Show that no ring of order 6 is an integral domain.

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.

• You can do $(1+1)(1+1+1)$ no matter what $R$ is (with identity). Nov 4, 2021 at 15:58
• 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. Nov 4, 2021 at 16:01
• @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. Nov 4, 2021 at 16:03
• @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 $6$ Nov 4, 2021 at 16:06
• $(1+1)(1+1+1)$ is six times $1\cdot1$ and therefore $0$, no matter what $1\cdot1$ is. Nov 4, 2021 at 16:11

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.