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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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$\mathbb{Z}$ is an integral domain of characteristic $0$ whose group of units is $\{1,-1\}$ , so it is not a field: $2$ is nonzero and not invertible

Example

Let $\mathbb{Z}$ be the integers with the commutative ring structure of $\mathbb{Z}$ is a commutative ring and an ordered ring, the published construction being an instance of the general definitions, and let $2 := 1 + 1$ . Then:

$\mathbb{Z}$ is an integral domain (Zero divisor, and integral domain: a commutative ring with $1 \ne 0$ and no zero divisors);
$\mathbb{Z}^{\times} = \{1,-1\}$ ( $(\mathbb{Z}, \cdot, 1)$ is a commutative monoid whose group of units is $\{1, -1\}$ ; equivalently $u \mid 1$ holds exactly for $u = 1$ and $u = -1$ , with the unit group structure from The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times}$ is a group under multiplication; $0 \in R^{\times}$ only in the zero ring);
$2 \ne 0$ and $2 \notin \mathbb{Z}^{\times}$ , so $\mathbb{Z}$ is not a field (Field);
$\operatorname{char}(\mathbb{Z}) = 0$ (The characteristic of a ring: the least $n \ge 1$ with $n \cdot 1_R = 0$ when one exists, and $0$ otherwise).

So an integral domain need not be a field: claim 1 with claim 3 is the witness that the two notions differ.

Facts & Assumptions

Given: The integers $\mathbb{Z}$ with the operations of Arithmetic on the integers and the order of Order on the integers, and the numeral $2 := 1 + 1$ (The integers as equivalence classes of pairs of naturals).

[L1]

$(\mathbb{Z},+,\cdot,0,1)$ is a commutative ring ( $\mathbb{Z}$ is a commutative ring and an ordered ring, the published construction being an instance of the general definitions, Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides, Commutative ring).

[L2]

If $x, y \in \mathbb{Z}$ are nonzero then $xy \ne 0$ (The integers have no zero divisors; multiplicative cancellation).

[L3]

$(\mathbb{Z},\cdot,1)$ is a commutative monoid whose group of units is $\{1,-1\}$ ( $(\mathbb{Z}, \cdot, 1)$ is a commutative monoid whose group of units is $\{1, -1\}$ ; equivalently $u \mid 1$ holds exactly for $u = 1$ and $u = -1$ , Left inverse, right inverse, and invertible element of a monoid, The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times}$ is a group under multiplication; $0 \in R^{\times}$ only in the zero ring).

[L4]

The order on $\mathbb{Z}$ is total and compatible with addition, and $\mathbb{Z}$ is a commutative ring (The integers form a totally ordered ring, The integers form a commutative ring, Order on the integers).

[L5]

$\iota : \mathbb{N} \to \mathbb{Z}$ , $\iota(n) = [(n,0)]$ , is injective and preserves addition, multiplication and order (The naturals embed in the integers); the formulas for $0_{\mathbb Z}=[(0,0)]$ and $1_{\mathbb Z}=[(1,0)]$ give $\iota(0)=0$ and $\iota(1)=1$ (Arithmetic on the integers).

[L6]

Induction on $\mathbb{N}$ , and $n + 1 = \sigma(n)$ on $\mathbb{N}$ (The principle of mathematical induction, Addition of natural numbers, The natural numbers $\mathbb{N}$ (von Neumann)).

[L7]

The additive multiples in a ring satisfy $0 \cdot 1 = 0$ and $\sigma(n) \cdot 1 = n \cdot 1 + 1$ for $n \in \mathbb{N}$ (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ).

[L8]

$\operatorname{char}(R)$ is the least $n \ge 1$ with $n \cdot 1_R = 0_R$ , or $0$ if there is none (The characteristic of a ring: the least $n \ge 1$ with $n \cdot 1_R = 0$ when one exists, and $0$ otherwise); a field is a commutative ring in which every nonzero element is a unit (Every field is a commutative ring with $1 \ne 0$ ; it is an integral domain, and it is a commutative division ring, Field).

Verification

technique · direct

1.1

Claim 1: $\mathbb{Z}$ is a commutative ring by [L1]; $1 \ne 0$ , since $1 = \iota(1)$ , $0 = \iota(0)$ and $\iota$ is injective while $1 \ne 0$ in $\mathbb{N}$ ; and $\mathbb{Z}$ has no zero divisors by [L2]. So $\mathbb{Z}$ is an integral domain.

