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 \(0 \ne r_i \in R\). Define \(h: M \to N\) by \( \begin{cases} h(r_1b_1 + \cdots + r_nb_n) = r_1j(b_1) + · · · + r_nj(b_n) & \text{if} r_1b_1 + · · · + r_nb_n \ne 0 \\ h(0_M) = 0_N \end{cases} \)

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 \(h(b_i) = j(b_i)\). Then in particular \(h: (M, +) → (N, +)\) is a group homomorphism and therefore \(h(0_M) = 0_N\) by properties of group homomorphisms. Furthermore, if \(m = r_1b_1 + · · · + r_nb_n \) then \( h(m) = h(r_1b_1 + · · · + r_nb_n) = r_1h(b_1) + · · · + r_nh(b_n) = r_1j(b_1) + · · · + r_nj(b_n) \) by the definition of homomorphism, and because \(h(b_i) = j(b_i)\).