Abstract Algebra - Field

发表于 2020-10-17 分类于 Math Disqus：本文字数： 2.4k 阅读时长 ≈ 9 分钟

Definition

If $A$ is a communicative ring with unity in which every nonzero element is invertible, $A$ is called a field. $\mathbb{Q}$, $\mathbb{R}$ and $\mathbb{C}$ are infinite fields.

Substitution function

Let $E$ be a field, $F$ a subfield of $E$, and $c$ any element of $E$.

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The substitution function $\sigma_c$ is defined as follows:

For every polynomial $a(x)$ in $F[x]$,

\[ \sigma_c(a(x)) = a(c) \]

Thus, $\sigma_c$ is the function "substitute $c$ for $x$."

Algebraic and Transcendental

An element $c$ in $E$ is called algebraic over $F$ if it is the root of some nonzero polynomial $a(x)$ in $F[x]$. Otherwise, $c$ is called transcendental over $F$.

Let $J_c$ denote the kernel of $\sigma_c$. Obviously $c$ is algebraic over $F$ iff $J_c$ contains nonzero polynomials, and transcendental over $F$ iff $J_c = \{0\}$.

Range

The range of $\sigma_c$ is a subfield of $E$, which is a set of all the elements $a(c)$, for all $a(x)$ in $F[x]$.

\[ \text{Range } \sigma_c = \{a(c): a(x) \in F[x]\} \]

The range of $\sigma_c$ is the smallest field containing $F$ and $c$. This field can also be called as the field generated by $F$ and $c$, and is denoted by the symbol

\[ F(c) \]

In a nutshell: $\sigma_c$ is a homomorphism with domain $F[x]$, range $F(c)$, and kernel $J_c = \langle p(x) \rangle$. Thus, by the FHT,

\[ F(c) \cong F[x]/ \langle p(x) \rangle \]

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Field Extension

Why should we be interested in looking at the extension of fields? There are several reasons, but one is very special. If $F$ is an arbitrary field, there are , in general, polynomials over $F$ which have no roots in $F$. But every polynomials over $F$ has roots. If those roots are not already in $F$, they are in a suitable extension of $F$.

Basic Theorem

Let $F$ be a field and $a(x)$ a non-constant polynomial in $F[x]$. There exists an extension field $E$ of $F$ and an element $c$ in $E$ such that $c$ is a root of $a(x)$.

The Idea

Let $F$ and $K$ be fields. If $K$ is an extension of $F$, we may regard $K$ as being a vector space over $F$. We may treat the elements in $K$ as "vectors" and the elements in $F$ as "scalars".

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Adding elements in $K$ can be regarded as vector addition.
Add and multiply elements in $F$ can be regarded as scalar addition and multiplication.
Multiply an element of $F$ by an element of $K$ can be regarded as scalar multiplication.

If $K$, as a vector space over $F$, is of finite dimension, we call $K$ a finite extension of $F$. If the dimension of the vector space $K$ is $n$, we say that $K$ is an extension of degree $n$ over $F$. This is symbolized by writing

\[ [K:F] = n \]

Theorem 1

The degree of $F(c)$ over $F$ is equal to the degree of the minimum polynomial of $c$ over $F$.

Theorem 2

Suppose $F \subseteq K \subseteq E$ where $E$ is a finite extension of $K$ and $K$ is a finite extension of $F$. Then $E$ is a finite extension of $F$, and

\[ [E:F] = [E:K][K:F] \]

An extension $F(c)$ formed by adjoining a single element to $F$ is called a simple extension of $F$. And extension $F(c_1, \dots, c_n)$, formed by adjoining a finite number of elements $c_1, \dots, c_n$, is called an iterated extension. It is called "iterated" because it can be formed step by step, one simple extension at a time:

\[ F \subseteq F(c_1) \subseteq F(c_1, c_2) \subseteq \cdots \subseteq F(c_1, c_2, \dots, c_n) \]

If $c_1, \dots, c_n$ are algebraic over $F$, then $F(c_1, c_2, \dots, c_n)$ is a finite extension of $F$.

Theorem 3

If $K$ is a finite extension of $F$, every element of $K$ is algebraic over $F$.

Ruler and Compass

If $\mathscr{A}$ is any set of points in the plane, consider operations of the following two kinds:

Ruler operation: Through any two points in $\mathscr{A}$, draw a straight line.
Compass operation: Given three points $A, B$ and $C$ in $\mathscr{A}$, draw a circle with center $C$ and radius equal in length to the segment $AB$.

