TITLE: A subset of a field that is a subfield
QUESTION [6 upvotes]: It can be verified that the following assertion is true: a subset $S$ of a field $F$ is a subfield if $S$ contains the additive and multiplicative identities 0 and 1, if $S$ is closed under addition, multiplication, additive inverses, and $S-\{0\}$ is closed under multiplicative inverses.  An exercise asks to show that the condition $0,1 \in S$ can be replaced by the condition that ''$S$ contains at least two elements''.  The hint given is ''Consider $ax=a$.'' 
Suppose $S$ contains at least two distinct elements, say $a,b$.  By the hypotheses, $-a \in S$, so $a+(-a) =0 \in S$.  At least one of $a,b$ is nonzero, say $a \ne 0$. Then its inverse $a^{-1} \in S$ and so the product $a a^{-1}=1 \in S$.  Thus, $0,1 \in S$.  This solves the exercise.  My question is whether there's another solution that uses the hint of considering $ax=a$.

REPLY [2 votes]: One possible use of the hint is to show that $1 \in S$ using the following:
Assume $S$ has at least two elements. Pick one which isn't zero and call it $a$. Consider the mapping from $S\rightarrow S$ which sends $x$ to $ax$. Since we work in a field $F$, the function is a bijection since it has an inverse. (namely the function that sends $x$ to $a^{-1}x$.) Thus the equation $ax=a$ has a solution in $S$. This is also a solution in $F$, so $1=x \in S$.
I think that this is probably what the author had in mind, especially since the hint was changed to provide a different (probably cleaner) way to show that $1 \in S$. I guess they might have had this approach in mind since considering the same function is a good way to show that any finite integral domain is a field.