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

2.4. SUM AND PRODUCT 29
2.4.4. Definition. A family of submodules Mα ⊂ M constitutes a direct sum if
any element x ∈
α Mα has a unique finite representation
x =
xα, xα ∈ Mα
2.4.5. Proposition. The following conditions are equivalent
(1) The family Mα ⊂ M constitutes a direct sum.
(2) The natural homomorphism
Mα →
is an isomorphism.
(3) For all β
α
α
α
Mα
Mβ ∩
Mα = 0
α=β
Proof. (1) and (2) are equivalent. If (1) is true and xβ =
α=β xα ∈ Mβ ∩
α=β Mα, then xβ −
α=β xα = 0. Therefore by uniqueness xβ = 0 and (3)
is true. Conversely if (3) is true and
α xα = 0, then xβ = −
α=β xα ∈
Mβ ∩
α=β Mα = 0. for any β. This shows uniqueness and therefore (1) is
true.
2.4.6. Definition. Let R be a ring. A module isomorphic to a direct sum
α
R of
copies of the ring R is a free module.
A basis of a module, is a subset Y such that any element admits a unique finite
representation
α aαyα, where aα ∈ R, yα ∈ Y .
The standard basis of
α R consists of the elements eα, where each is a family
with 0 for β = α and exactly 1 at index α.
2.4.7. Proposition. A module is free if and only if it admits a basis. If yα is a basis
of F then there is an isomorphism
f :
R → F
where f( aα) = aαyα.
α
Proof. Given a free module f :
α R → F then yα = f(eα) is a basis. Conversely
given a basis yα ∈ F then 1yα : R → F 2.3.11 is a family of homomorphisms
giving a homomorphism f :
α R → F by 2.4.3. As f( aα) = aαyα
it follows that f is bijective and therefore by 2.1.5 an isomorphism.
2.4.8. Remark. The polynomial ring R[X1, . . . , Xn] is free as R-module.
2.4.9. Example. (1) A nonzero ideal is a free module if and only if it has a basis
consisting of a nonzero divisor. Namely if x1 = x2 where in a basis then the
product x1x2 has two different representations.
(2) Let I ⊂ R be an ideal. The module R/I is free if and only if I = 0 or I = R.
2.4.10. Proposition. Any module over a field is free. Conversely if any module
over a nonzero ring is free, then the ring is a field
Proof. By Zorn’s lemma any vector space admits a basis. If I ⊂ R is an ideal and
R/I is free, then I = Ann(R/I) is either 0 or R. So R is a field.