Abstract Algebra — 04 · Rings, Ideals, Integral Domains and Fields

Groups have one operation. Most of the algebra you already use has two: integers, polynomials and matrices can be added and multiplied, and the distributive law ties the two together. A structure with both is a ring. This post builds ring theory the same way posts 01–03 built group theory: definitions and examples, then substructures (ideals play the role of normal subgroups), quotients and a First Isomorphism Theorem. Two questions then drive the rest. Which rings behave like $\mathbb{Z}$, with unique factorisation? And how do you manufacture new fields, such as $\mathbb{C}$ or a field with 4 elements? The answer to the second, "take polynomials modulo an irreducible polynomial", is the starting point of Galois theory in post 05.

Series map: 00 Integers, Relations & Modular Arithmetic → 01 Groups & Subgroups → 02 Cosets, Lagrange & Quotient Groups → 03 Homomorphisms, Group Actions & Sylow → 04 Rings, Ideals & Fields (this post) → 05 Field Extensions & Galois Theory.

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Learning Outcome

• State the ring axioms (with identity), and recognise units, zero divisors, integral domains and fields.
• Prove that a finite integral domain is a field and that the characteristic of a domain is 0 or a prime.
• Define ring homomorphisms, ideals and quotient rings, and prove the First Isomorphism Theorem for rings.
• Prove that $R/P$ is a domain exactly when $P$ is prime, and $R/M$ is a field exactly when $M$ is maximal.
• Use the division algorithm in $F[x]$ and prove that $F[x]$ is a principal ideal domain.
• Test polynomials for irreducibility: roots, the rational root test, Eisenstein's criterion, reduction mod $p$.
• Place a ring in the chain Euclidean domain ⊂ PID ⊂ UFD ⊂ integral domain, and show that $\mathbb{Z}[\sqrt{-5}]$ is not a UFD.
• Build $\mathbb{C}$ as $\mathbb{R}[x]/(x^2+1)$ and the finite fields $\mathbb{F}_4$ and $\mathbb{F}_8$, and compute in them.

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Chapter 1 — Rings

Terminology

• Ring — a set $R$ with two operations $+$ and $\cdot$ such that:
  1. $(R, +)$ is an abelian group, with identity $0$ and inverses $-a$;
  2. multiplication is associative: $(ab)c = a(bc)$;
  3. there is a multiplicative identity $1$ with $1a = a1 = a$;
  4. the distributive laws hold: $a(b + c) = ab + ac$ and $(a + b)c = ac + bc$.
• Convention. In this series every ring has a $1$ (some books call these "rings with unity" and allow rings without one). Multiplication need not be commutative.
• Commutative ring — a ring with $ab = ba$ for all $a, b$.
• Zero ring — $\{0\}$ with $1 = 0$. It is the only ring in which $1 = 0$ (if $1 = 0$ then $a = 1a = 0a = 0$).
• Unit — an element $u$ with a two-sided inverse: $uv = vu = 1$. The units form a group $R^\times$ under multiplication.
• Zero divisor — a nonzero $a$ for which some nonzero $b$ has $ab = 0$ or $ba = 0$.
• Integral domain — a commutative ring with $1 \ne 0$ and no zero divisors.
• Field — a commutative ring with $1 \ne 0$ in which every nonzero element is a unit.
• Division ring — like a field but not necessarily commutative (the quaternions are the standard example).

Examples

Ring	Commutative?	Units	Zero divisors?
$\mathbb{Z}$	yes	$\pm 1$	none (domain)
$\mathbb{Q}, \mathbb{R}, \mathbb{C}$	yes	all nonzero	none (fields)
$\mathbb{Z}/n\mathbb{Z}$	yes	classes with $\gcd(a, n) = 1$	yes, unless $n$ is prime
$M_2(\mathbb{R})$, $2\times 2$ real matrices	no	matrices with $\det \ne 0$	yes
$\mathbb{R}[x]$, real polynomials	yes	nonzero constants	none
$\mathbb{Z}[i] = \{a + bi : a, b \in \mathbb{Z}\}$	yes	$\pm 1, \pm i$	none
$C[0,1]$, continuous functions, pointwise operations	yes	functions with no zero	yes

A few of the entries, worked out:

• $\mathbb{Z}/12\mathbb{Z}$. The units are $1, 5, 7, 11$ (coprime to 12; each is its own inverse, e.g. $5 \cdot 5 = 25 \equiv 1$). Every other nonzero class is a zero divisor: $2 \cdot 6 = 12 \equiv 0$, $3 \cdot 4 \equiv 0$, $8 \cdot 3 \equiv 0$, and so on, giving $2, 3, 4, 6, 8, 9, 10$. In $\mathbb{Z}/n\mathbb{Z}$ every nonzero element is either a unit or a zero divisor: if $d = \gcd(a, n) > 1$, then $a \cdot (n/d) \equiv 0$ with $n/d \not\equiv 0$.
• Matrices. $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$, although neither factor is zero.
• Gaussian integers. Use the norm $N(a + bi) = a^2 + b^2 = (a+bi)(a-bi)$, which is multiplicative: $N(zw) = N(z)N(w)$. If $uv = 1$ then $N(u)N(v) = 1$ in the positive integers, so $N(u) = 1$, and $a^2 + b^2 = 1$ leaves $\pm 1, \pm i$.
• Functions. Let $f$ be zero on $[0, \tfrac12]$ and $g$ zero on $[\tfrac12, 1]$, both continuous and not identically zero (for example $f(t) = \max(0, t - \tfrac12)$, $g(t) = \max(0, \tfrac12 - t)$). Then $fg = 0$.

