Chapter 17

Medium

Fields — Definition and Basic Properties

00 · Symbol Glossary

$F$A field

A commutative ring with unity in which every nonzero element is a unit. The strongest, most restrictive kind of ring studied in this course — division is always available except by zero.

$F^\times$Nonzero elements of a field

$F^\times = F\setminus\{0\}$ . In a field (unlike a general ring), this coincides exactly with the group of units from Chapter 11 — every nonzero element is invertible.

$\mathrm{char}(F)$Characteristic of F

The smallest positive integer $n$ with $\underbrace{1+1+\cdots+1}_{n} = 0$ in $F$ , or $0$ if no such $n$ exists. A single number capturing deep structural information about $F$ .

$n\cdot1$n copies of the unity, added

Shorthand for $1+1+\cdots+1$ ( $n$ times), the image of the integer $n$ under the unique ring homomorphism $\mathbb{Z}\to F$ sending $1\mapsto1_F$ .

$\mathrm{Frac}(R)$Field of fractions

The smallest field containing an integral domain $R$ , built from formal fractions $a/b$ ( $b\neq0$ ) exactly the way $\mathbb{Q}$ is built from $\mathbb{Z}$ .

01 · The Definition

Definition — Field

A field is a commutative ring with unity ( $1\neq0$ ) in which every nonzero element is a unit: $F^\times = F\setminus\{0\}$ .

The strongest ring in this course's hierarchy

Field $\implies$ integral domain $\implies$ commutative ring with unity $\implies$ ring — a strict hierarchy of increasingly demanding structures, mirroring group $\implies$ monoid $\implies$ semigroup $\implies$ magma from Chapter 01. Every field is automatically a domain (proven next), but the reverse fails ( $\mathbb{Z}$ is a domain, not a field).

02 · Fields Are Domains, and Finite Domains Are Fields

Theorem 17.1 — Every Field Is an Integral Domain

Every field is an integral domain.

Proof

Let $F$ be a field and suppose $ab=0$ with $a\neq0$ . Since $F$ is a field, $a^{-1}$ exists. Multiply: $a^{-1}(ab) = a^{-1}\cdot0=0$ , and the left side is $(a^{-1}a)b = 1\cdot b=b$ . So $b=0$ : there are no zero divisors.

Theorem 17.2 — Every Finite Integral Domain Is a Field

If $R$ is a finite integral domain, then $R$ is a field.

Proof

Let $a\in R$ , $a\neq0$ . Consider the map $\lambda_a: R\to R$ , $\lambda_a(x) = ax$ . Injective: if $ax_1=ax_2$ , then $a(x_1-x_2)=0$ ; since $a\neq0$ and $R$ has no zero divisors, $x_1-x_2=0$ , i.e. $x_1=x_2$ . Since $R$ is finite and $\lambda_a$ is an injective map from $R$ to itself, $\lambda_a$ is also surjective (an injective self-map of a finite set is automatically a bijection — the pigeonhole principle). In particular, $1\in R$ is hit: $\lambda_a(x)=ax=1$ for some $x\in R$ . So $a$ has a multiplicative inverse. Since $a$ was an arbitrary nonzero element, every nonzero element of $R$ is a unit: $R$ is a field.

Example — Z_p is a field exactly when p is prime

For prime $p$ : $\mathbb{Z}_p$ is a finite commutative ring with unity; it's a domain because $ab\equiv0\pmod p$ forces $p\mid ab$ , and since $p$ is prime, $p\mid a$ or $p\mid b$ (Theorem 15.3-style reasoning, or directly Euclid's lemma), i.e. $a\equiv0$ or $b\equiv0\pmod p$ — no zero divisors. By Theorem 17.2, $\mathbb{Z}_p$ is a field. (This matches Theorem 11.4 + Theorem 12.6 from Chapters 11–12: every nonzero class is coprime to the prime $p$ , hence a unit.)

For composite $n=ab$ with $1<a,b<n$ : $[a][b]=[n]=[0]$ with $[a],[b]\neq[0]$ — zero divisors, so $\mathbb{Z}_n$ is not even a domain, let alone a field.

03 · A Gallery of Fields

Example — The familiar number fields

$\mathbb{Q}$ , $\mathbb{R}$ , $\mathbb{C}$ : each is a commutative ring with unity in which every nonzero element has a multiplicative inverse (reciprocals for $\mathbb{Q},\mathbb{R}$ ; $\frac{\bar z}{\lvert z\rvert^2}$ for nonzero $z\in\mathbb{C}$ ). All three are fields. $\mathbb{Z}$ is not a field — it's a domain (Chapter 11) but $2$ , for instance, has no integer reciprocal.

Example — $\mathbb{Z}_5$'s field structure, recomputed

From Exercise 1.5: every nonzero element of $\mathbb{Z}_5$ has an inverse ( $1^{-1}=1$ , $2^{-1}=3$ , $3^{-1}=2$ , $4^{-1}=4$ ). Theorem 17.2 now explains why this had to work out, before any computation: $5$ is prime, $\mathbb{Z}_5$ is a finite domain, hence automatically a field.

