For other kinds of direct sums see Direct sum
Definition
Given two rings (R,+R,×R) and (S,+S,×S) their direct sum is defined on the set R×S (where × is the Cartesian product), that is:
and is denoted:[1]
where the operation + and × are defined as follows:
- Given (x,y), (x′,y′)∈R⊕S we define:
- Addition as: (x,y)+(x′,y′)=(x+x′,y+y′) or more formally (x,y)+(x′,y′)=(x+Rx′,y+Sy′)
- Multiplication as: (x,y)(x′,y)=(xx′,yy′) or more formally (x,y)(x′,y′)=(x×Rx′,y×Sy′)
Other group properties
Unity
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Theorem: The ring R⊕S has unity if and only if both R and S have unity[2]
R⊕S has unity ⟹ both R and S have unity.
- Let (α,β)=e∈R⊕S be that unity. Suppose that neither, or just one of R and S have unity.
- We know that ∀(x,y)∈R⊕S that e(x,y)=(x,y)e=(x,y)
- This means that:
- ∀x∈R[αx=xα=x]
- ∀y∈S[βy=yβ=y]
- However then α is the unity of R and β is the unity of S
- This contradicts that one or both of them didn't have unity!
- So this half of the proof is complete
R and S have unity ⟹ R⊕S has unity
- Let eR be the unity of R and eS be the unity of S
- I claim that (eR,eS) is the unity of R⊕S
- Let (x,y)∈R⊕S be given, then:
- (eR,es)(x,y)=(eRx,eSy)=(x,y)
- (x,y)(eR,eS)=(xeR,yeS)=(x,y)
- We have shown ∀(x,y)∈R⊕S[(eR,eS)(x,y)=(x,y)(eR,eS)=(x,y)]
- That is the definition of R⊕S 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 e=(eS,eR) explicitly.
Commutative
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Theorem: R⊕S is a commutative ring if and only if both R and S are commutative rings
See also
References
- Jump up ↑ Fundamentals of Abstract Algebra - Neal H. McCoy - An Expanded Version
- Jump up ↑ My (Alec's) own work
