Rings, Domains and Fields: Examples and Counterexamples

Let $\mathbb{Z}$ be the integers (The integers as equivalence classes of pairs of naturals) with the operations of Arithmetic on the integers and the order of Order on the integers. Then:

1. $(\mathbb{Z}, +, \cdot, 0, 1)$ is a ring in the sense of Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides, and it is commutative (Commutative ring);
2. $\mathbb{Z}$ with the order $\le$ is an ordered ring in the sense of Ordered ring: a ring with a total order compatible with addition and with positives closed under multiplication;
3. its positive cone $P = \{\, x \in \mathbb{Z} : 0 < x \,\}$ satisfies trichotomy and closure in the sense of The order presentation and the positive-cone presentation of an ordered ring determine each other: $P = \{\, x : 0 < x \,\}$ satisfies trichotomy and closure, and $a < b :\iff b - a \in P$ recovers the order, and the order it induces is the published order of Order on the integers.

The point of the example is that nothing is being built: the published The integers form a commutative ring and The integers form a totally ordered ring were proved before rings were defined and used the words "commutative ring" and "totally ordered ring" informally. This item records that those words, as used there, mean exactly what Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides and Ordered ring: a ring with a total order compatible with addition and with positives closed under multiplication mean, so that a statement about rings may be applied to $\mathbb{Z}$ without translation.

• The multiplicative axiom matches on the nose. The integers form a totally ordered ring proves the strict statement "$0 < x$ and $0 < y$ imply $0 < xy$", which is exactly axiom (OR2) of Ordered ring: a ring with a total order compatible with addition and with positives closed under multiplication as that definition states it. That is why the step establishing claim 2 above is a citation and not an argument, and it is the reason Ordered ring: a ring with a total order compatible with addition and with positives closed under multiplication adopts the strict form.

• $\mathbb{Z}$ is not a field, and it is an integral domain; both are recorded separately in $\mathbb{Z}$ 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, for which this item supplies the ring structure.

• A later page that needs "$\mathbb{Z}$ is a ring" on its own spine must re-derive it there rather than cite this item, since examples pages are leaves in the reading order. The derivation is the three lines above.

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:

1. $\mathbb{Z}$ is an integral domain (Zero divisor, and integral domain: a commutative ring with $1 \ne 0$ and no zero divisors);
2. $\mathbb{Z}^{\times} = \{1,-1\}$ ($(\mathbb{Z}, \cdot, 1)$ 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);
3. $2 \ne 0$ and $2 \notin \mathbb{Z}^{\times}$, so $\mathbb{Z}$ is not a field (Field);
4. $\operatorname{char}(\mathbb{Z}) = 0$ (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.

• 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.

Let $F$ be either $\mathbb{Q}$ (The rationals form a field) or $\mathbb{R}$ (The reals form a field), with its published order (The rationals form a totally ordered field, The reals form a totally ordered field). Then:

1. $F$ is a commutative ring with $1 \ne 0$ and an integral domain (Every field is a commutative ring with $1 \ne 0$; it is an integral domain, and it is a commutative division ring, Zero divisor, and integral domain: a commutative ring with $1 \ne 0$ and no zero divisors);
2. $F$ with its order is an ordered ring (Ordered ring: a ring with a total order compatible with addition and with positives closed under multiplication), and the set $P = \{\, x \in F : 0 < x \,\}$ is a positive cone making $F$ an ordered field in the sense of Ordered field, whose induced order is the published one;
3. $\operatorname{char}(F) = 0$ (The characteristic of a ring: the least $n \ge 1$ with $n \cdot 1_R = 0$ when one exists, and $0$ otherwise).

• Two presentations of one order, reconciled here rather than assumed. The published The rationals form a totally ordered field and The reals form a totally ordered field state the order form; the published Ordered field states the positive-cone form. The step establishing claim 2 above passes between them using The order presentation and the positive-cone presentation of an ordered ring determine each other: $P = \{\, x : 0 < x \,\}$ satisfies trichotomy and closure, and $a < b :\iff b - a \in P$ recovers the order, and that is the only reason Canonical naturals are positive and strictly increasing, which is stated for an ordered field, may be applied to $\mathbb{Q}$ and $\mathbb{R}$ here.

• The multiples agree with the canonical naturals. By [L6] the element $n \cdot 1_F$ appearing in The characteristic of a ring: the least $n \ge 1$ with $n \cdot 1_R = 0$ when one exists, and $0$ otherwise is the $\iota(n)$ of The canonical natural $\iota(n) = n \cdot 1_F$ of a field, so claim 3 is also the statement that $\iota$ never takes the value $0_F$ on $n \ge 1$ in these two fields. More is true and is quoted rather than proved here: Canonical naturals are positive and strictly increasing shows $\iota$ is strictly increasing on $n \ge 1$, hence injective.

