Field Theory

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. $\blacksquare$

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 $f(\alpha) = 0$. $\blacksquare$

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 $f$. $\blacksquare$

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 $m = p^n - 1$. $\blacksquare$

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 $\mathbb{Q}$.

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

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

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

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