Commutative algebra - Department of Mathematical Sciences - old ...

10
Dedekind rings
10.1. Principal ideal domains
10.1.1. Lemma. Let R be a domain. The set of elements x ∈ M in a module with
Ann(x) = 0 is a submodule.
Proof. The product of two nonzero elements is nonzero.
10.1.2. Definition. Let R be a domain. An element x ∈ M in a module is a torsion
element if Ann(x) = 0. By 10.1.1 the set of torsion elements is a submodule
T (M) ⊂ M, the torsion submodule. If T (M) = 0 then M is a torsion free
module.
10.1.3. Lemma. Let R be a domain.
(1) A flat module is torsion free.
(2) M/T (M) is torsion free.
(3) Let K be the fraction field. Then T (M) = Ker(M → M ⊗R K).
(4) If M → N → L is exact, then T (M) → T (N) → T (L) is exact.
(5) If U ⊂ R is multiplicative, then U −1 T (M) = T (U −1 M).
Proof. (1) If 0 = a ∈ R and F flat, then aF is injective. (2), (4), (5) These are
clear. (3) This follows from 4.4.1.
10.1.4. Corollary. Let R be a domain and M a module. The following are equivalent.
(1) M is torsion free.
(2) MP is torsion free for all prime ideals P .
(3) MP is torsion free for all maximal ideals P .
10.1.5. Lemma. Let R be a principal ideal domain. A submodule of a finite free
module is free.
Proof. Let F ⊂ R n be a submodule and p : R n → R the last projection. Then
p(F ) is a principal ideal and free. By induction F ∩ Ker p is free, so F F ∩
Ker p ⊕ p(F ) is free.
10.1.6. Proposition. Let R be a principal ideal domain.
(1) A torsion free module is flat.
(2) A finite torsion free module is free.
Proof. (1) For a nonzero ideal (a) ⊂ R the composite R (a) → R is aM.
For a torsion free F the homomorphism aF : F (a) ⊗R F → F is injective.
So F is flat by 3.7.12. (2) Let K be the fraction field. Let F ⊂ F ⊗R K be a
torsion free submodule and suppose x1, . . . , xn ∈ F give a basis for F ⊗R K.
Then F ′ = Rx1 ⊕ · · · ⊕ Rxn ⊂ F is a free submodule such that F/F ′ is a finite
111