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Definition: Zero-Divisors.

A nonzero element $a$ in a commutative ring $R$ is called a zero divisor if there is a non zero element $b\in R$ such that $ab=0$.

Consider the set $\mathbb Z$ with the operations $\oplus$ and $\otimes$ defined for $a, b \in \mathbb Z$ by $a\oplus b=a+b−1$ and $a \otimes b = ab − (a + b) + 2$.

My question

Are there any zero divisors in this ring?

Do we need to check zero divisor in case of non-commutative ring.

The solution I have for zero divisor part is

Solution: No, there are no zero divisors in this ring. It is crucial to remember that the “zero element” in this ring is the additive identity, namely, 1. So, we need to check whether there are elements a = 1 and b = 1 such that a ⊗ b = 1, i.e. ab − (a + b) + 2 = 1 (a − 1)(b − 1) = 0 The only way for this equation to hold is that either a = 1 or b = 1. Thus there can be no zero divisors in this ring.

Why we are checking a ⊗ b = 1 ,we should check a ⊗ b = 0 where $a,b \neq 0$

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why group theory when you are actually talking about rings? – Mohan Apr 11 '13 at 7:58
This is ring theory, not group theory. Note that this Z is isomorphic to the usual integers (with the usual addition and multiplication operations), via the map $x\mapsto x-1$. Or you can check for zero divisors directly. Noncommutative rings may have zero divisors, whether you need to check for them depends on whether or not what you desire to accomplish requires knowing if there are zero divisors. – anon Apr 11 '13 at 8:00
Note that $a,b\in R$ are zero divisors iff $ab=0_R$. Now the additive identity of $R$ is $0_R=1$, so you need to check that $a\otimes b=1$ for any $a,b\in R$. This was even explained right in the solution; what were your thoughts when you were reading the explanation about how $1$ is the additive identity and that $a,b$ are zero divisors iff their product is the additive identity, i.e. $1$? – anon Apr 11 '13 at 8:05
Improve your title please – Alexander Gruber Apr 11 '13 at 8:13

As suggested by anon in his comment to OP, note that the map $\varphi: \Bbb{Z} \to \Bbb{Z}$ given by $\varphi(x) = x - 1$ is an isomorphism between the two rings $$ \varphi: (\Bbb{Z}, \oplus, \otimes) \to (\Bbb{Z}, +, \cdot). $$

In fact $$\varphi(x) + \varphi( y) = x - 1 + y - 1 = \varphi(x \oplus y),$$ and $$\varphi(x) \cdot \varphi( y) = (x - 1) \cdot (y - 1) = xy - (x+y) + 1 = \varphi(x \otimes y).$$

Everything becomes easy then, because it's the usual ring structure on the integers we are talking about, only in disguise.

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I didn't understand how Isomorphism and zero divisor are related. – TLE Apr 11 '13 at 8:13
@TLE, once you know there is an isomorphism, look at $\Bbb{Z}$ with the usual operations. There are no (non-trivial) zero divisors here, of course. So there are no zero divisors in the isomorphic ring. Simply because a zero divisor in one ring maps under isomorphism onto a zero divisor in the other ring. – Andreas Caranti Apr 11 '13 at 8:17
@TLE: If two rings are isomorphic, then (roughly speaking), anything that is true about the first is also true about the second. In particular, the usual integers don't have any zero divisors, so neither does this ring. – anon Apr 11 '13 at 8:18

$$ab - (a + b) + 2 = (a - 1)(b - 1) + 1$$ so for $a \otimes b = 0$ you would need $(a - 1)(b - 1) = -1$. The only ways to do that with integers is $1\cdot(-1)$ or $(-1)\cdot 1$. What does that say about $a$ and $b$?

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You're checking the wrong condition. $0$ is not the additive identity of $Z$ under the addition operation $\oplus$, so we should not be checking $a\otimes b=0$. As $1$ is the additive identity, we should be checking $a\otimes b=1$. This is the entire substance of the question. – anon Apr 11 '13 at 8:11

$ab-(a+b)+2=1\Leftrightarrow (a-1)(b-1)=0\Leftrightarrow a=1$ or $b=1$, which are the additive identity.

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Certainly that's possible. Take $b=2$ and $a=0$, or $b=0$ and $a=2$. But $ab-(a+b)+2=0$ is the wrong condition to check, since $0$ is not the additive identity of $Z$ in this situation, and checking that $a,b$ are zero divisors amounts to checking if $a\otimes b$ is equal to the additive identity. – anon Apr 11 '13 at 8:08
@anon, you are right, so now it becomes easier. – Easy Apr 11 '13 at 8:15

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