Subring Test

Theorem
Let $$S$$ be a subset of a ring $$\left({R, +, \circ}\right)$$

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


 * 1) $$S \ne \varnothing$$;
 * 2) $$\forall x, y \in S: x + \left({-y}\right) \in S$$;
 * 3) $$\forall x, y \in S: x \circ y \in S$$.

Proof

 * If $$S$$ is a subring of $$\left({R, +, \circ}\right)$$, the conditions hold by virtue of the ring axioms as applied to $$S$$.


 * Conversely, suppose the conditions hold. We check that the ring axioms hold for $$S$$.


 * 1) A: Addition forms a Group: By 1 and 2 above, it follows from the One-step Subgroup Test that $$\left({S, +}\right)$$ is a subgroup of $$\left({R, +}\right)$$, and therefore a group.
 * 2) M0: Closure of Ring Product: From 3, $$\left({S, \circ}\right)$$ is closed.
 * 3) M1: Associativity: From Restriction of Operation Associativity, $$\circ$$ is associative on $$R$$, therefore also associative on $$S$$.
 * 4) D: Distributivity: From Restriction of Operation Distributivity, $$\circ$$ distributes over $$+$$ for the whole of $$R$$, therefore for $$S$$ also.

So $$\left({S, +, \circ}\right)$$ is a ring, and therefore a subring of $$\left({R, +, \circ}\right)$$.