# Direct sum (ring)

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:

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

[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 $\forall (x,y)\in R\oplus S$ that $e(x,y)=(x,y)e=(x,y)$
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 $\forall (x,y)\in R\oplus S[(e_R,e_S)(x,y)=(x,y)(e_R,e_S)=(x,y)]$
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)][/itex] 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: