Subfield Tests

Proof

The main part of this proof follows from the subgroup tests which we apply to the whole set $K$ as an additive group, and then to $K-\{0\}$ as a multiplicative group.

Let $a\in K$ noting that $K\ne \varnothing$. Then $a-a=0\in K$ from property (2). Then with $a=0$ we have that $a-b=0-b=-b\in K$ for arbitrary $b\in K$, hence we have closure under inverses in $(K,+)$. As such we have that $a-(-b)=a+b\in K$ for arbitrary $a,b\in K$. Hence $(K,+)$ is an abelian group, noting that commutativity is inherited from the parent group.

Now, let $K^{\ast }=K-\{0\}$. We will prove that this set is an abelian group under multiplication. Let $a,b\in K^{\ast }$, which we note is non-empty by assumption (1). Then from property (3), we have that $b\times b^{-1}=1\in K$. This means that $1\times b^{-1}=b^{-1}\in K$ and therefore that $a\times (b^{-1}{)}^{-1}=a\times b\in K$. Again, commutativity follows from the parent group.

The distributive law then follows from the parent group, and closure as established above.