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:


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

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

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


 * $(1): \quad$ 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.
 * $(2): \quad$ M0: Closure of Ring Product: From $3$, $\left({S, \circ}\right)$ is closed.
 * $(3): \quad$ M1: Associativity: From Restriction of Associative Operation is Associative, $\circ$ is associative on $R$, therefore also associative on $S$.
 * $(4): \quad$ 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 $\left({R, +, \circ}\right)$.

Also defined as
Some sources insist that for $\left({S, +, \circ}\right)$ to be a subring of $\left({R, +, \circ}\right)$, the unity (if there is one) must be the same for both, but this is an extra condition as this is not necessarily the general case.