First consequences of the axioms

For all $a, b$ in a ring:

1. $0 \cdot a = a \cdot 0 = 0$. Proof: $0a = (0 + 0)a = 0a + 0a$; subtract $0a$ from both sides.
2. $(-a)b = a(-b) = -(ab)$. Proof: $ab + (-a)b = (a + (-a))b = 0b = 0$.
3. $(-a)(-b) = ab$, and in particular $(-1)^2 = 1$.

Cancellation. In an integral domain, if $a \ne 0$ and $ab = ac$ then $b = c$, because $a(b - c) = 0$ forces $b - c = 0$. This is exactly what fails in $\mathbb{Z}/12\mathbb{Z}$: $2 \cdot 1 = 2 \cdot 7$ but $1 \ne 7$.

Every field is an integral domain: if $a \ne 0$ and $ab = 0$, then $b = a^{-1}ab = 0$. The converse fails ($\mathbb{Z}$), except for finite rings:

Theorem. A finite integral domain is a field.

Proof. Let $R$ be a finite domain and $a \ne 0$. The map $\mu_a : R \to R$, $x \mapsto ax$, is injective by cancellation. An injective map from a finite set to itself is surjective, so $ax = 1$ for some $x$. Since $R$ is commutative, $x$ is a two-sided inverse. ∎

Corollary. $\mathbb{Z}/n\mathbb{Z}$ is a field if and only if $n$ is prime. (If $n = ab$ with $1 < a, b < n$, then $a$ and $b$ are zero divisors. If $n = p$ is prime, $\mathbb{Z}/p\mathbb{Z}$ has no zero divisors because $p \mid ab$ implies $p \mid a$ or $p \mid b$, post 00; so it is a finite domain.) This field is written $\mathbb{F}_p$.

Characteristic

The characteristic of a ring $R$ is the least $n \ge 1$ with $n \cdot 1 = 1 + 1 + \cdots + 1 = 0$ ($n$ terms), or $0$ if there is no such $n$. So $\operatorname{char}\mathbb{Z} = \operatorname{char}\mathbb{Q} = 0$ and $\operatorname{char}\mathbb{Z}/n\mathbb{Z} = n$.

Theorem. The characteristic of an integral domain is $0$ or a prime.

Proof. Suppose $\operatorname{char} R = n > 0$ and $n = ab$ with $1 \le a, b < n$. Then $0 = n \cdot 1 = (a \cdot 1)(b \cdot 1)$. In a domain one factor is $0$, contradicting the minimality of $n$. And $n \ne 1$ because $1 \ne 0$. ∎

Watch out: characteristic is about $n \cdot 1$, not about the size of the ring. The infinite field $\mathbb{F}_p(t)$ of rational functions over $\mathbb{F}_p$ has characteristic $p$.

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Chapter 2 — Homomorphisms, ideals and quotient rings

Terminology

• Subring — a subset $S \subseteq R$ containing $1$ that is closed under $+$, $-$ and $\cdot$. Example: $\mathbb{Z} \subset \mathbb{Z}[i] \subset \mathbb{C}$.
• Ring homomorphism — a map $\varphi : R \to S$ with $\varphi(a + b) = \varphi(a) + \varphi(b)$, $\varphi(ab) = \varphi(a)\varphi(b)$ and $\varphi(1) = 1$.
• Kernel — $\ker\varphi = \{a : \varphi(a) = 0\}$.
• Ideal — an additive subgroup $I \subseteq R$ that absorbs multiplication: $r a \in I$ and $a r \in I$ for all $r \in R$, $a \in I$. (For commutative rings one condition suffices.)
• Principal ideal — in a commutative ring, $(a) = aR = \{ra : r \in R\}$, the smallest ideal containing $a$. More generally $(a_1, \dots, a_k) = \{r_1a_1 + \cdots + r_ka_k\}$.

Examples of homomorphisms: reduction $\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$; evaluation $\mathbb{R}[x] \to \mathbb{C}$, $f \mapsto f(i)$; complex conjugation $\mathbb{C} \to \mathbb{C}$; the inclusion of a subring.

Ideals are the kernels

Theorem. The kernel of a ring homomorphism is an ideal.

Proof. It is an additive subgroup, being the kernel of a homomorphism of additive groups (post 03). If $\varphi(a) = 0$ then $\varphi(ra) = \varphi(r) \cdot 0 = 0$ and $\varphi(ar) = 0$. ∎

Ideals are to rings what normal subgroups are to groups: exactly the things you can take a quotient by. Two quick facts:

• If an ideal $I$ contains a unit $u$, then $I = R$: for any $r$, $r = (ru^{-1})u \in I$. So a proper ideal contains no units, and in particular does not contain $1$.
• A commutative ring $R$ with $1 \ne 0$ is a field if and only if its only ideals are $\{0\}$ and $R$. (If $R$ is a field, any nonzero ideal contains a unit. Conversely, for $a \ne 0$, $(a) = R \ni 1$, so $ra = 1$ for some $r$.) Consequently every ring homomorphism from a field to a nonzero ring is injective: its kernel is a proper ideal (it misses 1), hence $\{0\}$.

