Contents / Abstract Algebra / Integral Domains and Fields

Chapter 6

Integral Domains and Fields

Cancellation as the defining feature of a domain; every field is a domain and every finite domain is a field, by a pigeonhole argument; characteristic 0 or prime, the prime subfield, and finite fields of order p^n; the Frobenius endomorphism; and the field of fractions constructed in full, with the well-definedness checks that are where the domain hypothesis is spent.

Introduction

A commutative ring is a very permissive object. The axioms ask that addition be an abelian group, that multiplication be associative and commutative with an identity, and that the two distribute; they ask nothing at all about what happens when a product comes out zero. In $Z /6 Z$ the product $2 \cdot 3$ is $0$ even though neither factor is, and the consequences are immediate and destructive: one cannot cancel, the equation $2 x = 4$ has two solutions, a linear polynomial can have more roots than its degree, and the whole edifice of divisibility theory collapses before it is built. This chapter is about the two conditions that repair the damage.

The weaker condition is no zero-divisors, and a commutative ring satisfying it is an integral domain. It is exactly strong enough to restore cancellation and to make the degree of a product the sum of the degrees, which is why $Z$ , $Z [i]$ , $Z [- 5]$ and every polynomial ring $k [x]$ live in this chapter rather than outside it. The stronger condition is that every nonzero element be invertible, and a commutative ring satisfying it is a field. Every field is a domain, and the converse fails for the most familiar reason in algebra: $2$ has no inverse in $Z$ .

Between those two statements sits the prettiest theorem in the chapter, and the one worth reading it for: every finite integral domain is a field. The proof occupies four lines and uses no ring theory at all beyond cancellation — it is the pigeonhole principle, applied to a map that cancellation forces to be injective. Finiteness closes the gap between the two conditions completely, so that among finite commutative rings there is nothing strictly between "no zero-divisors" and "a field". That single theorem explains at a stroke why $Z / p Z$ is a field for $p$ prime, and it is the seed from which the classification of finite fields grows.

The other two movements of the chapter are invariants and constructions. The characteristic measures how many times $1$ must be added to itself to reach $0$ , and in a domain it is forced to be $0$ or a prime — a constraint with no analogue in general rings, where $Z /12 Z$ happily has characteristic $12$ . Characteristic $p$ brings with it the Frobenius map $x \mapsto x^{p}$ , which is a ring homomorphism because the binomial coefficients in between are all divisible by $p$ ; a fact that is a student's error in characteristic $0$ and a theorem in characteristic $p$ . Finally, the field of fractions runs the passage from $Z$ to $Q$ in general: every integral domain embeds in a field, canonically and minimally, and the construction is worth doing once in full because the verifications — that the relation is transitive, that addition is well defined on classes — are where the hypothesis "domain" is actually spent.

Throughout, ring means commutative ring with multiplicative identity $1$ , and a ring homomorphism is required to send $1$ to $1$ . We write $R^{\times}$ for the group of units of $R$ , as in Rings and subrings.

6.1Zero-divisors, domains and cancellation

The defect in $Z /6 Z$ has a name.

Definition 6.1 (Zero-divisor). Let $R$ be a ring. A nonzero element $a \in R$ is a zero-divisor if there exists a nonzero $b \in R$ with $ab = 0$ .

The insistence that both $a$ and $b$ be nonzero is what gives the notion content. In every ring $0 \cdot b = 0$ , so if $0$ were allowed to count, every ring with more than one element would have a zero-divisor and the definition would classify nothing.

Definition 6.2 (Integral domain). An integral domain (or simply a domain) is a ring $R$ with $1 \neq = 0$ and no zero-divisors; equivalently, for all $a, b \in R$ ,

ab = 0 ⟹ a = 0 or b = 0.

Two hypotheses are doing work here and both are easy to skip over. The first is commutativity, which we have folded into the standing convention; a noncommutative ring with no zero-divisors is called a domain but not an integral one, and the theory diverges. The second is $1 \neq = 0$ , which excludes the zero ring ${0}$ . In the zero ring the implication $ab = 0 \Rightarrow a = 0$ holds vacuously, since every element is $0$ , so without the extra clause the zero ring would be a domain — and then also, by the definition below, a field, and every theorem about fields would have a one-element counterexample. Excluding it costs nothing and is standard.

Intuition. A domain is a ring in which multiplication cannot destroy information. If $a \neq = 0$ , then multiplying by $a$ sends distinct elements to distinct elements: no two inputs get crushed together, and in particular nothing but $0$ is crushed onto $0$ . In $Z /6 Z$ , multiplication by $2$ crushes the six elements down onto only three, and once that has happened you cannot read $b$ off from $2 b$ .

That description is not a metaphor; it is a theorem, and it is the reason domains are the right setting for divisibility.

Proposition 6.3 (Cancellation characterises domains). Let $R$ be a ring with $1 \neq = 0$ . Then $R$ is an integral domain if and only if the cancellation law holds: for all $a, b, c \in R$ with $a \neq = 0$ ,

ab = a c ⟹ b = c .

Proof. Suppose $R$ is a domain and $ab = a c$ with $a \neq = 0$ . Then $a (b - c) = ab - a c = 0$ . Since $R$ has no zero-divisors and $a \neq = 0$ , the second factor must vanish: $b - c = 0$ , that is $b = c$ .

Conversely, suppose cancellation holds and let $ab = 0$ with $a \neq = 0$ . Write the hypothesis as $ab = a \cdot 0$ , which is legitimate because $a \cdot 0 = 0$ in any ring. Cancelling the nonzero $a$ gives $b = 0$ . Hence no zero-divisor exists and $R$ is a domain.∎

The proposition says the two properties are the same property wearing different clothes, and the "only if" direction is the one used daily: it licenses every step of the form "divide both sides by $a$ " in a ring where division is not available. Note carefully what cancellation does not assert. It does not produce an element $a^{- 1}$ ; in $Z$ we may cancel the $3$ in $3 b = 3 c$ without $\frac{1}{3}$ existing anywhere in sight. Cancellation is a statement about the injectivity of a map, not about the existence of an inverse, and keeping those apart is the whole distinction between a domain and a field.

Example 6.4 (Cancellation fails exactly where zero-divisors live). In $Z /6 Z$ , exhibit $b \neq = c$ with $2 b = 2 c$ , and explain which hypothesis of Cancellation characterises domains has failed.

Solution.

Take $b = 1$ and $c = 4$ . Then $2 b = 2$ and $2 c = 8 \equiv 2 (mod 6)$ , so $2 b = 2 c$ while $b \neq = c$ .
The multiplier $2$ is nonzero, so the failure is not in the hypothesis $a \neq = 0$ . What fails is the conclusion, and by the proposition that forces $Z /6 Z$ to have a zero-divisor.
Indeed $b - c = 1 - 4 = - 3 \equiv 3$ , and $2 \cdot 3 = 6 \equiv 0$ : the witnesses are $2$ and $3$ .
The moral is that the counterexample is not an accident of the numbers $1$ and $4$ . Any zero-divisor $a$ with partner $b$ gives $a \cdot b = a \cdot 0$ with $b \neq = 0$ , so cancellation fails at $a$ immediately.

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Example 6.5 (The Gaussian integers form a domain). Let $Z [i] = {a + bi : a, b \in Z} \subseteq C$ . Show that $Z [i]$ is an integral domain, and compute the norm $N (z) = z \overset{z}{ˉ}$ of $3 + 2 i$ .

Solution.

$Z [i]$ is closed under addition and under multiplication, since $(a + bi) (c + d i) = (a c - b d) + (a d + b c) i$ has integer coefficients, and it contains $0$ and $1$ . So it is a subring of $C$ .
$C$ is a field and therefore has no zero-divisors. A subring of a ring without zero-divisors has none either: if $z, w \in Z [i]$ are nonzero, they are nonzero in $C$ , so $z w \neq = 0$ in $C$ , hence in $Z [i]$ .
Since $1 \neq = 0$ in $Z [i]$ , it is a domain.
$N (3 + 2 i) = (3 + 2 i) (3 - 2 i) = 9 + 4 = 13$ . In general $N (a + bi) = a^{2} + b^{2}$ , a non-negative integer, and $N$ is multiplicative: $N (z w) = N (z) N (w)$ , because $\overline{z w} = \overset{z}{ˉ} \overset{w}{ˉ}$ .

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The argument in step 2 is worth extracting, because it disposes of a large supply of examples at once.

Proposition 6.6 (Subrings inherit domainhood). Let $R$ be an integral domain and let $S \subseteq R$ be a subring containing $1$ . Then $S$ is an integral domain.

Proof. $S$ is commutative and has $1 \neq = 0$ because those statements are inherited from $R$ . If $a, b \in S$ satisfy $ab = 0$ , then the same equation holds in $R$ , where the absence of zero-divisors forces $a = 0$ or $b = 0$ .∎

So every subring of $C$ containing $1$ is a domain: $Z$ , $Z [i]$ , $Z [2]$ , $Z [- 5] = {a + b - 5 : a, b \in Z}$ , the rationals, the reals. The last of these will matter later precisely because it is a domain in which factorisation into irreducibles is not unique — $6 = 2 \cdot 3 = (1 + - 5) (1 - - 5)$ — which shows that "domain" is a genuinely weaker condition than the unique-factorisation property studied in Polynomial rings, and that the containments in the figure below are all strict.

The other standard source of domains is polynomials, and there the no-zero-divisor hypothesis is not merely inherited but actively used.

Theorem 6.7 (Degrees add over a domain). Let $D$ be an integral domain and let $f, g \in D [x]$ be nonzero. Then $f g \neq = 0$ and

de g (f g) = de g f + de g g .

Consequently $D [x]$ is an integral domain, and $D [x]^{\times} = D^{\times}$ .

Proof. Write $f = a_{m} x^{m} + \dots + a_{0}$ with $a_{m} \neq = 0$ , so $de g f = m$ , and $g = b_{n} x^{n} + \dots + b_{0}$ with $b_{n} \neq = 0$ , so $de g g = n$ . Multiplying out, the coefficient of $x^{m + n}$ in $f g$ is $a_{m} b_{n}$ , and no higher power of $x$ occurs. Since $D$ is a domain and $a_{m}, b_{n}$ are both nonzero, $a_{m} b_{n} \neq = 0$ . Hence $f g$ has a nonzero coefficient in degree $m + n$ and none above it, which is both $f g \neq = 0$ and the degree formula.

$D [x]$ is commutative with $1 \neq = 0$ , so the first part makes it a domain. For the units, suppose $f g = 1$ . Then $de g f + de g g = de g 1 = 0$ , and since degrees of nonzero polynomials are non-negative, $de g f = de g g = 0$ . So $f$ and $g$ are constants, and $f \in D^{\times}$ . Conversely every unit of $D$ is a unit of $D [x]$ .∎

Both conclusions fail without the hypothesis. Over $Z /4 Z$ , which is not a domain, $(2 x + 1)^{2} = 4 x^{2} + 4 x + 1 = 1$ , so $2 x + 1$ is a unit of degree $1$ ; and $(2 x) (2 x) = 0$ , so $D [x]$ has zero-divisors and the degree formula is meaningless. Every pathology traces back to the single fact that $2 \cdot 2 = 0$ in the coefficient ring.

Figure 6.8. The standard chain of commutative-ring conditions. Each containment is strict:

Z /6 Z

is a ring that is not a domain,

Z [- 5]

a domain that is not a UFD,

Z [x]

a UFD that is not a PID,

Z

a Euclidean domain that is not a field.

Example 6.9 (Which of these rings are domains?). Decide, with reasons, whether each of $Z /15 Z$ , $Z /17 Z$ , $Z \times Z$ , and $M_{2} (R)$ is an integral domain.

Solution.

$Z /15 Z$ : not a domain. $3 \cdot 5 = 15 \equiv 0$ with both factors nonzero.
$Z /17 Z$ : a domain. If $ab \equiv 0 (mod 17)$ then $17 ∣ ab$ , and since $17$ is prime Euclid's lemma gives $17 ∣ a$ or $17 ∣ b$ , that is $a \equiv 0$ or $b \equiv 0$ .
$Z \times Z$ with componentwise operations: not a domain, since $(1, 0) \cdot (0, 1) = (0, 0)$ while neither factor is the zero element $(0, 0)$ . Every nontrivial direct product of rings fails for exactly this reason.
$M_{2} (R)$ : not a domain, and for two independent reasons — it is not commutative, and $(0010)^{2} = 0$ exhibits a nonzero element whose square vanishes.

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Item 2 generalises, and the generalisation is one of the two halves of the central classification of the next section.

Proposition 6.10 (Zero-divisors in $Z / n Z$ ). Let $n \geq 2$ and let $\overset{a}{ˉ} \in Z / n Z$ be nonzero. Then $\overset{a}{ˉ}$ is a unit if $g cd (a, n) = 1$ , and a zero-divisor if $g cd (a, n) > 1$ . Consequently $Z / n Z$ is an integral domain if and only if $n$ is prime, and when $n$ is composite the number of zero-divisors is $n - 1 - φ (n)$ .

Proof. If $g cd (a, n) = 1$ , Bézout's identity supplies integers $u, v$ with $a u + n v = 1$ ; reducing mod $n$ gives $\overset{a}{ˉ} \overset{u}{ˉ} = \overset{ˉ}{1}$ , so $\overset{a}{ˉ}$ is a unit.

If $d = g cd (a, n) > 1$ , put $b = n / d$ . Then $0 < b < n$ , so $\overset{ˉ}{b} \neq = 0$ , and $ab = a (n / d) = n (a / d)$ is a multiple of $n$ because $d ∣ a$ . Hence $\overset{a}{ˉ} \overset{ˉ}{b} = 0$ with both factors nonzero, so $\overset{a}{ˉ}$ is a zero-divisor.

Every nonzero class therefore falls into exactly one of the two cases, since the two conditions on $g cd (a, n)$ are exclusive and exhaustive, and a unit is never a zero-divisor by A unit is never a zero-divisor below. If $n$ is composite, write $n = d e$ with $1 < d, e < n$ ; then $\overset{ˉ}{d}$ is a zero-divisor and $Z / n Z$ is not a domain. If $n$ is prime, every nonzero class is coprime to $n$ , hence a unit, hence not a zero-divisor. Finally there are $n - 1$ nonzero classes of which $φ (n)$ are units, leaving $n - 1 - φ (n)$ zero-divisors.∎

Example 6.11 (Counting zero-divisors in $Z /36 Z$ ). How many zero-divisors does $Z /36 Z$ have, and which classes are they?

Solution.

$36 = 2^{2} \cdot 3^{2}$ , so $φ (36) = 36 (1 - \frac{1}{2}) (1 - \frac{1}{3}) = 36 \cdot \frac{1}{2} \cdot \frac{2}{3} = 12$ .
By Zero-divisors in $Z / n Z$ the count is $36 - 1 - 12 = 23$ .
Concretely these are the nonzero classes sharing a factor with $36$ : every nonzero even class and every nonzero multiple of $3$ . There are $17$ nonzero even classes and $11$ nonzero multiples of $3$ , overlapping in the $5$ nonzero multiples of $6$ , giving $17 + 11 - 5 = 23$ . ✓
The units are the $12$ classes $1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35$ .

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Exercises for §6.1 →

6.2Fields

Cancellation lets us undo multiplication by a nonzero element in the sense of recovering $b$ from $ab$ . A field lets us undo it in the stronger sense of performing the division.

Definition 6.12 (Unit). An element $u$ of a ring $R$ is a unit if there is $v \in R$ with $uv = 1$ . The units form an abelian group $R^{\times}$ under multiplication, and the inverse $v = u^{- 1}$ is unique.

Definition 6.13 (Field). A field is a ring $F$ with $1 \neq = 0$ in which every nonzero element is a unit; equivalently, $F^{\times} = F ∖ {0}$ , so that the nonzero elements form an abelian group under multiplication.

$Q$ , $R$ , $C$ are fields; so is $Q (i) = {p + q i : p, q \in Q}$ , and so is the field of rational functions $k (x)$ . Not a field: $Z$ , since $2^{- 1} \in / Z$ ; nor $k [x]$ , since $x^{- 1}$ is not a polynomial; nor $Z /6 Z$ .

Proposition 6.14 (A unit is never a zero-divisor). Let $R$ be a ring and $u \in R^{\times}$ . Then $u$ is not a zero-divisor.

Proof. Suppose $u b = 0$ . Multiplying on the left by $u^{- 1}$ gives $b = 1 \cdot b = (u^{- 1} u) b = u^{- 1} (u b) = u^{- 1} \cdot 0 = 0$ . So no nonzero $b$ annihilates $u$ .∎

Theorem 6.15 (Every field is an integral domain). Every field is an integral domain.

Proof. Let $F$ be a field. It is a commutative ring with $1 \neq = 0$ by definition. If $ab = 0$ and $a \neq = 0$ , then $a$ is a unit, so by A unit is never a zero-divisor we get $b = 0$ . Hence $F$ has no zero-divisors.∎

Pitfall. The converse is false, and $Z$ is the counterexample to remember. $Z$ has no zero-divisors whatever, so cancellation holds throughout; yet $Z^{\times} = {1, - 1}$ , so every integer other than $\pm 1$ fails to be invertible. "Can cancel" and "can divide" are different statements, and only one of them follows from the other.

The distance between the two notions is measured by the ideal structure, which is the point of view taken in Ideals and quotient rings: a field has only the two trivial ideals, whereas $Z$ has one ideal $(n)$ for every $n \geq 0$ .

Proposition 6.16 (Fields have no proper nonzero ideals). A ring $R$ with $1 \neq = 0$ is a field if and only if its only ideals are ${0}$ and $R$ .

Proof. Let $R$ be a field and $I \neq = {0}$ an ideal. Choose $a \in I$ with $a \neq = 0$ . Then $a^{- 1}$ exists and $1 = a^{- 1} a \in I$ because ideals absorb multiplication by ring elements. Hence for every $r \in R$ we get $r = r \cdot 1 \in I$ , so $I = R$ .

Conversely suppose the only ideals are ${0}$ and $R$ , and let $a \neq = 0$ . The principal ideal $(a) = {r a : r \in R}$ contains $a \neq = 0$ , so it is not ${0}$ , and therefore $(a) = R$ . In particular $1 \in (a)$ , say $1 = r a$ , which exhibits $r$ as an inverse of $a$ .∎

Example 6.17 (Inverting an element of $F_{13}$ ). Find $7^{- 1}$ in $Z /13 Z$ , both by search and by the Euclidean algorithm.

Solution.

By search: $7 \cdot 2 = 14 \equiv 1 (mod 13)$ . So $7^{- 1} = 2$ .
By the Euclidean algorithm, which is the method that scales: $13 = 1 \cdot 7 + 6$ , then $7 = 1 \cdot 6 + 1$ . Back-substituting, $1 = 7 - 6 = 7 - (13 - 7) = 2 \cdot 7 - 13$ .
Reducing mod $13$ gives $2 \cdot 7 \equiv 1$ , confirming $7^{- 1} = 2$ .
The computation used only $g cd (7, 13) = 1$ , which is exactly the hypothesis of Zero-divisors in $Z / n Z$ ; the same run of the algorithm on $g cd (6, 15) = 3$ would fail to reach $1$ and instead exhibit $6$ as a zero-divisor mod $15$ .

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Example 6.18 (The Gaussian integers are not a field). Determine $Z [i]^{\times}$ , and deduce that $Z [i]$ is a domain that is not a field.

Solution.

If $z w = 1$ then applying the multiplicative norm $N (a + bi) = a^{2} + b^{2}$ gives $N (z) N (w) = N (1) = 1$ . Both factors are non-negative integers, so $N (z) = 1$ .
$a^{2} + b^{2} = 1$ with $a, b \in Z$ forces $(a, b) \in {(\pm 1, 0), (0, \pm 1)}$ , so $z \in {1, - 1, i, - i}$ .
Each of these four is indeed a unit: $i \cdot (- i) = 1$ and $(- 1)^{2} = 1$ . Hence $Z [i]^{\times} = {\pm 1, \pm i}$ , a cyclic group of order $4$ .
Since $N (1 + i) = 2 \neq = 1$ , the element $1 + i$ is nonzero and not a unit, so $Z [i]$ is not a field — though by The Gaussian integers form a domain it is a domain.

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Finite domains

Everything so far has kept fields strictly stronger than domains. Finiteness collapses the distinction.

Theorem 6.19 (Every finite integral domain is a field). Let $D$ be a finite integral domain. Then $D$ is a field.

Proof. Let $a \in D$ with $a \neq = 0$ ; we must produce an inverse. Consider the multiplication map

μ_{a} : D \to D, μ_{a} (x) = a x .

$μ_{a}$ is injective: if $μ_{a} (x) = μ_{a} (y)$ , that is $a x = a y$ , then cancelling the nonzero $a$ using Cancellation characterises domains gives $x = y$ .

Now $μ_{a}$ is an injective map from the finite set $D$ to itself, so by the pigeonhole principle it is surjective: an injection cannot miss a point of a finite set of the same size without two points colliding elsewhere. Formally, $∣ μ_{a} (D) ∣ = ∣ D ∣$ by injectivity and $μ_{a} (D) \subseteq D$ , so $μ_{a} (D) = D$ .

In particular $1$ is in the image, so there is $b \in D$ with $ab = 1$ . Thus $a$ is a unit. Since $a \neq = 0$ was arbitrary and $1 \neq = 0$ holds in a domain, $D$ is a field.∎

Three remarks on that proof, because it is short enough that its content can slide past.

First, finiteness is used exactly once, and it is indispensable: $Z$ is an infinite domain in which $μ_{2}$ is injective and very far from surjective, missing every odd number. Second, no inverse is ever constructed. The argument is purely a counting argument; it tells you $a^{- 1}$ exists and tells you nothing about how to find it, and indeed finding inverses in a large finite field is a computation (the Euclidean algorithm, or $a^{- 1} = a^{q - 2}$ in a field of order $q$ ) that the theorem does not perform. Third, the same pigeonhole idea with $μ_{a}$ replaced by $x \mapsto a x$ on the nonzero elements shows that $D ∖ {0}$ is a finite cancellative commutative monoid, hence a group — which is the same theorem phrased in the language of Groups and their axioms.

Corollary 6.20 (Classification of the rings $Z / n Z$ ). For $n \geq 2$ , the ring $Z / n Z$ is a field if and only if $n$ is prime. In that case we write it $F_{p}$ .

Proof. If $n$ is prime then $Z / n Z$ is a domain by Zero-divisors in $Z / n Z$ , and it is finite, so Every finite integral domain is a field makes it a field.

If $n$ is composite then $Z / n Z$ has a zero-divisor, again by Zero-divisors in $Z / n Z$ , and a field has none by Every field is an integral domain.∎

It is instructive that the forward direction can also be run without the pigeonhole, by Bézout: $g cd (a, p) = 1$ for every $a \neq \equiv 0$ , so $a u + p v = 1$ gives the inverse explicitly. The two proofs are genuinely different — one is effective and special to $Z$ , the other is ineffective and applies to every finite domain, including the ones we build by hand in the next section.

Example 6.21 (The smallest field above a bound). What is the smallest $n > 100$ for which $Z / n Z$ is a field?

Solution.

By Classification of the rings $Z / n Z$ we need the least prime exceeding $100$ .
$101$ is not divisible by $2, 3, 5$ or $7$ , and $1 1^{2} = 121 > 101$ , so trial division by primes up to $101 < 11$ suffices and $101$ is prime.
Final answer: $n = 101$ .
Note that $121 = 1 1^{2}$ is not an answer to a similar-looking question: $Z /121 Z$ is not a field, since $11 \cdot 11 = 0$ there. There is a field of order $121$ , but it is not $Z /121 Z$ — see A finite field has prime-power order.

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Exercises for §6.2 →

6.3Characteristic and the prime subfield

Every ring receives a canonical map from $Z$ : there is exactly one ring homomorphism $χ : Z \to R$ , forced by $χ (1) = 1$ and additivity to satisfy $χ (n) = n \cdot 1 = 1 + \dots + 1$ . What that map does is the first thing to know about $R$ .

Definition 6.22 (Characteristic). The characteristic $char R$ of a ring $R$ is the least positive integer $n$ with $n \cdot 1 = 0$ , if such an $n$ exists, and $0$ otherwise. Equivalently, $ker χ = (char R) \subseteq Z$ .

The equivalence is worth unwinding, since it is what makes the definition natural rather than arbitrary. $ker χ$ is an ideal of $Z$ , and every ideal of $Z$ is $(m)$ for a unique $m \geq 0$ ; the characteristic is that $m$ . Characteristic $0$ means $χ$ is injective, so $R$ contains an isomorphic copy of $Z$ .

$char Z = char Q = char R = char C = 0$ , while $char (Z / n Z) = n$ for every $n \geq 1$ — including composite $n$ , so $char (Z /12 Z) = 12$ .

Proposition 6.23 (The characteristic annihilates everything). If $char R = n$ , then $n r = 0$ for every $r \in R$ . Moreover $m r = 0$ for all $r$ if and only if $n ∣ m$ .

Proof. Write $n r = r + \dots + r = (1 + \dots + 1) r = (n \cdot 1) r = 0 \cdot r = 0$ , using distributivity for the middle step.

For the second claim, if $n ∣ m$ write $m = nk$ ; then $m r = k (n r) = 0$ . Conversely if $m r = 0$ for all $r$ , take $r = 1$ to get $m \cdot 1 = 0$ , so $m \in ker χ = (n)$ , that is $n ∣ m$ .∎

So the additive group of a ring of characteristic $n > 0$ has exponent dividing $n$ : every element is killed by $n$ , and $n$ is the least positive integer that kills them all. In a domain the possible values of $n$ are severely restricted.

Theorem 6.24 (The characteristic of a domain is $0$ or prime). Let $D$ be an integral domain. Then $char D$ is either $0$ or a prime number.

Proof. Suppose $n = char D > 0$ . First, $n \neq = 1$ , since $1 \cdot 1 = 1 \neq = 0$ in a domain.

Suppose $n$ were composite, say $n = ab$ with $1 < a, b < n$ . In $D$ we then have

(a \cdot 1) (b \cdot 1) = (ab) \cdot 1 = n \cdot 1 = 0,

where the first equality is the distributive expansion of a product of two repeated sums. Since $D$ is a domain, $a \cdot 1 = 0$ or $b \cdot 1 = 0$ . Either conclusion contradicts the minimality of $n$ , because $0 < a < n$ and $0 < b < n$ .

Hence $n$ has no factorisation with both factors strictly between $1$ and $n$ , and $n \neq = 1$ , so $n$ is prime.∎

Pitfall. The theorem is about domains, not about finite rings. $Z /12 Z$ is finite with characteristic $12$ ; the hypothesis it fails is having no zero-divisors, and indeed $3 \cdot 4 = 0$ . Conversely the theorem allows characteristic $0$ for a domain of any size, and $Z$ takes that option.

Example 6.25 (Characteristic of a product and of a polynomial ring). Compute $char (Z /4 Z \times Z /6 Z)$ and $char (F_{5} [x])$ .

Solution.

The identity of the product is $(1, 1)$ , so $n \cdot (1, 1) = (n \cdot 1, n \cdot 1) = (\overset{n}{ˉ}, \overset{n}{ˉ})$ , which is zero exactly when $4 ∣ n$ and $6 ∣ n$ .
The least such positive $n$ is $lcm (4, 6) = 12$ , so the characteristic is $12$ . In general $char (R \times S) = lcm (char R, char S)$ , with the convention that an lcm involving $0$ is $0$ .
For $F_{5} [x]$ , the identity is the constant polynomial $1$ , and $n \cdot 1$ is the constant polynomial $\overset{n}{ˉ} \in F_{5}$ , zero exactly when $5 ∣ n$ . So $char (F_{5} [x]) = 5$ .
Consistency check: $F_{5} [x]$ is a domain by Degrees add over a domain, and $5$ is prime, as The characteristic of a domain is $0$ or prime demands. The product ring is not a domain, and $12$ is not prime. ✓

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Characteristic tells us not merely a number but which field sits inside, at the bottom of everything.

Definition 6.26 (Prime subfield). The prime subfield of a field $F$ is the intersection of all subfields of $F$ ; equivalently, the smallest subfield of $F$ .

