UMP for Free Modules - Revision history

Kfagerstrom at 21:56, 9 March 2023

2023-03-09T21:56:24Z

Kfagerstrom at 22:07, 8 March 2023

2023-03-08T22:07:56Z

Kfagerstrom: /* UMP for Free Modules */

2023-03-08T22:04:21Z

UMP for Free Modules

Kfagerstrom at 18:47, 8 March 2023

2023-03-08T18:47:29Z

Kfagerstrom at 17:23, 8 March 2023

2023-03-08T17:23:30Z

Kfagerstrom: Created page with "UMP for free modules Let $R$ be a ring, $M$ a free $R$ -module with basis $B$ , $N$ an $R$ -module, $j:B\..."$

2023-03-08T17:17:48Z

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\..."

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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>.

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

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 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>.
 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
@@ Line 22: / Line 24: @@
 </math> by the definition of homomorphism, and because <math>h(b_i) = j(b_i)</math>.
-[[Category:Modules]]
+[[Category:Modules]] [[Category: Free Modules]] [[Category: Proofs]]

← Older revision		Revision as of 22:07, 8 March 2023
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	</math> by the definition of homomorphism, and because <math>h(b_i) = j(b_i)</math>.		</math> by the definition of homomorphism, and because <math>h(b_i) = j(b_i)</math>.

−	[[Category\|Modules]]	+	[[Category:Modules]]

← Older revision		Revision as of 22:04, 8 March 2023
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	<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> 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>.		</math> by the definition of homomorphism, and because <math>h(b_i) = j(b_i)</math>.
		+
		+	[[Category\|Modules]]

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-UMP for free modules
+==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>.
@@ Line 6: / Line 6: @@
 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
+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(r1b1 + · · · + rnbn) = r1j(b1) + · · · + rnj(bn) if r1b1 + · · · + rnbn 6= 0
+h(0_M) = 0_N
-h(0M) = 0N
+\end{cases} </math>
 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
+''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 = r1b1 + · · · + rnbn then
+Then in particular <math>h: (M, +) → (N, +)</math> is a group homomorphism and therefore <math>h(0_M) = 0_N</math>
-h(m) = h(r1b1 + · · · + rnbn) = r1h(b1) + · · · + rnh(bn) = r1j(b1) + · · · + rnj(bn)
+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(bi) = j(bi).
+<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)

← Older revision		Revision as of 17:23, 8 March 2023
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	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).