Subfield Test

Theorem
Let $$\left({F, +, \circ}\right)$$ be a field, and let $$K$$ be a subset of $$F$$.

Then $$\left({K, +, \circ}\right)$$ is a subfield of $$\left({F, +, \circ}\right)$$ iff these all hold:


 * 1) The subset $$K^* \ne \varnothing$$;
 * 2) $$\forall x, y \in K: x + \left({-y}\right) \in K$$;
 * 3) $$\forall x, y \in K: x \circ y \in K$$;
 * 4) $$x \in K^* \implies x^{-1} \in K^*$$.

Proof

 * If $$\left({K, +, \circ}\right)$$ is a subfield of $$\left({F, +, \circ}\right)$$, the conditions hold by virtue of the field axioms.


 * Conversely, suppose the conditions hold.

By 1 to 3, it follows from Subring Test that $$\left({K, +, \circ}\right)$$ is a subring of $$\left({F, +, \circ}\right)$$.

By 4, every element of $$K^*$$ has a product inverse (and incidentally, from the Two-step Subgroup Test, that $$\left({K^*, \circ}\right)$$ is a group).

Therefore, $$\left({K, +, \circ}\right)$$ is a ring in which every element has a product inverse, which makes $$\left({K, +, \circ}\right)$$ a division ring.

As $$\left({F, +, \circ}\right)$$ is a field, then $$\circ$$ is commutative on all of $$F$$, and therefore also on $$K$$ by Restriction of Operation Commutativity.

Thus $$\left({K, +, \circ}\right)$$ is a field.