Integers form Commutative Ring

Theorem

The set of integers $\Z$ forms a commutative ring under addition and multiplication.


Proof

We have that:

From Integers under Addition form Abelian Group, the algebraic structure $\struct {\Z, +}$ is an abelian group.
From Integers under Multiplication form Countably Infinite Commutative Monoid, the algebraic structure $\struct {\Z, \times}$ is a commutative monoid and therefore a commutative semigroup.
Integer Multiplication Distributes over Addition.

Thus all the ring axioms are fulfilled, and $\struct {\Z, +, \times}$ is a commutative ring.


By Integer Multiplication has Zero, the zero is $0$.

$\blacksquare$


Sources

  • 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains: Example $21.1$
  • 1970: B. Hartley and T.O. Hawkes: Rings, Modules and Linear Algebra ... (previous) ... (next): Chapter $1$: Rings - Definitions and Examples: $1$: The definition of a ring: Definitions $1.1 \ \text{(c)}$
  • 1972: A.G. Howson: A Handbook of Terms used in Algebra and Analysis ... (previous) ... (next): $\S 4$: Number systems $\text{I}$: The rational integers
  • 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 55$. Special types of ring and ring elements: $(1)$
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