6
$\begingroup$

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?

$\endgroup$
9
  • 2
    $\begingroup$ Sometimes $0$ is considered a zero divisor, and sometimes it is not. $\endgroup$ Commented Mar 2, 2013 at 19:54
  • 2
    $\begingroup$ 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 $\endgroup$ Commented Mar 2, 2013 at 19:57
  • $\begingroup$ Essentially, a number is "not a zero divisor" if you can always cancel it from an equation in the ring. $\endgroup$ Commented Mar 2, 2013 at 20:00
  • $\begingroup$ 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 ? $\endgroup$ Commented Mar 2, 2013 at 20:05
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Mar 2, 2013 at 20:08

3 Answers 3

4
$\begingroup$

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.

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

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?

$\endgroup$
3
  • $\begingroup$ Since $a\neq 0$, in order for $a(b-c) = 0$, it must be the case that $b-c = 0$ right? $\endgroup$
    – Student
    Commented Mar 2, 2013 at 19:52
  • 1
    $\begingroup$ Because $a$ is not a zero divisor, not because $a\neq 0$. $0$ is the prototypical zero-divisor. @Student $\endgroup$ Commented Mar 2, 2013 at 19:53
  • $\begingroup$ @ThomasAndrews: I see now. Thanks! $\endgroup$
    – Student
    Commented Mar 2, 2013 at 19:55
-3
$\begingroup$

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

$\endgroup$
2
  • 3
    $\begingroup$ 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. $\endgroup$ Commented Mar 2, 2013 at 19:53
  • $\begingroup$ 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. $\endgroup$ Commented Mar 2, 2013 at 23:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .