# Why is a finite integral domain always field?

This is how I'm approaching it: let $R$ be a finite integral domain and I'm trying to show every element in $R$ has an inverse:

• let $R-\{0\}=\{x_1,x_2,\ldots,x_k\}$,
• then as $R$ is closed under multiplication $\prod_{n=1}^k\ x_i=x_j$,
• therefore by canceling $x_j$ we get $x_1x_2\cdots x_{j-1}x_{j+1}\cdots x_k=1$,
• by commuting any of these elements to the front we find an inverse for first term, e.g. for $x_m$ we have $x_m(x_1\cdots x_{m-1}x_{m+1}\cdots x_{j-1}x_{j+1}\cdots x_k)=1$, where $(x_m)^{-1}=x_1\cdots x_{m-1}x_{m+1}\cdots x_{j-1}x_{j+1}\cdots x_k$.

As far as I can see this is correct, so we have found inverses for all $x_i\in R$ apart from $x_j$ if I am right so far. How would we find $(x_{j})^{-1}$?

• The multiplication by a nonzero element, being injectve, is surjective. Sep 7, 2011 at 13:39
• So for $0 \neq a \in R$ the map $T:x\to ax$ is injective as $ker(T)=\{0\}$ (no zero divisors) and as R is finite it must be surjective. Therefore T is invertible, and every a must have an inverse? Sep 7, 2011 at 13:49
• – lhf
Sep 7, 2011 at 16:00
• It follows from math.stackexchange.com/questions/1428221 Jan 19, 2017 at 13:04

Remember that cancellation holds in domains. That is, if $c \neq 0$, then $ac = bc$ implies $a=b$. So, given $x$, consider $x, x^2, x^3,......$. Out of finiteness there would be a repetition sometime: $x^n = x^m$ for some $n >m$. Then, by cancellation, $x^{n-m} =1$, and $x$ has an inverse.

• Unless, of course, $x$ is zero. Sep 7, 2011 at 13:41
• Indeed, as there are no zero divisors. So could we equally say for some $n$, $x=x^n$, or do we need to specify the m? Sep 7, 2011 at 13:43
• @Myerson: Well, of course. :) LHS: There is no need to specify an $m$. It is enough that some such $n$ and $m$ exist. Sep 7, 2011 at 13:43
• @George: sorry I meant do we know that for some product of $x$, say $x^n$, $x^n$ will equal $x$ Sep 7, 2011 at 13:51
• If you need to identify an $N$ for which you get a repetition for some $n<N$, the number of elements of the domain (being finite) will suffice (as it is the number of non-zero elements plus 1, so in any collection there must be two the same). Sep 7, 2011 at 14:41

While the common simple pigeonhole-based proof has already been given, it is worth emphasis that your proof is completable. Put $$\rm\:u = x_j\ne 0.\:$$ Either $$\rm\:u^2 = u\:\ (so\:\ u = 1)\:$$ or $$\rm\: u^2 = x_{\:k}\mid 1,\,$$ thus $$\rm\:u\mid 1.\:$$ Therefore all nonzero elements of $$\rm\:R\:$$ are units  (note $$\rm\ u^2 \ne 0\:$$ by $$\rm\:u\ne 0)$$ $$\$$ QED

In fact we can generalize such pigeonhole-based ideas. The Theorem below is one simple way. Note that the above proof is just the special case when $$\rm\:R\:$$ is a domain and $$\rm\:|\cal N|$$ $$= 1\:.$$

Theorem $$\$$ If all but finitely many elements of a ring $$\rm\:R\:$$ are units or zero-divisors (incuding $$0$$), then all elements of $$\rm\:R\:$$ are units or zero-divisors.

