# 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 $ab^{-1}\in H$.

Can someone help me?

Thanks.

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These conditions simply make $H$ closed under the operations of $K$, so $H$ is a subring of $K$, hence a subfield. – Dane Dec 30 '11 at 20:38
@Dane: Being a subring of a field does not imply being a field, of course; one has $\mathbb{Z}\subset\mathbb{Q}$ for example. However the condition that $a,b\in H\implies ab^{-1}\in H$ implies that $H$ has multiplicative inverses of its non-zero elements (which one must assume exist...) – Zev Chonoles Dec 30 '11 at 20:40
@Zev: You need to include the clause that $b\neq 0$, and you also need to include a clause specifying that $H$ does not consist only of the element $0$. Otherwise, the statement is incorrect. – Arturo Magidin Dec 30 '11 at 21:00
@Arturo: You're correct, of course; I was assuming the condition was meant to be interpreted as "$ab^{-1}\in H$ when this expression makes sense" but nevertheless it's best to point out the problem. – Zev Chonoles Dec 30 '11 at 21:06
@Zev: Even if you interpret it as "provided it makes sense", you run into the issue of $H=\emptyset$ or $H=\{0\}$, which satisfy the given conditions (with your reading for the second one) but are not subfields. The emtpy set strikes again! (-: – Arturo Magidin Dec 30 '11 at 21:16

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:

1. $H\neq\emptyset$ and $H\neq\{0\}$.
2. If $a,b\in H$, then $a-b\in H$.
3. 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 $K$. $\Box$

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Thank you Arturo. – Aslan986 Dec 30 '11 at 20:57

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.

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