Theorem. Every ideal of $\mathbb{Z}$ is $n\mathbb{Z} = (n)$ for a unique $n \ge 0$.

Proof. Let $I \ne \{0\}$ and let $n$ be the least positive element of $I$. For $a \in I$ divide: $a = qn + r$ with $0 \le r < n$. Then $r = a - qn \in I$, so $r = 0$ by minimality, and $a \in (n)$. ∎

In $\mathbb{Z}$, $(a, b) = (\gcd(a, b))$: for example $(4, 6) = (2)$. This is Bézout's identity from post 00 in ideal language.

Quotient rings

Let $I$ be an ideal of $R$. The additive cosets $a + I$ form an abelian group $R/I$ (post 02). Define multiplication by $(a + I)(b + I) = ab + I$.

Well defined. If $a' = a + i$ and $b' = b + j$ with $i, j \in I$, then $a'b' = ab + (aj + ib + ij)$, and the bracket lies in $I$ because $I$ absorbs multiplication. This is precisely where "ideal" (rather than just "subring") is needed. The ring axioms for $R/I$ follow from those of $R$, with $1 + I$ as identity, and $\pi : R \to R/I$, $a \mapsto a + I$, is a surjective homomorphism with kernel $I$.

Theorem (First Isomorphism Theorem for rings). If $\varphi : R \to S$ is a ring homomorphism, then $R/\ker\varphi \cong \operatorname{im}\varphi$ via $a + \ker\varphi \mapsto \varphi(a)$.

Proof. By the group version (post 03) the map $\bar\varphi$ is a well-defined isomorphism of additive groups. It is multiplicative: $\bar\varphi((a + K)(b + K)) = \bar\varphi(ab + K) = \varphi(ab) = \varphi(a)\varphi(b)$; and $\bar\varphi(1 + K) = \varphi(1) = 1$. ∎

Worked applications:

• $\mathbb{Z}[x]/(x) \cong \mathbb{Z}$. Evaluation at 0, $f \mapsto f(0)$, is surjective, and its kernel is the polynomials with zero constant term, which is $(x)$.
• $\mathbb{R}[x]/(x^2 + 1) \cong \mathbb{C}$. Evaluation at $i$ is surjective ($a + bx \mapsto a + bi$). Its kernel contains $x^2 + 1$; Chapter 6 shows that $(x^2 + 1)$ is a maximal ideal, so the kernel (a proper ideal containing it) is exactly $(x^2 + 1)$.
• $\mathbb{Z}[x]/(2, x) \cong \mathbb{Z}/2\mathbb{Z}$. The map $f \mapsto f(0) \bmod 2$ is surjective with kernel $\{f : f(0) \text{ even}\} = (2, x)$.
• Chinese Remainder Theorem. If $\gcd(m, n) = 1$, the map $\mathbb{Z} \to \mathbb{Z}/m \times \mathbb{Z}/n$, $a \mapsto (a \bmod m, a \bmod n)$, is a surjective ring homomorphism (surjective by post 00's CRT) with kernel $(m) \cap (n) = (mn)$. So $\mathbb{Z}/mn\mathbb{Z} \cong \mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$ as rings, and taking units gives $\phi(mn) = \phi(m)\phi(n)$ for Euler's function.

Trick: the fastest way to understand $R/I$ is again to find a surjective homomorphism out of $R$ whose kernel is $I$. Evaluation maps are the usual candidates for polynomial rings.

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Chapter 3 — Prime and maximal ideals

From here on, "ring" means commutative ring with $1$.

Terminology

• Prime ideal — a proper ideal $P$ such that $ab \in P$ implies $a \in P$ or $b \in P$.
• Maximal ideal — a proper ideal $M$ such that no ideal lies strictly between $M$ and $R$: if $M \subseteq J \subseteq R$ then $J = M$ or $J = R$.

In $\mathbb{Z}$: $(p)$ for a prime $p$ is prime (this is Euclid's lemma) and maximal; $(0)$ is prime (because $\mathbb{Z}$ is a domain) but not maximal, since $(0) \subsetneq (2) \subsetneq \mathbb{Z}$; $(6)$ is neither, since $2 \cdot 3 \in (6)$ but $2, 3 \notin (6)$.

The two quotient theorems

Theorem. $P$ is a prime ideal of $R$ if and only if $R/P$ is an integral domain.

Proof. $R/P$ is not the zero ring exactly when $1 \notin P$, i.e. when $P$ is proper. In $R/P$, the product $(a + P)(b + P) = ab + P$ is zero exactly when $ab \in P$, and $a + P$ is zero exactly when $a \in P$. So "no zero divisors in $R/P$" says word for word "$ab \in P \Rightarrow a \in P$ or $b \in P$". ∎

Theorem. $M$ is a maximal ideal of $R$ if and only if $R/M$ is a field.

Proof. ($\Rightarrow$) Let $a + M \ne 0$, i.e. $a \notin M$. The ideal $M + (a) = \{m + ra\}$ strictly contains $M$, so by maximality it is $R$ and contains $1$: $1 = m + ra$ for some $m \in M$, $r \in R$. Then $(r + M)(a + M) = ra + M = 1 - m + M = 1 + M$, so $a + M$ is a unit. Also $1 + M \ne 0$ because $M$ is proper.

