Difference between revisions of "UMP for Free Modules"
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| − | UMP for | + | ==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\to N</math> a function. Then there exists a unique <math>R</math>-module homomorphism <math>h:M\to N</math> such that <math>h(b)=j(b)\ \forall b\in B</math>. | 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\to N</math> a function. Then there exists a unique <math>R</math>-module homomorphism <math>h:M\to N</math> such that <math>h(b)=j(b)\ \forall b\in B</math>. | ||
| + | |||
| + | ==Proof== | ||
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. | 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 | + | |
| + | ''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> | <math>m = r_1b_1 + \cdots + r_nb_n</math> | ||
| − | with <math>b_i\in B </math> distinct and 0 | + | with <math>b_i\in B </math> distinct and <math>0 \ne r_i \in R</math>. Define <math>h: M \to N</math> by <math> \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( | + | h(0_M) = 0_N |
| − | h( | + | \end{cases} </math> |
| + | |||
One can check that this satisfies the conditions to be an R-module homomorphism (exercise!). | 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( | + | |
| − | Then in particular h: (M, +) → (N, +) is a group homomorphism and therefore h( | + | ''Uniqueness:'' Let <math>h : M → N</math> be an <math>R</math>-module homomorphism such that <math>h(b_i) = j(b_i)</math>. |
| − | by properties of group homomorphisms. Furthermore, if m = | + | Then in particular <math>h: (M, +) → (N, +)</math> is a group homomorphism and therefore <math>h(0_M) = 0_N</math> |
| − | h(m) = h( | + | by properties of group homomorphisms. Furthermore, if <math>m = r_1b_1 + · · · + r_nb_n </math> then |
| − | by the definition of homomorphism, and because h( | + | <math> 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) |
| + | </math> by the definition of homomorphism, and because <math>h(b_i) = j(b_i)</math>. | ||
| + | |||
| + | [[Category:Modules]] [[Category: Free Modules]] [[Category: Proofs]] | ||
Latest revision as of 21:56, 9 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 \(h(b)=j(b)\ \forall b\in B\).
Proof
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 \(h(b_i) = j(b_i)\).
