Field Theory, Part 1: Basic Theory and Algebraic Extensions

October 18th, 2017

Introduction

Recall that a field is a commutative ring in which every nonzero element has a multiplicative inverse.

Definition: The characteristic of a field is the additive order of $1$. For example, if $1+1+1=0$, then we say the field has characteristic $3$. If $1+ 1 + \dots$ is never equal to $0$, we say the field has characteristic $0$. The characteristic of a field is either $0$ or a prime.

Denote $1+ 1 + \dots + 1$, added $n$ times, we denote this element $n\cdot 1$. For each field $F$, we have a natural homomorphism $\mathbb{Z}\rightarrow F$, which maps $n$ to $n\cdot 1$. Note that a homomomorphism into a field is either zero identically or an isomorphism; thus the image of this map can be realized as a subfield of $F$.

The kernel of this homomorphism is exactly $(\text{char} F)\mathbb{Z}$. By the isomorphism theorems, then, $F$ contains either a subring isomorphic to $\mathbb{Z}$ (in which case $F$ contains $\mathbb{Q}$) or else $F$ contains a subring isomorphic to $\mathbb{Z} / p\mathbb{Z}$ (in which case $\mathbb{F}_p$, the finite field of $p$ elements, is a subfield).

Definition: The prime subfield of a field $F$ is the subfield generated by $1$ additively. It is either $\mathbb{Q}$ or $\mathbb{F}_p$, the finite field of $p$ elements.

Definition: If $K$ is a field containing a subfield $F$, then $K$ is an extension of $F$. The prime subfield is called the base field of an extension.

Definition: The degree of $K/F$, the extension $K$ over $F$, is the dimension of $K$ as a vector space over $F$.

Definition: Let $K$ be an extension of $F$. Then for $\alpha \in K$, $F(\alpha)$ denotes the smallest subfield of $K$ which contains $F$ and $\alpha$. This is called a simple extension of $F$; a simple extension is not, in general, simply an extension of degree $2$ over $F$.

Theorem 1

Let $F$ be a field and $p(x) \in F[x]$ an irreducible polynomial. Then there exists a field $K$ containing $F$ such that $p(x)$ has a root.

We can prove this by considering the field:

$$ K = \frac{F[x]}{(p(x))} $$

Since $p$ is irreducible, and $F[x]$ is a PID, $p$ spans a maximal ideal, and thus $K$ is indeed a field. Furthermore, we have the canonical projection:

$$ \pi: F[x] \rightarrow K $$

When restricted to $F$, this map is an isomorphism. Since it sends $1$ to $1$, it is an isomorphism and therefore an image of $F$ lies in $K$. Thus, since $\pi$ is a homomorphism, denoting the image in the quotient with a bar, we have:

$$ \overline{p(x)} = p(\overline{x}) = 0 $$

And thus, $\bar{x}$ is a root of $p$. In particular, let:

$$ p(x) = a_nx^n + \dots + a_1x + a_0 $$

Then if $\theta = \bar{x}$, then the above proof gives us a basis for $K$:

$$ 1, \theta, \dots, \theta^{n-1} $$

And thus, $[K:F]=n$, i.e. $K$ is a vector space over $F$ of dimension $n$. It remains to check that this is indeed a basis, i.e. that it is linearly independent; this follows from the fact that $p$ is irreducible.

Theorem 2

Let $F$ be a field and $p(x)\in F[x]$ an irreducible polynomial. Suppose that $K$ is an extension of $F$ containing a root $\alpha$ of $p(x)$ such that $p(\alpha)=0$. Let $F(\alpha)$ denote the subfield of $K$ generated over $F$ by $\alpha$. Then:

$$ F(\alpha) \cong \frac{F[x]}{(p(x))} $$

This theorem tells us that any field over $F$ in which $p(x)$ contains a root contains a subfield isomorphic to the extension we considered in Theorem 1. The natural homomorphism that allows us to prove this identity is:

$$ \varphi: F[x] \rightarrow F(\alpha)\subseteq K \\ f(x) \mapsto f(\alpha) $$

This homomorphism is exactly evaluation. With some work, we can prove that this is a nontrivial ring homomorphism; thus the quotient ring is indeed a field.