• $\mathbb{Q}$ and $\mathbb{R}$ are domains for a reason stronger than necessary. They have no zero divisors because every nonzero element is invertible, not because of any cancellation argument; the same reasoning gives nothing about $\mathbb{Z}$, whose domain property is recorded separately in $\mathbb{Z}$ 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.

Let $X := \{\, n \in \mathbb{N} : n \ge 1 \,\}$ (The natural numbers $\mathbb{N}$ (von Neumann), Order on the natural numbers) and let $\mathcal{C}$ be the set of Cauchy sequences of rationals, that is, the set of those functions $X \to \mathbb{Q}$ that satisfy the condition of Cauchy sequence of rationals; that definition indexes its sequences $(a_n)_{n \ge 1}$, so $X$, and not $\mathbb{N}$, is the index set. Give $\mathcal{C}$ the termwise operations and the constant sequences $0$ and $1$. Then:

1. $\mathcal{C}$ is a commutative ring in the sense of Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides and Commutative ring, with $1 \ne 0$;

2. $\mathcal{C}$ is a subring (Subring: a subset containing $1_R$ and closed under addition, additive inverses and multiplication) of the ring $\mathbb{Q}^{X}$ of all functions $X \to \mathbb{Q}$ with pointwise operations (The ring $R^{X}$ of all functions from a set $X$ into a ring, with pointwise operations);

3. $\mathcal{C}$ is not an integral domain (Zero divisor, and integral domain: a commutative ring with $1 \ne 0$ and no zero divisors): the sequences

   $a = (1,0,0,0,\dots), \qquad b = (0,1,1,1,\dots)$

   defined by $a_1 = 1$, $a_n = 0$ for $n \ge 2$, and $b_1 = 0$, $b_n = 1$ for $n \ge 2$, are both Cauchy, both nonzero, and satisfy $ab = 0$; so each is a zero divisor.

• The index set is stated because it is not $\mathbb{N}$. Cauchy sequence of rationals indexes its sequences from $n = 1$, while The ring $R^{X}$ of all functions from a set $X$ into a ring, with pointwise operations takes an arbitrary index set; the ambient ring in claim 2 is therefore $\mathbb{Q}^{X}$ with $X = \{\, n \in \mathbb{N} : n \ge 1 \,\}$, not $\mathbb{Q}^{\mathbb{N}}$. Since $\mathbb{N}$ contains $0$ (The natural numbers $\mathbb{N}$ (von Neumann)) the two are different sets of functions, and the subring claim would be false as stated about the second.

• Why this matters for the construction of $\mathbb{R}$. The real numbers are defined as the quotient $\mathcal{C}/\mathcal{N}$ by the null sequences (The real numbers), and The reals form a field proves that quotient is a field. The present example shows the field property cannot come from $\mathcal{C}$ alone: $\mathcal{C}$ is not even a domain. What The reals form a field actually uses is The null ideal is maximal, a property of $\mathcal{N}$ inside $\mathcal{C}$. The two zero divisors above cause no trouble in the quotient: $a$ is a null sequence (Null sequence), so its class is $0$, and $b$ differs from the constant sequence $1$ by a null sequence, so its class is $1$.

• The witnesses are chosen to keep the verification short. Both are eventually constant, hence Cauchy with no $\varepsilon$ bookkeeping, and their supports are disjoint, which makes the product zero at every index.

Let $R$ and $S$ be rings (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides) with $1_R \ne 0_R$ and $1_S \ne 0_S$, and let $R \times S$ be the product ring (The product ring $R \times S$ with componentwise operations, its identity $(1_R, 1_S)$ and its units $R^{\times} \times S^{\times}$). Then:

1. $(1_R, 0_S)$ and $(0_R, 1_S)$ are nonzero elements of $R \times S$ whose product is $(0_R, 0_S)$; each is therefore a zero divisor (Zero divisor, and integral domain: a commutative ring with $1 \ne 0$ and no zero divisors);
2. consequently $R \times S$ is never an integral domain, whatever $R$ and $S$ are; in particular a product of two integral domains is not one;
3. the concrete instance is $\mathbb{Z} \times \mathbb{Z}$, where $(1,0)(0,1) = (0,0)$ while $\mathbb{Z}$ itself is an integral domain ($\mathbb{Z}$ 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).

• This is the cheapest source of zero divisors available. No arithmetic beyond $x \cdot 0 = 0$ is used, and the two witnesses are written down rather than found. The other standard source is a ring of functions from a set with at least two points to a coefficient ring with $1\ne0$ (The ring $R^{X}$ of all functions from a set $X$ into a ring, with pointwise operations), and the two constructions are the same phenomenon: a point where one factor vanishes and the other does not.

