UMP for Free Modules

UMP for free modules

Let \(R\) be a ring, \(M\) a free \(R\)-module with basis \(B\), \(N\) an \(R\)-module, \(j:B\to N\) a function. Then there exists a unique \(R\)-module homomorphism \(h:M\to N\) such that \(h(b)=j(b)\ \forall b\in B\).

As \(m=\sum_{i=1}^n r_ib_i,\ b_i\in B\) unique, imples \(h(m):=\sum_{i=1}^n r_ij(b_i)\) is well defined

We have two things to prove: existence and uniqueness. Existence: By Lemma 1.59, any \(0 \ne m ∈ M\) can be written uniquely as \(m = r_1b_1 + \cdots + r_nb_n\) with \(b_i\in B \) distinct and 0 6= ri ∈ R. Define h: M → N by ( h(r1b1 + · · · + rnbn) = r1j(b1) + · · · + rnj(bn) if r1b1 + · · · + rnbn 6= 0 h(0M) = 0N One can check that this satisfies the conditions to be an R-module homomorphism (exercise!). Uniqueness: Let h : M → N be an R-module homomorphism such that h(bi) = j(bi). Then in particular h: (M, +) → (N, +) is a group homomorphism and therefore h(0m) = 0N by properties of group homomorphisms. Furthermore, if m = r1b1 + · · · + rnbn then h(m) = h(r1b1 + · · · + rnbn) = r1h(b1) + · · · + rnh(bn) = r1j(b1) + · · · + rnj(bn) by the definition of homomorphism, and because h(bi) = j(bi).