현대대수학II 중간고사 대비 요약
내일 현대대수학II 중간고사를 앞두고 있는 만큼, 다시 한 번 내용들을 복습하면서 정리해보았다. 미처 진도를 다 나가지 못한 Galois Theory 부분들은 기말고사 때 정리해 보겠다!
PART VI: Extension Fields
Chp 29: Introduction to Extension Fields
Definition:
A field $E$ is an extension field of a field $E$ if $F \le E$.
Theorem 29.3$($Kronecker's Theorem$)$:
Let $F$ be a field and let $f(x)$ be a nonconstant polynomial in $F[x]$. Then there exist an extension field $E=F[x]/\left< p(x) \right> $ of $F$ and an $\alpha=x+ \left< p(x) \right> \in E$ such that $f(\alpha)=0$.
Definition:
An element $\alpha$ of an extension field $E$ of a field $F$ is algebraic over $F$ if $f(\alpha)=0$ for some nonzero $f(x) \in F[x].$ If $\alpha$ is not algebraic over $F$, then $\alpha$ is transcendental over $F$.
Definition:
An element of $\mathbb{C}$ that is algebraic over $\mathbb{Q}$ is an algebraic number. A transcendental number is an element $\mathbb{C}$ that is transcendental over $\mathbb{Q}$.
Definition:
Let $E$ be an extension field of a field $F$, and let $\alpha \in E$ be algebraic over $F$. The unique monic polynomial $p(x)$ having the property described in Theorem 29.13 is the irreducible polynomial for $\alpha$ over $F$ and will be denoted by $irr(\alpha, F)$. The degree of $irr(\alpha,F)$ is the degree of $\alpha$ over $F$, denote by $deg(\alpha,F)$.
Definition:
- Case I: $\alpha$ is algebraic over $F$
Then as in Theorem 29.13, $ker(\phi_{\alpha})=\left< irr(\alpha,F) \right>$ and by Theorem 27.25, $\left<irr(\alpha,F\right>$ is a maximal ideal of $F[x]$. Therefore, $ F[x] / \left< irr(\alpha,F) \right>$ is a field and is isomorphic to the image $\phi_{\alpha} [F[x]]$ in $E$. THis subfield $\phi_{\alpha} [F[x]]$ of $E$ is then smallest subfield of $E$ containing $F$ and $\alpha$. We shall denote this field by $F(\alpha)$.
- Case II: $\alpha$ is transcendental over $F$
Then by Theorem 29.12, $\phi_{\alpha}$ gives an isomorphism of $F[x]$ with a subdomain of $E$. Thus in this case $\phi_{\alpha} [F[x]]$ is not a field but an integral domain that we shall denote by $F[\alpha]$. $( \because ^\nexists irr(\alpha,F). )$ By Corollary 21.8, $E$ contains a field of quotients of $F[\alpha]$, which is this the smallest subfield of $E$ containing $F$ and $\alpha$. As in Case I, we denote this field by $F(\alpha)$.
Definition:
An extension field $E$ of a field $F$ is a simple extension of $F$ if $E=F(\alpha)$ for some $\alpha in E$.
Theorem 29.18:
Let $E$ be a simple extension $F(\alpha)$ of a field $F$, and let $\alpha$ be algebraic over $F$. Let the $deg(\alpha,F) \ge 1$. Then every element $\beta \in E = F(\alpha)$ can be uniquely expressed in form
$$ \beta = b_{0} + b_{1} \alpha + \cdots b_{n-1} \alpha^{n-1}, $$
where the $ b_{i}$ are in $F$.
Proved easily by evaluation homomorphism $\phi_{\alpha}$.
Chp 30: Vector Spaces
Definition:
Let $F$ be a field. A vector space over $F$ $($ or $F$-vector space $)$ consists of an abelian group $V$ under addition together with an operation of scalar multiplication of each element of $V$ by each element of $F$ on the left, such that for all $a, b \in F$ and $\alpha, \beta \in V$ the following conditions are satisfied:
- $a \alpha \in V$.
- $a (b\alpha) = (ab) \alpha$.
- $(a+b) \alpha = (a\alpha) + (b\alpha)$.
- $a(\alpha + \beta) = (a\alpha) + (b\alpha)$.
- $1\alpha = \alpha$.
The elements of $V$ are vectors and the elements of $F$ are scalars.
Definition:
A vector space $V$ over a field $F$ is finite-dimensional if there is a finite subset of $V$ whose vectors span $V$.
Definition:
If $V$ is a vector space over a field $F$, the vectors in a subset $ B = \{ \beta_{i} | i \in I \} $ of $V$ form a basis for $V$ over $F$ if they span $V$ and are linearly independent.
Definition:
If $V$ is a finite-dimensional vector space over a field $F$, the number of elements in a basis is the dimension of $V$ over $F$.
Theorem 30.23:
Let $E$ be an extension field of F, and let $\alpha \in E$ be algebraic over $F$. If $deg(\alpha,F)=n$, then $F(\alpha)$ is n-dimensional vector space over $F$ with basis $ \{ 1, \alpha, \ldots, \alpha^{n-1} \} $. Furthermore, every element $\beta$ of $F(\alpha)$ is algebraic over $F$, and $deg(\beta, F) \le deg(\alpha,F)$.
