A false proof for the theorem about finite integral domains and fields

Question

Related to another post of mine, right now I am looking at a proof for the theorem:

A finite integral domain $R$ is a field.

In short, the proof uses a map $f_a:R\to R$ defined for some $a \in R$ by $f_a(x)=ax$ for all $x \in R$. Injectivity is easy to show and surjectivity is attained from injectivity and pigeonhole considerations. So now we know that $1_R \in Im(f)$ and that we have a unique element $r$ in the function domain $R$ such that $f_a(r)=1_R = ar$. That is how we show that every element has exactly one inverse and thus our integral domain $R$ satisfies field properties.

My question is why can't the proof work when $R$ is infinite? $f_a(x)$ might as well still be a bijection and we're done. When does infinity get in the way here?

Thanks.

Answer 1

The argument doesn't work for an infinite $R$ because for infinite sets injectivity does not imply surjectivity.

Think about the function $x \to 2x$ on the integers.

Answer 2

It can be made to work for infinite domains but you need to use finiteness in some other form.

One generalization is this

A finite-dimensional commutative $K$-algebra without zero divisors is a field

The proof uses exactly the same map $f_a$, which is $K$-linear and injective, hence surjective, by the rank–nullity theorem.