Proof that a finite domain is a 'body', a little question In my country a 'body' is defined as $F,+,\cdot$ such that

*

*$F,+$ is a commutative group

*$F\setminus\{0\}, \cdot$ is a group

*$\cdot$ is distributive to $+$
(while in my country a field is a 'body' where $F,\cdot$ also has the commutative property)
I have a question regarding the proof that a finite domain is a 'body'. (this seems to be know as Wedderburn's little theorem?)
Anyhow the textbook proof goes as follows:
Proof
Say $R$ is a finite domain, this means that it doesn't contain any zero divisors.
Since a finite domain is a ring, $R\setminus \{0\},\cdot$ has the (A) associative property. To prove it is a 'body' we need to prove the (N) (neutral) and (I) (inversive) properties in $R\setminus \{0\}, \cdot$
We choose $a\in R\setminus \{ 0\}$ en define $f_a: R\setminus \{0\} \to R\setminus\{0\}; x \mapsto ax$.
We prove that $f_a$ is bijective. First of all it's injective since if $f_a(x) = f_a(y)$ then $ax = ay$. And since $F,+$ is a group we can conclude $ax-ay=0$.
The distributive property is valid so $a(x-y)=0$. Because $a\not = 0$ and there aren't any zero divisors this means $x=y$.

The following is less clear.

*

*Why does this imply surjectivity?

and especially:

*

*Why does $R$ contain the 1 (neutral element)? Since we don't know yet that $R,\cdot$ is a group.


Since $f_a$ is an injective mapping of a finite set in itself it must be surjective.
This means that for any given $f(x) = ax \in R\setminus\{0\}$ there is an $x$ in $R\setminus\{ 0\}$. So $ax = 1$ is valid and $x$ is an inverse of $a$.
Since we can setup the proof for any $a$ in $R\setminus\{0\}$ this proves the neutral and inverse properties.
 A: Usually, a domain is assumed to have an identity.  Since $R$ is finite, surjectivity of a map $R \to R$ follows from injectivity: if $x \in R$ were not in the image, then the function would map $n$ elements to at most $n-1$ elements, which means that function is not injective.
Even without assuming the existence of an identity, you can prove it as follows.  We have already proved that both left multiplication by a fixed non-zero element is injective and surjective.  Hence for every nonzero $a \in R$ there exists $x_a \in R$ such that $a x_a = a$.  Multiply on the left by any nonzero $b$ to get $ba x_a = ba$.  But also $ba x_{ba} = ba$ (by definition).  Since $a$ and $b$ are nonzero, so is $ba$, and hence left multiplication by $ba$ is injective, so we conclude that $x_a = x_{ba}$.  But for a fixed nonzero $a$, the elements $ba$ range over all nonzero elements of $R$.  Hence $x_a$'s are all equal, independent of $a$, to some fixed element $x_0$.  We conclude that $a x_0 = a$ for all nonzero $a$, and of course for 0 as well, so this proves that $x_0$ is a right identity for $R$.  By similar arguments (reversing left and right), there is a left identity for $R$, and by a simple argument these two identities must then be equal, so there is an identity.
