Chapter 18

Rigorous

Field Extensions

00 · Symbol Glossary

$K/F$K is an extension of F

Read " $K$ over $F$ ." Means $F$ is a subfield of $K$ . The slash is notation, not division.

$[K:F]$Degree of the extension

The dimension of $K$ as a vector space over $F$ (using $F$ 's scalars, $K$ 's vectors, and $K$ 's own addition and multiplication-by- $F$ -elements). Finite or infinite.

$F(\alpha)$Simple extension generated by α

The smallest field containing both $F$ and $\alpha$ . Read " $F$ adjoin alpha."

$m_\alpha(x)$Minimal polynomial of α over F

The monic polynomial of least degree in $F[x]$ having $\alpha$ as a root, when $\alpha$ is algebraic over $F$ .

$F[\alpha]$Subring generated by α

The smallest subring of $K$ containing $F$ and $\alpha$ — polynomial expressions in $\alpha$ with coefficients in $F$ , before knowing whether inverses are needed.

01 · Extensions and Degree

Definition — Field Extension and Degree

If $F$ is a subfield of $K$ , $K$ is an extension of $F$ , written $K/F$ . Since $F$ 's scalars act on $K$ 's elements via $K$ 's own multiplication, $K$ is naturally a vector space over $F$ ; its dimension is the degree $[K:F]$ .

Example — [C:R] = 2

Every complex number is uniquely $a+bi$ with $a,b\in\mathbb{R}$ — exactly the statement that $\{1,i\}$ is a basis for $\mathbb{C}$ as an $\mathbb{R}$ -vector space. So $[\mathbb{C}:\mathbb{R}]=2$ .

Example — [R:Q] = ∞

$\mathbb{R}$ contains uncountably many elements but $\mathbb{Q}$ -linear combinations of any finite list of reals form only a countable set — so no finite basis can span $\mathbb{R}$ over $\mathbb{Q}$ . $[\mathbb{R}:\mathbb{Q}]=\infty$ .

02 · The Tower Law

Theorem 18.1 — Tower Law

If $F\subseteq K\subseteq L$ are fields with $[K:F]$ and $[L:K]$ both finite, then $[L:F]$ is finite and:

[L:F] = [L:K]\cdot[K:F]

Proof

Let $\{a_1,\ldots,a_m\}$ be a basis for $K$ over $F$ ( $m=[K:F]$ ) and $\{b_1,\ldots,b_n\}$ a basis for $L$ over $K$ ( $n=[L:K]$ ). Claim: $\{a_ib_j : 1\le i\le m,\,1\le j\le n\}$ ( $mn$ elements) is a basis for $L$ over $F$ .

Spanning. Any $\ell\in L$ is a $K$ -linear combination $\ell = \sum_j c_jb_j$ ( $c_j\in K$ ). Each $c_j$ , being in $K$ , is an $F$ -linear combination $c_j=\sum_i d_{ij}a_i$ ( $d_{ij}\in F$ ). Substituting: $\ell = \sum_j\big(\sum_i d_{ij}a_i\big)b_j = \sum_{i,j}d_{ij}(a_ib_j)$ — an $F$ -linear combination of the $a_ib_j$ .

Independence. Suppose $\sum_{i,j}d_{ij}a_ib_j=0$ with $d_{ij}\in F$ . Regroup as $\sum_j\big(\sum_i d_{ij}a_i\big)b_j=0$ . Each coefficient $\sum_i d_{ij}a_i$ lies in $K$ ; since $\{b_j\}$ is independent over $K$ , every one of these $K$ -coefficients must be $0$ : $\sum_i d_{ij}a_i=0$ for each $j$ . Since $\{a_i\}$ is independent over $F$ , this forces $d_{ij}=0$ for every $i$ (for each fixed $j$ ). So all $d_{ij}=0$ .

The $mn$ products $a_ib_j$ span $L$ and are $F$ -linearly independent: a basis. So $[L:F]=mn=[L:K][K:F]$ .

Example — Applying the Tower Law

$\mathbb{Q}\subseteq\mathbb{Q}(\sqrt2)\subseteq\mathbb{Q}(\sqrt2,\sqrt3)$ . If $[\mathbb{Q}(\sqrt2):\mathbb{Q}]=2$ and $[\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}(\sqrt2)]=2$ (verified in Section 05), Theorem 18.1 gives $[\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}] = 2\times2=4$ — instantly, without constructing a basis of the full four-dimensional space by hand.