($\Leftarrow$) If $R/M$ is a field, $M$ is proper. Let $M \subsetneq J$ and pick $a \in J \setminus M$. Then $a + M$ has an inverse $r + M$, so $ra - 1 \in M \subseteq J$. Since $ra \in J$, also $1 \in J$ and $J = R$. ∎

Corollary. Every maximal ideal is prime (a field is a domain).

Examples in $\mathbb{Z}[x]$.

• $(x)$ is prime but not maximal: $\mathbb{Z}[x]/(x) \cong \mathbb{Z}$ is a domain but not a field. Indeed $(x) \subsetneq (2, x) \subsetneq \mathbb{Z}[x]$.
• $(2, x)$ is maximal: $\mathbb{Z}[x]/(2, x) \cong \mathbb{Z}/2\mathbb{Z}$, a field.
• $(x^2 + 1)$ is prime but not maximal: $\mathbb{Z}[x]/(x^2+1) \cong \mathbb{Z}[i]$, a domain that is not a field.

A standard existence result (it needs Zorn's lemma): every proper ideal of a ring with $1$ is contained in a maximal ideal.

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Chapter 4 — Polynomial rings over a field

Let $F$ be a field. $F[x]$ is the ring of polynomials $a_0 + a_1x + \cdots + a_nx^n$ with $a_i \in F$. If $a_n \ne 0$, the degree is $n$ and $a_n$ is the leading coefficient; the polynomial is monic if $a_n = 1$. The zero polynomial has no degree (some books use $-\infty$).

Degree, units, roots

If $R$ is a domain, the leading coefficient of $fg$ is the product of the leading coefficients, which is nonzero. Hence, for nonzero $f, g$:

$$\deg(fg) = \deg f + \deg g,$$

$R[x]$ is again a domain, and the units of $F[x]$ are the nonzero constants (a unit $f$ has $\deg f + \deg f^{-1} = 0$).

Theorem (division algorithm). For $f, g \in F[x]$ with $g \ne 0$ there are unique $q, r \in F[x]$ with $f = qg + r$ and either $r = 0$ or $\deg r < \deg g$.

Proof sketch. Existence by induction on $\deg f$: if $\deg f \ge \deg g = m$, subtract $(a_n/b_m)\,x^{n-m}g$ to kill the leading term of $f$ and apply induction to the result. Dividing by $b_m$ is where a field is used. Uniqueness: if $qg + r = q'g + r'$ then $(q - q')g = r' - r$; the right side has degree $< m$, so $q = q'$ by the degree formula, and then $r = r'$. ∎

Worked example. Divide $f = x^4 + 3x^3 - 2x + 5$ by $g = x^2 + x + 1$ in $\mathbb{Q}[x]$:

$$x^4 + 3x^3 - 2x + 5 = (x^2 + x + 1)(x^2 + 2x - 3) + (8 - x).$$

(Check: $(x^2 + x + 1)(x^2 + 2x - 3) = x^4 + 3x^3 - x - 3$, and adding $8 - x$ gives $f$.)

Factor theorem. $a$ is a root of $f$ if and only if $(x - a) \mid f$ (divide by $x - a$; the remainder is the constant $f(a)$). So a nonzero polynomial of degree $n$ over a domain has at most $n$ roots.

Common pitfall: over a ring with zero divisors this fails. In $\mathbb{Z}/8\mathbb{Z}$, $x^2 - 1$ has four roots: $1, 3, 5, 7$ (each squares to $1$ mod 8).

F[x] is a principal ideal domain

A principal ideal domain (PID) is an integral domain in which every ideal is principal.

Theorem. If $F$ is a field, $F[x]$ is a PID.

Proof. Let $I \ne \{0\}$ be an ideal and let $g \in I$ be nonzero of least degree. For $f \in I$ write $f = qg + r$. Then $r = f - qg \in I$ and, if $r \ne 0$, $\deg r < \deg g$, contradicting minimality. So $r = 0$ and $I = (g)$. ∎

The proof is the one for $\mathbb{Z}$ with "size" replaced by "degree". As in $\mathbb{Z}$, $(f, g) = (\gcd(f, g))$, and the Euclidean algorithm computes the gcd:

$$\begin{aligned} x^6 - 1 &= x^2\,(x^4 - 1) + (x^2 - 1), \\ x^4 - 1 &= (x^2 + 1)(x^2 - 1) + 0, \end{aligned}$$

so $\gcd(x^6 - 1, x^4 - 1) = x^2 - 1$.

Watch out: $\mathbb{Z}[x]$ is not a PID, although $\mathbb{Z}$ is. Suppose $(2, x) = (h)$. Then $h \mid 2$, so $h$ is a constant $\pm 1$ or $\pm 2$. $h = \pm 1$ is impossible because $(2, x)$ is proper (its quotient is $\mathbb{Z}/2$). $h = \pm 2$ is impossible because $x \notin (2)$.

