Invertible Integers under Multiplication
Theorem
The only invertible elements of $\Z$ for multiplication (that is, units of $\Z$) are $1$ and $-1$.
Corollary 1
Let $a, b \in \Z$ such that $a b = 1$.
Then $a = b = \pm 1$.
Corollary 2
The integers $\struct {\Z, +, \times}$ do not form a field.
Proof
Let $x > 0$ and $x y > 0$.
Aiming for a contradiction, suppose first that $y \le 0$.
Then from Multiplicative Ordering on Integers and Ring Product with Zero:
- $x y \le x \, 0 = 0$
From this contradiction we deduce that $y > 0$.
Let $x > 0$ and $x y = 1$.
Then:
- $y > 0$
and by Natural Numbers are Non-Negative Integers:
- $y \in \N$
Hence by Invertible Elements under Natural Number Multiplication:
- $x = 1$
Thus $1$ is the only element of $\N$ that is invertible for multiplication.
Therefore by Natural Numbers are Non-Negative Integers and Product with Ring Negative, the result follows.
$\blacksquare$
Sources
- 1964: Iain T. Adamson: Introduction to Field Theory ... (previous) ... (next): Chapter $\text {I}$: Elementary Definitions: $\S 1$. Rings and Fields: Example $1$
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): Chapter $4$. Groups: Exercise $13$
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $20$. The Integers: Theorem $20.10$
- 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $6$: Polynomials and Euclidean Rings: $\S 26$. Divisibility: Example $47$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 55$. Special types of ring and ring elements: $(6)$
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers