# Prove that a ring with 48 elements is not an integral domain

I am trying to show that a ring with 48 elements is not an integral domain.

Let $$R$$ be a ring with 48 elements. I know I need to show that $$ab = 0$$ for some nonzero elements $$a , b \in R$$ in order to conclude that $$R$$ cannot be an integral domain. But I'm not seeing how to use the fact that the ring has 48 elements to make progress towards this.

Am I supposed to identity $$R$$ with some other ring 48-element ring that I can actually make algebraic calculations with? That would be a big help. Otherwise, I don't know what the elements of $$R$$ are, and so I can't begin to try to find the appropriate elements $$a, b \in R$$.

I don't know of any results that can help me identity a 48-element ring $$R$$ with another ring. I only know of such results with fields (the classification of finite fields, for example.)

Thanks!

## 3 Answers

A finite integral domain must be a field. Take $$x \in R$$ with $$x \neq 0$$. Then, because its an integral domain, you have cancellation. Thus, the maps $$r \to x r$$ is injective. But, since this is a finite set it must also be surjective. So, there is an $$r \in R$$ such that $$xr =1$$.

Thus, $$R$$ is a field. But, then, you know that it must have prime power order. Since $$48$$ is not a prime power, it can't be a field - and so can't be an integral domain.

Let $$(R,+,*)$$ be the ring of $$48$$ elements. Consider the group $$(R,+)$$ of order $$3\times 2^4$$. By Sylow's theorems $$(R,+)$$ has an element $$a$$ of order $$3$$ and an element $$b$$ of order $$2$$. Now, look at $$3\cdot (a*b)=(3\cdot a)*b=0*b=0$$$$\text{and}$$$$2\cdot (a*b)=a*(2\cdot b)=a*0=0$$$$\implies a*b=3\cdot (a*b)-2\cdot (a*b)=0.$$ So we have two non-zero elements $$a,b$$ such that, $$a*b=0$$. That's $$(R,+,*)$$ can't be integral domain.

It is easier to prove by contradiction. Multiplication by a fixed element in an integral domain is always injective.