In a finite integral domain, is the inverse of every element a positive power of that element? Let $D$ be a finite integral domain with unity and let $x \in D$. Is it true that then $x^{-1} = x^n$ for some $n \in N$, and thus every element has an inverse. This was my solution to proving that $D$ is also a field. I am thinking of the analogue proof in groups (See Grillet's abstract algebra corollary 2.4) and since cancellation laws hold and there cannot be a zero divisor the prerequisites seem to hold. I am not sure however, could somebody confirm?
 A: Yes, it's true: since $D$ is finite, the set of positive powers of a non-zero element $a$ is finite, hence there exist distinct positive powers $r$ and $s$ (say $s>r$ such that $a^r=a^s$. We can simplify both sides by $a^r$, since we're in a domain, hence $a^{s-r}=1$.
To prove a finite integral domain is field, there's a simpler proof: for any $a\ne 0$, mutltiplication by $a$ in $D$ is injective, and in a finite set, for a map from $D$ to $D$, injective, surjective and bijective are equivalent properties. Hence $1$ is attained, i.e. there exists $xin D$  such that $ax=1$.
The same type of argument is used to prove that, if $A\hookrightarrow B$ is an injective morphism of integral domains such that $B$ is a finite $A$-module, then $A$ is a field if and only if $B$ is a field.
A: An alternative proof: The invertible elements in any ring  form what is called the multiplicative group under multiplication. Since we are now in a finite group we have that $a^{-1}=a^k$ for some $k$ (by your previous result for groups).
