10.3 Rings, Ideals, Modules, Polynomial Rings, and Fields

10.3 Rings, Ideals, Polynomial Rings, and Field Extensions

Ring and field theory expand algebraic study to systems equipped with two binary operations. On the GRE Mathematics Subject Test, questions test ring axioms, zero divisors, integral domains, prime versus maximal ideals, polynomial factorization in PIDs, and the construction and properties of finite fields.

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Rings, Zero Divisors, and Integral Domains

A ring $(R, +, \cdot)$ is a set $R$ equipped with addition and multiplication such that:

1. $(R, +)$ is an abelian group (additive identity $0$, additive inverse $-a$).
2. Multiplication is associative: $(a \cdot b) \cdot c = a \cdot (b \cdot c)$.
3. Multiplication distributes over addition: $a(b + c) = ab + ac$ and $(a + b)c = ac + bc$.

Special Ring Classifications

• Commutative Ring: $ab = ba$ for all $a, b \in R$.
• Ring with Unity: There exists an element $1 \neq 0$ such that $1 \cdot a = a \cdot 1 = a$ for all $a \in R$.
• Units: An element $u \in R$ is a unit if there exists $v \in R$ such that $uv = vu = 1$. The units form a multiplicative group $R^*$.
• Zero Divisors: A non-zero element $a \in R$ is a zero divisor if there exists a non-zero $b \in R$ such that $ab = 0$ or $ba = 0$.
• Integral Domain: A commutative ring with unity $1 \neq 0$ that has no zero divisors. Equivalently, the cancellation law holds: if $ab = ac$ and $a \neq 0$, then $b = c$.
• Division Ring (Skew Field): A ring with unity in which every non-zero element is a unit (multiplicative inverses exist, but multiplication may be non-commutative).
• Field: A commutative division ring. Every non-zero element is a unit.

Structural Theorems

• Wedderburn's Little Theorem: Every finite division ring is a field.
• Finite Domain Theorem: Every finite integral domain is a field. (Proof: For non-zero $a \in D$, the map $x \mapsto ax$ is injective by cancellation; by finiteness, it is surjective, so $ax = 1$ has a solution).

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Ideals, Quotient Rings, and Homomorphisms

An ideal $I$ of a ring $R$ is an additive subgroup ($I \le (R, +)$) that satisfies the absorption property: $r \in R, x \in I \implies rx \in I \text{ and } xr \in I$

• If $R$ is commutative, the ideal generated by a single element $a$ is the principal ideal $\langle a \rangle = aR = {ar : r \in R}$.
• An ideal $I$ contains a unit if and only if $I = R$. Consequently, the only ideals of a field $F$ are ${0}$ and $F$.

Quotient Ring $R/I$

For an ideal $I \subseteq R$, the set of cosets $R/I = {r + I : r \in R}$ forms a ring under: $(r + I) + (s + I) = (r + s) + I, \qquad (r + I)(s + I) = rs + I$ The absorption property of ideals guarantees that multiplication of cosets is well-defined.

First Isomorphism Theorem for Rings

If $\phi: R \to S$ is a ring homomorphism (preserving addition and multiplication), then:

• The kernel $\ker(\phi) = {r \in R : \phi(r) = 0}$ is an ideal of $R$.
• The quotient ring $R / \ker(\phi)$ is isomorphic to the image $\operatorname{im}(\phi) \subseteq S$: $R / \ker(\phi) \cong \operatorname{im}(\phi)$

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Prime Ideals vs. Maximal Ideals

Let $R$ be a commutative ring with unity and $I \subsetneq R$ a proper ideal:

Characterizations via Quotients

• Prime Ideal: $P$ is prime if whenever $ab \in P$, either $a \in P$ or $b \in P$. $\mathbf{P \text{ is a prime ideal}} \iff R/P \text{ is an integral domain}$
• Maximal Ideal: $M$ is maximal if no proper ideal strictly contains $M$ (if $M \subseteq J \subseteq R$ for an ideal $J$, then $J = M$ or $J = R$). $\mathbf{M \text{ is a maximal ideal}} \iff R/M \text{ is a field}$

Core Implications

Because every field is an integral domain, every maximal ideal is prime.

• The converse is false in general: In $\mathbb{Z}[x]$, the ideal $\langle x \rangle$ is prime because $\mathbb{Z}[x]/\langle x \rangle \cong \mathbb{Z}$ (an integral domain), but it is not maximal because $\mathbb{Z}$ is not a field. The ideal $\langle 2, x \rangle$ strictly contains $\langle x \rangle$.
• In a Principal Ideal Domain (PID), every non-zero prime ideal is maximal.

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Polynomial Rings $F[x]$, PIDs, and UFDs

When $F$ is a field, the polynomial ring $F[x]$ possesses rich algebraic structure:

• Division Algorithm: For $f(x), g(x) \in F[x]$ with $g(x) \neq 0$, there exist unique $q(x), r(x) \in F[x]$ such that $f(x) = g(x)q(x) + r(x)$ with $\deg(r) < \deg(g)$ or $r = 0$.
• Euclidean Domain (ED): The degree function $\deg$ serves as a Euclidean valuation.
• Principal Ideal Domain (PID): Every ideal in $F[x]$ is principal: $I = \langle p(x) \rangle$.
• Unique Factorization Domain (UFD): Every non-constant polynomial factors uniquely into irreducible polynomials up to order and unit associates.