Proof $$\$$ Suppose the finite set $$\,\cal N$$ of nonunit non-zero-divisors is nonempty. Let $$\rm\: r\in \cal N.\,$$ Then all positive powers $$\rm\:r^n\:$$ are also in $$\rm\,\cal N$$ since powering preserves the property of being a nonunit and non-zero-divisor (if $$\rm\ a\,r^n = 0\:$$ then, since non-zerodivisors are cancellable, we deduce $$\rm\:a = 0\:$$ by cancelling the $$\rm\:n\:$$ factors of $$\rm\:r).\,$$ So pigeonholing the powers $$\rm\:r^n\:$$ into the finite set $$\rm\,\cal N$$ yields $$\rm\:m>n\:$$ such that $$\rm\:r^m = r^n,\$$ so $$\,\rm\:r^n(r^{m-n} - 1) = 0\:.\:$$ As $$\,\rm r^n\in\cal N$$ it is not a zero-divisor so we can cancel it, which, finally, yields that $$\rm\:r^{m-n}=1,\:$$ so $$\rm\:r\:$$ is a unit, contradiction. $$\$$ QED

Corollary $$\$$ Every element of a finite ring is either a unit or a zero-divisor (including $$0$$).
Therefore a finite integral domain is a field.

Remrk  For a bit deeper example see my proof here that generalizes (to "fewunit" rings) Euclid's classic constructive proof that there are infinitely many primes. Such ideas generalize to monoids and will come to the fore when one learns algebraic local-global methods, esp. localization of rings.

• Thanks! that's satisfying to know Sep 7, 2011 at 15:17
• @LHS Kudos to you for devising your own proof. It's a nice alternative to the frequently regurgitated well-known proofs in the other answers (which, alas, do not even address your question, i.e. your method of proof). Please see my comment posted below your question. Sep 7, 2011 at 16:22
• Thanks Bill! I only post here if i'm completely stuck -and it is much more satisfying to work out your own way. See response comment. Sep 7, 2011 at 16:45
• @LHS I wish that we had many more questions like yours above. I actually learned a little something by reflecting on your method of proof - which happens only very rarely here. Keep striving to find your own proofs before consulting the "standard" proofs. That is one of the best ways to learn mathematics. Then, someday, one of your own proofs may be the standard proof. Sep 7, 2011 at 18:14
• That means a lot! Our lecturers tell us the same thing, I always try, and if I get something that looks promising I will post it along with the question :) Haha possibly, lets hope! I do need to coauthor a paper with my Erdős number 1 tutor at some point! Sep 7, 2011 at 18:34

It sufficies to prove that there exists $1\in R$ such that $a1=1a=a$ for any $a\in R$, and that every $a\neq 0$ is invertible in $R$. So let $R=\{a_1,\dots,a_n\}$ with the $a_i$'s pairwise distinct. Let $a=a_k\neq 0$. Then the elements $$aa_1,aa_2,\dots, aa_n$$ are also pairwise distinct (if $aa_i=aa_j$ with $i\neq j$ then $a(a_i-a_j)=0$ wich forces $a_i=a_j$ since we are in an integral domain and $a\neq 0$). But then the map $\Psi:R\to R$ defined by $$\Psi(a_i)=aa_i$$ is injective by what we have proved before. Since $R$ is finite it is also surjective, then it is a bijection. This means that every element of $R$ can be written as $aa_i$ for some element $a_i\in R$. In particular $a$ itself can be written in this way: there exixsts $a_{i_0}\in R$ such that $a=aa_{i_0}=a_{i_0}a$.

Now we claim that $a_{i_0}$ is the unit element of $R$: indeed let $x=aa_i$ any element in $R$. Then $$x=aa_i=(aa_{i_0})a_i=(a_{i_0}a)a_i=a_{i_0}(aa_i)=a_{i_0}x$$ and also $$x=a_ia=a_i(aa_{i_0})=(a_ia)a_{i_0}=xa_{i_0}.$$ We shall denote this element $a_{i_0}$ with $1$. Now, from the fact that $1$ is in $R$, $1$ can be written as $1=aa_j$ for some $a_j\in R$. But then $a$ is invertible in $R$.

