Direct sum (ring)

For other kinds of direct sums see Direct sum

Definition

Given two rings $(R,+_R,\times_R)$ and $(S,+_S,\times_S)$ their direct sum is defined on the set $R\times S$ (where $\times$ is the Cartesian product), that is:

• $R\times S=\{(x,y)\vert\ x\in R\wedge y\in S\}$

and is denoted:

• $R\oplus S=(R\times S,+,\times)$ or simply $R\oplus S$ as Mathematicians are lazy

where the operation $+$ and $\times$ are defined as follows:

• Given $(x,y),\ (x',y')\in R\oplus S$ 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 $(x,y)(x',y')=(x\times_Rx',y\times_Sy')$

Other group properties

Unity

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Theorem: The ring $R\oplus S$ has unity if and only if both $R$ and $S$ have unity

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$R\oplus S$ has unity $\implies$ both $R$ and $S$ have unity.

Let $(\alpha,\beta)=e\in R\oplus S$ be that unity. Suppose that neither, or just one of $R$ and $S$ have unity.

We know that $$\forall (x,y)\in R\oplus S$$ that $$e(x,y)=(x,y)e=(x,y)$$

This means that:

• $\forall x\in R[\alpha x=x\alpha=x]$
• $\forall y\in S[\beta y=y\beta=y]$

However then $\alpha$ is the unity of $R$ and $\beta$ 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 $\implies$ $R\oplus S$ has unity

Let $e_R$ be the unity of $R$ and $e_S$ be the unity of $S$

I claim that $(e_R,e_S)$ is the unity of $R\oplus S$

Let $(x,y)\in R\oplus S$ be given, then:

• $(e_R,e_s)(x,y)=(e_Rx,e_Sy)=(x,y)$
• $(x,y)(e_R,e_S)=(xe_R,ye_S)=(x,y)$

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 $R\oplus S$ 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 $e=(e_S,e_R)$ explicitly.

Commutative

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Theorem: $R\oplus S$ is a commutative ring if and only if both $R$ and $S$ are commutative rings

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TODO:

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See also

• Ring
• Homomorphism (ring)