Identity and zero in a Ring Given a ring R with with addition and multiplication on $\mathbb Z$ defined by $a\oplus b = a+b-4$ and $a\otimes b = ab-4a-4b+20$, what is the zero and what is the identity?
My thoughts: are they just o and 1? or do I have to use these special rules to find them? I am sometimes confused how to find these by the rules given.
 A: If the ring was the integers with normal addition and multiplication, then the zero and identity would be 0 and 1.
But your ring has different definitions for addition and multiplication, so the zero and identity are likely to be different.
The 'zero' is the value $z$ such that $z \oplus b = b$ and $a \oplus z = a$ and $z \otimes b = z$ and $a \otimes z = z$
The 'identity' is the value $i$ such that $i \otimes b  = b$ and $a \otimes i = a$
Very simple to work out the zero.
Slightly harder to get the identity, but not rocket science.
A: I'm going to write $0_R$ and $1_R$ to denote the additive and multiplicative identities for $R$; this notational convention should help stop you from confusing $0_R$ with the number $0$ and $1_R$ with the number $1$. It may be the case that $0_R=0$ or $1_R=1$, but these equations do not hold in general and probably don't hold for your problem.
To figure out what $0_R$ is, recall that $a\oplus 0_R=a$ for any $a\in\mathbb Z$; that equation defines $0_R$. Now expand out $\oplus$ using its definition and solve for $0_R$.
Similarly, to figure out what $1_R$ is, recall that $a\otimes 1_R=a$ for any $a\in\mathbb Z$. Now expand out $\otimes$ using its definition and solve for $1_R$.
A: Perhaps I should mention this spoiler. 
This structure is obtained via transport of structure from the usual operations on integers via the function
$$
f : \mathbb{Z} \to \mathbb{Z}
\qquad
x \mapsto x - 4
$$
That is
$$
f(x \oplus y) = f(x + y - 4) = x + y - 8 = f(x) + f(y)
$$
and
$$
f(x \otimes y) = f(xy-4x-4y+20) = xy -4x -4y + 20 - 4 = (x-4)(y-4) = f(x) f(y). 
$$