Irreducibility tests

A nonzero non-unit $f \in F[x]$ is irreducible if it is not a product of two polynomials of smaller degree. Irreducible polynomials are the "primes" of $F[x]$, and whether $f$ is irreducible depends on the field: $x^2 - 2$ is irreducible over $\mathbb{Q}$ but factors as $(x - \sqrt2)(x + \sqrt2)$ over $\mathbb{R}$; $x^2 + 1$ is irreducible over $\mathbb{R}$ but not over $\mathbb{C}$.

1. Degree 2 or 3. A polynomial of degree 2 or 3 over a field is irreducible if and only if it has no root in the field (any factorisation includes a linear factor).
2. Rational root test. If $f = a_nx^n + \cdots + a_0 \in \mathbb{Z}[x]$ and $p/q$ in lowest terms is a root, then $p \mid a_0$ and $q \mid a_n$.
3. Gauss's lemma. A polynomial in $\mathbb{Z}[x]$ that factors in $\mathbb{Q}[x]$ into factors of lower degree already factors in $\mathbb{Z}[x]$ into factors of those same degrees. So it suffices to look for integer factorisations.
4. Eisenstein's criterion. Let $f = a_nx^n + \cdots + a_0 \in \mathbb{Z}[x]$ and $p$ a prime with $p \nmid a_n$, $p \mid a_i$ for all $i < n$, and $p^2 \nmid a_0$. Then $f$ is irreducible in $\mathbb{Q}[x]$.
5. Reduction mod $p$. Let $f \in \mathbb{Z}[x]$ and let $p$ be a prime not dividing the leading coefficient. If the reduction $\bar f \in \mathbb{F}_p[x]$ is irreducible (of the same degree), then $f$ is irreducible in $\mathbb{Q}[x]$.

Worked examples:

• $x^3 - 2$: Eisenstein with $p = 2$. Also, the rational root test leaves only $\pm 1, \pm 2$, none of which is a root.
• $x^3 + 2x + 1$: candidates $\pm 1$ give $4$ and $-2$, so no rational root; degree 3, so irreducible over $\mathbb{Q}$.
• $x^5 - 4x + 2$: Eisenstein with $p = 2$ ($2 \mid -4$, $2 \mid 2$, $4 \nmid 2$). This polynomial returns in post 05.
• Cyclotomic polynomial $\Phi_5(x) = x^4 + x^3 + x^2 + x + 1 = (x^5 - 1)/(x - 1)$. Eisenstein does not apply directly, but $f(x)$ is irreducible exactly when $f(x+1)$ is, and
  $\Phi_5(x + 1) = x^4 + 5x^3 + 10x^2 + 10x + 5$, which is Eisenstein at $5$.
• $x^4 + 5x^3 + 3x^2 + 7x + 11$: reduce mod 2 to $x^4 + x^3 + x^2 + x + 1$, which is irreducible in $\mathbb{F}_2[x]$. So the original is irreducible over $\mathbb{Q}$.

Common pitfall: "no rational roots" proves irreducibility only in degree 2 and 3. $x^4 + 4$ has no rational root, yet $x^4 + 4 = (x^2 + 2x + 2)(x^2 - 2x + 2)$.

Common pitfall: the reduction test only works one way. $x^2 + 1$ is irreducible over $\mathbb{Q}$, but mod 2 it is $(x + 1)^2$. A reducible reduction proves nothing; so does a failed Eisenstein check.

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Chapter 5 — Unique factorisation

Terminology

In an integral domain $R$:

• $a$ divides $b$, written $a \mid b$, if $b = ac$ for some $c \in R$.
• $a$ and $b$ are associates if $a = ub$ for a unit $u$ (for example $2$ and $-2$ in $\mathbb{Z}$, or $1 + i$ and $i(1+i) = -1 + i$ in $\mathbb{Z}[i]$).
• A nonzero non-unit $p$ is irreducible if $p = ab$ forces $a$ or $b$ to be a unit.
• A nonzero non-unit $p$ is prime if $p \mid ab$ forces $p \mid a$ or $p \mid b$ (equivalently, $(p)$ is a prime ideal).
• Unique factorisation domain (UFD) — a domain in which every nonzero non-unit is a product of irreducibles, uniquely up to order and associates.
• Euclidean domain — a domain with a function $N : R \setminus \{0\} \to \mathbb{Z}_{\ge 0}$ such that for all $a$ and $b \ne 0$ there are $q, r$ with $a = qb + r$ and $r = 0$ or $N(r) < N(b)$.

Prime implies irreducible in any domain: if $p$ is prime and $p = ab$, then $p \mid ab$, say $p \mid a$, so $a = pc$ and $p = pcb$; cancelling gives $cb = 1$, so $b$ is a unit. The converse holds in a UFD (and so in a PID) but fails in general, which is exactly how unique factorisation breaks.

The hierarchy

$$\text{fields} \subset \text{Euclidean domains} \subset \text{PIDs} \subset \text{UFDs} \subset \text{integral domains}.$$

• Euclidean ⇒ PID: the proof that $F[x]$ is a PID works verbatim with $N$ in place of degree.
• PID ⇒ UFD: a standard theorem (existence uses the fact that increasing chains of ideals in a PID stop; uniqueness uses "irreducible ⇒ prime" in a PID).
• Every inclusion is strict, as the diagram's examples show.

