13

The problem is to prove that, in a commutative ring with identity, the set of ideals in which every element is a zero divisor has a maximal element with respect the order of inclusion, and that every maximal element is prime. But I´m thinking* that considering the set of all zero-divisors, this set as I see it´s an ideal, and thus must be the only maximal element.

An ideal is defined as a subset J of the ring R , such that for every $ x\in R$ we have $ xJ \subset J $

I think that I´m wrong, only because it´s rare.

An extra question but related, there exist rings with element x, such that x is not a zero divisor nor a unit?

Zach Boyd
  • 908
Susuk
  • 445
  • Careful: an (two sided) ideal is usually defined as a subgroup of the additive group of the ring $R$, such that for any $x\in R$ and $xJ,Jx\subseteq J$. A left ideal is one for which $xJ\subseteq J$, and a right ideal is one for which $Jx\subseteq J$. – Pedro Dec 19 '13 at 22:56
  • Related: https://math.stackexchange.com/questions/44481/showing-the-set-of-zero-divisors-is-a-union-of-prime-ideals – Smiley1000 Aug 15 '23 at 21:59

2 Answers2

15

Your definition of an ideal seems nonstandard: for most folks, an ideal must also be a subgroup of the additive group of the ring. So, for counterexample to your conjecture, in the ring $\mathbf Z/6\mathbf Z$ there are zero-divisors $2$ and $3$, the sum of which is a unit.

To the second question: yes! As Alex notes in the comments, $2 \in \mathbf Z$ is an example. Any integral domain which is not a field also provides examples. But it's a good pigeonhole exercise to show that you cannot find such an element in a finite ring.

Dylan Moreland
  • 19,819
  • His definition is fine if he meant subgroup instead of subset, which seems likely, and you didn't answer his second question--he asked for a non-unit nonzero divisor, like $2\in\mathbb{Z}$. – Alex Youcis Jan 21 '12 at 05:49
  • @Alex Of course! I think that's what I've said. I didn't notice the extra question, but I'll address it now. – Dylan Moreland Jan 21 '12 at 05:51
  • @Alex Also, my thought was that if you only use the definition as written, then it seems possible to "prove" that the zerodivisors form an ideal. – Dylan Moreland Jan 21 '12 at 15:39
10

It is not an ideal, as the next example show. Consider the ring $\mathbb R \times \mathbb R$. The elements $(1,0)$ and $(0,1)$ are zero-divisors but their sum is not. Thus, the set of zero-divisors is not closed under sums.

For your second question, consider the polynomial ring in one variable $\mathbb R[x]$, clearly $x$ is neither a zero-divisor or a unit.

azarel
  • 13,120