If a subring $B$ of a field $F$ is closed with respect to multiplicative inverses then $B$ is a field. ($B$ is then called a subfield of $F$)
I started out my proof like this:
Assume a subring $B$ of a field $F$ is closed with respect to multiplicative inverses. By definition, a subring is closed with respect to addition, negative (additive) and multiplication. Therefore $B$ contains all these properties.
To show that $B$ is a field then we must show that $B$ is a commutative ring with unity in which every nonzero element is invertible
$B$ is a ring since every subring is also a ring.
Since we are given that $B$ is closed with respect to multiplicative inverses, then it must contain a unity which shall be denoted $1$.
Also, since it is now a subring with unity and multiplicative inverse means that every element in $B$ is invertible or :
where $x$ is the multiplicative inverse and $b$ is an element in $B$.
Now the only problem I have is how do I show that $B$ is commutative, the only thing I was thinking is to use the invertible part but i'm not sure.