The intersection of any family of subfields is a subfield — it contains $0$ and $1$ and is closed under the four operations because each member is — so the definition is legitimate, and the intersection is contained in every subfield by construction.

Theorem 6.27 (Identification of the prime subfield). Let $F$ be a field.

If $char F = p > 0$ , the prime subfield of $F$ is isomorphic to $F_{p}$ .
If $char F = 0$ , the prime subfield of $F$ is isomorphic to $Q$ .

Proof. Let $χ : Z \to F$ be the canonical homomorphism, with image $χ (Z) = {n \cdot 1 : n \in Z}$ . By the first isomorphism theorem of Ring homomorphisms and ideals, $χ (Z) ≅ Z / ker χ = Z / (char F)$ .

(1) If $char F = p$ , then $χ (Z) ≅ Z / p Z = F_{p}$ , which is a field by Classification of the rings $Z / n Z$ . So $χ (Z)$ is a subfield. Any subfield $K \subseteq F$ contains $1$ and is closed under addition and negation, hence contains every $n \cdot 1$ , hence contains $χ (Z)$ . Therefore $χ (Z)$ is the smallest subfield, that is the prime subfield, and it is isomorphic to $F_{p}$ .

(2) If $char F = 0$ , then $χ$ is injective and $χ (Z) ≅ Z$ , which is not a field. But $F$ is a field, so every nonzero $χ (n)$ is invertible in $F$ , and the set

K = {χ (m) χ (n)^{- 1} : m, n \in Z, n \neq = 0}

is closed under the four operations and is therefore a subfield. The map $m / n \mapsto χ (m) χ (n)^{- 1}$ is well defined — if $m / n = m^{'} / n^{'}$ then $m n^{'} = m^{'} n$ , so $χ (m) χ (n^{'}) = χ (m^{'}) χ (n)$ and the two products agree — and it is a ring homomorphism $Q \to F$ , injective because $Q$ is a field and its kernel is a proper ideal, hence ${0}$ by Fields have no proper nonzero ideals. Its image is $K$ , so $K ≅ Q$ . As in (1), any subfield contains every $χ (n)$ and, being closed under inverses, every $χ (m) χ (n)^{- 1}$ ; so $K$ is the prime subfield.∎

The construction in part (2) is the field of fractions of $Z$ , built inside $F$ , and it will reappear in full generality in the last section. The upshot is a clean dichotomy: every field contains $Q$ or contains some $F_{p}$ , and never both. In particular no field of characteristic $p$ contains $Q$ , so a field containing $Q$ is automatically infinite, and every finite field has prime characteristic.

Theorem 6.28 (A finite field has prime-power order). Let $F$ be a finite field. Then $char F = p$ for some prime $p$ , and $∣ F ∣ = p^{n}$ where $n = [F : F_{p}]$ is the dimension of $F$ as a vector space over its prime subfield. In particular no field has $6$ , $10$ or $12$ elements.

Proof. $F$ is a field, hence a domain, so $char F$ is $0$ or prime by The characteristic of a domain is $0$ or prime. It cannot be $0$ : characteristic $0$ would make $χ : Z \to F$ injective and $F$ infinite. So $char F = p$ , and by Identification of the prime subfield the prime subfield is a copy of $F_{p}$ .

$F$ is a vector space over $F_{p}$ : the field axioms of $F$ give exactly the vector-space axioms, with scalar multiplication the restriction of the multiplication of $F$ . Since $F$ is finite it is finite-dimensional; let $n$ be its dimension and $e_{1}, \dots, e_{n}$ a basis. Every element of $F$ is uniquely $λ_{1} e_{1} + \dots + λ_{n} e_{n}$ with $λ_{i} \in F_{p}$ , and there are $p$ independent choices for each of the $n$ coordinates. Hence $∣ F ∣ = p^{n}$ .

Since $6$ , $10$ and $12$ are not prime powers, no field has that many elements.∎

The converse — that for every prime power $q = p^{n}$ there is a field of order $q$ , unique up to isomorphism — is true and is proved in Galois theory using splitting fields of $x^{q} - x$ . Here we shall build the smallest non-prime case by hand.

The Frobenius endomorphism

In characteristic $p$ the binomial theorem degenerates in a way that is enormously useful.

Theorem 6.29 (Frobenius endomorphism). Let $R$ be a commutative ring with $char R = p$ prime. Then the map

φ : R \to R, φ (x) = x^{p}

is a ring homomorphism: $φ (1) = 1$ , $φ (x y) = φ (x) φ (y)$ , and

(x + y)^{p} = x^{p} + y^{p} .

Proof. $φ (1) = 1^{p} = 1$ , and $φ (x y) = (x y)^{p} = x^{p} y^{p} = φ (x) φ (y)$ by commutativity.

For additivity, the binomial theorem — valid in any commutative ring — gives

(x + y)^{p} = k = 0 \sum p (k p) x^{k} y^{p - k} .

Fix $0 < k < p$ . Then

(k p) = \frac{p !}{k ! ( p - k )!},

so $k! (p - k)! (k p) = p!$ . The prime $p$ divides the right-hand side. It divides none of the factors of $k!$ or of $(p - k)!$ , since all of those are positive integers strictly less than $p$ and $p$ is prime. By Euclid's lemma, therefore, $p ∣ (k p)$ .

By The characteristic annihilates everything, every term with $0 < k < p$ has coefficient a multiple of $p$ and so vanishes in $R$ . Only $k = 0$ and $k = p$ survive, leaving $(x + y)^{p} = x^{p} + y^{p}$ .∎

Pitfall. $(x + y)^{p} = x^{p} + y^{p}$ is the "freshman's dream", and it is a genuine theorem in characteristic $p$ and a genuine error everywhere else. The proof uses the prime hypothesis twice over: Euclid's lemma needs $p$ prime, and the conclusion needs the characteristic to be that same $p$ . In $Z /4 Z$ , which has characteristic $4$ , $(1 + 1)^{4} = 16 \equiv 0$ but $1^{4} + 1^{4} = 2 \neq = 0$ .

Corollary 6.30 (Frobenius is an automorphism of a finite field). If $F$ is a finite field of characteristic $p$ , then $φ (x) = x^{p}$ is an automorphism of $F$ , and $φ$ fixes $F_{p}$ pointwise. If $∣ F ∣ = p^{n}$ then $φ^{n} = id$ .

Proof. $φ$ is a ring homomorphism by Frobenius endomorphism. Its kernel is an ideal of $F$ not containing $1$ , hence ${0}$ by Fields have no proper nonzero ideals; so $φ$ is injective. Being an injective map from the finite set $F$ to itself it is surjective, by the same pigeonhole step used in Every finite integral domain is a field. Hence $φ$ is a bijective ring homomorphism, that is an automorphism.

For $a \in F_{p}$ , Fermat's little theorem gives $a^{p} = a$ , so $φ$ fixes the prime subfield. Finally, $F^{\times}$ is a group of order $p^{n} - 1$ , so $x^{p^{n} - 1} = 1$ for every $x \neq = 0$ by Lagrange's theorem, whence $x^{p^{n}} = x$ for every $x \in F$ including $0$ . Since $φ^{n} (x) = x^{p^{n}}$ , we get $φ^{n} = id$ .∎

The group generated by $φ$ is therefore cyclic of order dividing $n$ — in fact exactly $n$ — and it is the Galois group of $F$ over $F_{p}$ . That identification is the entry point to Galois theory; here we note only the computational payoff, that raising to the $p$ -th power in characteristic $p$ is a linear operation.

