Created the Abstract Algebra theorems and definitions cheat sheet

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+\chapter{Vector Spaces}
+\subimport{./}{definition-and-examples.tex}
+\subimport{./}{subspaces.tex}
+\subimport{./}{linear-independence.tex}
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+\section{Definition and Examples}
+
+\begin{definition}[Vector Space]
+	A set $V$ is said to be a \textit{vector space} over a field $\F$ if $V$ is an Abelian group under addition (denoted by $+$) and, if for each $a \in \F$ and $v \in V$, there is an element $av \in V$ such that the following conditions hold for all $a,b \in \F$ and all $u,v \in V$.
+	\begin{enumerate}
+		\item $a(v + u) = av + au$
+		\item $(a + b)v = av + bv$
+		\item $a(bv)=(ab)v$
+		\item $1v=v$
+	\end{enumerate}
+\end{definition}
+
+\begin{remark}
+	The members of a vector space are called \textit{vectors}. The members of the field are called \textit{scalars}. The operation that combines a scalar $a$ and a vector $v$ to form the vector $av$ is called \textit{scalar multiplication}. In general, we will denote vectors by letters from the end of the alphabet, such as $u,v,w$, and scalars by letters from the beginning of the alphabet, such as $a,b,c$.
+\end{remark}
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+\section{Linear Independence}
+
+\begin{definition}[Linearly Dependent, Linearly Independent]
+	A set $S$ of vectors is said to be \textit{linearly dependent} over a field $\F$ if there are vectors $v_1,v_2,\dots,v_n$ from $S$ and elements $a_1,a_2,\dots,a_n$ from $\F$, not all zero, such that $a_1v_1+a_2v_2+\dots+a_nv_n = 0$. A set of vectors that is not linearly dependent over $\F$ is called \textit{linearly independent} over $\F$.
+\end{definition}
+
+\begin{definition}[Basis]
+	Let $V$ be a vector space over $\F$. A subset $B$ of $V$ is called a \textit{basis} for $V$ if $B$ is linearly independent over $\F$ and every element of $V$ is a linear combination of elements of $B$.
+\end{definition}
+
+\begin{theorem}[Invariance of Basis Size]
+	If $\{u_1,u_2,\dots,u_m\}$ and $\{w_1,w_2,\dots,w_n\}$ are both bases of a vector space $V$ over a field $\F$, then $m=n$.
+\end{theorem}
+
+\begin{definition}[Dimension]
+	A vector space that has a basis consisting of $n$ elements is said to have \textit{dimension $n$}. For completeness, the trivial vector space $\{0\}$ is said to be spanned by the empty set and to have dimension 0.
+
+	\noindent A vector space that has a finite basis is called \textit{finite dimensional}; otherwise, it is called \textit{infinite dimensional}.
+\end{definition}
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+\section{Subspaces}
+
+\begin{definition}[Subspace]
+	Let $V$ be a vector space over a field $\F$ and let $U$ be a subset of $V$. We say that $U$ is a \textit{subspace} of $V$ if $U$ is also a vector space over $\F$ under the operations of $V$.
+\end{definition}
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+\chapter{Extension Fields}
+\subimport{./}{the-fundamental-theorem-of-field-theory.tex}
+\subimport{./}{splitting-fields.tex}
+\subimport{./}{zeros-of-an-irreducible-polynomial.tex}
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+\section{Splitting Fields}
+
+\begin{definition}[Splitting Field]
+	Let $\E$ be an extension field of $\F$ and let $f(x) \in \F[x]$ with degree at least 1. We say that $f(x)$ \textit{splits} in $\E$ if there are elements $a \in \F$ and $a_1,a_2,\dots,a_n \in \E$ such that
+	\[ f(x) = a(x-a_1)(x-a_2)\dots(x-a_n) \]
+	We call $\E$ a \textit{splitting field for $f(x)$ over $\F$} if
+	\[ \E = \F(a_1,a_2,\dots,a_n) \]
+\end{definition}
+
+\begin{theorem}[Existence of Splitting Fields]
+	Let $\F$ be a field and let $f(x)$ be a nonconstant element of $\F[x]$. Then there exists a splitting field $\E$ for $f(x)$ over $\F$.
