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 $\struct {R, +, \circ}$.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old
• 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $5$: Rings: $\S 22$. Quotient Rings: Example $40$
• 1970: B. Hartley and T.O. Hawkes: Rings, Modules and Linear Algebra ... (previous) ... (next): $\S 2.2$: Homomorphisms: Definition $2.5$
• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 58.1$ Ideals