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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!

testguy807
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3 Answers3

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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.

Rushabh Mehta
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msm
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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.

Sumanta
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  • Finite integral domains are fields and finite fields are of the form $\Bbb Z_p,$ for $p$ a prime. So all the finite integral domains are prime ordered. Since $48$ is not a prime it follows that no ring of order $48$ can be an integral domain. – Anacardium Nov 09 '20 at 16:53
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It is easier to prove by contradiction. Multiplication by a fixed element in an integral domain is always injective.

lestudent
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