Null Ring is Ideal

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

Then the null ring $\struct {\set {0_R}, +, \circ}$ is an ideal of $\struct {R, +, \circ}$.

Proof
From Null Ring is Subring of Ring, $\struct {\set {0_R}, +, \circ}$ is a subring of $\struct {R, +, \circ}$.

Also, from Ring Product with Zero:
 * $\forall x \in R: x \circ 0_R = 0_R = 0_R \circ x \in \set {0_R}$

thus fulfilling the condition for $\struct {\set {0_R}, +, \circ}$ to be an ideal of $\struct {R, +, \circ}$.