m an ideal in A. Then A is a local ring with maximal ideal m if and only if all elements in A \ m are invertible in A. It is clear that if all the elements of A \ m are invertible then m is maximal and A can not have other maximal ideals than m.
1) Describe Spec(A). (2) What is the dimension of Spec(A)? 7. Let A = Zn be the cartesian product of the ring Z with itself n times. (1) Describe Spec(A). (2) What is the dimension of Spec(A)? 8. A topological space X is connected if there is no non-empty subset of X different from X. Let A be a ring and let X = Spec(A). Show that the following assertions are equivalent: (1) The space X = Spec(A) is not connected. (2) There are elements f and g in A such that f g = 0, f 2 = f , g 2 = g, and f + g = 1.