Problem Statement:
Show that no ring of order $6$ is an integral domain.

Some Definitions:
Integral Domain: a commutative ring with identity and no zero divisors.
$\mathbb{Z}_n$: group of the elements ${0, 1, 2,...,n−1}$ under addition modulo n.

My Attempt:
Let $(R, +, \cdot)$ be a ring of order $6$.

Clearly, $(R, +)$ is $\mathbb{Z}_6$ (upto isomorphism), since $R$ must be an abelian group to be a ring.
I recognize that if I can show $1\in\mathbb{Z}_6$ is the identity of $R$, then I am done [$\because (1+1)(1+1+1)=6=0$]. However, if this statement is true, I am unable to figure out a way to show this.

How do I proceed?

I am aware that this can be proved in better ways [like by using the fact that every finite integral domain is a field and that every finite field has a prime power order], however, I am interested to know if this can be proved the way I am going.

  • 3
    $\begingroup$ You can do $(1+1)(1+1+1)$ no matter what $R$ is (with identity). $\endgroup$
    – Randall
    Nov 4, 2021 at 15:58
  • $\begingroup$ You know the additive structure is $\mathbb{Z}_6$, but you do not know that the multiplicative structure is such that $1\cdot 1=1$, so you can't do what you are doing in brackets. For a very simple example when it isn't, take any nontrivial additive automorphism $\phi$ of $\mathbb{Z}_6$, and use transport of structure to define a new multiplication where $\phi(1)$ is the multiplicative identity. $\endgroup$ Nov 4, 2021 at 16:01
  • $\begingroup$ @Randall: You don't know if the additive generator $1$ is the multiplicative identity, though. Consider the additive automorphism $\varphi\colon\mathbb{Z}_6\to\mathbb{Z}_6$ given by $1\mapsto 5$. Then define $a\bullet b = \phi^{-1}(\phi(a)\phi(b))$. Under this multiplication, $1\bullet 1 =5$, so $1$ is not the multiplicative identity. $\endgroup$ Nov 4, 2021 at 16:03
  • 1
    $\begingroup$ @ArturoMagidin That's not the point of Randall's comment. Even if you don't know that $1$ is the multiplicative identity, you still can do $(1+1)(1+1+1)$ to get $1\cdot 1 +1\cdot 1 +1\cdot 1 +1\cdot 1 +1\cdot 1 +1\cdot 1 = 0$ because every element has additive order $6$ $\endgroup$
    – jjagmath
    Nov 4, 2021 at 16:06
  • 3
    $\begingroup$ $(1+1)(1+1+1)$ is six times $1\cdot1$ and therefore $0$, no matter what $1\cdot1$ is. $\endgroup$
    – Magma
    Nov 4, 2021 at 16:11

2 Answers 2


You are aware that this collection has the additive structure of $\mathbb{Z}_6$, but you are wondering if a multiplicative identity exists. But you don't need that. You can do it just using the additive generator of $\mathbb{Z}_6$. I think in the comments some users are writing $1$ for the additive generator of $\mathbb{Z}_6$, and that's fine, but it also comes with the connotation of being the multiplicative identity which is what you are wondering about. So use something else like $a$ instead.

As an additive group, you have $\{0,a,2a,3a,4a,5a\}$. Now what is $a\cdot a$? It is in there somewhere: $a\cdot a=ka$.

So what is $(2a)(3a)$? That's $(a+a)(a+a+a)=\cdots=6a^2=6(ka)=(6k)a=0$.


The possible additive orders of elements in the underlying abelian group are $1, 2,3, 6$.

Cauchy's theorem says there must be elements of additive order $2$ and $3$. The sum of two such elements must have additive order $6$. Call such an element $c$.

Then $2c$ and $3c$ are nonzero and $(2c)(3c)=(6c)c=0$, showing the ring is not a domain.


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