Difference between revisions of "Direct sum (ring)"
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Latest revision as of 14:32, 8 June 2015
For other kinds of direct sums see Direct sum
Definition
Given two rings [ilmath](R,+_R,\times_R)[/ilmath] and [ilmath](S,+_S,\times_S)[/ilmath] their direct sum is defined on the set [ilmath]R\times S[/ilmath] (where [ilmath]\times[/ilmath] is the Cartesian product), that is:
- [ilmath]R\times S=\{(x,y)\vert\ x\in R\wedge y\in S\}[/ilmath]
and is denoted:[1]
- [ilmath]R\oplus S=(R\times S,+,\times)[/ilmath] or simply [ilmath]R\oplus S[/ilmath] as Mathematicians are lazy
where the operation [ilmath]+[/ilmath] and [ilmath]\times[/ilmath] are defined as follows:
- Given [ilmath](x,y),\ (x',y')\in R\oplus S[/ilmath] we define:
- Addition as: [ilmath](x,y)+(x',y')=(x+x',y+y')[/ilmath] or more formally [ilmath](x,y)+(x',y')=(x+_Rx',y+_Sy')[/ilmath]
- Multiplication as: [ilmath](x,y)(x',y)=(xx',yy')[/ilmath] or more formally [ilmath](x,y)(x',y')=(x\times_Rx',y\times_Sy')[/ilmath]
Other group properties
Unity
Theorem: The ring [ilmath]R\oplus S[/ilmath] has unity if and only if both [ilmath]R[/ilmath] and [ilmath]S[/ilmath] have unity[2]
[ilmath]R\oplus S[/ilmath] has unity [ilmath]\implies[/ilmath] both [ilmath]R[/ilmath] and [ilmath]S[/ilmath] have unity.
- Let [ilmath](\alpha,\beta)=e\in R\oplus S[/ilmath] be that unity. Suppose that neither, or just one of [ilmath]R[/ilmath] and [ilmath]S[/ilmath] have unity.
- We know that [math]\forall (x,y)\in R\oplus S[/math] that [math]e(x,y)=(x,y)e=(x,y)[/math]
- This means that:
- [ilmath]\forall x\in R[\alpha x=x\alpha=x][/ilmath]
- [ilmath]\forall y\in S[\beta y=y\beta=y][/ilmath]
- However then [ilmath]\alpha[/ilmath] is the unity of [ilmath]R[/ilmath] and [ilmath]\beta[/ilmath] is the unity of [ilmath]S[/ilmath]
- This means that:
- This contradicts that one or both of them didn't have unity!
- We know that [math]\forall (x,y)\in R\oplus S[/math] that [math]e(x,y)=(x,y)e=(x,y)[/math]
- So this half of the proof is complete
[ilmath]R[/ilmath] and [ilmath]S[/ilmath] have unity [ilmath]\implies[/ilmath] [ilmath]R\oplus S[/ilmath] has unity
- Let [ilmath]e_R[/ilmath] be the unity of [ilmath]R[/ilmath] and [ilmath]e_S[/ilmath] be the unity of [ilmath]S[/ilmath]
- I claim that [ilmath](e_R,e_S)[/ilmath] is the unity of [ilmath]R\oplus S[/ilmath]
- Let [ilmath](x,y)\in R\oplus S[/ilmath] be given, then:
- [ilmath](e_R,e_s)(x,y)=(e_Rx,e_Sy)=(x,y)[/ilmath]
- [ilmath](x,y)(e_R,e_S)=(xe_R,ye_S)=(x,y)[/ilmath]
- Let [ilmath](x,y)\in R\oplus S[/ilmath] be given, then:
- We have shown [math]\forall (x,y)\in R\oplus S[(e_R,e_S)(x,y)=(x,y)(e_R,e_S)=(x,y)][/math]
- That is the definition of [ilmath]R\oplus S[/ilmath] having a unity.
- I claim that [ilmath](e_R,e_S)[/ilmath] is the unity of [ilmath]R\oplus S[/ilmath]
- This completes the proof.
- We have shown {{M|1=\exists e\in R\oplus S\forall (x,y)\in R\oplus S[e(x,y)=(x,y)e=(x,y)]</math> is true (the exact definition)
- Where [ilmath]e=(e_S,e_R)[/ilmath] explicitly.
Commutative
Theorem: [ilmath]R\oplus S[/ilmath] is a commutative ring if and only if both [ilmath]R[/ilmath] and [ilmath]S[/ilmath] are commutative rings
TODO:
