Extended Field Trip

Having explored the realms of groups and rings, we can now explore fields!

For any group $G$ with identity $0$ and set $K \subseteq G$ , define $K^* := K \backslash \{0 \}$ .

Definition 1. A field is a ring $\mathbb F$ with multiplicative identity $1$ such that $(\mathbb F, +)$ and $(\mathbb F^*, \cdot)$ form Abelian groups. Given two fields $\mathbb F, \mathbb K$ , we call $\mathbb K$ a field extension of $\mathbb F$ if $\mathbb F \subseteq \mathbb K$ .

Our discussion on rings built up to our discussion on polynomials, because irreducible polynomials help us generate new fields. Let $f \in \mathbb F[x]$ denote an irreducible polynomial.

Example 1. The ring $\mathbb F[x]/\langle f \rangle$ forms a field extension of $\mathbb F$ . In particular, $\mathbb C \cong \mathbb R[x]/\langle x^2 + 1 \rangle$ is a field extension of $\mathbb R$ . Denote $i \equiv x + \langle x^2 + 1 \rangle$ .

Example 2. Slightly more generally, for any prime number $p$ , $\mathbb Q[x]/\langle x^2 - p \rangle$ is a field extension of $\mathbb Q$ . Denote $\sqrt{p} \equiv x + \langle x^2 - p \rangle$ .

Observe that in both examples, the root $\alpha := x + \langle f \rangle$ satisfies $f(\alpha) = 0$ . Therefore, $\mathbb F[x]/\langle f \rangle$ contains both $\mathbb F$ and $\alpha$ . Fix $\mathbb F \subseteq \mathbb K$ as fields.

Definition 2. For any $K \subseteq \mathbb K \backslash \mathbb F$ , denote the smallest sub-field of $\mathbb K$ containing $\mathbb F$ and $K$ by $\mathbb F(K)$ , so that $\mathbb F \subseteq \mathbb F(K) \subseteq \mathbb K$ .

In the case $K = \{\alpha\}$ , denote $\mathbb F(\alpha) := \mathbb F(\{\alpha\})$ , which we call a simple extension of $\mathbb F$ .

Theorem 1. $\mathbb F[x]/\langle f \rangle \cong \mathbb F(\alpha)$ .

Proof. Define the map $\phi : \mathbb F[x] \to \mathbb F(\alpha)$ by $\phi(g) = g(\alpha)$ . We remark that $\ker(\phi) = \langle f \rangle$ , and by construction,

$\phi(\mathbb F[x]) \subseteq \mathbb F(\alpha).$

By the first isomorphism theorem for rings, $\phi(\mathbb F[x]) \cong \mathbb F[x]/\langle f \rangle$ is a field that contains $\mathbb F \cup \{\alpha\}$ . By Definition 2,

$\mathbb F(\alpha) \subseteq \phi(\mathbb F[x]).$

Therefore, $\mathbb F[x]/\langle f \rangle \cong \phi(\mathbb F[x]) = \mathbb F(\alpha)$ .

Since $\alpha$ is constructed as a root of $f$ in the field extension $\mathbb K = \mathbb F[x]/\langle f \rangle$ , we call $\mathbb F(\alpha)$ an algebraic extension of $\mathbb F$ .

Definition 2. An element $\alpha \in \mathbb K$ is algebraic (over $\mathbb F$ ) if there exists a polynomial $f \in \mathbb F[x]$ such that $f(\alpha) = 0$ .

The element $\alpha$ is transcendental over $\mathbb F$ if it is not algebraic over $\mathbb F$ .

A field extension $\mathbb K \supseteq \mathbb F$ is algebraic if every element in $\mathbb K$ is algebraic over $\mathbb F$ .

Example 3. Using Examples 1 and 2, $\mathbb C = \mathbb R(i) \supseteq \mathbb R$ and $\mathbb Q(\sqrt p) \supseteq \mathbb Q$ are an algebraic extensions.