Example 6.31 (Frobenius in $F_{7}$ and in $F_{9}$ ). Compute $φ (x) = x^{7}$ for every $x \in F_{7}$ . Then, in a field $F$ of order $9$ , explain why $φ (x) = x^{3}$ is not the identity.

Solution.

In $F_{7}$ , Fermat's little theorem gives $x^{7} = x$ for all $x$ , so $φ = id$ . This is Frobenius is an automorphism of a finite field with $n = 1$ : the whole field is its own prime subfield, and Frobenius fixes the prime subfield.
In $F$ of order $9 = 3^{2}$ we have $n = 2$ , so $φ^{2} = id$ but $φ$ itself need not be. The fixed points of $φ$ are the solutions of $x^{3} = x$ , that is $x (x - 1) (x + 1) = 0$ .
A polynomial of degree $3$ over a field has at most $3$ roots, so $φ$ fixes at most $3$ of the $9$ elements — precisely the prime subfield $F_{3}$ .
Hence $φ \neq = id$ on $F$ , and $φ$ has order exactly $2$ . The general statement is that the fixed field of $φ^{d}$ inside $F_{p^{n}}$ is $F_{p^{g c d (d, n)}}$ .

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Building $F_{4}$ by hand

$Z /4 Z$ is not a field; a field of order $4$ exists nonetheless, and constructing it shows exactly how the two differ.

Example 6.32 (The field with four elements). Construct a field $F_{4}$ with four elements and write out its addition and multiplication tables.

Solution.

By A finite field has prime-power order, a field of order $4 = 2^{2}$ has characteristic $2$ and is a $2$ -dimensional vector space over $F_{2}$ . So its elements are ${0, 1, a, a + 1}$ for a basis ${1, a}$ , and the addition table is forced: it is the Klein four-group, with $x + x = 0$ for every $x$ .
Only the value of $a^{2}$ remains to be chosen. It must lie in the field, and the possibilities are $0, 1, a, a + 1$ . If $a^{2} = 0$ then $a$ is a zero-divisor, impossible. If $a^{2} = 1$ then $(a + 1)^{2} = a^{2} + 1 = 0$ by Frobenius endomorphism in characteristic $2$ , again a zero-divisor. If $a^{2} = a$ then $a (a - 1) = 0$ with $a \neq = 0, 1$ , impossible.
So $a^{2} = a + 1$ , which is to say $a$ is a root of $x^{2} + x + 1$ , the unique irreducible quadratic over $F_{2}$ . Concretely, $F_{4} = F_{2} [x] / (x^{2} + x + 1)$ with $a = \overset{x}{ˉ}$ .
Multiplication now follows: $a \cdot a = a + 1$ ; $a (a + 1) = a^{2} + a = (a + 1) + a = 1$ ; and $(a + 1)^{2} = a^{2} + 1 = (a + 1) + 1 = a$ .
In particular $a^{- 1} = a + 1$ , every nonzero element is invertible, and $F_{4}^{\times} = {1, a, a + 1}$ is cyclic of order $3$ generated by $a$ , since $a^{1} = a$ , $a^{2} = a + 1$ , $a^{3} = 1$ .

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Figure 6.33. The multiplication table of

F_{4} = {0, 1, a, a + 1}

, where

a^{2} = a + 1

. Every nonzero row is a permutation of the nonzero elements — the visible signature of *Every finite integral domain is a field*.

Example 6.34 ( $F_{4}$ is not $Z /4 Z$ ). Show that $F_{4}$ and $Z /4 Z$ are not isomorphic as rings, in two ways.

Solution.

By characteristic: $char F_{4} = 2$ , since $1 + 1 = 0$ there, whereas $char (Z /4 Z) = 4$ . Characteristic is preserved by ring isomorphism, because an isomorphism sends $1$ to $1$ and is additive.
By structure: $F_{4}$ is a field, while $Z /4 Z$ has the zero-divisor $2$ .
Additively they already differ: $(F_{4}, +)$ is the Klein four-group $C_{2} \times C_{2}$ , and $(Z /4 Z, +)$ is cyclic of order $4$ .
The lesson is that "the ring of order $4$ " is not a well-defined phrase, and that $Z / n Z$ is only one of several rings of order $n$ .

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Exercises for §6.3 →

6.4The field of fractions

$Z$ is a domain and not a field, but it sits inside $Q$ , which is a field; and $Q$ is not merely some field containing $Z$ but the smallest one. This section carries out that construction for an arbitrary integral domain, and the point of doing it in full is to see where the hypothesis is spent.

Fix an integral domain $D$ and set

S = D \times (D ∖ {0}) = {(a, b) : a, b \in D, b \neq = 0} .

Think of $(a, b)$ as the formal fraction $a / b$ . Distinct pairs must be identified, since $1/2$ and $2/4$ are to be the same, and the identification is cross-multiplication.

Definition 6.35 (The fraction relation). On $S$ define $(a, b) \sim (c, d)$ if and only if $a d = b c$ .

Lemma 6.36 ( $\sim$ is an equivalence relation). The relation $\sim$ on $S$ is reflexive, symmetric and transitive.

Proof. Reflexive. $ab = ba$ by commutativity, so $(a, b) \sim (a, b)$ .

Symmetric. If $a d = b c$ then $c b = d a$ , so $(c, d) \sim (a, b)$ .

Transitive. Suppose $(a, b) \sim (c, d)$ and $(c, d) \sim (e, f)$ , that is $a d = b c$ and $c f = d e$ . Multiply the first equation by $f$ and the second by $b$ :

a df = b c f, b c f = b d e .

Hence $a df = b d e$ , that is $d (a f - b e) = 0$ . Now $d \neq = 0$ by the definition of $S$ , and $D$ is a domain, so $a f - b e = 0$ , that is $a f = b e$ and $(a, b) \sim (e, f)$ .∎

Transitivity is the step that needs $D$ to be a domain, and it is the only place in the whole construction where cancellation is unavoidable. Over $Z /6 Z$ the relation is not transitive: $(1, 1) \sim (3, 3)$ since $1 \cdot 3 = 1 \cdot 3$ , and $(3, 3) \sim (2, 4)$ since $3 \cdot 4 = 12 \equiv 0$ and $3 \cdot 2 = 6 \equiv 0$ ; but $(1, 1) \neq \sim (2, 4)$ , since $1 \cdot 4 = 4 \neq = 2 = 1 \cdot 2$ . A relation that is not transitive has no quotient, so the construction stops before it begins.

Write $\frac{a}{b}$ for the equivalence class of $(a, b)$ , and let $Frac (D) = S / \sim$ be the set of classes.

Definition 6.37 (Operations on fractions). On $Frac (D)$ define

\frac{a}{b} + \frac{c}{d} = \frac{a d + b c}{b d}, \frac{a}{b} \cdot \frac{c}{d} = \frac{a c}{b d} .

Lemma 6.38 (The operations are well defined). The formulas above do not depend on the chosen representatives, and the results lie in $S$ .

