Here's how Fraleigh proves: Every finite integral domain is a field in his book:
Let
\begin{equation*}
0, 1, a_1, \dots, a_n
\end{equation*}
be all the elements of the finite domain $D$. Now, consider
\begin{equation*}
a1, aa_1, \dots, aa_n
\end{equation*}
Since the multiplicative cancellation laws hold in $D$, it means that each of $a1, aa_1, \dots, aa_n$ are distinct from each other since $aa_i = aa_j \implies a_i = a_j$. Also, since $D$ has no divisors of $0$, neither of $a1, aa_1, \dots, aa_n$ can be zero. Hence, $a1, aa_1, \dots, aa_n$ are elements $1, a_1, \dots, a_n$ in some order. So, either $a1 = 1 \implies a = 1$ or $aa_i = 1$ for some $i$.
My addition: If $a = 1$, then the conditional in question is trivially satisfied and there is nothing to prove. So, without loss of generality, assume $aa_i = 1$.
This shows that $a$ has a multiplicative inverse, $a_i$. $\square$
I have two questions: firstly, is my addition to the proof valid? Secondly, how does $D$ has no divisors of $0$ imply "neither of $a1, aa_1, \dots, aa_n$ can be zero" (in bold above). The definition of 0 divisors that Fraleigh has given is:
If $a$ and $b$ are two nonzero elements of a ring $R$ s.t. $ab = 0$, then $a$ and $b$ are divisors of 0.
To conclude that "neither of $a1, aa_1, \dots, aa_n$ can be zero" from this definition, I think we would need to know that the product of any two terms from $a1, aa_1, \dots, aa_n$ is zero but we don't know this. What am I missing? Thanks!
Best Answer
The whole point is to show that none of the products $a1,aa_1,\ldots,aa_n$ is $0$. Suppose that some $aa_k$ were $0$. We know that $a$ and $a_k$ are not $0$; if $aa_k$ were $0$, $a$ and $a_k$ would by definition be divisors of $0$, but we know that $D$ has no divisors of $0$. Thus, $aa_k$ cannot be $0$. The same argument shows that $a1$ cannot be $0$, though in that case it’s even easier, since $a1=a$, and we know that $a\ne 0$.
Your addition is correct but not really necessary: one would hope that the reader can be trusted to recognize that if $a=1$, we already know that it has a multiplicative inverse, so we’re really interested in the other cases.