Gaussian integers are Euclidean

Take $N(a + bi) = a^2 + b^2$. To divide $\alpha$ by $\beta$, compute $\alpha/\beta$ in $\mathbb{C}$ and round each coordinate to the nearest integer to get $q$; the rounding error has both coordinates at most $\tfrac12$, so $N(\alpha/\beta - q) \le \tfrac14 + \tfrac14 < 1$ and $r = \alpha - q\beta$ has $N(r) < N(\beta)$.

Worked example. Divide $7 + 2i$ by $2 - i$:

$$\frac{7 + 2i}{2 - i} = \frac{(7 + 2i)(2 + i)}{5} = \frac{12 + 11i}{5} = 2.4 + 2.2i.$$

Round to $q = 2 + 2i$. Then $r = (7 + 2i) - (2 + 2i)(2 - i) = (7 + 2i) - (6 + 2i) = 1$, and $N(1) = 1 < N(2 - i) = 5$. So $\mathbb{Z}[i]$ is a PID and a UFD.

A domain without unique factorisation

Let $R = \mathbb{Z}[\sqrt{-5}] = \{a + b\sqrt{-5} : a, b \in \mathbb{Z}\}$ with norm $N(a + b\sqrt{-5}) = a^2 + 5b^2$, again multiplicative. In $R$,

$$6 = 2 \cdot 3 = (1 + \sqrt{-5})(1 - \sqrt{-5}).$$

• Norms. $N(2) = 4$, $N(3) = 9$, $N(1 \pm \sqrt{-5}) = 1 + 5 = 6$.
• Units. $N(u) = 1$ forces $a^2 + 5b^2 = 1$, so $u = \pm 1$.
• All four factors are irreducible. A proper factor of an element of norm 4, 9 or 6 would have norm 2 or 3. But $a^2 + 5b^2 = 2$ or $3$ has no integer solutions ($b \ne 0$ gives at least 5; $b = 0$ needs $a^2 \in \{2, 3\}$).
• The factorisations really differ. $2$ is not an associate of $1 \pm \sqrt{-5}$ (the only units are $\pm 1$, and the norms 4 and 6 differ).

So $R$ is not a UFD. Equivalently, $2$ is irreducible but not prime: $2 \mid (1 + \sqrt{-5})(1 - \sqrt{-5})$, but $2$ divides neither factor, since $(1 \pm \sqrt{-5})/2 \notin R$.

The other separating examples, stated without proof: $\mathbb{Z}[x]$ and $F[x, y]$ are UFDs (Gauss: $R$ a UFD implies $R[x]$ a UFD) but not PIDs, as shown for $(2, x)$ above; and $\mathbb{Z}\big[\tfrac{1 + \sqrt{-19}}{2}\big]$ is a PID that is not Euclidean (Dummit & Foote, §8.1–8.2).

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Chapter 6 — Building fields from polynomials

Theorem. Let $F$ be a field and $p \in F[x]$ nonconstant. Then $F[x]/(p)$ is a field if and only if $p$ is irreducible.

Proof. In a PID, $(p)$ is maximal exactly when $p$ is irreducible. If $p$ is irreducible and $(p) \subseteq (h)$, then $h \mid p$, so $h$ is a unit (and $(h)$ is everything) or an associate of $p$ (and $(h) = (p)$). If $p = ab$ with $a, b$ of lower degree, then $(p) \subsetneq (a) \subsetneq F[x]$. Now apply "maximal ⇔ quotient is a field" from Chapter 3. ∎

What the elements look like. By the division algorithm every class has exactly one representative of degree $< n = \deg p$. So $F[x]/(p)$ is an $n$-dimensional vector space over $F$ with basis $1, \alpha, \dots, \alpha^{n-1}$, where $\alpha = x + (p)$ is a root of $p$. If $F = \mathbb{F}_q$ is finite, the field has $q^n$ elements. To multiply, multiply polynomials and reduce with $p(\alpha) = 0$; to invert, use the extended Euclidean algorithm (since $\gcd(f, p) = 1$, there are $s, t$ with $sf + tp = 1$, so $s = f^{-1}$).

The complex numbers

$x^2 + 1$ has no real root, so it is irreducible over $\mathbb{R}$ and $\mathbb{R}[x]/(x^2 + 1)$ is a field whose elements are $a + b\alpha$ with $\alpha^2 = -1$:

$$(a + b\alpha)(c + d\alpha) = ac + (ad + bc)\alpha + bd\,\alpha^2 = (ac - bd) + (ad + bc)\alpha.$$

That is exactly complex multiplication, and the evaluation map of Chapter 2 is the isomorphism with $\mathbb{C}$. Example: $(3 + 2\alpha)(1 - \alpha) = 3 - 3\alpha + 2\alpha - 2\alpha^2 = 5 - \alpha$, matching $(3 + 2i)(1 - i) = 5 - i$.

A field with 4 elements

Over $\mathbb{F}_2$, $x^2 + x + 1$ has no root ($0 \mapsto 1$, $1 \mapsto 1$), so it is irreducible and $\mathbb{F}_4 = \mathbb{F}_2[x]/(x^2 + x + 1)$ is a field with elements $0, 1, \alpha, \alpha + 1$ and the rule $\alpha^2 = \alpha + 1$ (since $-1 = 1$ in characteristic 2).