Theorem 2. For any $\alpha \in \mathbb K$ , exactly one of the following holds.

If $\alpha$ is algebraic, then there exists a unique monic irreducible polynomial $f_{\mathbb F}^{\alpha} \in \mathbb F[x]$ such that $f_{\mathbb F}^{\alpha}(\alpha) = 0$ and $\mathbb F[x]/\langle f_{\mathbb F}^{\alpha} \rangle \cong \mathbb F(\alpha)$ .
If $\alpha$ is transcendental, then $\mathbb F(\alpha) \cong Q(\mathbb F[x])$ , where the right-hand side denotes the fraction field of $\mathbb F[x]$ .

Proof. Define the evaluation map $\phi : \mathbb F[x] \to \mathbb F(u)$ by $\phi(g) = g(\alpha)$ , as per Theorem 1, which can be shown to be a ring homomorphism.

Suppose $\alpha$ is algebraic. Then there exists a monic polynomial $f \in \mathbb F[x]$ such that $f(\alpha) = 0$ . Since $\mathbb F[x]$ is a UFD, write $f$ as a product of monic irreducible polynomials. If $\alpha$ is not a root of any of these polynomials, then $f(\alpha) \neq 0$ , a contradiction. Therefore, there exists a monic irreducible polynomial $f_{\mathbb F}^{\alpha} \in \mathbb F[x]$ such that $f_{\mathbb F}^{\alpha}(\alpha) = 0$ . By the argument in Theorem 1, $\mathbb F[x] / \langle f_{\mathbb F}^{\alpha} \rangle \cong \mathbb F(\alpha)$ as fields. Furthermore, if there exists another monic irreducible polynomial $g \in \mathbb F[x]$ such that $g(\alpha) = 0$ , then

$\mathbb F[x] / \langle f_{\mathbb F}^{\alpha} \rangle \cong \mathbb F(\alpha) \cong \mathbb F[x] / \langle g \rangle$

implies that $\langle f_{\mathbb F}^{\alpha} \rangle = \langle g \rangle$ . Since both polynomials are monic and irreducible, we must have $f_{\mathbb F}^{\alpha} = g$ , establishing uniqueness.

Suppose $\alpha$ is transcendental. Then for any $f \in \mathbb F[x]$ , $f(\alpha) \neq 0$ . Therefore, we can extend $\phi$ to act on $\phi : Q(\mathbb F[x]) \to \mathbb F(\alpha)$ by defining

$\phi(f/g) := f(\alpha)/g(\alpha).$

The observation

$\mathbb F \cup \{\alpha \} \subseteq \phi(Q(\mathbb F[x])) \subseteq \mathbb F(\alpha)$

implies that $\phi(Q(\mathbb F[x])) = \mathbb F(\alpha)$ . However, note that $\ker(\phi) = \{ 0 \}$ , so that $\phi$ is injective. Therefore, $\phi$ is a bijective ring homomorphism, so that

$Q(\mathbb F[x]) \cong \phi( Q(\mathbb F[x]) ) = \mathbb F(\alpha).$

Definition 3. We can regard field extensions $\mathbb K \supseteq \mathbb F$ as vector spaces over the field $\mathbb F$ . Define the degree of the extension by $[\mathbb K : \mathbb F] := \dim_{\mathbb F}(\mathbb K)$ . If this number is finite, we say that $\mathbb K$ is a finite extension of $\mathbb F$ .

Theorem 3. If $\alpha \in \mathbb K$ is algebraic over $\mathbb F$ , then $[\mathbb F(\alpha) : \mathbb F] = \deg(f_{\mathbb F}^{\alpha})$ , so that $\mathbb F(\alpha) \supseteq \mathbb F$ is a finite extension.