Chp 31: Algebraic Extensions
Definition:
An extension field $E$ of a field $F$ is an algebraic extension of $F$ if every element in $E$ is algebraic over $F$.
Definition:
If an extension field $E$ of a field $F$ is of finite dimension $n$ as a vector space over $F$, then $E$ is a finite extension of degree $n$ over $F$. We shall let $ \left[ E : F \right] $ be the degree $n$ of $E$ over $F$.
Theorem 31.4:
If $E$ is a finite extension field of a field $F$, and $K$ is a finite extension field of $E$, then $K$ is a finite extension of $F$, and
$$ \left[ K : F \right] = \left[K : E \right] \left[ E : F \right]. $$
Theorem 31.12:
Let $E$ be an extension field of $F$. Then
$ \bar{F}_{E} = \{ \alpha \in E | \alpha $ is algebraic over $F \} $
is a subfield of $E$, the algebraic closure of $F$ in $E$.
Definition:
A field $F$ is algebraically closed if every nonconstant polynomial in $F[x]$ has a zero in $F$.
$e.g. \mathbb{C}$
Theorem 31.17:
Every field $F$ has an algebraic closure, that is, an algebraic extension $\bar{F}$ that is algebraically closed.
Chp 32: Geometric Constructions
Definition:
A real number $\alpha$ is constructible if we can construct a line segment of length $|\alpha|$ in a finite number of steps from this given segment of unit length by using a straightedge and a compass.
Theorem 32.1:
If $\alpha, \beta$ are constructible real numbers, then so are $ \alpha + \beta, \alpha - \beta, \alpha\beta, \alpha/\beta$, if $\beta \neq 0$.
$\Rightarrow$ The set of all constructible real numbers forms a subfield $F$ of the field of real numbers.
Chp 33: Finite Fields
Theorem 33.5:
The multiplicative group $\left< F^{*},\cdot \right>$ of nonzero elements of a finite field $F$ is cyclic.
Lemma 33.8:
If $F$ is a field of prime characteristic $p$ with algebraic closure $\bar{F}$, then $x^{p^{n}}-x$ has $p^{n}$ distinct zeros in $\bar{F}$.
Theorem 33.10:
A finite field $GF(p^{n})$ of $p^{n}$ elements exists for every prime power $p^{n}$.
Corollary 33.11:
If $F$ is any finite field, then for every positive integer $n$, there is an irreducible polynomial in $F[x]$ of degree $n$.
Theorem 33.12:
Let $p$ be a prime and let $n \in \mathbb{Z}^{+}$. If $E$ and $E'$ are fields of order $p^{n}$, then $E \cong E'$.
PART X: Automorphism and Galois Theory
Chp 48: Automorphisms of FIelds
Definition:
Let $E$ be an algebraic extension of a field $F$. Two elements $\alpha, \beta \in E$ are conjugate over $F$ if $irr(\alpha, F)=irr(\beta,F)$.
Theorem 48.3$($The Conjugation Isomorphisms$)$:
Let $F$ be a field, and let $\alpha, \beta$ be algebraic over $F$ with $deg(\alpha,F)=n$. The map $\psi_{\alpha, \beta}: F(\alpha) \rightarrow F(\beta)$ defined by
$$ \psi_{\alpha, \beta} (c_{0} + c_{1} \alpha + \cdots + c_{n-1} \alpha^{n-1}) = c_{0} + c_{1} \beta + \cdots + c_{n-1} \beta^{n-1}$$
for $c_{i} \in F$ is an isomorphism of $F(\alpha)$ onto $F(\beta)$ if and only if $\alpha$ and $\beta$ are conjugate over $F$.
Definition:
An isomorphism of a field onto itself is an automorphism of the field.
Definition:
If $\sigma$ is an automorphism of a field $E$ onto some field, then an element $a$ of $E$ is left fixed by $\sigma$ if $\sigma(a)=a$. A collection $S$ of isomorphisms of $E$ leaves a subfield $F$ of $E$ fixed if each $a \in F$ is left fixed by every $\sigma \in S$. If $\{\sigma\}$ leaves $F$ fixed, then $\sigma$ leaves $F$ fixed.
Definition:
The field $E_{\{\sigma_{i}\}}$ of Theorem 48.11 is the fixed field of $\{\sigma_{i} | i \in I \}$. For a single automorphism $\sigma$, we shall refer to $E_{\sigma}$ as the fixed field of $\sigma$.
Theorem 48.14:
The set of all automorphisms of a field $E$ is a group$(Aut(E))$ under function composition.
Definition:
The group $G(E/F)$ of the preceding theorem is the group of automorphisms of $E$ leaving $F$ fixed, or, more briefly, the group of $E$ over $F$.
Theorem 48.19:
Let $F$ be a finite field of characteristic $p$. Then the map $\sigma_{p}: F \rightarrow F$ defined by $\sigma_{p} (a)=a^{p}$ for $a\in F$ is an automorphism, the Frobenius Automorphism, of $F$. Also, $F_{\{\sigma_{p}\}} \cong \mathbb(Z)_{p}$.
