Null Ring is Ideal
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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