Difference between revisions of "UMP for Free Modules"
Kfagerstrom (talk | contribs) (Created page with "UMP for free modules Let <math>R</math> be a ring, <math>M</math> a free <math>R</math>-module with basis <math>B</math>, <math>N</math> an <math>R</math>-module, <math>j:B\...") |
Kfagerstrom (talk | contribs) |
||
| Line 4: | Line 4: | ||
As <math>m=\sum_{i=1}^n r_ib_i,\ b_i\in B</math> unique, imples <math>h(m):=\sum_{i=1}^n r_ij(b_i)</math> is well defined | As <math>m=\sum_{i=1}^n r_ib_i,\ b_i\in B</math> unique, imples <math>h(m):=\sum_{i=1}^n r_ij(b_i)</math> is well defined | ||
| + | |||
| + | We have two things to prove: existence and uniqueness. | ||
| + | Existence: By Lemma 1.59, any <math>0 \ne m ∈ M</math> can be written uniquely as | ||
| + | <math>m = r_1b_1 + \cdots + r_nb_n</math> | ||
| + | with <math>b_i\in B </math> 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). | ||
Revision as of 17:23, 8 March 2023
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).