+\end{theorem}
+
+\begin{theorem}[$\mathbf{\F(a) \approx \F[x]/\lr{p(x)}}$]
+	Let $\F$ be a field and let $p(x) \in \F[x]$ be irreducible over $\F$. If $a$ is a zero of $p(x)$ in some extension $\E$ of $\F$, then $\F(a)$ is isomorphic to $\F[x]/\lr{p(x)}$. Furthermore, if $\deg p(x) = n$, then every member of $\F(a)$ can be uniquely expressed in the form
+	\[ c_{n-1}a^{n-1}+c_{n-2}a^{n-2}+\dots+c_1a+c_0 \]
+	where $c_0,c_1,\dots,c_{n-1} \in \F$.
+\end{theorem}
+
+\begin{corollary}[$\mathbf{\F(a) \approx \F(b)}$]
+	Let $\F$ be a field and let $p(x) \in \F[x]$ be irreducible over $\F$. If $a$ is a zero of $p(x)$ in some extension $\E$ of $\F$ and $b$ is a zero of $p(x)$ in some extension $\E'$ of $\F$, then the fields $\F(a)$ and $\F(b)$ are isomorphic.
+\end{corollary}
+
+\begin{lemma}
+	Let $\F$ be a field, let $p(x) \in \F[x]$ be irreducible over $\F$, and let $a$ be a zero of $p(x)$ in some extension of $\F$. If $\phi$ is a field isomorphism from $\F$ to $\F'$ and $b$ is a zero of $\phi(p(x))$ in some extension of $\F'$, then there is an isomorphism from $\F(a)$ to $\F'(b)$ that agrees with $\phi$ on $\F$ and carries $a$ to $b$.
+\end{lemma}
+
+\begin{theorem}[Extending $\mathbf{\phi: \F \to \F'}$]
+	Let $\phi$ be an isomorphism from a field $\F$ to a field $\F'$ and let $f(x) \in \F[x]$. If $\E$ is a splitting field for $f(x)$ over $\F$ and $\E'$ is a splitting field for $\phi(f(x))$ over $\F'$, then there is an isomorphism from $\E$ to $\E'$ that agrees with $\phi$ on $\F$.
+\end{theorem}
+
+\begin{corollary}[Splitting Fields Are Unique]
+	Let $\F$ be a field and let $f(x) \in \F[x]$. Then any two splitting fields of $f(x)$ over $\F$ are isomorphic.
+\end{corollary}
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+\section{The Fundamental Theorem of Field Theory}
+
+\begin{definition}[Extension Field]
+	A field $\E$ is an \textit{extension field} of a field $\F$ if $\F \subseteq \E$ and the operations of $\F$ are those of $\E$ restricted to $\F$.
+\end{definition}
+
+\begin{theorem}[Fundamental Theorem of Field Theory (Kronecker's Theorem, 1887)]
+	Let $\F$ be a field and let $f(x)$ be a nonconstant polynomial in $\F[x]$. Then there is an extension field $\E$ of $\F$ in which $f(x)$ has a zero.
+\end{theorem}
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+\section{Zeros of an Irreducible Polynomial}
+
+\begin{definition}[Derivative]
+	Let $f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0$ belong to $\F[x]$. The \textit{derivative} of $f(x)$, denoted by $f'(x)$, is the polynomial $na_nx^{x-1} + (n-1)a_{n-1}x^{n-2} + \dots + a_1$ in $\F[x]$.
+\end{definition}
+
+\begin{lemma}[Properties of the Derivative]
+	Let $f(x)$ and $g(x) \in \F[x]$ and let $a \in \F$. Then
+	\begin{enumerate}
+		\item $(f(x) + g(x))' = f'(x) + g'(x)$.
+		\item $(af(x))' = af'(x)$.
+		\item $(f(x)g(x))' = f(x)g'(x) + g(x)f'(x)$.
+	\end{enumerate}
+\end{lemma}
+
+\begin{theorem}[Criterion for Multiple Zeros]
+	A polynomial $f(x)$ over a field $\F$ has a multiple zero in some extension $\E$ if and only if $f(x)$ and $f'(x)$ have a common factor of positive degree in $\F[x]$.