Proof. First, the denominators. If $b \neq = 0$ and $d \neq = 0$ then $b d \neq = 0$ , because $D$ is a domain. So both formulas produce elements of $S$ ; this is the second place the hypothesis is spent.

Now suppose $(a, b) \sim (a^{'}, b^{'})$ , so $a b^{'} = a^{'} b$ . It suffices to check independence in the first argument, since the formulas are symmetric in the two fractions.

Multiplication. We must show $(a c, b d) \sim (a^{'} c, b^{'} d)$ , that is $a c b^{'} d = b d a^{'} c$ . Rearranging by commutativity, the left side is $(a b^{'}) (c d)$ and the right side is $(a^{'} b) (c d)$ ; these agree because $a b^{'} = a^{'} b$ .

Addition. We must show $(a d + b c, b d) \sim (a^{'} d + b^{'} c, b^{'} d)$ , that is

(a d + b c) b^{'} d = b d (a^{'} d + b^{'} c) .

Expanding, the left side is $a b^{'} d^{2} + b b^{'} c d$ and the right side is $a^{'} b d^{2} + b b^{'} c d$ . The second terms agree, and the first agree because $a b^{'} = a^{'} b$ .∎

Theorem 6.39 (The field of fractions). Let $D$ be an integral domain. Then $Frac (D)$ , with the operations above, is a field with zero $\frac{0}{1}$ and identity $\frac{1}{1}$ , and

ι : D \to Frac (D), ι (a) = \frac{a}{1}

is an injective ring homomorphism. Every element of $Frac (D)$ has the form $ι (a) ι (b)^{- 1}$ with $b \neq = 0$ .

Proof. The ring axioms are routine verifications from the formulas; we check the ones that are not pure symbol-pushing.

Zero and identity. $\frac{a}{b} + \frac{0}{1} = \frac{a \cdot 1 + b \cdot 0}{b} = \frac{a}{b}$ , and $\frac{a}{b} \cdot \frac{1}{1} = \frac{a}{b}$ . Note $\frac{0}{1} \neq = \frac{1}{1}$ , since $0 \cdot 1 = 0 \neq = 1 = 1 \cdot 1$ in $D$ ; so the ring is not the zero ring.

Negatives. $\frac{a}{b} + \frac{- a}{b} = \frac{ab - ab}{b ^{2}} = \frac{0}{b ^{2}} = \frac{0}{1}$ , the last equality because $0 \cdot 1 = b^{2} \cdot 0$ .

Inverses. Suppose $\frac{a}{b} \neq = \frac{0}{1}$ , which says $a \cdot 1 \neq = b \cdot 0$ , that is $a \neq = 0$ . Then $(b, a) \in S$ , and

\frac{a}{b} \cdot \frac{b}{a} = \frac{ab}{ba} = \frac{1}{1}

since $ab \cdot 1 = ba \cdot 1$ . So every nonzero element is invertible and $Frac (D)$ is a field.

The embedding. $ι (a + a^{'}) = \frac{a + a ^{'}}{1} = \frac{a}{1} + \frac{a ^{'}}{1}$ because $\frac{a}{1} + \frac{a ^{'}}{1} = \frac{a \cdot 1 + 1 \cdot a ^{'}}{1}$ , and $ι (a a^{'}) = \frac{a a ^{'}}{1} = \frac{a}{1} \cdot \frac{a ^{'}}{1}$ , and $ι (1) = \frac{1}{1}$ . Injectivity: $ι (a) = ι (a^{'})$ means $a \cdot 1 = 1 \cdot a^{'}$ , that is $a = a^{'}$ .

The last claim. For $b \neq = 0$ , $ι (b)^{- 1} = \frac{1}{b}$ by the inverse computation, so $ι (a) ι (b)^{- 1} = \frac{a}{1} \cdot \frac{1}{b} = \frac{a}{b}$ .∎

We may therefore regard $D$ as a subring of $Frac (D)$ and drop $ι$ from the notation, writing $a$ for $\frac{a}{1}$ . The precise sense in which $Frac (D)$ is the smallest such field is the following, which is the property that pins it down up to unique isomorphism.

Theorem 6.40 (Universal property of the field of fractions). Let $D$ be an integral domain and $K$ a field. For every injective ring homomorphism $f : D \to K$ there is a unique ring homomorphism $\tilde{f} : Frac (D) \to K$ with $\tilde{f} \circ ι = f$ , namely

\tilde{f} (\frac{a}{b}) = f (a) f (b)^{- 1},

and $\tilde{f}$ is injective.

Proof. Existence. Since $f$ is injective and $b \neq = 0$ , we have $f (b) \neq = 0$ , so $f (b)^{- 1}$ exists in the field $K$ and the formula makes sense. It is independent of representatives: if $(a, b) \sim (a^{'}, b^{'})$ then $a b^{'} = a^{'} b$ , so $f (a) f (b^{'}) = f (a^{'}) f (b)$ , and multiplying by $f (b)^{- 1} f (b^{'})^{- 1}$ gives $f (a) f (b)^{- 1} = f (a^{'}) f (b^{'})^{- 1}$ .

That $\tilde{f}$ is a ring homomorphism is a direct computation:

\tilde{f} (\frac{a}{b} \cdot \frac{c}{d}) = f (a c) f (b d)^{- 1} = f (a) f (c) f (b)^{- 1} f (d)^{- 1} = \tilde{f} (\frac{a}{b}) \tilde{f} (\frac{c}{d}),

and similarly $\tilde{f} (\frac{a d + b c}{b d}) = (f (a) f (d) + f (b) f (c)) f (b)^{- 1} f (d)^{- 1} = f (a) f (b)^{- 1} + f (c) f (d)^{- 1}$ . Also $\tilde{f} (\frac{1}{1}) = f (1) f (1)^{- 1} = 1$ , and $\tilde{f} (ι (a)) = f (a) f (1)^{- 1} = f (a)$ .

Uniqueness. Let $g$ be any homomorphism with $g \circ ι = f$ . For $b \neq = 0$ , applying $g$ to $ι (b) \cdot \frac{1}{b} = \frac{1}{1}$ gives $f (b) g (\frac{1}{b}) = 1$ , so $g (\frac{1}{b}) = f (b)^{- 1}$ is forced. Hence $g (\frac{a}{b}) = g (ι (a)) g (\frac{1}{b}) = f (a) f (b)^{- 1} = \tilde{f} (\frac{a}{b})$ .

Injectivity. $ker \tilde{f}$ is an ideal of the field $Frac (D)$ not containing $1$ , hence ${0}$ by Fields have no proper nonzero ideals.∎

Corollary 6.41 ( $Frac (D)$ is the smallest field containing $D$ ). If $K$ is a field and $D \subseteq K$ is a subring, then $K$ contains a subfield isomorphic to $Frac (D)$ , namely ${a b^{- 1} : a, b \in D, b \neq = 0}$ . Moreover $Frac (D)$ is determined up to a unique isomorphism fixing $D$ .

Proof. Apply Universal property of the field of fractions to the inclusion $f : D ↪ K$ , which is an injective homomorphism. The induced $\tilde{f}$ is injective, so $Frac (D) ≅ \tilde{f} (Frac (D)) = {a b^{- 1}} \subseteq K$ .