Addition (coefficients add mod 2):

$+$	$0$	$1$	$\alpha$	$\alpha+1$
$0$	$0$	$1$	$\alpha$	$\alpha+1$
$1$	$1$	$0$	$\alpha+1$	$\alpha$
$\alpha$	$\alpha$	$\alpha+1$	$0$	$1$
$\alpha+1$	$\alpha+1$	$\alpha$	$1$	$0$

Multiplication:

$\times$	$0$	$1$	$\alpha$	$\alpha+1$
$0$	$0$	$0$	$0$	$0$
$1$	$0$	$1$	$\alpha$	$\alpha+1$
$\alpha$	$0$	$\alpha$	$\alpha+1$	$1$
$\alpha+1$	$0$	$\alpha+1$	$1$	$\alpha$

Two entries in detail: $\alpha(\alpha + 1) = \alpha^2 + \alpha = (\alpha + 1) + \alpha = 1$, and $(\alpha + 1)^2 = \alpha^2 + 2\alpha + 1 = \alpha^2 + 1 = \alpha$. Every nonzero element has an inverse, and the multiplicative group $\{1, \alpha, \alpha + 1\}$ is cyclic of order 3.

Common pitfall: $\mathbb{F}_4$ is not $\mathbb{Z}/4\mathbb{Z}$. In $\mathbb{Z}/4\mathbb{Z}$, $2 \cdot 2 = 0$, so it is not even a domain. The additive group of $\mathbb{F}_4$ is $\mathbb{Z}/2 \times \mathbb{Z}/2$, not $\mathbb{Z}/4$.

A field with 8 elements

$x^3 + x + 1$ has no root in $\mathbb{F}_2$ ($0 \mapsto 1$, $1 \mapsto 1$), so, being cubic, it is irreducible, and $\mathbb{F}_8 = \mathbb{F}_2[x]/(x^3 + x + 1)$ has $2^3 = 8$ elements $c_0 + c_1\alpha + c_2\alpha^2$ with $\alpha^3 = \alpha + 1$. The powers of $\alpha$ run through all 7 nonzero elements:

$k$	0	1	2	3	4	5	6
$\alpha^k$	$1$	$\alpha$	$\alpha^2$	$\alpha + 1$	$\alpha^2 + \alpha$	$\alpha^2 + \alpha + 1$	$\alpha^2 + 1$

and $\alpha^7 = \alpha \cdot \alpha^6 = \alpha^3 + \alpha = 1$. This "log table" makes multiplication easy: $\alpha^i \cdot \alpha^j = \alpha^{(i + j) \bmod 7}$. For example $(1 + \alpha)^{-1} = (\alpha^3)^{-1} = \alpha^4 = \alpha^2 + \alpha$; check: $(1 + \alpha)(\alpha^2 + \alpha) = \alpha^3 + \alpha = 1$.

Post 05 proves that for every prime power $p^n$ there is exactly one field of that size, up to isomorphism.

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Practice Exercises

1. List the units and the zero divisors of $\mathbb{Z}/15\mathbb{Z}$.

Solution

Units are the classes coprime to 15: $1, 2, 4, 7, 8, 11, 13, 14$ (there are $\phi(15) = 8$). The other nonzero classes, $3, 5, 6, 9, 10, 12$, are zero divisors: for example $3 \cdot 5 = 15 \equiv 0$ and $6 \cdot 5 = 30 \equiv 0$.

2. Show that $(x)$ is a prime ideal of $\mathbb{Q}[x]$ that is also maximal, while in $\mathbb{Z}[x]$ it is prime but not maximal. What changes?

Solution

Evaluation at 0 gives $\mathbb{Q}[x]/(x) \cong \mathbb{Q}$, a field, so $(x)$ is maximal in $\mathbb{Q}[x]$. The same map gives $\mathbb{Z}[x]/(x) \cong \mathbb{Z}$, a domain but not a field, so $(x)$ is prime but not maximal in $\mathbb{Z}[x]$; indeed $(x) \subsetneq (2, x)$. What changed is the coefficient ring: the quotient is the coefficient ring itself.

3. Decide whether $x^3 + 2x + 1$ is irreducible over $\mathbb{Q}$ and over $\mathbb{F}_3$.

Solution

Over $\mathbb{Q}$: the rational root test leaves $\pm 1$, giving values $4$ and $-2$; no root, degree 3, so irreducible. Over $\mathbb{F}_3$: $f(0) = 1$, $f(1) = 4 \equiv 1$, $f(2) = 13 \equiv 1$; no root, so irreducible, and $\mathbb{F}_3[x]/(x^3 + 2x + 1)$ is a field with $27$ elements.

4. Use Eisenstein's criterion to show $x^4 + 10x + 5$ is irreducible over $\mathbb{Q}$. Why does "it has no rational root" not suffice on its own?

Solution

With $p = 5$: $5 \nmid 1$ (leading coefficient), $5$ divides $0, 0, 10, 5$, and $25 \nmid 5$. So it is irreducible. Absence of roots only rules out linear factors; a quartic could still split into two quadratics, as $x^4 + 4$ does.

5. In $\mathbb{F}_4 = \mathbb{F}_2[x]/(x^2 + x + 1)$, compute $\alpha^3$, $(\alpha + 1)^2$ and $\alpha^{-1}$.

Solution

$\alpha^3 = \alpha \cdot \alpha^2 = \alpha(\alpha + 1) = \alpha^2 + \alpha = 1$. $(\alpha + 1)^2 = \alpha^2 + 1 = \alpha$ (the cross term $2\alpha$ is $0$). Since $\alpha(\alpha + 1) = 1$, $\alpha^{-1} = \alpha + 1$.

