Direct sum (ring)

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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]

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 contradicts that one or both of them didn't have unity!
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]
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.
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:



See also

References

  1. Fundamentals of Abstract Algebra - Neal H. McCoy - An Expanded Version
  2. My (Alec's) own work