For uniqueness, suppose $L$ is a field with an injection $j : D \to L$ having the same universal property. The two properties give homomorphisms $u : Frac (D) \to L$ and $v : L \to Frac (D)$ compatible with $ι$ and $j$ . Then $v \circ u$ and the identity are both homomorphisms $Frac (D) \to Frac (D)$ compatible with $ι$ , so by the uniqueness clause they agree; symmetrically $u \circ v = id_{L}$ . Hence $u$ is an isomorphism, and it is the unique one fixing $D$ .∎

Example 6.42 ( $Frac (Z) = Q$ ). Verify that the construction applied to $D = Z$ reproduces $Q$ .

Solution.

$S = {(a, b) : a, b \in Z, b \neq = 0}$ and $(a, b) \sim (c, d)$ iff $a d = b c$ — which is exactly the schoolroom test for two fractions being equal.
The operations are the schoolroom rules for adding and multiplying fractions, and The operations are well defined is the proof, usually skipped, that those rules do not depend on whether one writes $\frac{1}{2}$ or $\frac{2}{4}$ .
$Q$ is a field containing $Z$ , so by $Frac (D)$ is the smallest field containing $D$ it contains a copy of $Frac (Z)$ , namely ${a b^{- 1}} = Q$ itself. Hence $Frac (Z) ≅ Q$ .
The same corollary, applied inside an arbitrary field of characteristic $0$ , is precisely part (2) of Identification of the prime subfield: the prime subfield is $Frac (Z)$ built inside $F$ .

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Example 6.43 (The fraction field of the Gaussian integers). Identify $Frac (Z [i])$ concretely.

Solution.

$Z [i]$ is a subring of the field $C$ , so by $Frac (D)$ is the smallest field containing $D$ its fraction field is ${z w^{- 1} : z, w \in Z [i], w \neq = 0} \subseteq C$ .
Rationalise: for $w = c + d i \neq = 0$ ,
$\frac{a + bi}{c + d i} = \frac{( a + bi ) ( c - d i )}{c ^{2} + d ^{2}} = \frac{a c + b d}{c ^{2} + d ^{2}} + \frac{b c - a d}{c ^{2} + d ^{2}} i,$
and $c^{2} + d^{2} = N (w)$ is a positive integer, so the real and imaginary parts are rational.
Conversely every $p + q i$ with $p, q \in Q$ is such a quotient: write $p = a / n$ , $q = b / n$ over a common denominator $n$ , so $p + q i = (a + bi) / n$ .
Hence $Frac (Z [i]) = Q (i)$ , the Gaussian rationals. For instance $\frac{1}{1 + i} = \frac{1 - i}{2} = \frac{1}{2} - \frac{1}{2} i$ . ✓

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Figure 6.44. The Gaussian integers, with the circles

N (z) = 1, 2, 5

. The four points on the innermost circle are exactly the units

\pm 1, \pm i

; the multiplicativity of

N

is what forces a unit to have norm

1

Example 6.45 (Norms in $Z [- 5]$ and the failure of unique factorisation). In $D = Z [- 5]$ define $N (a + b - 5) = a^{2} + 5 b^{2}$ . Compute $N (1 + - 5)$ , determine $D^{\times}$ , and show that $6$ has two genuinely different factorisations.

Solution.

$N (1 + - 5) = 1^{2} + 5 \cdot 1^{2} = 6$ . The norm is multiplicative, since $N (z) = z \overset{z}{ˉ}$ for the complex conjugate.
If $u \in D^{\times}$ then $N (u) N (u^{- 1}) = 1$ with both factors positive integers, so $N (u) = 1$ : $a^{2} + 5 b^{2} = 1$ forces $b = 0$ and $a = \pm 1$ . Hence $D^{\times} = {1, - 1}$ .
$6 = 2 \cdot 3 = (1 + - 5) (1 - - 5)$ , and $N$ of the four factors is $4, 9, 6, 6$ respectively.
Each factor is irreducible: a proper factorisation of an element of norm $4$ , $9$ or $6$ would need a factor of norm $2$ or $3$ , and $a^{2} + 5 b^{2} \in {2, 3}$ has no integer solutions ( $b = 0$ leaves $a^{2} \in {2, 3}$ ).
The two factorisations are not rearrangements of one another up to units, since $N (2) = 4$ differs from $N (1 \pm - 5) = 6$ and the only units are $\pm 1$ . So $D$ is a domain that is not a UFD, which is the strictness of the second containment in the figure The standard chain of commutative-ring conditions.

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Summary. The chapter in six lines. A domain is a ring where cancellation works; a field is a ring where division works; every field is a domain and $Z$ shows the converse fails. Finiteness repairs the failure: every finite integral domain is a field, by pigeonhole applied to $x \mapsto a x$ . The characteristic of a domain is $0$ or prime, and it determines the prime subfield — $Q$ or $F_{p}$ — so a finite field has order $p^{n}$ . In characteristic $p$ , $x \mapsto x^{p}$ is a homomorphism, and an automorphism when the field is finite. Every domain embeds in a smallest field, its field of fractions, whose construction needs "domain" exactly twice: for transitivity of $\sim$ and for $b d \neq = 0$ .

Exercises for §6.4 →

Common Mistakes

Assuming a domain is a field. $Z$ is the standing counterexample: no zero-divisors at all, yet $Z^{\times} = {\pm 1}$ . Cancellation is not division.
Assuming a field is only a domain when it is finite. Every field is an integral domain has no finiteness hypothesis; it is the converse that needs finiteness.
Forgetting $1 \neq = 0$ . Without it the zero ring qualifies as a domain and as a field, and the two words stop meaning anything.
Calling $0$ a zero-divisor. By convention a zero-divisor is nonzero. Otherwise every ring but the zero ring has one.
Believing the characteristic of a finite ring must be prime. It must be prime in a domain. $Z /12 Z$ is finite with characteristic $12$ , and $Z /4 Z \times Z /6 Z$ has characteristic $12$ too.
Confusing $F_{p^{n}}$ with $Z / p^{n} Z$ . These agree only when $n = 1$ . $F_{4}$ is a field of characteristic $2$ ; $Z /4 Z$ has characteristic $4$ and a zero-divisor, and they are not even isomorphic as additive groups.
Expecting a field of every order. By A finite field has prime-power order there is no field with $6$ , $10$ , $12$ or $15$ elements.
Using $(x + y)^{p} = x^{p} + y^{p}$ outside characteristic $p$ . It is a theorem exactly when the characteristic is the prime $p$ being used as the exponent, and false otherwise — including in $Z /4 Z$ with $p = 4$ , which is not prime.
Building fractions over a ring with zero-divisors. The relation $a d = b c$ is not transitive without the domain hypothesis, so there is no quotient set to put operations on.
Reducing a fraction without justification. $\frac{a}{b} = \frac{a c}{b c}$ requires $c \neq = 0$ , and the equality is a statement about equivalence classes, not about the chosen representatives.

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

Practice this chapter

Introduction

6.1Zero-divisors, domains and cancellation

6.2Fields

Finite domains

6.3Characteristic and the prime subfield

The Frobenius endomorphism

Building F4​ by hand

6.4The field of fractions

Building $F_{4}$ by hand