6. Show that $3$ is irreducible but not prime in $\mathbb{Z}[\sqrt{-5}]$, using $9 = (2 + \sqrt{-5})(2 - \sqrt{-5})$.

Solution

$N(3) = 9$. A proper factor would have norm 3, and $a^2 + 5b^2 = 3$ has no integer solution, so $3$ is irreducible. Next, $(2 + \sqrt{-5})(2 - \sqrt{-5}) = 4 + 5 = 9$, so $3$ divides this product. But $3$ divides neither factor, since $(2 \pm \sqrt{-5})/3 \notin \mathbb{Z}[\sqrt{-5}]$. So $3$ is not prime.

7. Show that $\mathbb{Z}[i]/(2 + i) \cong \mathbb{Z}/5\mathbb{Z}$.

Solution

Let $\psi : \mathbb{Z} \to \mathbb{Z}[i]/(2 + i)$ be the natural map. It is surjective: $i \equiv -2$ modulo $(2 + i)$, so $a + bi \equiv a - 2b$, an integer. Its kernel contains $5 = (2 + i)(2 - i)$. If an integer $n$ lies in $(2 + i)$, then $n = (2 + i)\gamma$ and $n^2 = N(n) = 5N(\gamma)$, so $5 \mid n^2$ and $5 \mid n$. So the kernel is $5\mathbb{Z}$, and the First Isomorphism Theorem gives $\mathbb{Z}/5\mathbb{Z} \cong \mathbb{Z}[i]/(2 + i)$.

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Cheat-sheet

Concept	Key fact
Ring (this series)	abelian group under $+$, associative $\cdot$ with $1$, distributive
Unit / zero divisor	invertible / nonzero with $ab = 0$ for some nonzero $b$
Integral domain	commutative, $1 \ne 0$, no zero divisors; cancellation holds
Field	every nonzero element a unit; finite domain is a field
Characteristic	least $n$ with $n \cdot 1 = 0$; a domain has 0 or a prime
Ideal	additive subgroup absorbing multiplication; exactly the kernels
First Isomorphism Theorem	$R/\ker\varphi \cong \operatorname{im}\varphi$
Prime ideal	$R/P$ is a domain
Maximal ideal	$R/M$ is a field; maximal implies prime
$F[x]$	division algorithm; PID; degree-$n$ polynomial has at most $n$ roots
Irreducibility	deg 2–3: no roots; rational roots; Eisenstein; reduce mod $p$
Hierarchy	field ⊂ Euclidean ⊂ PID ⊂ UFD ⊂ domain, all strict
Non-UFD	$6 = 2 \cdot 3 = (1+\sqrt{-5})(1-\sqrt{-5})$ in $\mathbb{Z}[\sqrt{-5}]$
New fields	$F[x]/(p)$ is a field exactly when $p$ is irreducible; size $q^n$
$\mathbb{F}_4$	$\mathbb{F}_2[x]/(x^2+x+1)$, $\alpha^2 = \alpha + 1$
$\mathbb{F}_8$	$\mathbb{F}_2[x]/(x^3+x+1)$, $\alpha^3 = \alpha + 1$, $\alpha$ has order 7

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What's next

Post 05 turns the construction of Chapter 6 into a theory. If $p$ is irreducible over $F$, the field $F[x]/(p)$ contains a root of $p$, so it is an extension of $F$ of degree $\deg p$. Degrees multiply in towers, which already settles the Greek problems of doubling the cube and trisecting an angle. Adding all the roots of a polynomial gives its splitting field, and the automorphisms of that field form its Galois group. The Fundamental Theorem of Galois Theory matches subfields with subgroups, and it explains why there is no formula in radicals for the roots of a general quintic.

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References

• D. S. Dummit and R. M. Foote, Abstract Algebra, 3rd edition, Wiley, 2004 — Chapters 7 (rings, ideals, quotients), 8 (Euclidean domains, PIDs, UFDs; the example $\mathbb{Z}[(1+\sqrt{-19})/2]$) and 9 (polynomial rings, irreducibility criteria).
• M. Artin, Algebra, 2nd edition, Pearson, 2011 — Chapters 11 (rings) and 12 (factoring, Gauss's lemma, Gaussian integers).
• J. A. Gallian, Contemporary Abstract Algebra, 10th edition, CRC Press, 2021 — rings, integral domains, ideals and factor rings, divisibility in integral domains.
• T. W. Hungerford, Algebra, Springer GTM 73, 1974 — Chapter III for a more formal treatment (rings without 1, Zorn's lemma for maximal ideals).
• MIT OpenCourseWare, 18.702 Algebra II (Spring 2011) — lecture notes on rings, factoring and finite fields.
• K. Conrad, Expository papers ("blurbs") — short notes on ring theory, Eisenstein's criterion and finite fields.

Note: this post is in progress. Planned additions: a full proof that PIDs are UFDs, Gauss's lemma with proof, localisation and fields of fractions, and a short section on noncommutative examples (quaternions, matrix rings and their two-sided ideals).

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