L1L2L5

1.2

Claim 2 is [L3]: the multiplicative monoid of the ring $\mathbb{Z}$ is $(\mathbb{Z},\cdot,1)$ , and its group of units is $\{1,-1\}$ .

L1L3

1.3

$0 < 1 < 2$ : $1 = \iota(1)$ lies in the image of $\iota$ , so $0 \le 1$ , and $1 \ne 0$ by injectivity of $\iota$ ; adding $1$ to $0 < 1$ gives $1 < 1 + 1 = 2$ .

L4L5

1.4

The map $n \mapsto n \cdot 1$ from $\mathbb{N}$ to $\mathbb{Z}$ is $\iota$ . Both send $0$ to $0$ , since $0 \cdot 1 = 0$ and $\iota(0) = 0$ ; and if $n \cdot 1 = \iota(n)$ then $\sigma(n)\cdot 1 = n \cdot 1 + 1 = \iota(n) + \iota(1) = \iota(n+1) = \iota(\sigma(n))$ , because $\iota$ preserves addition. Induction on $\mathbb{N}$ gives the claim.

L5L6L7

2.1

Claim 3: by step 1.3, $0 < 1 < 2$ , so $2 \ne 0$ and $2 \ne 1$ ; and $-1 < 0 < 2$ by adding $-1$ to $0 < 1$ and using transitivity, so $2 \ne -1$ . Hence $2 \notin \{1,-1\} = \mathbb{Z}^{\times}$ by step 1.2. A field has every nonzero element a unit, so $\mathbb{Z}$ is not a field.

step 1.2step 1.3L4L8

2.2

Claim 4: for $n \in \mathbb{N}$ with $n \ge 1$ we have $n \cdot 1 = \iota(n)$ by step 1.4, and $\iota(n) \ne \iota(0) = 0$ because $\iota$ is injective and $n \ne 0$ . So no $n \ge 1$ satisfies $n \cdot 1 = 0$ , and $\operatorname{char}(\mathbb{Z}) = 0$ .

step 1.4L5L8

3.1

Claims 1 to 4 are established in steps 1.1, 1.2, 2.1 and 2.2.

step 1.1step 1.2step 2.1step 2.2∎

Remarks

This is the standard witness that "integral domain" is strictly weaker than "field". $\mathbb{Z}$ satisfies every clause of Zero divisor, and integral domain: a commutative ring with $1 \ne 0$ and no zero divisors and fails the one extra clause a field asks for. The gap is exactly the failure of $2$ to be invertible, and $(\mathbb{Z}, \cdot, 1)$ is a commutative monoid whose group of units is $\{1, -1\}$ ; equivalently $u \mid 1$ holds exactly for $u = 1$ and $u = -1$ is what pins the units down.
Characteristic $0$ and infinite additive order are the same statement here. By The characteristic of a ring is the additive order of $1_R$ , with $0$ recording infinite order; $n \cdot 1_R = 0$ holds exactly when $\operatorname{char}(R) \mid n$ ; and in an integral domain every nonzero element has the same additive order as $1_R$ the characteristic is the additive order of $1$ , so claim 4 says that $1$ has infinite additive order in $(\mathbb{Z},+)$ ; and since $\mathbb{Z}$ is a domain, the same lemma says every nonzero integer has infinite additive order too.
The units are read off $(\mathbb{Z}, \cdot, 1)$ is a commutative monoid whose group of units is $\{1, -1\}$ ; equivalently $u \mid 1$ holds exactly for $u = 1$ and $u = -1$ , not off the order. That lemma proves $\mathbb{Z}^{\times} = \{1,-1\}$ from the divisibility relation and the bound on divisors; nothing on this page reproves it, and no group-theoretic example page is cited for it.

Depends on

Used by

A product of two rings with 1 ≠ 0 always has zero divisors: (1,0)(0,1) = (0,0) in ℤ × ℤ, so a product of integral domains is never an integral domain Example

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 89 results over 21 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Integral domain (Wikipedia) (standard reference, not scraped)
Unit (ring theory) (Wikipedia) (standard reference, not scraped)
Thomas W. Judson, Abstract Algebra: Theory and Applications, §16.4: Integral Domains and Fields (standard reference, not scraped)

Z\mathbb{Z}Z is an integral domain of characteristic 000 whose group of units is {1,−1}\{1,-1\}{1,−1}, so it is not a field: 222 is nonzero and not invertible