04 · Characteristic

Definition — Characteristic

$\mathrm{char}(F)$ is the smallest positive integer $n$ with $n\cdot1 = \underbrace{1+\cdots+1}_n = 0$ in $F$ , if such $n$ exists; otherwise $\mathrm{char}(F):=0$ .

Example — Characteristics of familiar fields

$\mathrm{char}(\mathbb{Q})=\mathrm{char}(\mathbb{R})=\mathrm{char}(\mathbb{C})=0$ : no finite sum of $1$ 's is ever $0$ in these fields (they all contain the ordinary positive integers as a genuinely infinite, nonzero set).

$\mathrm{char}(\mathbb{Z}_5)=5$ : $1+1+1+1+1 = 5 = [0]$ in $\mathbb{Z}_5$ , and no smaller positive sum of $1$ 's vanishes (checking: $1,2,3,4\neq0$ in $\mathbb{Z}_5$ ).

Theorem 17.3 — Characteristic Is Zero or Prime

For any field $F$ , $\mathrm{char}(F)$ is either $0$ or a prime number.

Proof

Suppose $\mathrm{char}(F)=n>0$ and $n$ is not prime, so $n=ab$ with $1<a,b<n$ . Then:

(a\cdot1)(b\cdot1) = (ab)\cdot1 = n\cdot1 = 0

using distributivity to expand $(a\cdot1)(b\cdot1)$ as $ab$ copies of $1\cdot1=1$ added together. Since $F$ is a field (Theorem 17.1, no zero divisors), $a\cdot1=0$ or $b\cdot1=0$ . But $1\le a,b<n$ , so $a\cdot1=0$ (or $b\cdot1=0$ ) contradicts $n$ being the smallest positive integer with $n\cdot1=0$ . So $n$ cannot have a nontrivial factorization: $n$ is prime.

Common mistake — Characteristic 4 is impossible for a field

Could a field have characteristic $4$ ? By Theorem 17.3, no — $4=2\times2$ is not prime, so no field can have characteristic exactly $4$ . (Indeed, $\mathbb{Z}_4$ itself is not a field, consistent with this — it's exactly the kind of ring the theorem rules out as a field's characteristic, though $\mathbb{Z}_4$ still makes sense as a ring with "additive characteristic 4" in a looser sense outside field theory.)

05 · The Prime Subfield

Theorem 17.4 — The Prime Subfield

Let $F$ be a field. There is a unique ring homomorphism $\psi:\mathbb{Z}\to F$ with $\psi(1)=1_F$ (given by $\psi(n)=n\cdot1_F$ ). If $\mathrm{char}(F)=p>0$ , $\mathrm{im}\,\psi \cong \mathbb{Z}_p$ ; if $\mathrm{char}(F)=0$ , $\psi$ is injective and extends to a copy of $\mathbb{Q}$ inside $F$ .

Proof

Existence and uniqueness of $\psi$ follow from the universal property of $\mathbb{Z}$ as the free ring on no generators with unity forced to $1_F$ — concretely, any ring homomorphism from $\mathbb{Z}$ is completely determined by where $1$ goes (Theorem 6.1-style forcing: $\psi(n)=\psi(1+\cdots+1)=n\psi(1)$ ), so $\psi(n):=n\cdot1_F$ is the only candidate, and it's directly checked to be a homomorphism using distributivity in $F$ .

$\ker\psi = \{n\in\mathbb{Z} : n\cdot1_F=0\}$ . If $\mathrm{char}(F)=p>0$ : the smallest positive element of $\ker\psi$ is $p$ (by definition of characteristic), and $\ker\psi$ , being an ideal of $\mathbb{Z}$ (Theorem 13.3), must be $p\mathbb{Z}$ (Theorem 3.6-style: the ideal generated by its smallest positive element). By the First Isomorphism Theorem (Theorem 13.6), $\mathbb{Z}/p\mathbb{Z} \cong \mathrm{im}\,\psi$ , i.e. $\mathbb{Z}_p \cong \mathrm{im}\,\psi \subseteq F$ .

If $\mathrm{char}(F)=0$ : no positive integer is in $\ker\psi$ , so $\ker\psi=\{0\}$ , and by Theorem 13.5, $\psi$ is injective — giving a copy of $\mathbb{Z}$ inside $F$ . Since $F$ is a field, every nonzero image $\psi(m)$ has an inverse in $F$ , allowing every "fraction" $\psi(m)^{-1}\psi(n)$ to be formed inside $F$ ; this produces a copy of $\mathbb{Q}$ sitting inside $F$ (made precise as the field of fractions construction, Section 06).

Every field contains a copy of Q or some Z_p

Theorem 17.4 says the characteristic completely determines a field's "smallest possible piece": either $\mathbb{Q}$ (characteristic 0) or $\mathbb{Z}_p$ (characteristic $p$ ), sitting inside every field as its prime subfield. This single number is the first invariant checked when studying any new field, and it governs enormous amounts of behavior in Chapter 18's field extensions (for instance, whether $x^p-a$ can have a repeated root).

06 · The Field of Fractions

Section 05 promised that characteristic-0 fields contain a copy of $\mathbb{Q}$ . The general construction behind this — building the smallest field containing a given integral domain — is worth naming even without full proof.