+\end{theorem}
+
+\begin{theorem}[Zeros of an Irreducible]
+	Let $f(x)$ be an irreducible polynomial over a field $\F$. If $\F$ has characteristic 0, then $f(x)$ has no multiple zeros. If $\F$ has characteristic $p \neq 0$, then $f(x)$ has a multiple zero if it is of the form $f(x) = g(x^p)$ for some $g(x)$ in $\F[x]$.
+\end{theorem}
+
+\begin{definition}[Perfect Field]
+	A field $\F$ is called \textit{perfect} if $\F$ has characteristic 0 or if $\F$ has characteristic $p$ and $\F^p=\{a^p\ \vert\ a \in \F\} = \F$.
+\end{definition}
+
+\begin{theorem}[Finite Fields Are Perfect]
+	Every finite field is perfect.
+\end{theorem}
+
+\begin{theorem}[Criterion for No Multiple Zeros]
+	If $f(x)$ is an irreducible polynomial over a perfect field $\F$, then $f(x)$ has no multiple zeros.
+\end{theorem}
+
+\begin{theorem}[Zeros of an Irreducible over a Splitting Field]
+	Let $f(x)$ be an irreducible polynomial over a field $\F$ and let $\E$ be a splitting field of $f(x)$ over $\F$. Then all the zeros of $f(x)$ in $\E$ have the same multiplicity.
+\end{theorem}
+
+\begin{corollary}[Factorization of an Irreducible over a Splitting Field]
+	Let $f(x)$ be an irreducible polynomial over a field $\F$ and let $\E$ be a splitting field of $f(x)$. Then $f(x)$ has the form
+	\[ a(x-a_1)^n(x-a_2)^n\dots(x-a_t)^n \]
+	where $a_1,a_2,\dots,a_t$ are distinct elements of $\E$ and $a \in \F$.
+\end{corollary}
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+\chapter{Algebraic Extensions}
+\subimport{./}{characterization-of-extensions.tex}
+\subimport{./}{finite-extensions.tex}
+\subimport{./}{properties-of-algebraic-extensions.tex}
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+\section{Characterization of Extensions}
+
+\begin{definition}[Types of Extensions]
+	Let $\E$ be an extension field of a field $\F$ and let $a \in \E$. We call $a$ \textit{algebraic over $\F$} if $a$ is the zero of some nonzero polynomial in $\F[x]$. If $a$ is not algebraic over $\F$, it is called \textit{transcendental over $\F$}. An extension $\E$ of $\F$ is called an \textit{algebraic} extension of $\F$ if every element of $\E$ is algebraic over $\F$. If $\E$ is not an algebraic extension of $\F$, it is called a \textit{transcendental} extension of $\F$. An extension of $\F$ of the form $\F(a)$ is called a \textit{simple} extension of $\F$.
+\end{definition}
+
+\begin{theorem}[Characterization of Extensions]
+	Let $\E$ be an extension field of the field $\F$ and let $a \in \E$. If $a$ is transcendental over $\F$, then $\F(a) \approx \F(x)$. If $a$ is algebraic over $\F$, then $\F(a) \approx \F[x]/\lr{p(x)}$, where $p(x)$ is a polynomial in $\F[x]$ of minimum degree such that $p(a) = 0$. Moreover, $p(x)$ is irreducible over $\F$.
+\end{theorem}
+
+\begin{theorem}[Uniqueness Property]
+	If $a$ is algebraic over a field $\F$, then there is a unique monic irreducible polynomial $p(x)$ in $\F[x]$ such that $p(a)=0$. The polynomial with this property is called the \textit{minimal polynomial for $a$ over $\F$}.
+\end{theorem}
+
+\begin{theorem}[Divisibility Property]
+	Let $a$ be algebraic over $\F$, and let $p(x)$ be the minimal polynomial for $a$ over $\F$. If $f(x) \in \F[x]$ and $f(a) = 0$, then $p(x)$ divides $f(x)$ in $\F[x]$.
