Necessary and Sufficient Condition for a sub-field Is there any  necessary and sufficient condition to determine whether a subset $H$ of a given field $K$ is a subfield? 
In some paper I have found something like that: $H$ is a field if for all $a, b\in H$, we have $a-b\in H$ and $ab^{-1}\in H$.
But I'm not sure about this property..
Can someone help me? 
Thanks.
 A: HINT $\rm\ \exists\: a\ne 0\in H\ \Rightarrow\ 0 = a-a\in H\ \Rightarrow 1 = a\:a^{-1}\in H\ \Rightarrow 1\cdot a^{-1} = a^{-1}\in H\ $ hence $\rm\ b\cdot(a^{-1})^{-1} = b\cdot a\in H\:.\:$ Thus $\rm\:H\:$ is a subring of $\rm\:K\:$ by the subring test. Being a nontrivial ring whose nonzero elements are invertible, $\rm\:H\:$ is a field.
A: The statement is false as written, since you could have $b=0$.
A subset $H$ of $K$ is a subfield if and only if $H$ is a subgroup of $K$ under addition, and the nonzero elements of $H$ are a subgroup of the multiplicative group of nonzero elements of $K$. 
Thus, $H\subseteq K$ is a subfield of $K$ if and only if:


*

*$H\neq\emptyset$ and $H\neq\{0\}$. 

*If $a,b\in H$, then $a-b\in H$.

*If $a,b\in H$, $a\neq 0$, $b\neq 0$, then $ab^{-1}\in H$.


Proof. The conditions clearly hold if $H$ is a subfield. Conversely, if $H$ satisfies $H\neq\emptyset$ and 2, then $H$ is a subgroup of $K$. Taking $r\in H-\{0\}$ (possible since $H\neq\{0\}$ and $H\neq\emptyset$), setting $a=b=r$ gives $1\in H$, and then condition 3 shows that $H-\{0\}$ is a (multiplicative) subgroup of $K-\{0\}$. Thus, $H$ is closed under addition, products, additive inverses, nonzero multiplicative inverses, and every nonzero element has an inverse. Thus, $H$ is a field, hence a subfield of $K$. $\Box$