Construction — Field of Fractions (taken as given)

For an integral domain $R$ , define $\mathrm{Frac}(R)$ as the set of formal fractions $a/b$ ( $a,b\in R$ , $b\neq0$ ), where $a/b$ and $c/d$ are identified whenever $ad=bc$ (exactly how $\tfrac12=\tfrac24$ is justified for ordinary fractions), with addition and multiplication defined by the usual fraction rules: $\tfrac{a}{b}+\tfrac{c}{d} := \tfrac{ad+bc}{bd}$ , $\tfrac{a}{b}\cdot\tfrac{c}{d} := \tfrac{ac}{bd}$ . Verifying these are well-defined and satisfy the field axioms is routine (each step mirrors why ordinary fraction arithmetic works) but tedious; we take it as given. $R$ embeds into $\mathrm{Frac}(R)$ via $a\mapsto a/1$ , and $\mathrm{Frac}(\mathbb{Z})=\mathbb{Q}$ recovers the familiar construction of the rationals.

Example — The field of rational functions

$\mathrm{Frac}(F[x])$ (for a field $F$ ) is the field of rational functions $F(x) = \{p(x)/q(x) : p,q\in F[x],\,q\neq0\}$ — built from the polynomial ring $F[x]$ exactly the way $\mathbb{Q}$ is built from $\mathbb{Z}$ . This field will reappear when field extensions need a "generic" transcendental element.

07 · Exercises

EXERCISE 17.1

Check whether $\mathbb{Z}_{11}$ is a finite domain, then apply Theorem 17.2 directly.

$11$ is prime, so $\mathbb{Z}_{11}$ has no zero divisors (Euclid's lemma reasoning, as in Section 03). It's finite. By Theorem 17.2, $\mathbb{Z}_{11}$ is a field — every nonzero element among $\{1,\ldots,10\}$ has a multiplicative inverse mod 11, without needing to compute each one individually.

Using Theorem 17.2, explain why $\mathbb{Z}_{11}$ is a field without computing any inverses directly.

EXERCISE 17.2

Apply Theorem 17.3: what are the possible characteristics for a field of exactly 4 elements?

By Theorem 17.3, $\mathrm{char}(F) \in \{0\} \cup \{\text{primes}\}$ . Since $F$ is finite (order 4), it cannot have characteristic $0$ (that would force $\mathbb{Z}$ , an infinite set, to embed inside it — impossible in a finite field, since $1,2\cdot1,3\cdot1,\ldots$ would need to be all distinct forever). So $\mathrm{char}(F)$ is a prime; and since $n\cdot1_F=0$ must occur within the field's own $4$ elements, $\mathrm{char}(F) \le4$ . The only prime $\le4$ dividing into a consistent structure of size 4 is $\mathrm{char}(F)=2$ (a field of order $4=2^2$ always has characteristic $2$ , matching that its prime subfield $\mathbb{Z}_2$ must have order dividing $4$ ).

A field $F$ has exactly 4 elements. Using Theorem 17.3, what must $\mathrm{char}(F)$ be, and why can't it be $0$ ?

EXERCISE 17.3

Compute $1+1+1+1+1+1$ inside $\mathbb{Z}_6$ and see when it first hits zero — but first check whether $\mathbb{Z}_6$ even qualifies as a field.

$\mathbb{Z}_6$ is not a field (6 is composite, $[2][3]=[0]$ are zero divisors, Chapter 11), so "characteristic" in the field-theoretic sense of Theorem 17.3 doesn't strictly apply to it as a guarantee of primality — but the raw definition ( $n\cdot1=0$ ) still makes sense for any ring: $1+1+1+1+1+1=6\equiv0\pmod6$ , and no smaller positive sum vanishes, so this ring has "characteristic" 6 in the general ring sense. This is consistent with Theorem 17.3 only applying to fields — $\mathbb{Z}_6$ being a counterexample-in-waiting is exactly why the theorem needs the field hypothesis (no zero divisors) to rule out composite characteristics.

Does $\mathbb{Z}_6$ contradict Theorem 17.3, given that $1+1+1+1+1+1=0$ in $\mathbb{Z}_6$ but $6$ is not prime? Explain.

08 · Chapter Summary

Concept	Statement
Field	Commutative ring with unity where every nonzero element is a unit
Fields are domains	Thm 17.1
Finite domains are fields	Thm 17.2 (via injective-hence-surjective multiplication map)
$\mathbb{Z}_p$ is a field	Exactly when $p$ is prime
Characteristic	Smallest $n>0$ with $n\cdot1=0$ , or $0$
Characteristic is 0 or prime	Thm 17.3
Prime subfield	$\mathbb{Z}_p$ (char $p$ ) or $\mathbb{Q}$ (char 0), inside every field (Thm 17.4)
Field of fractions	$\mathrm{Frac}(R)$ , generalizing $\mathbb{Q}=\mathrm{Frac}(\mathbb{Z})$

Next: Chapter 18 — Field Extensions studies fields built on top of a smaller field, developing the algebraic/transcendental distinction and the degree of an extension needed for Chapter 20's Galois theory.