Algebra (Unknown 27). - Index of

26 Algebra 2
It remains to show that maps α: P ↦→ S −1 P for P ∩ S = ∅ and β: Q ↦→ φ −1 (Q)
are inverse to each other. From 9.5 we already know that α ◦ β is the indentity map.
To show β ◦ α is the indentity map let a ∈ φ −1 (S −1 P ). Then a/1 = p/s for some
p ∈ P, s ∈ S. Hence there is u ∈ S such that u(p − as) = 0, i.e. a(us) ∈ P .
Since us ∈ S, we have us ∉ P . Since P is prime we deduce a ∈ P . Thus
φ −1 (S −1 P ) = P .
9.7. Corollary. Let P be a prime ideal of A. Then the localization A P has only one
maximal ideal, namely M P = P A P . Thus, Spec (A P ) = {QA P : Q ∈ Spec (A), Q ⊂
P }, m-Spec (A P ) = {P A P }.
Proof. Indeed, prime ideals of A P correspond to prime ideals of A which are contained
in P . Hence the only maximal ideal of A P corresponds to P . All other maximal
ideals of A dissappear in A P .
Definition. Let P be a prime ideal of a ring A. The residue field of A at P is
k(P ) = A P /M P .
Example. pZ (p) is the only maximal ideal of Z (p) . Note that Z (p) isn’t a field ( p
isn’t invertible in Z (p) ). The residue field of Z at (p) is
k( (p) ) = Z (p) /pZ (p) = F p .
9.8. Definition. A ring A is called a local ring if m-Spec (A) consists of one element,
i.e. A has exactly one maximal ideal.
Example. The localization A P is a local ring. Every field is a local ring.
9.9. If A is a local ring and M is its maximal ideal, then 1 + M ⊂ A × . Indeed,
if for some m ∈ M 1 + m were not a unit of A, then the ideal (1 + m) is a proper
ideal of A, hence it is contained in the maximal ideal M ; since m ∈ M we would get
1 = 1 + m − m ∈ M , a contradiction.
9.10. Nakayama’s Lemma. Let N be a module over a ring A generated by k
elements n 1 , . . . , n k . Let I be an ideal of A.
a) If N = IN = { ∑ k
j=1 c jn j : c j ∈ I} then there is an element a ∈ 1 + I such that
aN = 0.
b) If M is the maximal ideal of a local ring A and MN = N , then N = 0.
Proof. a) Let N = IN . Assume that N ≠ 0. Let n 1 , . . . , n k ∈ N be the minimal set
of generators of the A-module N with the minimal possible k. Then n 1 ∈ N = IN ,
so n 1 = ∑ k
j=1 b 1jn j with b 1j ∈ I. Then (1 − b 11 )n 1 − b 12 n 2 − · · · − b 1k n k = 0.
Similarly we find b ij for i = 2, . . . , n. So for the matrix C = E − B, B = (b ij ) we