Indeed, we can totally describe $F(\alpha)$ using this theorem:

Corollary

Suppose that $p(x)$ has degree $n$. Then:

$$ F(\alpha) = \{a_0 + a_1\alpha + \dots + a_{n-1}\alpha^{n-1}\} \subseteq K $$

Where $a_i \in F$.

Describing the fields which are generated by more than element is a little more complicated.

Note that Theorem 2 tells us that the roots of an irreducible polynomial are, in a sense, indistinguishable; adjoining any root of an irreducible polynomial yields an isomorphic field. We extend this result.

Theorem 3

Let $\varphi: F\rightarrow \tilde{F}$ be an isomorphism of fields. Let $p(x) \in F[x]$ be an irreducible polynomial and let $p’(x) = \varphi(p(x))$ (we simply map each coefficient under $\varphi$). Then $p’(x)$ is irreducible.

Let $\alpha$ be a root of $p(x)$ and $\beta$ be a root of $p’(x)$ in some extension of $F’$. Then there is an isomorphism:

$$ \sigma: F(\alpha) \rightarrow F'(\beta) $$

$$ \sigma: a \mapsto \beta $$

And such that $\sigma$ restricted to $F$ is exactly $\varphi$.

Thus, we can extend any isomorphism of fields to an isomorphism of simple extensions which maps roots to roots. In particular, if $F=F’$ and $\varphi$ is the identity, then this tells us that $F(\alpha)\cong F(\beta)$, where $\beta$ is another root of $p(x)$. This will be vital to understanding Galois theory.

Theorem 4 (Eisenstein’s Criterion)

Suppose that we have a polymial in $\mathbb{Q}[x]$ given by:

$$ a_nx^n + \dots + a_1x+a_0 $$

Then if there exists a prime $p$ such that:

Algebraic Extensions

Definition: The element $\alpha \in K$ is said to be algebraic over $F$ if $\alpha$ is a root of some nonzero polynomial with coefficients in $F$. Otherwise, $\alpha$ is said to be transcendental over $F$. The extension $K/F$ is algebraic if every element of $K$ is algebraic over $F$.

From the Euclidean algorithm, we get:

Definition: Let $\alpha$ be algebraic over $F$. Then there exists a unique monic irreducible polynomial $m_{\alpha, F}(x)\in F[x]$ which has $\alpha$ as a root. This polynomial is called the minimal polynomial of $\alpha$ and we say the degree of $\alpha$ is the degree of this polynomial.

Proposition 1

Let $\alpha$ be algebraic over $F$, and let $F(\alpha)$ be the field generated by $\alpha$ over $F$. Then:

$$ F(\alpha) \cong \frac{F[x]}{(m_\alpha(x))} $$

This proves that in particular:

$$ [F(\alpha):F] = \text{deg }\alpha $$

Thus, the degree of a simple extension is exactly the degree of the minimal polynomial, and we have an explicit way of computing simple extensons corresponding to algebraic elements.

Proposition 2

The element $\alpha$ is algebraic over $F$ iff the simple extension $F(\alpha)/F$ is finite.

This tells us that the property that $\alpha$ is algebraic over $F$ is equivalent to the property that $[F(\alpha):F]$ is finite. In particular, we have the corollary:

Proposition 3

If an extension $K/F$ is finite, then it is algebraic.

A simple algebraic extension is finite, but in general the converse is not true, since there are infinite algebraic extensions.