Proof. Denote $n:= \deg(f_{\mathbb F}^\alpha)$ . We claim that $K := \{1,\alpha,\dots, \alpha^{n-1}\}$ form a basis for $\mathbb F(\alpha)$ . For linear independence, the existence of nonzero scalars $c_k \in \mathbb F$ such that

$\displaystyle \sum_{k=0}^{n-1} c_k \alpha^k = 0$

imply the existence of a polynomial $g(x) := \sum_{k=0}^{n-1} c_k x^k \in \mathbb F[x]$ such that $g(\alpha) = 0$ . By Theorem 2, $g \in \langle f_{\mathbb F^{\alpha}} \rangle$ so that

$n-1 = \deg(g) \geq \deg(f_{\mathbb F}^{\alpha}) = n,$

a contradiction. Therefore, $K$ must be linearly independent. For the spanning set, write $f_{\mathbb F}^{\alpha}(x) = \sum_{k=0}^{n-1} d_kx^k + x^n$ . Since $f_{\mathbb F}^{\alpha}(\alpha) = 0$ , we must have

$\displaystyle \alpha^n = \sum_{k=0}^{n-1} (-d_k)\alpha^k \in \mathrm{span}_{\mathbb F}(K).$

Inductively, $\alpha^m \in \mathrm{span}_{\mathbb F}(K)$ for $m \geq n$ , since $\alpha^{m-1} \in \mathrm{span}_{\mathbb F}(K)$ implies

$\begin{aligned} \alpha^m &= \alpha \cdot \alpha^{m-1} = \alpha \cdot \sum_{k=0}^{n-1} d_k \alpha^k \\ &= \sum_{k=1}^{n-1} d_{k-1} \alpha^{k} + d_{n-1} \alpha^n \in \mathrm{span}_{\mathbb F}(K).\end{aligned}$

Then

$\mathbb F( \alpha ) \cong \mathbb F[x]/\langle f_{\mathbb F}^{\alpha} \rangle \cong \mathrm{span}_{\mathbb F}(K).$

Theorem 4. If $\mathbb K \supseteq \mathbb F$ is a finite extension, then it is an algebraic extension.

Proof. Suppose $\mathbb K \supseteq \mathbb F$ is a finite extension so that $[K : F] =: n < \infty$ . Fix $\alpha \in K$ . The set $\{1,\alpha, \dots, \alpha^n\}$ has $n+1 > n$ elements, and thus must be linearly dependent. Therefore, there exists scalars $c_i \in \mathbb F$ , not all $0$ , such that

$\displaystyle \sum_{i=0}^n c_i \alpha^i = 0.$

Defining the polynomial $f := \sum_{i=0}^n c_i \mu_i \in \mathbb F[x]$ , we have $f(\alpha) = 0$ . Therefore, $\alpha$ is algebraic.

Can we measure “stacks” or “towers” of field extensions? Of course!

Lemma 1. Every vector space $V$ over $\mathbb F$ has a basis.

Proof. Assume $V$ is not finite-dimensional for the sake of non-triviality. Let $\mathcal S$ denote the collection of linearly independent subsets of $V$ , and partially order its elements via $\subseteq$ . For any chain $\mathcal C \subseteq \mathcal S$ , the linearly independent subset $K := \bigcup_{C \in \mathcal C} C$ is an upper bound for every subset in $\mathcal C$ . By Zorn’s lemma, $\mathcal S$ contains a maximal element $B$ , that is, a linearly independent subset of $V$ . It suffices to check that $V = \mathrm{span}(B)$ . To that end, the existence of $\mathbf v \in V\backslash \mathrm{span}(B)$ implies $B \subsetneq B \cup \{\mathbf v \}$ , so that by the maximality of $B$ , $B \cup \{\mathbf v \} \subseteq B$ , a contradiction.

Remark 1. We establish detailed proofs for other domain-specific results out of necessity, rather than curiosity. This is an example of an applied mathematics motivation combined with pure mathematical techniques.

