TITLE: Prove that R is a ring under 'special' definitions of multiplication and addition
QUESTION [7 upvotes]: 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:

R is an Abelian group under multiplication
R must be closed and associative under multiplication
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?

REPLY [2 votes]: Let (R,+,x) be a ring.
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: $M(a,0)=(a)(0)+a+0=a$