TITLE: Understanding Fraleigh's proof of: Every finite integral domain is a field
QUESTION [8 upvotes]: 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!

REPLY [2 votes]: Treating $0$ and $1$ separately from the rest of the elements of $D$ makes the proof unnecessarily complicated and seems to have produced some confusion. So here's a shorter and possibly less confusing version of the proof.
Apparently Fraleigh has already established the cancellation law for integral domains: If $ab=ac$ and $a\neq0$, then $b=c$. Now, to show that $D$ is a field, I'll consider an arbitrary non-zero element $a\in D$ and show that it has a multiplicative inverse. Multiplication by $a$ is a function $f:D\to D$, namely $f(x)=ax$ for all $x\in D$. Cancellation says exactly that this $f$  is one-to-one. From elementary set theory, we know that a one-to-one function from a finite set $D$ into itself must be onto. So $1$ is in the image of $f$, i.e., there is $x\in D$ with $f(x)=1$. That means $ax=1$, and so $x$ is the desired inverse of $a$.