Division Subring Test

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

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


 * $$(1) \quad L^* \ne \varnothing$$


 * $$(2) \quad \forall x, y \in L: x + \left({-y}\right) \in L$$


 * $$(3) \quad \forall x, y \in L: x \circ y \in L$$


 * $$(4) \quad x \in L^* \implies x^{-1} \in L^*$$

Necessary Condition
Suppose $$\left({L, +, \circ}\right)$$ is a division subring of $$\left({K, +, \circ}\right)$$.

The conditions $$(1)$$ to $$(3)$$ hold by virtue of the Subring Test.

Then $$(4)$$ also holds by the definition of a division ring:
 * $$\forall x \in L^*: \exists ! x^{-1} \in L^*: x^{-1} \circ x = x \circ x^{-1} = 1_L$$

Sufficient Condition
Suppose the conditions $$(1)$$ to $$(4)$$ hold.

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

By $$(4)$$, every element of $$L^*$$ has a product inverse.

Thus, from the Two-step Subgroup Test, $$\left({L^*, \circ}\right)$$ is a group.

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