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

Question

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

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?

Answer 1

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.

Answer 2

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!