• The hypothesis is on both factors. If $S$ is the one-element ring then $(0_R,1_S) = (0_R,0_S)$ is the zero of $R \times S$ and the argument collapses; indeed $R \times S$ is then essentially $R$ again, and may well be a domain. That is why both $1_R \ne 0_R$ and $1_S \ne 0_S$ are assumed.

• The failure is not about commutativity. $R$ and $S$ need not be commutative for claim 1; commutativity enters only in claim 2, because Zero divisor, and integral domain: a commutative ring with $1 \ne 0$ and no zero divisors requires a domain to be commutative and the product of two commutative rings is commutative (The product ring $R \times S$ with componentwise operations, its identity $(1_R, 1_S)$ and its units $R^{\times} \times S^{\times}$).

Let $Z = \{z\}$ be a one-element set, and define $z + z := z$, $z \cdot z := z$, $0_Z := z$ and $1_Z := z$. Then:

1. $Z$ is a commutative ring (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides, Commutative ring), the zero ring, and $1_Z = 0_Z$;
2. up to the choice of the single element, it is the only ring in which $1 = 0$: any ring $R$ with $1_R = 0_R$ has $R = \{0_R\}$ (In any ring $0 \cdot a = a \cdot 0 = 0$, $(-a)b = a(-b) = -(ab)$, $(-a)(-b) = ab$, $(-1)a = -a$ and $a(b - c) = ab - ac$);
3. $Z$ has no zero divisors, and is nevertheless not an integral domain (Zero divisor, and integral domain: a commutative ring with $1 \ne 0$ and no zero divisors), because it fails $1 \ne 0$;
4. $Z$ is not a division ring (Division ring: a ring with $1 \ne 0$ in which every nonzero element is a unit) and not a field (Field), for the same reason;
5. $\operatorname{char}(Z) = 1$ (The characteristic of a ring: the least $n \ge 1$ with $n \cdot 1_R = 0$ when one exists, and $0$ otherwise).

• This is the item that makes $1 \ne 0$ a visible hypothesis. Zero divisor, and integral domain: a commutative ring with $1 \ne 0$ and no zero divisors, Division ring: a ring with $1 \ne 0$ in which every nonzero element is a unit and Field each carry that clause, and the zero ring is what each of them excludes. It satisfies every other clause of the first two: it is a commutative ring, it has no zero divisors, and every nonzero element of it is vacuously a unit. For Field the clause $0 \ne 1$ is not the only one that fails, and this is worth saying rather than glossing: axiom (M) there asks that $F \setminus \{0\}$ be an abelian group with identity $1$, and in the zero ring that set is empty, so it has no identity at all. The two failures are the same phenomenon, since $1 = 0$ is exactly what empties $F \setminus \{0\}$.

• Characteristic $1$ occurs exactly here. By claim 2 a ring has $1 = 0$ only if it is the zero ring, and by The characteristic of a ring: the least $n \ge 1$ with $n \cdot 1_R = 0$ when one exists, and $0$ otherwise the characteristic is $1$ exactly when $1 \cdot 1_R = 1_R = 0_R$. So the zero ring is the only ring of characteristic $1$, and every other ring has characteristic $0$ or a characteristic that is at least $2$.

• The zero ring is not excluded from being a ring. Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides does not require $1 \ne 0$, deliberately. The ring of all functions from the empty set into a ring has exactly one element, the empty function (The ring $R^{X}$ of all functions from a set $X$ into a ring, with pointwise operations), so it is the zero ring; requiring $1 \ne 0$ in the definition of a ring would make that construction partial.

The integers are not literally a subset of the rationals in this library: $\mathbb{Q}$ is a set of equivalence classes of pairs of integers, so "$\mathbb{Z}$ inside $\mathbb{Q}$" means the image of the embedding $j : \mathbb{Z} \to \mathbb{Q}$, $j(k) = [(k,1)]$, of The integers embed in the rationals. Write $K := j(\mathbb{Z}) \subseteq \mathbb{Q}$ and let $2 := 1 + 1$ in $\mathbb{Z}$. Then:

1. $K$ is a subring of the ring $\mathbb{Q}$ (Subring: a subset containing $1_R$ and closed under addition, additive inverses and multiplication);
2. $K$ is not a subfield of $\mathbb{Q}$ (Subfield: a subring of a field closed under inverses of its nonzero elements, and therefore a field with the restricted operations): the element $j(2)$ is a nonzero member of $K$ whose inverse in $\mathbb{Q}$ does not lie in $K$;
3. so the inverse-closure clause (K2) of Subfield: a subring of a field closed under inverses of its nonzero elements, and therefore a field with the restricted operations is not implied by being a subring.