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

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.
