Zero of Subring is Zero of Ring

Theorem

Let $\struct {R, +, \times}$ be a ring whose zero is $0$.

Let $\struct {S, +, \times}$ be a subring of $\struct {R, +, \times}$.

The zero of $\struct {S, +, \times}$ is also $0$.

Proof

By definition, $\struct {S, +, \times}$ is a subset of $R$ which is a ring.

By definition of ring, $\struct {S, +}$ and $\struct {R, +}$ are groups such that $S \subseteq R$.

So, by definition, $\struct {S, +}$ is a subgroup of $\struct {R, +}$.

By Identity of Subgroup, the identity of $\struct {S, +}$, which is $0$, is also the identity of $\struct {R, +}$.

$\blacksquare$