03 · Algebraic Elements and Minimal Polynomials

Definition — Algebraic and Transcendental

Let $K/F$ be an extension. $\alpha\in K$ is algebraic over $F$ if $f(\alpha)=0$ for some nonzero $f\in F[x]$ ; otherwise $\alpha$ is transcendental over $F$ .

Definition — Minimal Polynomial

If $\alpha$ is algebraic over $F$ , its minimal polynomial $m_\alpha(x)$ is the monic polynomial of least degree in $F[x]$ with $m_\alpha(\alpha)=0$ .

Theorem 18.2 — The Minimal Polynomial Is Irreducible and Unique

$m_\alpha(x)$ is irreducible over $F$ , and it is the unique monic polynomial of its degree vanishing at $\alpha$ .

Proof

Irreducible. Suppose $m_\alpha = fg$ with $\deg f,\deg g < \deg m_\alpha$ . Then $f(\alpha)g(\alpha) = m_\alpha(\alpha)=0$ ; since $K$ is a field (no zero divisors), $f(\alpha)=0$ or $g(\alpha)=0$ . Either way, a polynomial of degree strictly less than $\deg m_\alpha$ vanishes at $\alpha$ — contradicting minimality (after scaling to make it monic, using that $F$ is a field so any nonzero leading coefficient is invertible). So no such factorization exists (beyond trivial unit factors): $m_\alpha$ is irreducible.

Unique. Suppose $g$ is also monic of degree $\deg m_\alpha$ with $g(\alpha)=0$ . Then $(m_\alpha-g)(\alpha) = 0-0=0$ , and $\deg(m_\alpha-g) < \deg m_\alpha$ (the leading terms, both monic of the same degree, cancel). If $m_\alpha-g\neq0$ , scaling to monic gives a smaller-degree polynomial vanishing at $\alpha$ , contradicting minimality. So $m_\alpha-g=0$ : $g=m_\alpha$ .

04 · Simple Extensions

The minimal polynomial completely determines the structure of the smallest field containing $\alpha$ .

Lemma 18.3 — Nonzero Prime Ideals in a PID Are Maximal

In a PID $R$ , every nonzero prime ideal $(p)$ is maximal.

Proof

Suppose $(p)\subseteq(d)\subseteq R$ for some $d\in R$ . Then $d\mid p$ , so $p=dc$ for some $c\in R$ . Since $(p)$ is prime, $p$ is a prime element (Chapter 15's convention), hence irreducible (Theorem 15.3). So in $p=dc$ , either $d$ or $c$ is a unit. If $d$ is a unit, $(d)=R$ . If $c$ is a unit, $d=pc^{-1}$ is an associate of $p$ , so $(d)=(p)$ . Either way, no ideal sits strictly between $(p)$ and $R$ : $(p)$ is maximal.

Theorem 18.4 — Structure of F(α) for Algebraic α

Let $\alpha\in K$ be algebraic over $F$ with minimal polynomial $m_\alpha$ of degree $n$ . Then:

F(\alpha) \;\cong\; F[x]/(m_\alpha), \qquad [F(\alpha):F] = n

with $\{1,\alpha,\alpha^2,\ldots,\alpha^{n-1}\}$ a basis for $F(\alpha)$ over $F$ .

Proof

Consider the evaluation homomorphism $\mathrm{ev}_\alpha: F[x]\to K$ , $f\mapsto f(\alpha)$ (a ring homomorphism, Chapter 13's evaluation example specialized to polynomials). $\ker(\mathrm{ev}_\alpha) = \{f\in F[x] : f(\alpha)=0\}$ is an ideal (Theorem 13.3); since $F[x]$ is Euclidean (Chapter 15), hence a PID, this ideal is $(g)$ for some $g$ , and $g$ has the minimal degree among nonzero elements of the ideal (the proof of Theorem 15.1 identifies the generator exactly this way) — so $g$ is (up to a unit scalar) the minimal polynomial: $\ker(\mathrm{ev}_\alpha) = (m_\alpha)$ .

$(m_\alpha)$ is a nonzero prime ideal: $m_\alpha$ is irreducible (Theorem 18.2), hence prime in the PID $F[x]$ (Theorem 15.4). By Lemma 18.3, $(m_\alpha)$ is maximal. By Theorem 12.6, $F[x]/(m_\alpha)$ is a field.