Domain Hierarchy

$\text{Fields} \subset \text{Euclidean Domains} \subset \text{Principal Ideal Domains} \subset \text{Unique Factorization Domains} \subset \text{Integral Domains}$ Note: $\mathbb{Z}[x]$ is a UFD, but not a PID (the ideal $\langle 2, x \rangle$ is not principal).

Irreducibility and Maximality in $F[x]$

Let $p(x) \in F[x]$ with $\deg(p) \ge 1$: $\langle p(x) \rangle \text{ is maximal} \iff \langle p(x) \rangle \text{ is prime} \iff p(x) \text{ is irreducible over } F$ If $p(x)$ is irreducible over $F$, then the quotient ring $E = F[x]/\langle p(x) \rangle$ is a field.

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Field Extensions and Finite Fields $\mathbb{F}_{p^n}$

Constructing Extension Fields

Let $p(x) \in F[x]$ be irreducible with $\deg(p(x)) = n$. The quotient $E = F[x]/\langle p(x) \rangle$ is an extension field of $F$ with degree $[E : F] = n$.

• The element $\alpha = x + \langle p(x) \rangle$ satisfies $p(\alpha) = 0$, meaning $\alpha$ is a root of $p(x)$ in $E$.
• Every element in $E$ has a unique representation as: $c_0 + c_1\alpha + c_2\alpha^2 + \dots + c_{n-1}\alpha^{n-1} \quad (c_i \in F)$

Structure of Finite Fields (Galois Fields $\mathbb{F}_q$)

Let $\mathbb{F}_q$ be a finite field:

1. Prime Power Cardinality: The characteristic of $\mathbb{F}_q$ must be a prime $p$. The field contains $\mathbb{Z}_p$ as its prime subfield, forming a vector space over $\mathbb{Z}_p$ of dimension $n$. Therefore, its order must be a prime power: $q = p^n$
2. Uniqueness: For every prime power $p^n$, there exists a unique field of order $p^n$ up to isomorphism, denoted $\mathbb{F}_{p^n}$ or $\operatorname{GF}(p^n)$.
3. Cyclic Multiplicative Group: The non-zero elements form a cyclic group under multiplication: $(\mathbb{F}_{p^n}^*, \cdot) \cong (\mathbb{Z}_{p^n - 1}, +)$
4. Subfield Criterion: $\mathbb{F}{p^d}$ is a subfield of $\mathbb{F}{p^n}$ if and only if $d$ divides $n$.

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Ring Classification and Ideal Summary

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Step-by-Step Worked Problem

Problem: Construct the finite field $\mathbb{F}_8 = \operatorname{GF}(2^3)$ as a quotient ring $\mathbb{Z}_2[x] / \langle p(x) \rangle$. Verify irreducibility of $p(x) = x^3 + x + 1$, compute the multiplicative inverse of $\alpha + 1$ (where $\alpha = x + \langle p(x) \rangle$), and find the order of $\alpha$ in $\mathbb{F}_8^*$.

Solution:

• Step 1: Verify Irreducibility. For a polynomial of degree 2 or 3 over a field, it is reducible if and only if it has a root in the base field. Testing elements of $\mathbb{Z}_2$ in $p(x) = x^3 + x + 1$:
• $p(0) = 0 + 0 + 1 = 1 \neq 0$.
• $p(1) = 1 + 1 + 1 = 3 \equiv 1 \neq 0$.

Since $p(x)$ has no roots in $\mathbb{Z}_2$, it is irreducible. Hence $\langle x^3 + x + 1 \rangle$ is a maximal ideal in $\mathbb{Z}_2[x]$, and $\mathbb{F}_8 = \mathbb{Z}_2[x]/\langle x^3 + x + 1 \rangle$ is a field with $2^3 = 8$ elements.

• Step 2: Multiplicative Inverse of $\alpha + 1$. In $\mathbb{F}_8$, $\alpha^3 + \alpha + 1 = 0 \implies \alpha^3 = \alpha + 1$ (since $-1 \equiv +1 \pmod 2$). We seek $\beta = a_2\alpha^2 + a_1\alpha + a_0$ such that $(\alpha + 1)\beta = 1$. Multiply $(\alpha + 1)(\alpha^2 + \alpha)$: $(\alpha + 1)(\alpha^2 + \alpha) = \alpha^3 + \alpha^2 + \alpha^2 + \alpha = \alpha^3 + 2\alpha^2 + \alpha = (\alpha + 1) + 0 + \alpha = 2\alpha + 1 \equiv 1 \pmod 2$ Thus, $(\alpha + 1)^{-1} = \alpha^2 + \alpha$.

• Step 3: Multiplicative Order of $\alpha$. The group $\mathbb{F}_8^$ has order $8 - 1 = 7$. By Lagrange's theorem, the order of $\alpha$ must divide 7. Since $7$ is prime and $\alpha \neq 1$, $|\alpha| = 7$. The element $\alpha$ is a primitive element (generator) of $\mathbb{F}_8^$.

