# Prove that R is a ring under 'special' definitions of multiplication and addition

## Question:

Let R be a ring with a 1. Define $\bar R$ to have the same elements of R with addition $$\oplus: a \oplus b = a +b +1$$ andmultiplication $$\otimes: a \otimes b = ab + a +b$$ Prove that $\bar R$ is a ring under $\oplus$ and $\otimes$ and has a 1.

## My Attempt:

My understanding of a ring is it should be a non-empty set together with operations addition and multiplication satisfying the following conditions:

1. R is an Abelian group under multiplication
2. R must be closed and associative under multiplication
3. Multiplication must be distributive over addition i.e. $a(b+c) = ab + ac$ and $(a+b)c= ac +bc$

So I try to show that all those conditions are satisfied:

$$1$$

$$a \otimes b = ab + a + b$$ $$b \otimes a = ba + b +a$$ In order for $a \otimes b$ to be equal to $b \otimes a$ it must be that $ba = ab$ (how do I prove this? - do I even need to prove this?)

$$2$$

Closed: $$a \otimes b = ab + a + b$$ if $a, b \in \bar R$ then $ab, \in \bar R$, $(a + b) \in \bar R$ and therefore $ab + a + b$ must be in $\bar R$. (Do I need to prove this? Or is that pattern of thought good enough?)

Associative: let's introduce an element $c \in \bar R$ and examine the associativity property.

\begin{align} (a \otimes b) \otimes c & = \ (ab + a + b) \otimes c\\ & = \ abc + ac + bc + ab + a + b + c \\ & = \ a \otimes (bc + b + c) \\ & = \ a \otimes (b \otimes c) \\ & = \ a \otimes b \otimes c \\ \end{align}

So it is indeed associative under multiplication.

$$3$$

\begin{align} a \otimes (b \oplus c) & = \ a \otimes (b + c + 1) \\ & = \ ab + ac + a + a + b + c + 1 \\ & = \ ab + b + a + ac + a + c + 1 \\ & = \ a\otimes b \oplus \ a \otimes c \\ \end{align}

And I'm sure if I tinker around with the $(a + b)c = ac + bc$ equation I will find that it is also true so it does satisfy the distributive law.

The second part of the question demands that I show that $\bar R$ has a 1. I understand a 1 to have the property: $a\times1= a$.

So considering it in this ring we're looking for a $x$ that does this:

$$a \otimes x = a$$ $$a \otimes x = ax + x + a = a$$ Just from looking at the equation and using my intuition one can see that when $x=0$ the equation is true. So I would say that x=0 is the 1 of this ring. 0 is the multiplicative identity of $\bar R$

## My Concerns:

Can I conclude after having done all this that $\bar R$ is in fact a Ring? Have I answered the question sufficiently enough?

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The keyword to look up here is "transport of structure." –  Qiaochu Yuan Nov 18 '12 at 11:24

It seems you are mistaking the axioms in your definition : the first axiom should state that $\bar{R}$ is an abelian group under addition, not multiplication. So you need to prove that addition is closed, associative and commutative (this is trivial), find an identity for addition, and prove that every element has an inverse.

Then, with your proof for axioms 2 and 3, you'll have proved that $\bar{R}$ is indeed a ring.

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Thank you! Was using my notes from class. I must have made a mistake while typing it out. –  Siyanda Nov 18 '12 at 10:50
In general, let $p:R→R$ be a bijection, then define $A(m,n)=p^{−1}(p(m)+p(n))$ and define $M(m,n)=p^{−1}(p(m)p(n))$. The structure (R,A,M) is a ring. Moreover the bijection $p:R→R$ is a ring isomorphism from (R,A,M) to (R,+,x).
In fact, by Letting p(n)=n+1 we find that $M(m,n)=p^{−1}(p(m)p(n))=mn+m+n$ and $A(m,n)=m+n+1$. Thus, $1_{(R,A,M)}=p^{-1}(1_{(R,+,x)})=1-1=0$ .
Check: $M(a,0)=(a)(0)+a+0=a$