By the First Isomorphism Theorem (Theorem 13.6): $F[x]/\ker(\mathrm{ev}_\alpha) \cong \mathrm{im}(\mathrm{ev}_\alpha)$ , i.e. $F[x]/(m_\alpha) \cong F[\alpha]$ (the image is exactly the subring generated by $\alpha$ over $F$ , since every polynomial expression in $\alpha$ is hit). Since the left side is a field, $F[\alpha]$ is already a field — meaning $F[\alpha]=F(\alpha)$ (a field containing $F$ and $\alpha$ that's already closed under inverses needs nothing more adjoined).

Finally, by the division algorithm (Theorem 14.5), every coset in $F[x]/(m_\alpha)$ has a unique representative of degree $<n$ (divide by $m_\alpha$ and keep the remainder), so $\{1,x,\ldots,x^{n-1}\}+(m_\alpha)$ is a basis for $F[x]/(m_\alpha)$ over $F$ , corresponding to $\{1,\alpha,\ldots,\alpha^{n-1}\}$ under the isomorphism. So $[F(\alpha):F]=n$ .

Example — Q(√2)

$\sqrt2$ is a root of $x^2-2$ , irreducible over $\mathbb{Q}$ (no rational root, by the classical proof that $\sqrt2$ is irrational, together with Corollary 14.7 — a degree-2 polynomial with no root can't factor into two degree-1 pieces). So $m_{\sqrt2}(x)=x^2-2$ , and by Theorem 18.4:

\mathbb{Q}(\sqrt2) \cong \mathbb{Q}[x]/(x^2-2), \qquad [\mathbb{Q}(\sqrt2):\mathbb{Q}]=2

with basis $\{1,\sqrt2\}$ — matching the familiar description $\mathbb{Q}(\sqrt2) = \{a+b\sqrt2 : a,b\in\mathbb{Q}\}$ .

Example — Constructing C purely algebraically

$x^2+1$ is irreducible over $\mathbb{R}$ (Chapter 15, Section 06's example: no real root). By Theorem 18.4, $\mathbb{R}[x]/(x^2+1)$ is a field of degree 2 over $\mathbb{R}$ , with basis $\{1,\bar x\}$ (writing $\bar x$ for the coset $x+(x^2+1)$ ) satisfying $\bar x^2 = -1$ (since $x^2\equiv-1$ modulo $x^2+1$ ). This field is exactly $\mathbb{C}$ , with $\bar x$ playing the role of $i$ — the complex numbers, built with no appeal to square roots of negative numbers, purely as a quotient of a polynomial ring.

05 · A Degree-4 Extension

Example — [Q(√2,√3):Q] = 4

$[\mathbb{Q}(\sqrt2):\mathbb{Q}]=2$ (Section 04). Working inside $K=\mathbb{Q}(\sqrt2,\sqrt3)$ , consider $\sqrt3$ over the base field $\mathbb{Q}(\sqrt2)$ : $\sqrt3$ satisfies $x^2-3$ , and this remains irreducible over $\mathbb{Q}(\sqrt2)$ (if $\sqrt3=a+b\sqrt2$ for $a,b\in\mathbb{Q}$ , squaring gives $3=a^2+2b^2+2ab\sqrt2$ ; since $\sqrt2\notin\mathbb{Q}$ , this forces $ab=0$ , and checking $b=0$ gives $3=a^2$ with no rational solution, while $a=0$ gives $3=2b^2$ , also no rational solution — so no such $a,b$ exist, and $\sqrt3\notin\mathbb{Q}(\sqrt2)$ ). So $[\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}(\sqrt2)]=2$ by Theorem 18.4 applied over the base field $\mathbb{Q}(\sqrt2)$ .

By the Tower Law (Theorem 18.1): $[\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}] = 2\times2=4$ , with basis $\{1,\sqrt2,\sqrt3,\sqrt6\}$ (products of the two individual bases, as in Theorem 18.1's proof).

06 · Finite Extensions Are Algebraic

Theorem 18.5 — Finite Extensions Are Algebraic

If $[K:F]$ is finite, every $\alpha\in K$ is algebraic over $F$ .

Proof

Let $n=[K:F]$ . The $n+1$ elements $1,\alpha,\alpha^2,\ldots,\alpha^n$ all live in the $n$ -dimensional $F$ -vector space $K$ , so they are linearly dependent over $F$ (more vectors than the dimension): there exist $c_0,\ldots,c_n\in F$ , not all zero, with $c_0+c_1\alpha+\cdots+c_n\alpha^n=0$ . This is exactly the statement that $\alpha$ is a root of the nonzero polynomial $c_0+c_1x+\cdots+c_nx^n\in F[x]$ : $\alpha$ is algebraic over $F$ .

