# 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.

• Is $b=0\oplus b=b-4$? Is $b=1\otimes b=b-4-4b+20$? Commented May 7, 2014 at 2:43
• the way you put it there I dont bieleve so Commented May 7, 2014 at 2:47

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.

• thanks this helped a lot. zero was 4 and identity was 5. Commented May 7, 2014 at 5:22

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 $1_R$.

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 $$f(x \otimes y) = f(xy-4x-4y+20) = xy -4x -4y + 20 - 4 = (x-4)(y-4) = f(x) f(y).$$