The points of intersection of any two of these figures (line-line, line-circle, circle-circle) are said to be constructible in one step from $\mathscr{A}$. A point $P$ is called constructible from $\mathscr{A}$ if there are points $P_1, P_2, \dots, P_n = P$ such that $P_1$ is constructible in one step from $\mathscr{A}$, $P_2$ is constructible in one step from $\mathscr{A} \cup \{P_1\}$, and so on, so that $P_i$ is constructible in one step from $\mathscr{A} \cup \{P_1, P_2, \dots, P_{i-1}\}$

Suppose $P$ has coordinates $(a, b)$ and is constructed from $\mathbb{Q} \times \mathbb{Q}$ in one step. We associate with $P$ the field $\mathbb{Q}(a, b)$, obtained by adjoining to $\mathbb{Q}$ the coordinates of $P$. More generally, suppose $P$ is constructible from $\mathbb{Q}\times\mathbb{Q}$ in $n$ steps: there are then $n$ points $P_1, P_2, \dots, P_n = P$ such that each $P_i$ is constructible in one step from $\mathbb{Q} \times \mathbb{Q} \cup \{P_1, P_2, \dots, P_{i-1}\}$. Let the corrdinates of $P_1, P_2, \dots, P_n$ be $(a_1, b_1), \dots, (a_n, b_n)$ respectively. With the points $P_1, P_2, \dots, P_n$ we associate fields $K_1, K_2, \dots, K_n$ where $K_1 = \mathbb{Q}(a_1, b_1)$, and for each $i > 1$,

\[ K_i = K_{i-1}(a_i, b_i) \]

Thus, beginning with $\mathbb{Q}$, we adjoin first the coordinates of $P_1$, then the coordinates of $P_2$, and so on successively, yielding the sequence of extensions

\[ \mathbb{Q} \subseteq K_1 \subseteq K_2 \subseteq \cdots \subseteq K_n = K \]

We call $**K$ the field extension associated with the point $P$**.

Lemma

If $K_1, \dots, K_n$ are as defined previously, then $[K_i: K_{i-1}] = 1, 2$ or $4$.

Basic theorem on constructible points

If the point with coordinates $(a, b)$ is constructible, then the degree of $\mathbb{Q}(a)$ over $\mathbb{Q}$ is a power of $2$, and likewise for the degree of $\mathbb{Q}(b)$ over $\mathbb{Q}$.

Galois Theory

We will be concerned not so much with finding solutions as with the nature and properties of these solutions. As we shall discover, these properties turn out to depend less on the polynomials themselves than on the fields which contain their solutions. We will be speaking of field extensions, but polynomials will always be lurking in the background. Every extension will be generated by roots of a polynomial, and every theorem about these extensions will actually be saying something about the polynomials.

Root field

If $a(x)$ is a polynomial of degree $n$ in $F[x]$, let its roots be $c_1, c_2, \dots, c_n$. Then $F(c_1, c_2, \dots, c_n)$ is clearly the smallest extension of $F$ containing all the roots of $a(x)$. $F(c_1, c_2, \dots, c_n)$ is called the root field of $a(x)$ over $F$.

Basic theorems

Theorem 1

If $F$ has characteristic 0, irreducible polynomials over $F$ can never have multiple roots.

Theorem 2

Every finite extension of $F$ is a simple extension $F(c)$.

Isomorphism extension

Suppose $F_1$ and $F_2$ are fields, and $h:F_1 \to F_2$ is an isomorphism. Let $K_1$ and $K_2$ be extensions of $F_1$ and $F_2$, and let $\bar{h}: K_1 \to K_2$ also be an isomorphism. We call $\bar{h}$ an extension of $h$ if $\bar{h}(x) = h(x)$ for every $x$ in $F_1$, that is, if $h$ and $\bar{h}$ are the same on $F_1$.

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Theorem 1

Let $h:F_1 \to F_2$ be an isomorphism, and let $p(x)$ be irreducible in $F_1[x]$. Suppose $a$ is a root of $p(x)$, and $b$ a root of $hp(x)$. Then $h$ can be extended to an isomorphism

\[ \bar{h}: F_1(a) \to F_2(b) \]

Furthermore, $\bar{h}(a) = b$.

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Theorem 2

Suppose $a$ and $b$ are roots of the same irreducible polynomial $p(x)$ in $F[x]$. Then there is an isomorphism $g:F(a) \to F(b)$ such that $g(x) = x$ for every $x$ in $F$, and $g(a) = b$.

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Theorem 3

Let $K$ and $K'$ be finite extensions of $F$. Assume $K$ is the root field of some polynomial over $F$. If $h: K \to K'$ is an isomorphism which fiexes $F$, then $K = K'$.

Theorem 4

Let $L$ and $L'$ be finite extensions of $F$. Let $K$ be an extension of $L$ such that $K$ is a root field over $F$. Any isomorphism $h: L \to L'$ which fixes $F$ can be extended to an isomorphism $\bar{h}: K \to K$.

