Let $F$ be a field and let $K$ be a subset of $F$ with at least two elements. Prove that $K$ is a subfield of $F$ if, for any $a, b\in K$ with $b \neq 0$, $a-b$ and $ab^{-1}$ belong to $K$.
A field $F$ is a commutative ring with unity (1, that is, multiplicative identity) in which every nonzero element is unit (has a multiplicative inverse).
The way I understand this is that $$ (F,+) \ \ \text{ is an abelian group under addition} $$ $$ (F-\{ 0 \},*) \ \ \text{ is an abelian group under multiplication } $$
A nonempty $K \subseteq F $ is a subfield if it is a field with the same operations of $F$.
In group theory we have a test to easily verify that $H\subseteq G$ is a subgroup of a group $(G,*)$, that is, if $a*b^{-1} \in H, \forall a,b \in H$
An intuitive way to see subfield test is to think that we have to prove that $(K,+)$ is an subgroup of $(F,+)$ and that $(K-\{0\},*)$ is a subgroup of $(F-\{0\},*)$ ... In other words:
$$ a+(-b) = a-b \in K \ \ \text{...$(K,+)$ is a subgroup under addition}$$ and $$ a*b^{-1} \in K \ \ \text{...$(K - \{0\},*)$ is a subgroup under multiplication}$$
if these conditions hold, we can say that $K$ is a subfield. Is this correct?...if it's not...can you help me? thanks!