Extensions built from algebraic elements are automatically finite too

The converse direction — combining Theorem 18.4's degree formula with the Tower Law — shows that adjoining finitely many algebraic elements always produces a finite extension, exactly as Section 05's example demonstrated for $\sqrt2,\sqrt3$ . Finiteness and "built from algebraic elements" turn out to be two sides of the same coin, a theme Chapter 20's Galois correspondence relies on constantly.

07 · Exercises

EXERCISE 18.1

Apply the Tower Law directly, using the given intermediate degrees.

By Theorem 18.1: $[L:F] = [L:K][K:F] = 3\times5=15$ .

If $[K:F]=5$ and $[L:K]=3$ , find $[L:F]$ using the Tower Law.

EXERCISE 18.2

Find a polynomial with rational coefficients having $\sqrt[3]2$ as a root, and check irreducibility (no rational root, and it can't factor as degree 1 times degree 2 without a rational root).

$\sqrt[3]2$ is a root of $x^3-2$ . Rational root test: any rational root would be $\pm1,\pm2$ (divisors of the constant term), and none satisfy $x^3=2$ . Since a degree-3 polynomial with no rational root cannot factor into a linear times a quadratic (the linear factor would supply a rational root, by the Factor Theorem, Theorem 14.6), $x^3-2$ is irreducible over $\mathbb{Q}$ .

By Theorem 18.4, $m_{\sqrt[3]2}(x)=x^3-2$ and $[\mathbb{Q}(\sqrt[3]2):\mathbb{Q}]=3$ .

Find the minimal polynomial of $\sqrt[3]{2}$ over $\mathbb{Q}$ , and state $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]$ .

EXERCISE 18.3

Apply Theorem 18.5 directly — you don't need to find the polynomial explicitly.

Since $[\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}]=4$ (Section 05) is finite, Theorem 18.5 guarantees $\sqrt2+\sqrt3$ (an element of this field) is algebraic over $\mathbb{Q}$ , without needing to exhibit the polynomial. (It happens to satisfy $x^4-10x^2+1=0$ , found by repeatedly squaring $x=\sqrt2+\sqrt3$ and eliminating radicals — but Theorem 18.5 guarantees algebraicity before any such computation.)

Using Theorem 18.5 alone (without finding an explicit polynomial), explain why $\sqrt2+\sqrt3$ must be algebraic over $\mathbb{Q}$ .

EXERCISE 18.4

Apply Theorem 18.4 with the given irreducible polynomial directly.

$x^2+x+1$ is irreducible over $\mathbb{Z}_2$ (checking both elements of $\mathbb{Z}_2$ : $0^2+0+1=1\neq0$ ; $1^2+1+1=1\neq0$ — no root, and a degree-2 polynomial with no root over a field is irreducible). By Theorem 18.4, $\mathbb{Z}_2[x]/(x^2+x+1)$ is a field with $[\,\cdot\,:\mathbb{Z}_2]=2$ , hence $2^2=4$ elements: basis $\{1,\bar x\}$ over $\mathbb{Z}_2$ , giving elements $\{0,1,\bar x,1+\bar x\}$ . This is a field with exactly 4 elements — resolving Exercise 17.2's question by explicit construction.

Using Theorem 18.4 and the irreducible polynomial $x^2+x+1$ over $\mathbb{Z}_2$ , construct a field with exactly 4 elements.

08 · Chapter Summary

Concept	Statement
Extension, degree	$K/F$ ; $[K:F] = \dim_F K$
Tower Law	$[L:F]=[L:K][K:F]$ (Thm 18.1)
Algebraic / transcendental	Root of some / no nonzero $f\in F[x]$
Minimal polynomial	Least-degree monic vanishing at $\alpha$ ; irreducible and unique (Thm 18.2)
Nonzero prime ideals in a PID	Always maximal (Lemma 18.3)
$F(\alpha)$ structure	$F(\alpha)\cong F[x]/(m_\alpha)$ ; $[F(\alpha):F]=\deg m_\alpha$ (Thm 18.4)
$\mathbb{C}$ from $\mathbb{R}$	$\mathbb{R}[x]/(x^2+1) \cong \mathbb{C}$
Finite $\implies$ algebraic	$[K:F]$ finite $\implies$ every element of $K$ algebraic over $F$ (Thm 18.5)

Next: Chapter 19 — Splitting Fields & Algebraic Closures builds the smallest extension in which a given polynomial factors completely into linear pieces, the essential setting for Galois theory.