Prove that a ring of order $6$ can never be an integral domain.

Question

My solution:

Let $R$ be a ring of order $6$ which is an integral domain. This means, that $1+1\neq 0\in R$ and we note that, $(1+1)(1+1+1)=0,$ a contradiction as $1+1,1+1+1\neq 0.$ So, $R$ is not an integral domain.

However, I feel that my solution is not justified because, we can not claim that $1+1,1+1+1\neq 0$ if $1\neq 0.$ For if, $1+1=0$ it means $1=-1$ which seems to be nothing absurd apparently. Is there any way to solve the problem?

You need to explain why $1+1=0$ or $1+1+1=0$ lead to problems as well. — Arturo Magidin, Oct 31 '23 at 15:49
@ArturoMagidin Yes, that's what I have been trying to do right now, but till now, I seem to find no problems. — Thomas Finley, Oct 31 '23 at 15:50
If by $1$ you mean the generator of the additive group of $R$, then yes, the solution is correct. We have $1+1\neq 0$ and $1+1+1\neq 0$, but $(1+1)(1+1+1) = 0$ from distributivity (this shows that $R$, even if not unital, has to have zero divisors) — Jakobian, Oct 31 '23 at 15:51
@Jakobian Not necessarily a generator for $R.$ By $1$ I meant the unit element of $R$ which is not equal to the zero element of $R$. Precisely, $1\neq 0\in R$ is such that, $r.1=r=1.r,\forall r\in R.$ — Thomas Finley, Oct 31 '23 at 15:55
Well, the replace it by some element $a$ which generates the additive group of $R$ and the argument will be fine. — Jakobian, Oct 31 '23 at 15:56
@Jakobian Ok, but then why will $a+a,a+a+a\neq 0$ if $R=$ ? What problem will we be getting if $a+a=0\iff a=-a$ or if $a+a+a=0$ ? — Thomas Finley, Oct 31 '23 at 15:58
@ThomasFinley Since $a$ generates $R$, $0, ..., 5a$ need to be the only elements of $R$, all of them distinct. — Jakobian, Oct 31 '23 at 16:04
@Jakobian why should the additive group have a single generator? — spaceisdarkgreen, Oct 31 '23 at 16:27
@spaceisdarkgreen because its a commutative group of $6$ elements, hence cyclic — Jakobian, Oct 31 '23 at 16:36
@Jakobian: You are assuming the additive group is cyclic. You can prove it is cyclic (otherwise there are zero divisors...) but you have to actually prove it is cyclic. — Arturo Magidin, Oct 31 '23 at 17:29
@ArturoMagidin I don't see how its adding to what I already said. — Jakobian, Oct 31 '23 at 17:34
@spaceisdarkgreen Any group of order 2p (p>2 is prime) is isomorphic either to the Dihedral group $D_{2p}$ of order 2p or to a cyclic group of order $2p$. $D_6$ is not Abelian so you get isomorphism with the cyclic group of order 6. — Koro, Nov 01 '23 at 05:29
@koro or to leave non-abelian groups out of it, there’s only one abelian group of any square-free order, by classification of finite abelian groups / Chinese remainder theorem / what-have-you — spaceisdarkgreen, Nov 01 '23 at 16:14

score 2 · Accepted Answer · answered Oct 31 '23 at 16:48

Take any ring $R$ consisting of $6$ elements. $R$ has an "additive' group structure. By Cauchy's theorem, $R$ has an element of order $2$, called $a$ and an element of order $3$ called $b$, i.e., $2a=0, 3b=0, 2b\ne 0,a\ne 0$.

$a (2b)=a(b+b)= ab+ab=(a+a)b=(2a)b=0$ which is not possible if $R$ were an integral domain.

score 1 · Answer 2 · answered Oct 31 '23 at 15:49

The easiest way to solve this is to appeal to the cardinality of $R$. Note that $R$ must be a field, since for every nonzero $x\in R$, the map $f\colon R\to R$ given by $f(y)=xy$ is injective (because $R$ is a domain), so it is bijective because $R$ is finite, and we can find an element $y\in R$ such that $xy=1$. So $R$ is a field, and the number of elements in a finite field is of the form $p^{k}$, where $p$ is a prime number. Thus, $R$ cannot be a field.

Hope this helps!

Prove that a ring of order $6$ can never be an integral domain.

2 Answers2