+\end{theorem}
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+\section{Finite Extensions}
+
+\begin{definition}[Degree of an Extension]
+	Let $\E$ be an extension field of a field $\F$. We say that $\E$ \textit{has degree $n$ over $\F$} and write $[\E:\F]=n$ if $\E$ has dimension $n$ as a vector space over $\F$. If $[\E:\F]$ is finite, $\E$ is called a \textit{finite extension} of $\F$; otherwise, we say that $\E$ is an \textit{infinite extension} of $\F$.
+\end{definition}
+
+\begin{theorem}[Finite Implies Algebraic]
+	If $\E$ is a finite extension of $\F$, then $\E$ is an algebraic extension of $\F$.
+\end{theorem}
+
+\begin{theorem}[$\mathbf{[\K:\F] = [\K:\E][\E:\F]}$]
+	Let $\K$ be a finite extension field of the field $\E$ and let $\E$ be a finite extension field of the field $\F$. Then $\K$ is a finite extension field of $\F$ and $[\K:\F] = [\K:\E][\E:\F]$.
+\end{theorem}
+
+\begin{theorem}[Primitive Element Theorem (Steinitz, 1910)]
+	If $\F$ is a field of characteristic 0, and $a$ and $b$ are algebraic over $\F$, then there is an element $c$ in $\F(a,b)$ such that $\F(a,b) = \F(c)$.
+\end{theorem}
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+\section{Properties of Algebraic Extensions}
+
+\begin{theorem}[Algebraic over Algebraic Is Algebraic]
+	If $\K$ is an algebraic extension of $\E$ and $\E$ is an algebraic extension of $\F$, then $\K$ is an algebraic extension of $\F$.
+\end{theorem}
+
+\begin{corollary}[Subfield of Algebraic Elements]
+	Let $\E$ be an extension field of the field $\F$. Then the set of all elements of $\E$ that are algebraic over $\F$ is a subfield of $\E$.
+\end{corollary}
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+\chapter{Finite Fields}
+\subimport{./}{classification-of-finite-fields.tex}
+\subimport{./}{structure-of-finite-fields.tex}
+\subimport{./}{subfields-of-a-finite-field.tex}
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+\section{Classification of Finite Fields}
+
+\begin{theorem}[Classification of Finite Fields]
+	For each prime $p$ and each positive integer $n$, there is, up to isomorphism, a unique finite field of order $p^n$.
+\end{theorem}
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+\section{Structure of Finite Fields}
+
+\begin{theorem}[Structure of Finite Fields]
+	As a group under addition, $\gf(p^n)$ is isomorphic to
+	\[ \underbrace{\Z_p \oplus \Z_p \oplus \dots \oplus \Z_p}_\text{$n$ factors} \]
+	As a group under multiplication, the set of nonzero elements of $\gf(p^n)$ is isomorphic to $\Z_{p^n-1}$ (and is, therefore, cyclic).
+\end{theorem}
+
+\begin{remark}
+	Because there is only one field for each prime-power $p^n$, we may unambiguously denote it by $\gf(p^n)$, in honor of Galois, and call it the \textit{Galois field of order $p^n$}.
+\end{remark}
+
+\begin{corollary}
+	\[ [\gf(p^n):\gf(p)]=n \]
+\end{corollary}
+
+\begin{corollary}[$\mathbf{\gf(p^n)}$ Contains an Element of Degree $\mathbf{n}$]
+	Let $a$ be a generator of the group of nonzero elements of $\gf(p^n)$ under multiplication. Then $a$ is algebraic over $\gf(p)$ of degree $n$.
+\end{corollary}
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+\section{Subfields of a Finite Field}
+
+\begin{theorem}[Subfields of a Finite Field]
+	For each divisor $m$ of $n$, $\gf(p^n)$ has a unique subfield of order $p^m$. Moreover, these are the only subfields of $\gf(p^n)$.
+\end{theorem}
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+\chapter{Geometric Constructions}
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+\part{Fields}
+\subimport{chapters/chapter-19/}{chapter-19.tex}
+\subimport{chapters/chapter-20/}{chapter-20.tex}
+\subimport{chapters/chapter-21/}{chapter-21.tex}
+\subimport{chapters/chapter-22/}{chapter-22.tex}
+\subimport{chapters/chapter-23/}{chapter-23.tex}