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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?

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    Yes, this is correct. – 2017-01-01
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    Something that is intimately related to this is that every finite cancelable semigroup is a group. (perhaps a generalization). – 2017-01-01
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    Ah lovely, thank you =) – 2017-01-01

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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.

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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).

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    this proof just shows that if $x$ has an inverse then it is a power of $x$, in any finite ring. The other result shows that every non-zero $x$ has an inverse inside any finite integral domain. – 2017-01-01