# Subring Test

## Theorem

Let $S$ be a subset of a ring $\struct {R, +, \circ}$.

Then $\struct {S, +, \circ}$ is a subring of $\struct {R, +, \circ}$ if and only if these all hold:

$(1): \quad S \ne \O$
$(2): \quad \forall x, y \in S: x + \paren {-y} \in S$
$(3): \quad \forall x, y \in S: x \circ y \in S$

## Proof

### Necessary Condition

If $S$ is a subring of $\struct {R, +, \circ}$, 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 $\struct {S, +}$ is a subgroup of $\struct {R, +}$, and therefore a group.
$(2): \quad$ M0: Closure of Ring Product: From $3$, $\struct {S, \circ}$ 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 $\struct {S, +, \circ}$ is a ring, and therefore a subring of $\struct {R, +, \circ}$.

$\blacksquare$

## Also defined as

Some sources insist that for $\struct {S, +, \circ}$ to be a subring of $\struct {R, +, \circ}$, 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.