Field Theory

12.1 Field Extensions

A field extension is an inclusion $F \subseteq E$ of fields. We write $E/F$ and call $E$ an extension field of $F$ .

The degree of the extension, denoted $[E : F]$ Is the dimension of $E$ as a vector space over $F$ .

Proposition 12.1. If $F \subseteq E \subseteq K$ are field extensions, then $[K : F] = [K : E][E : F]$ .

Proof. If $\{\alpha_i\}$ is a basis for $E/F$ and $\{\beta_j\}$ is a basis for $K/E$ Then $\{\alpha_i \beta_j\}$ is a basis for $K/F$ . Count dimensions. $\blacksquare$

12.2 Algebraic Extensions

An element $\alpha \in E$ is algebraic over $F$ if there exists a non-zero polynomial $f \in F[x]$ With $f(\alpha) = 0$ . Otherwise $\alpha$ is transcendental over $F$ .

The minimal polynomial of $\alpha$ over $F$ is the monic polynomial of smallest degree in $F[x]$ Having $\alpha$ as a root.

Proposition 12.2. The minimal polynomial of $\alpha$ over $F$ is irreducible in $F[x]$ .

Proof. If $m_\alpha = fg$ with $\deg(f), \deg(g) \lt \deg(m_\alpha)$ Then $f(\alpha)g(\alpha) = 0$ So either $f(\alpha) = 0$ or $g(\alpha) = 0$ Contradicting the minimality of $\deg(m_\alpha)$ . $\blacksquare$

Theorem 12.3. $\alpha$ is algebraic over $F$ if and only if $[F(\alpha) : F] \lt \infty$ . In this case, $[F(\alpha) : F] = \deg(m_\alpha)$ .

Proof. If $\alpha$ is algebraic with minimal polynomial $m_\alpha$ of degree $n$ Then $\{1, \alpha, \alpha^2, \ldots, \alpha^{n-1}\}$ is a basis for $F(\alpha)/F$ (every element can be Reduced modulo $m_\alpha$ ), so $[F(\alpha) : F] = n$ . Conversely, if $[F(\alpha) : F] = n \lt \infty$ Then $\{1, \alpha, \ldots, \alpha^n\}$ is linearly dependent, giving a polynomial relation $f(\alpha) = 0$ . $\blacksquare$

12.3 Constructing Extension Fields

Theorem 12.4 (Kronecker”s Theorem). If $F$ is a field and $f \in F[x]$ is irreducible, then $E = F[x] / (f)$ is a field extension of $F$ containing a root of $f$ .

Proof. Since $f$ is irreducible and $F[x]$ is a PID, $(f)$ is a maximal ideal, so $E = F[x]/(f)$ Is a field. The element $\alpha = x + (f) \in E$ satisfies $f(\alpha) = f(x + (f)) = f(x) + (f) = (f) = 0$ I.e., $\alpha$ is a root of $f$ . $\blacksquare$

12.4 Finite Fields

Theorem 12.5. For every prime $p$ and every $n \geq 1$ There exists a field of order $p^n$ Unique up to isomorphism.

Proof (existence). Consider the splitting field of $f(x) = x^{p^n} - x$ over $\mathbb{F}_p$ . The set of roots of $f$ in the splitting field forms a field (since roots are closed under addition, Multiplication, and taking inverses), and it has exactly $p^n$ elements. $\blacksquare$

Proposition 12.6. The multiplicative group $\mathbb{F}_{p^n}^*$ of a finite field is cyclic.

Proof. $\mathbb{F}_{p^n}^*$ is a finite abelian group of order $p^n - 1$ . Let $m$ be the largest order of any element. By Lagrange, every element’s order divides $m$ . So $x^m = 1$ for all $x \in \mathbb{F}_{p^n}^*$ Meaning every element is a root of $x^m - 1$ . Since $x^m - 1$ has at most $m$ roots in a field, $p^n - 1 \leq m$ . But $m$ divides $p^n - 1$ So $m = p^n - 1$ . $\blacksquare$

12.5 Algebraic Closure

A field $F$ is algebraically closed if every non-constant polynomial in $F[x]$ has a root in $F$ .

Theorem 12.7 (Fundamental Theorem of Algebra). $\mathbb{C}$ is algebraically closed.

Remark. Every field $F$ has an algebraic closure $\overline{F}$ : an algebraically closed field That is an algebraic extension of $F$ . The algebraic closure is unique up to $F$ -isomorphism. For example, $\overline{\mathbb{Q}}$ is the field of all algebraic numbers. It is countable and Infinite-dimensional over $\mathbb{Q}$ .