Theorem 5 (Tower Rule). Let $\mathbb F \subseteq \mathbb K \subseteq \mathbb L$ be field extensions. Then $\mathbb F \subseteq \mathbb L$ is finite if and only if $\mathbb F \subseteq \mathbb K$ and $\mathbb K \subseteq \mathbb L$ . In this case, we recover the tower rule:

$[\mathbb L : \mathbb F] = [\mathbb L : \mathbb K] \cdot [\mathbb K : \mathbb F].$

Proof. Using Lemma 1, let $\mathcal U = \{u_i : i \in I\} \subseteq \mathbb L$ be a basis of $\mathbb L$ over $\mathbb K$ and $\mathcal V = \{v_j : j \in J\} \subseteq \mathbb K$ be a basis of $\mathbb K$ over $\mathbb F$ . We claim that

$\mathcal W := \{u_i v_j : (i,j) \in I \times J\}$

forms a basis of $\mathbb L$ over $\mathbb F$ . For linear independence,

$\displaystyle \sum_{(i,j) \in I \times J} c_{ij} u_i v_j = \sum_{j \in J} \left( \sum_{i \in I} c_{ij} u_i \right) v_j.$

Since $\mathcal U, \mathcal V$ are linearly independent,

$\begin{aligned} \sum_{(i,j) \in I \times J} c_{ij} u_i v_j = 0 \quad &\Rightarrow \quad \sum_{j \in J} \left( \sum_{i \in I} c_{ij} u_i \right) v_j = 0 \\ &\Rightarrow \quad \sum_{i \in I} c_{ij} u_i = 0 \\ &\Rightarrow \quad c_{ij} = 0.\end{aligned}$

Therefore, $\mathcal W$ is linearly independent.

To show that $\mathbb L = \mathrm{span}_{\mathbb F}(\mathcal W)$ , use $\mathbb L = \mathrm{span}_{\mathbb K}(\mathcal U)$ and $\mathbb K = \mathrm{span}_{\mathbb F}(\mathcal V)$ to construct $c_i, d_{ij}$ such that for any $w \in \mathbb L$ ,

$\begin{aligned} w = \sum_{i \in I} c_i u_i &= \sum_{i \in I} \left( \sum_{j \in J} d_{ij} v_j \right) u_i \\ &= \sum_{(i,j) \in I \times J} d_{ij} u_i v_j \in \mathrm{span}_{\mathbb F}(\mathcal W).\end{aligned}$

We note, of course, that only finitely many terms here are nonzero, so that the sum is still finite, and hence, well-defined. Here, the finite-ness results hold, and in this case,

$\begin{aligned} [\mathbb L : \mathbb F] &= |\mathcal W| = |I \times J| \\ &= |I| \cdot |J| \\ &= |\mathcal U| \cdot |\mathcal V| \\ &= [\mathbb L : \mathbb K] \cdot [\mathbb K : \mathbb F]. \end{aligned}$

This “stacking” of field extensions is responsible for several impossibility results in constructions in Euclidean geometry.

For now, we want to recall the seemingly trivial field extension $\mathbb R \subseteq \mathbb C$ . Recall that $f_{ \mathbb R}^i(x) = x^2 + 1$ . By Theorem 3, $[\mathbb C : \mathbb R] = \deg(f_{ \mathbb R}^i) = 2$ . Yet, the fundamental theorem of algebra also tells us that for any $f \in \mathbb C[x]$ , there exists at least one $\alpha \in \mathbb C$ such that $f(\alpha) = 0$ . In this case, we call $\mathbb C$ algebraically closed.

It turns out that every field $\mathbb F$ has an (effectively unique) algebraic extension $\overline{\mathbb F}$ that is algebraically closed. By effectively unique, we mean that for any algebraic extension $\mathbb K \supseteq \mathbb F$ that is algebraically closed, $\mathbb K \cong \overline{\mathbb F}$ as fields. Using this uniqueness, we can conclude that $\overline{\mathbb R} = \mathbb C$ .

We will explore this notion of algebraic closure next time.

—Joel Kindiak, 5 Feb 26, 1334H

KindiakMath

Extended Field Trip

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Extended Field Trip

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