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Why is a finite integral domain always field?

How do you prove this;

Let R be a finite commutative domain. Prove that R is a field.

I need to know this because doing a course in non commutative rings. Yet, my algebra has gone from summer holidays. Also, another list of exercise I can't do.

I know it's simple.

I have a, b, 1, 0 in R. ab=ba. Might need to use it finite. So for groups if you have finite then you apply 1 element over and over again you get back to it. Just need to find the order of the group. Yet, I don't see how I can use it here without proving a lot of other stuff. $a^{n-1}a=1$ Where n is the order of the ring???

Community
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algebears
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3 Answers3

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Pick a nonzero $a\in R$ and look at the number of elements in the set $\{ax\mid x\in R\}$. Is it possible to have less elements than in the whole of $R$? That should give you a clue.

Every definition of a domain that I've seen does assume that it contains a unit element $1$. In this case it doesn't really matter as a field contains a unit, so if the domain didn't contain a unit, then it would necessarily not be a field. If such a definition was used, you would be trying to prove a false statement.

Edvard F
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One nice way to prove this theorem comes from examining left multiplication operators.

Define $L_x:R \rightarrow R$ by $L_x(y)=xy$. $L_x(y+z)=L_x(y)+L_x(z)$ by the distributive law so $L_x$ is a (group) homomorphism. Now since you have no zero divisors if $x \not= 0$, then $L_x(y)=0$ implies $y=0$. Thus $L_x$ is injective. Now since $R$ is finite, $L_x$ must also be surjective (a map from a finite set to itself is one-to-one iff onto). Now that we know $L_x$ is surjective there must be some $y \in R$ such that $L_x(y)=1$. Thus $xy=yx=1$ so $x$ is a unit and so $R$ is a field.

A careful modification of this argument will show that in general any finite ring with $1$ (commutative or not) contains only $0$, zero divisors, and units. So finite rings with 1 which have no zero divisors are fields (using Wedderburn's theorem which says finite skew fields are fields).

Bill Cook
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What's wrong with the usual (I think) proof? Hint: Show that the only ideals are 0 and $R$.

Spoiler: Let $I$ be any ideal of $R$, and suppose $I \neq 0$. Choose $a \in I$, $a \neq 0$. The map given by left multiplication by $a$ is injective ( because $R$ is a domain) hence it must be surjective (since $R$ is finite). On the other hand, the image of this map is contained in $I$. Therefore $I = R$. This characterizes fields (why?), so you're done

anon
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