Chp 49: The isomorphism Extension Theorem
Theorem 49.3$($Isomorphism Extension Theorem$)$:
Let $E$ be an algebraic extension of a field $F$. Let $\sigma$ be an isomorphism of $F$ onto a field $F'$. Let $\bar{F'}$ be an algebraic closure of $F'$.Then $\sigma$ can be extended to an isomorphism $\tau$ of $E$ onto a subfield of $\bar{F'}$ such that $\tau(a)=\sigma(a)$ for all $a \in F$.
Corollary 49.5:
Let $\bar{F}$ and $\bar{F'}$ be two algebraic closures of $F$. Then $\bar{F}$ is isomorphic to $\bar{F'}$ under an isomorphism leaving each element of $F$ fixed.
Definition:
Let $E$ be a finite extension of a field $F$. The number of isomorphisms of $E$ onto a subfield of $\bar{F}$ leaving $F$ fixed is the index $ \{ E:F \} $ of $E$ over $F$.
Corollary 49.10:
If $F \le E \le K$, where $K$ is a finite extension field of the field $F$, then
$$\{ K :F \} = \{ K:E \} \{ E:F \}.$$
Chp 50: Splitting Fields
Definition:
Let $F$ be a field with algebraic closure $\bar{F}$. Let $ \{ f_{i} (x) | i \in I \}$ be a collection of polynomials in $F[x]$.
A field $E \le \bar{F}$ is the splitting field of $ \{ f_{i} (x) | i \in I \}$ over $F$ if $E$ is the smallest subfield of $\bar{F}$ containing $F$ and all zeros in $\bar{F}$ of each of the $f_{i} (x)$ for $i \in I$.
A field $K$ is a splitting field over $F$ if it is the splitting field of some set of polynomials in $F[x]$.
Theorem 50.3:
A field $E$, where $F \le E \le \bar{F}$, is a splitting field over $F$ if and only if every automorphism of $\bar{F}$ leaving $F$ fixed maps $E$ onto itself and thus induces an automorphism of $E$ leaving $F$ fixed.
Definition:
Let $E$ be an extension field of a field $F$. A polynomial $f(x) \in F[x]$ splits in $E$ if it factors into a product of linear factor in $E[x]$.
Corollary 50.6:
If $E \le \bar{F}$ is a splitting field over $F$, then every irreducible polynomial in $F[x]$ having a zero in $E$ splits in $E$.
Corollary 50.7:
If $E \le \bar{F}$ os a splitting field over $F$, then every isomorphic mapping of $E$ onto a subfield of $\bar{F}$ and leaving $F$ fixed is actually an automorphism of $E$. In particular, if $E$ is a splitting field of finite degree over $F$, then
$$ \{ E:F \} = |G(E/F)|. $$
Chp 51: Separable Extensions
Definition:
Let $f(x) \in F[x]$. An element $\alpha$ of $\bar{F}$ such that $f(\alpha)=0$ is a zero of $f(x)$ of multiplicity $\nu$ if $\nu$ is the greatest integer such that $(x-\alpha)^{\nu}$ is a factor of $f(x)$ in $\bar{F} [x]$.
Theorem 51.2:
Let $f(x)$ be irreducible in $F[x]$. Then all zeros of $f(x)$ in $\bar{F}$ have the same multiplicity.
Theorem 51.6:
If $E$ is a finite extension of $F$, then $ \{ E:F \} $ divides $[ E:F ]$.
Definition:
A finite extension $E$ of $F$ is a separable extension of $F$ if $ \{ E:F \} = [E:F] $.
An element $\alpha$ of $\bar{F}$ is separable over $F$ if $F(\alpha)$ is a separable extension of $F$.
An irreducible polynomial $f(x) \in F[x]$ is separable over $F$ if every zero of $f(x)$ in $\bar{F}$ is separable over $F$.
$ \{ F(\alpha) : F \}$ is the number of distinct zeros of $irr(\alpha,F)$.
Thus $\alpha$ is separable over $F$ $\iff$ $irr(\alpha,F)$ has all zeros of multiplicity 1.
This tells us at once that an irreducible polynomial $f(x) \in F[x]$ is separable over $F$ $\iff$ $f(x)$ has all zeros of multiplicity 1.
Corollary 51.10:
If $E$ is a finite extension of $F$, then $E$ is separable over $F$ if and only if if each $\alpha$ in $E$ is separable over $F$.
Definition:
A field is perfect if every finite extension is a separable extension.
Theorem 51.13:
Every field of characteristic zero is perfect.
Theorem 51.14:
Every finite field is perfect.
Theorem 51.15$($Primitive ELement Theorem$)$
Let $E$ be a finite separable extension of a field $F$. Then there exists $\alpha \in E$ such that $E=F(\alpha)$.
Theorem 51.16:
A finite extension of a field of characteristic zero is a simple extension.
Reference:
A First Course in Abstract Algebra International Edition, PEARSON