How to show that $(S, \oplus, \otimes)$ is a ring? Let $(R,+,\times)$ be a ring with additive identity $0 \in R$. On the set $S = \left.\left\{ (a,b) \,\right|\, a,b \in R \right\}$ the binary operators $\oplus$ and $\otimes$ are defined by:


*

*$(a, b) \oplus (c, d) = (a + c, b + d)$

*$(a, b) \otimes (c, d) = (a \times c + b \times d, a \times d + b \times c)$


How do I show that $(S, \oplus, \otimes)$ is a ring? I understand the axioms which a set needs to satisfy, but I'm getting confused.
 A: You show that the axioms hold one by one. Let $(a,a'),(b,b'),(c,c') \in S$. 


*

*Closure of addition: $(a,a')\oplus (b,b')=(a+b,a'+b') \in S$. 

*Associativity of addition: $(a,a')\oplus ((b,b') \oplus (c,c'))=(a,a')\oplus (b+c,b'+c')=(a+b+c,a'+b'+c')=(a+b,a'+b')\oplus (c,c') = ((a,b)\oplus (a',b'))\oplus (c,c')$.

*Commutativity of addition: Try this one yourself. 

*Additive identity: Show that $(0_R,0_R)$ is the additive identity.

*Additive inverse: Show that for an element $(a,a')$ the additive inverse is $(-a,-a)$, where $-a$ is the additive inverse of $a$ in $R$. 

*Closure of multiplication: Try this yourself. [Hint: similar to 1. but using $\otimes$ instead of $\oplus$] 

*Associativity of multiplication: Try this yourself. [Hint: similar to 2. but using $\otimes$ instead of $\oplus$]

*Distributivity: Work out $(a,a')\otimes ((b,b')\oplus (c,c'))$. Then work out $((a,a')\otimes (b,b')) \oplus ((a,a')\otimes (c,c'))$. They are the same so distributivity holds from one side. Now you need to show similarly that $((a,a')\oplus (b,b'))\otimes (c,c') =((a,a')\otimes (c,c'))\oplus ((a,a')\otimes (b,b'))$. 


I hope that helps you. 