• I may not have made myself clear enough, but the definition of an Integral Domain I am using has $a1=1a=1$ $\forall a \in R$ as an axiom Sep 7, 2011 at 13:56
• ok.. so the proof is even simpler than mine. However as far as I remember from my first algebra course, an integral domain is a commutative ring without $0$ divisors, and a commutative ring need not have the unit element. But, again, this just make your life easier so my proof holds with weaker assumptions. Sep 7, 2011 at 14:07
• You're certainly right about the commutative ring definition, from what I can see integral domains don't have to defined with 1, but are normally assumed to have it. Thanks very much, it's very helpful! Sep 7, 2011 at 14:11
• In current usage, the phrase integral domain almost universally refers to a commutative ring with nonzero 1 (and without zero divisors, of course). Sep 7, 2011 at 16:49

Here is another proof.For any $a\in R$ with $a\neq0$ consider the function $f_a:R\longrightarrow R$ defined by $f_a(x)=ax$ it is injective because $R$ is a domain, now 'cause $R$ is finite then $f_a$ is surjective because is injective, there is an element $b\in R$ which $f_a(b)=1$, then $ab=1$. Also is important to mention the Wedderburn theorem that proves that the ring is commutative.

• Regarding the last bit: I think most people here require integral domains to be commutative, but that is a good theorem. Sep 8, 2011 at 3:13
• yes, this proof is given as a corollary on p228 in Dummit & Foote 3rd edition
– qwr
Nov 22, 2018 at 1:03

Simple arguments have already been given. Let us do a technological one.

Let $A$ be a finite integral commutative domain. It is an artinian, so its radical $\mathrm{rad}(A)$ is nilpotent—in particular, the non-zero elements of $\mathrm{rad}(A)$ are themselves nilpotent: since $A$ is a domain, this means that $\mathrm{rad}(A)=0$. It follows that $A$ is semisimple, so it is a direct product of matrix rings over division rings. It is a domain, so there can only be one factor; it is is commutative, so that factor must be a ring of $1\times 1$ matrices over a commutative division ring. In all, $A$ must be a field.

• In Germany you would call this "mit Kanonen auf Spatzen schießen" :-) Oct 31, 2011 at 15:15
• Actually, using the Artinian property, this is even easier: Since $A$ is Artinian (by finiteness), it has dimension 0. On the other hand, the 0 ideal is a prime ideal (by virtue of being an integral domain), so it is the maximal ideal, and $A$ is hence local. Thus, $A^\times = A \setminus \{ 0 \}$... Mar 14, 2012 at 16:40

In fact, we can go a bit farther, and say that if $R$ is a finite commutative ring that has elements that are not zero-divisors, then $R$ has an identity. Furthermore, every nonzero element of $R$ is either a unit or a zero-divisor.

To see why, pick $a\in R\setminus\{0\}$ with $a$ not a zero-divisor. As $R$ is finite, the set $\{a,a^2,a^3,...\}$ must also be finite, whence there exist $m,n\in \mathbb{N}$ with $m<n$ and $a^m=a^n$.

We will now show that $a^{n-m}$ serves as an identity for $R$. Pick any $x\in R$. Then $a^m=a^n$ implies $a^mx=a^nx$, whence $a^m(a^{n-m}x-x)=0$. Now, since $a$ is not a zero divisor, it is clear that $a^m$ is not a zero-divisor. Thus, the only way we can have $a^m(a^{n-m}x-x)=0$ is if $a^{n-m}x-x=0$ or $a^{n-m}x=x$. Therefore $a^{n-m}=1_R$, and $R$ has an identity.

In fact, the proof of why any nonzero zero-divisor is a unit essentially follows from the same argument as above (letting $x=1$ now that we know that $R$ has an identity): if $a\in R\setminus\{0\}$ is not a zero-divisor, then there exist $0<m<n$ with $a^m=a^n\,\Rightarrow\,a^m(a^{n-m}-1)=0\,\Rightarrow\,a^{n-m}=1$ (since, again, if $a$ is not a zero-divisor, then neither can $a^m$ be a zero-divisor). Therefore, every nonzero element of $R$ is either a zero-divisor or a unit.

From here, it directly follows that every finite integral domain is a field, since integral domains have no zero-divisors.