Example

Let $F$ be a field of characteristic $2$, and $K$ an extension of degree $2$ (called a quadratic extension). Let $\alpha \in K$ be an element not in $F$. It must be algebraic. Its minimal polynomial cannot be degree $1$ (since $\alpha \notin F$); and so it is quadratic. It looks like:

$$ m_\alpha(x) = x^2 + bx+c $$

For some $b,c \in F$. Furthermore, $K = F(\alpha)$. The roots are given by:

$$ \alpha = \frac{-b\pm \sqrt{b^2-4c}}{2} $$

And $b^2-4c$ is not a square in $F$, since if it were then $\alpha \in F$.

Now, $F(\alpha) \subset F(\sqrt{b^2-4c})$ since $\alpha$ is an element of the field on the right. Conversely, $\sqrt{b^2-4c} = \pm(b+2\alpha)$ so we have the reverse inclusion.

We have just shown that any quadratic extension is of the form $F(\sqrt{D})$ where $D$ is an element of $F$ which is not a square in $F$; conversely, every such extension has degree $2$.

Theorem 5

Let $F\subseteq K \subseteq L$ be fields. Then:

$$ [L:F] = [L:K][K:F] $$

This is an analogous theorem to the one for groups; indeed this connection is deeper than it appears.

Corollary

Suppose $L/F$ finite extension, and $K$ a subfield of $L$ containing $F$. Then $[K:F]$ divides $[L:F]$.

Definition: An extension $K/F$ is finitely generated if there are element $\alpha_1, \dots, \alpha_k$ in $K$ such that:

$$ K = F(\alpha_1, \dots, \alpha_k) $$

As expected, we can obtain this field by recursively compounding a series of simple extensions, i.e.:

$$ (F(\alpha))(\beta) = F(\alpha, \beta) $$

Where $F(\alpha, \beta)$ is the smallest field containing $F, \alpha$, and $\beta$.

Theorem 6

The extension $K/F$ is finite iff $K$ is generated by a finite number of algebraic elements over $F$. If these elements have degrees $n_1, \dots, n_k$ then, $K$ is algebraic of degree at most $n_1n_2\dots n_k$.

To see this, notice that if $K/F$ is finite of degree $n$, then say $\alpha_1, \dots, \alpha_n$ is a basis for $K$ as a vector space over $F$. Then:

$$ [F(\alpha_i):F]\mid [K:F] = n $$

Therefore, by Proposition 2 each $\alpha_i$ is algebraic. Conversely, if $K$ is generated by a finite number of algebraic elements, then it is generated as a vector space by polynomials of those elements.

Corollary

Let $L/F$ be an arbitrary extension. Then the collection of elements of $L$ that are algebraic over $F$ forms a subfield $K$ of $L$.

Suppose that $\alpha, \beta$ are algebraic over $F$. Then, note that $\alpha \pm \beta, \alpha\beta, \alpha/\beta, \alpha^{-1}$ are all algebraic, and lie in the finite extension $F(\alpha, \beta)$; and since this extension is finite, these elements are algebraic. Thus, the collection of algebraic elements is closed under addition, multiplication, and inverses.

Theorem 7

If $K$ is algebraic over $F$ and $L$ is algebraic over $K$, then $L$ is algebraic over $F$.

We also ask about “intersections” of fields.

Definition: Let $K_1, K_2$ be subfields of $K$. The composite field of $K_1, K_2$, denoted $K_1K_2$, is the smallest subfield of $K$ containing both $K_1, K_2$. It is equivalently the intersection of all subfields of $K$ containing both $K_1$ and $K_2$.

Indeed, if $K_1, K_2$ are finite extensions, then if we combine their bases, we can construct a set of generators for $K_1K_2$. From this discussion, we can see:

Proposition 4

Let $K_1, K_2$ be two finite extensions of a field $F$ contained in $K$. Then:

$$ [K_1K_2: F] \leq [K_1:F][K2:F] $$

Corollary

Suppose that $[K_1: F] = n$, and $[K_2: F]=m$, then if $n,m$ are relatively prime then:

$$ [K_1K_2: F] = [K_1:F][K_2:F] = nm $$

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