Question

Prove Theorem 3.1 .

Short answer

Theorem 3.1 is proved.

Step-by-step solution

Determine R×S  with addition

Determine that $R\times S$with addition and multiplication is a ring.

Addition: ((r,s)+(r',s')=(r+r',s+s'))

Multiplication: ((r,r')+(s,s')=(rr',ss'))

Here, R and S arerings.

Let’s consider that,

1. 2$\left(r,s\right)+\left(r\text{'},s\text{'}\right)=\left(r+r\text{'},s+s\text{'}\right)\in R\times S\left(r+r\text{'}\in R,s+s\text{'}\in S\right)$
2. Addition: $$\begin{gathered} \left(r,s\right)+\left(\left(r\text{'},s\text{'}\right)+\left(r",s"\right)\right)=\left(r,s\right)+\left(r\text{'}+r",s\text{'}+s"\right)=\left(r+r\text{'}+r",s+s\text{'}+s"\right) \\ \left(\left(r,s\right)+\left(r\text{'},s\text{'}\right)\right)+\left(r",s"\right)=\left(r+r\text{'},s+s\text{'}\right)+\left(r",s"\right)=\left(r+r\text{'}+r",s+s\text{'}+s"\right) \\ \left(r,s\right)+\left(\left(r\text{'},s\text{'}\right)+\left(r",s"\right)\right)=\left(\left(r,s\right)+\left(r\text{'},s\text{'}\right)\right)+\left(r",s"\right) \end{gathered}$$

3. Commutative addition:

$\left(r,s\right)+\left(r\text{'},s\text{'}\right)=\left(r+r\text{'},s+s\text{'}\right)=\left(r\text{'}+r,s\text{'}+s\right)=\left(r\text{'},s\text{'}\right)+\left(r,s\right)$

By definition, addition is commutative in rings $R$and $S$.

4. Let’s assume that,

$0_{R\times S}=\left(0_{R,}{0}_S\right)$then,

$\left(r,s\right)+\left(0_{R,}{0}_S\right)=\left(r+0_{R,}s+0_S\right)=\left(r,s\right)=\left(0_R+r,0_S+s\right)=\left(0_{R,}{0}_S\right)+\left(r,s\right)$

5. Let’s consider,$\left(-r,-s\right)\in R\times S$

Then,$\left(r,s\right)+\left(-r,-s\right)=\left(r+\left(-r\right),s+\left(-s\right)\right)=\left(0_R,0_S\right)=0_{R\times S}$

Determine R×S  with multiplication

6. Multiplication:

$\left(r,s\right)\left(r\text{'}.s\text{'}\right)=\left(rr\text{'},ss\text{'}\right)\in R\times S(rr\text{'}\in R,ss\text{'}\in S)$

It is closed with multiplication.

7. Commutative multiplication:

$$\begin{gathered} \left(r,s\right)\left(\left(r\text{'},s\text{'}\right)\left(r",s"\right)\right)=\left(r,s\right)\left(r\text{'}rr"\right)\left(s\text{'}ss"\right)=\left(rr\text{'}r",ss\text{'}s"\right) \\ \left(\left(r,s\right)\left(r\text{'},s\text{'}\right)\right)\left(r",s"\right)=\left(rr\text{'},ss\text{'}\right)\left(r",s"\right)=\left(rr\text{'}r",ss\text{'}s"\right) \\ \left(r,s\right)\left(\left(r\text{'},s\text{'}\right)\left(r",s"\right)\right)=\left(\left(r,s\right)\left(r\text{'},s\text{'}\right)\right)\left(r",s"\right) \end{gathered}$$

8. Associative multiplication:

$$\begin{gathered} \left(r,s\right)\left(\left(r\text{'},s\text{'}\right)+\left(r",s"\right)\right)=\left(r,s\right)\left(r\text{'}+r",s\text{'}+s"\right)=\left(rr\text{'}+rr",ss\text{'}+ss"\right) \\ =\left(rr\text{'},ss\text{'}\right)+\left(rr",ss"\right)=\left(r,s\right)\left(r\text{'},s\text{'}\right)+\left(r,s\right)\left(r",s"\right) \\ \left(\left(r,s\right)+\left(r\text{'},s\text{'}\right)\right)\left(r",s"\right)=\left(r+r\text{'},s+s\text{'}\right)\left(r",s"\right)=\left(rr"+r\text{'}r",ss"+s\text{'}s"\right) \\ =\left(rr",ss"\right)+\left(r\text{'}r",s\text{'}s"\right)=\left(r,s\right)\left(r",s"\right)+\left(r\text{'},s\text{'}\right)\left(r",s"\right) \end{gathered}$$

Therefore, $R\times S$is a Ring