• We say that X generates or spans the submodule hXi. • We say a module is finitely generated if there is a finite set which generates it. Definition 6.5. • If M is a module over R, we say a set S ⊆ M is linearly independent if whenever we have an equation r1 s1 + r2 s2 + · · · + rk sk = 0 for ri ∈ R and si ∈ S we have r1 = . . . = rk = 0. • A set S is a basis for M if it is linearly independent and it spans M . Any module which has a basis is called a free module.
6.3
Quotient Modules and Isomorphism Theorems
Remark 6.3. All rings R are integral domains from now on. Definition 6.6. • If N is a submodule of M , then we can define the quotient module M/N and the associated operations exactly as quotient spaces were defined for vector spaces. • If M1 , M2 are modules, we define a module homomorphism exactly as linear transformations were defined for vector spaces. • We define isomorphisms and the kernel and image of a module homomorphism exactly as linear isomorphisms, kernels and images were defined for vector spaces. It is easy to see that, for a module homomorphism φ, ker(φ) and im(φ) are both submodules. Example 6.7. When R is a field, module homomorphisms are exactly linear maps. When R is Z a Z-module homomorphism is just a homomorphism of the abelian groups. As another important example, it is easy to see that if M is an R-module and N is a submodule of M then the definition of the module structure on M/N ensures precisely that the map q : M → M/N given by q(m) = m + N is a surjective module homomorphism. Lemma 6.1. (Submodule Correspondence) Let M be an R-module and N a submodule. Let q : M → M/N be the quotient map. If S is a submodule of M then q(S) is a submodule of M/N , while if T is a submodule of M/N then q −1 (T ) is a submodule of M . Moreover the map T 7→ q −1 (T ) gives an injective map from submodules of M/N to the submodules of M which contain N , thus submodules of M/N correspond bijectively to submodules of M which contain N . Proof. To check that q(S) and q −1 (T ) are submodules of N and M respectively follows directly from the definitions, we give the argument for q −1 (T ), since the argument for q(S) follows exactly the same pattern. If m1 , m2 ∈ q −1 (T ) then q(m1 ), q(m2 ) ∈ T and it follows since T is a submodule that q(m1 ) + q(m2 ) = q(m1 + m2 ) ∈ T which says precisely that m1 + m2 ∈ q −1 (T ). Similarly if r ∈ R then q(r · m1 ) = r · q(m1 ) ∈ T since q(m1 ) ∈ T and T is a submodule, so that r · m1 ∈ q −1 (T ). Thus q −1 (T ) is a submodule of M as required. Now if T is any subset of M/N we have q(q −1 (T )) = T simply because q is surjective. Since we have just checked q −1 (T ) is always a submodule in M , this immediately implies that 23