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