# Not a Zero Divisor

Let $R$ be a commutative ring. Then we say $a \in R$ is a zero divisor if there exists $b \neq 0$ such that $ab = 0$.

I want to know what it means to not be a zero divisor. So I tried to negate the statement: $a$ is not a zero divisor if for every $b \neq 0$ we have $ab \neq 0$.

Also taking the contrapositive of the initial statement I got the following: If for every $b \neq 0$, $ab \neq 0$, then $a$ is not a zero divisor.

Have I negated the definition of a zero divisor and taken the contrapositive correctly?

My book has the following theorem: Suppose $a$ is not a zero-divisor. Then if $ab = ac$, we can conclude that $b = c$.

Proof: $ab - ac = a(b-c) = 0$. Since $a$ is not a zero-divisor, $b-c = 0$ so $b=c$.

I don't see why $b-c = 0$ because $a$ is not a zero-divisor. Could someone explain?

• Sometimes $0$ is considered a zero divisor, and sometimes it is not. Commented Mar 2, 2013 at 19:54
• When is $0$ not considered a zero-divisor? I suppose sometimes you'll want to reference a non-trivial zero divisor, but it seems like always want $0$ to be considered a zero-divisor. @IsaacSolomon Commented Mar 2, 2013 at 19:57
• Essentially, a number is "not a zero divisor" if you can always cancel it from an equation in the ring. Commented Mar 2, 2013 at 20:00
• Thomas is spot on: zero definitely is a zero divisor according to any reasonable definition of the term. Moreover, many well-known theorems would be false if one took the absurd position that zero is not a zero-divisor. For example the result that in a noetherian ring the zero divisors consist of the union of the minimal primes. Need I point out the dire consequences for the heretics believing that Bourbaki, Atiyah-Macdonald, Matsumura,... are wrong on this ? Commented Mar 2, 2013 at 20:05
• To answer Student's question, the element $a\in R$ is not a zero-divisor iff the multiplication map $R\to R:x\mapsto ax$ is injective. Commented Mar 2, 2013 at 20:08

Yes a zero divisor is an element $$a\neq 0$$ such that you can find some $$b\neq 0$$ with $$ab= 0$$. The existence of zero divisors in a ring just means that the product of two non-zero elements can be zero.

So indeed, as you write, $$a\neq 0$$ is not a zero divisor if one of the following equivalent statements are satisfied:

• There does not exist a $$b\neq 0$$ such that $$ab = 0$$.
• $$ab = 0$$ implies that $$b = 0$$.
• $$b\neq 0$$ implies $$ab \neq 0$$.

So indeed is given $$a\neq 0$$ satisfies that all $$b\neq 0$$ you have that $$ab\neq0$$ then $$a$$ is not a zero divisor.

• @Student: Glad to help Commented Mar 2, 2013 at 19:59
• You mean a non-zero zero divisor.
– john
Commented Mar 4, 2020 at 17:37

Yes, you have determined the correct formulation for what it means to be a non-zero-divisor.

If $a$ is not a zero-divisor, then for every $r\neq 0$, we have that $ar\neq 0$. But $ar=0$ when $r=b-c$. What does that tell you?

• Since $a\neq 0$, in order for $a(b-c) = 0$, it must be the case that $b-c = 0$ right? Commented Mar 2, 2013 at 19:52
• Because $a$ is not a zero divisor, not because $a\neq 0$. $0$ is the prototypical zero-divisor. @Student Commented Mar 2, 2013 at 19:53
• @ThomasAndrews: I see now. Thanks! Commented Mar 2, 2013 at 19:55

$b-c=0$ because any number - that number gives $0$. Else you can't get $0$ if $b>c$ or $c>b$.

• While it is true that $x-x=0$, this is not what was asked here. Also, note that we are talking about general rings and not numbers: The notion $b>c$ might not even make sense in the given ring. For example, in $\mathbb Z/2\mathbb Z$, there is no ordering compatible with addition. Commented Mar 2, 2013 at 19:53
• Oh.. I didn't understood his question then, as for the >, I made a mistake, I should have said if a is different than b. Commented Mar 2, 2013 at 23:59