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GRE Exam Traps & Pitfalls

Trap 1: Confusing Prime and Maximal Ideals in Non-PIDs An ideal $P$ being prime only guarantees that $R/P$ is an integral domain, not a field. In $\mathbb{Z}[x]$, $\langle x \rangle$ is prime because $\mathbb{Z}[x]/\langle x \rangle \cong \mathbb{Z}$, but it is not maximal because $\mathbb{Z}$ is not a field.

Trap 2: Assuming Finite Field $\mathbb{F}{p^n}$ Is Isomorphic to $\mathbb{Z}{p^n}$ The finite field $\mathbb{F}{p^n}$ has characteristic $p$, meaning $p \cdot 1 = 0$. Its additive group is isomorphic to $(\mathbb{Z}p)^n$. In contrast, $\mathbb{Z}{p^n}$ has characteristic $p^n$ and contains zero divisors (e.g., $p \cdot p^{n-1} = 0$), so $\mathbb{Z}{p^n}$ is never a field for $n > 1$.

Trap 3: Irreducibility Is Field-Dependent The polynomial $x^2 + 1$ is irreducible over $\mathbb{R}$ (so $\mathbb{R}[x]/\langle x^2 + 1 \rangle \cong \mathbb{C}$ is a field), but reducible over $\mathbb{C}$ as $(x - i)(x + i)$, and reducible over $\mathbb{Z}_2$ as $(x + 1)^2$.

Trap 4: Subfield Divisibility Condition $\mathbb{F}{p^d}$ is a subfield of $\mathbb{F}{p^n}$ if and only if $d \mid n$, not $p^d \mid p^n$. For instance, $\mathbb{F}{2^2}$ is a subfield of $\mathbb{F}{2^6}$ because $2 \mid 6$, but $\mathbb{F}{2^4}$ is not a subfield of $\mathbb{F}{2^6}$ because $4 \nmid 6$.

Modules: Vector Spaces over Rings

Let $R$ be a ring with identity. A left $R$-module $M$ is an abelian group under addition together with scalar multiplication $R\times M\to M$ satisfying distributivity, associativity $(rs)m=r(sm)$, and $1_Rm=m$. A vector space is exactly a module whose scalar ring is a field. The distinction matters: a nonzero scalar in a ring need not be invertible, so familiar vector-space conclusions can fail.

Examples organize the theory:

• Every abelian group is a $\mathbb Z$-module, with $n\cdot m$ defined by repeated addition and additive inverses.
• Every ideal $I$ of a commutative ring $R$ is an $R$-submodule of the regular module $R$.
• $R^n$ is a free $R$-module with standard generators, but not every module has a basis.
• The quotient $M/N$ by a submodule $N$ is an $R$-module with $(m+N)+(m'+N)=(m+m')+N$ and $r(m+N)=rm+N$.

A nonempty subset $N\subseteq M$ is a submodule when it is closed under subtraction and scalar multiplication: $x,y\in N$ and $r\in R$ imply $x-y\in N$ and $rx\in N$. An $R$-linear map $\phi:M\to N$ preserves addition and scalar multiplication. Its kernel and image are submodules, and the module form of the First Isomorphism Theorem is $M/\ker\phi\cong\operatorname{im}\phi.$

A module generated by one element $m$ is cyclic: $Rm=\{rm:r\in R\}.$ As a $\mathbb Z$-module, every cyclic group is either $\mathbb Z$ or $\mathbb Z/n\mathbb Z$. Unlike vector spaces, a linearly independent spanning set need not exist. For example, $\mathbb Z/6\mathbb Z$ is a cyclic $\mathbb Z$-module but cannot be a free $\mathbb Z$-module because the nonzero element $\bar1$ is annihilated by 6.

Linear operators as polynomial-module actions

Let $V$ be a vector space over a field $F$ and $T:V\to V$ a linear operator. Define an $F[x]$-module structure on $V$ by $p(x)\cdot v=p(T)v.$ Then an $F$-subspace $W\subseteq V$ is an $F[x]$-submodule exactly when it is $T$-invariant: $T(W)\subseteq W$. The annihilator of a vector, $\operatorname{Ann}(v)=\{p\in F[x]:p(T)v=0\},$ is an ideal. Since $F[x]$ is a principal ideal domain, it is generated by a single monic polynomial. For a cyclic vector whose iterates span $V$, that generator records the operator's minimal polynomial on the cyclic subspace.

This viewpoint explains why quotient modules $F[x]/(p)$ model cyclic linear operators. In the basis $1,x,\ldots,x^{n-1}$ modulo a monic polynomial $p$ of degree $n$, multiplication by $x$ is represented by the companion matrix of $p$.

Common traps

Submodules must contain zero and be closed under multiplication by every ring scalar. A module over a general ring need not have a basis or a well-defined dimension. Torsion—nonzero $m$ with $rm=0$ for nonzero $r$—cannot occur in a vector space over a field but is common in modules.