12.6 Worked Examples: Field Extensions

Problem. Compute $[\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}]$ and find the minimal polynomial of $\sqrt{2} + \sqrt{3}$ over $\mathbb{Q}$ .

Solution

Solution. First, $[\mathbb{Q}(\sqrt{2}) : \mathbb{Q}] = 2$ since $x^2 - 2$ is irreducible over $\mathbb{Q}$ (by Eisenstein with $p = 2$ ). Then $\sqrt{3} \notin \mathbb{Q}(\sqrt{2})$ : if $\sqrt{3} = a + b\sqrt{2}$ With $a, b \in \mathbb{Q}$ Squaring gives $3 = a^2 + 2b^2 + 2ab\sqrt{2}$ Forcing $ab = 0$ . If $b = 0$ : $a^2 = 3$ Impossible in $\mathbb{Q}$ . If $a = 0$ : $2b^2 = 3$ Impossible in $\mathbb{Q}$ . So $[\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}(\sqrt{2})] = 2$ .

By the tower law: $[\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}] = 2 \cdot 2 = 4$ .

For the minimal polynomial of $\alpha = \sqrt{2} + \sqrt{3}$ : compute powers. $\alpha^2 = 5 + 2\sqrt{6}$ So $\alpha^2 - 5 = 2\sqrt{6}$ Giving $\alpha^4 - 10\alpha^2 + 25 = 24$ Hence $\alpha^4 - 10\alpha^2 + 1 = 0$ . One checks that $f(x) = x^4 - 10x^2 + 1$ is irreducible over $\mathbb{Q}$ (no rational roots, no quadratic factor), so $m_\alpha = x^4 - 10x^2 + 1$ . $\blacksquare$

Problem. Show that $\mathbb{Q}(\sqrt[4]{2})$ is not a Galois extension of $\mathbb{Q}$ .

Solution

Solution. The minimal polynomial of $\sqrt[4]{2}$ is $x^4 - 2$ (irreducible by Eisenstein with $p = 2$ ), So $[\mathbb{Q}(\sqrt[4]{2}) : \mathbb{Q}] = 4$ . The roots of $x^4 - 2$ are $\sqrt[4]{2}$ , $i\sqrt[4]{2}$ $-\sqrt[4]{2}$ , $-i\sqrt[4]{2}$ . The root $i\sqrt[4]{2}$ is not in $\mathbb{Q}(\sqrt[4]{2}) \subset \mathbb{R}$ .

Therefore $\mathbb{Q}(\sqrt[4]{2})$ is not the splitting field of $x^4 - 2$ And $|\mathrm{Aut}(\mathbb{Q}(\sqrt[4]{2})/\mathbb{Q})| = 2 < 4$ . The extension is not Galois. $\blacksquare$

Problem. Construct $\mathbb{F}_9$ as a quotient of $\mathbb{F}_3[x]$ .

Solution

Solution. We need an irreducible polynomial of degree $2$ in $\mathbb{F}_3[x]$ . Check $x^2 + 1$ : $f(0) = 1$ , $f(1) = 2$ , $f(2) = 4 + 1 = 5 \equiv 2 \pmod{3}$ . No roots, so irreducible. Thus $\mathbb{F}_9 = \mathbb{F}_3[x]/(x^2 + 1)$ .

Let $\alpha = x + (x^2 + 1)$ So $\alpha^2 = -1 = 2$ in $\mathbb{F}_3$ . Then: $\mathbb{F}_9 = \{a + b\alpha : a, b \in \mathbb{F}_3\} = \{0, 1, 2, \alpha, 1+\alpha, 2+\alpha, 2\alpha, 1+2\alpha, 2+2\alpha\}$ .

Multiplication: $(a + b\alpha)(c + d\alpha) = (ac + 2bd) + (ad + bc)\alpha$ . $\blacksquare$

12.7 The Primitive Element Theorem

Theorem 12.8 (Primitive Element Theorem). Every finite separable extension $E/F$ is simple: There exists $\theta \in E$ such that $E = F(\theta)$ .

Proof (sketch). If $F$ is infinite, it suffices to find $\theta = \alpha + c\beta$ for suitable $c \in F$ When $E = F(\alpha, \beta)$ . Only finitely many values of $c$ fail to work. For $F$ of characteristic $0$ Every finite extension is separable, so every finite extension of $\mathbb{Q}$ is simple. $\blacksquare$

Corollary 12.9. Every finite extension of $\mathbb{Q}$ is simple.

Example. $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})$ .

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