Theorem 5

Let $K$ be the root field of some polynomial over $F$. For every irreducible polynomial $p(x)$ in $F[x]$, if $p(x)$ has one root in $K$, then $p(x)$ must have all of its roots in $K$.

Theorem 6

Suppose $I \subseteq E \subseteq K$, where $E$ is a finite extension of $I$ and $K$ is a finite extension of $E$. If $K$ is the root field of some polynomial over $I$, then $K$ is also the root field of some polynomial over $E$.

Galois Group

If $K$ is the root field of a polynomial $a(x)$ in $F[x]$, the group of all the automorphisms of $K$ which fix $F$ is called the Galois group of $a(x)$. We also call it the Galois group of $K$ over $F$, and designates it by the symbol

\[ Gal(K:F) \]

Theorem 1

Let $K$ be the root field of some polynomial over $F$. The number of automorphisms of $K$ fixing $F$ is equal to the degree of $K$ over $F$.

Fixer and Fixfield

For example, $\mathbb{Q}(\sqrt 2, \sqrt 3)$ is an extension of degree 4 over $\mathbb{Q}$. There are 4 automorphisms of $\mathbb{Q}(\sqrt 2, \sqrt 3)$ which fix $\mathbb{Q}$. These 4 automorphisms are

\[ a+b\sqrt 2 + c\sqrt 3 + d\sqrt 6 \xrightarrow{\varepsilon} a+b\sqrt 2 + c\sqrt 3 + d\sqrt 6 \\ a+b\sqrt 2 + c\sqrt 3 + d\sqrt 6 \xrightarrow{\alpha} a-b\sqrt 2 + c\sqrt 3 - d\sqrt 6 \\ a+b\sqrt 2 + c\sqrt 3 + d\sqrt 6 \xrightarrow{\beta} a+b\sqrt 2 - c\sqrt 3 - d\sqrt 6 \\ a+b\sqrt 2 + c\sqrt 3 + d\sqrt 6 \xrightarrow{\gamma} a-b\sqrt 2 - c\sqrt 3 + d\sqrt 6 \]

The group table of the above automorphism is

$$ \[\begin{array}{c|c c c c} \circ & \varepsilon & \alpha & \beta & \gamma \\ \hline \varepsilon & \varepsilon & \alpha & \beta & \gamma \\ \alpha & \alpha &\varepsilon & \gamma & \beta \\ \beta & \beta & \gamma & \varepsilon & \gamma \\ \gamma & \gamma & \beta & \alpha & \varepsilon \\ \end{array}\]

Inclusion diagram of the Galois group of $\mathbb{Q}(\sqrt 2, \sqrt 3)$ over $\mathbb{Q}$ is as follows:

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The fixed fields intermediate between $\mathbb{Q}(\sqrt 2, \sqrt 3)$ and $\mathbb{Q}$ is as follows:

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One to one correspondence between intermediate fields and subgroups of Galois groups.

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Definition

Let $F$ be an arbitrary field and $K$ is a root field of some polynomial $a(x)$ in $F[x]$. Let $I$ be an intermediate field between $F$ and $K$. Since $K$ is the root field of $a(x)$ over $F$, it is also the root field of $a(x)$ over $I$ for every intermediate field $I$.

The letter $\mathbf{G}$ will denote the Galois group of $K$ over $F$. With each intermediate field $I$, we associate the group

\[ I^* = Gal(K:I) \]

that is, the group of all the automorphisms of $K$ which fix $I$. It is obviously a subgroup of $\mathbf{G}$. We call $I^*$ the fixer of $I$.

Conversely, with each subgroup $H$ of $\mathbf{G}$ we associate the subfield of $K$ containing all the $a$ in $K$ which are not changed by any $\pi \in H$. That is,

\[ \{a \in K: \pi(a) = a, \forall \pi \in H\} \]

This is a subfield of $K$. It obviously contains $F$, and is therefore one of the intermediate fields. It is called the fixed field, or fixfield of $H$.

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Theorem 2

If $H$ is the fixer of $I$, then $I$ is the fixfield of $H$.

Lemma

Let $H$ be a subgroup of $\mathbf{G}$, and $I$ the fixfield of $H$. The number of elements in $H$ is equal to $[K:I]$.

Theorem 3

If $I$ is the fixfield of $H$, then $H$ is the fixer of $I$.

Theorem 4

Suppose $E \subseteq I \subseteq K$, where $I$ is a root field over $E$ and $K$ is a root field over $I$. Then

\[ Gal(I:E) \cong \frac{Gal(K:E)}{Gal(K:I)} \]