Why is a finite integral domain always field?

Question

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 $(x_{j})^{-1}$?

Answer 1

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.

Answer 2

Your proof is completable. Put $\rm\:u = x_j\ne 0.\:$ Either $\rm\:u^2 = u\:\ (so\:\ u = 1)\: $ or $\rm\: u^2 = x_{\:k}\mid 1\:$ so $\rm\:u\mid 1.\:$ Therefore all nonzero elements of $\rm\:R\:$ are units. $\:$ (note $\rm\ u^2 \ne 0\:$ by $\rm\:u\ne 0\:$). $\ $ QED

In fact one 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 $\rm\:\cal N\:$ 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.

For a less trivial 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.

Answer 3

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 $R$.

Answer 4

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.

Answer 5

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.

Answer 6

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.

Answer 7

That’s a corollary from a more general result:

Lemma. Let $R$ be a finite commutative ring. Then every element it either a unit or a zero divisor.

Proof. Let $r \in R$ and consider the mapping

$$\varphi_r: R \to R, \quad a \mapsto ra.$$

$\varphi_r$ can either be injective or not.

• If $\varphi_r$ is not injective, there are elements $a,b \in R, a \neq b$ such that $ra = rb$. That is, $ra - rb = 0$, and hence $$r (a - b) = 0.$$ Since $a \neq b$, it holds that $a - b \neq 0$, so $r$ is a zero divisor.

• Otherwise, if $\varphi_r$ is in fact injective, it’s also surjective since it’s a mapping from a finite set to itself. Let $a$ be a preimage of $1 \in R$, that is, $$1 = \varphi_r(a) = ra.$$ But that means that $r$ is a unit. $\blacksquare$

Corollary. Every finite integral domain is a field.

Proof. An integral domain has no zero divisors except $0$. Hence, by the lemma, all elements except $0$ are units, $R^\times = R \setminus \{ 0 \}$, so $R$ is a field. $\blacksquare$

Answer 8

Why to find the inverse of $xj$ Can't it be solved this way $(x1.x2...xk) - xj = 0$ Then $xj(x1..xj-1.xj+1..xk -1)=0$ As $xj$ is not a $0$ and $R$ is integral We get $x1..xj-1.xj+1..xk = 1$ Choosing an arbitrary $xm$ $Xm(x1..xm-1.xm+1..xj-1.xj+1..xk)=1$ So $xm$ has an inverse