Created the Abstract Algebra theorems and definitions cheat sheet

part-4/chapters/chapter-21/chapter-21.tex

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+\chapter{Algebraic Extensions}
+\subimport{./}{characterization-of-extensions.tex}
+\subimport{./}{finite-extensions.tex}
+\subimport{./}{properties-of-algebraic-extensions.tex}

part-4/chapters/chapter-21/characterization-of-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}

part-4/chapters/chapter-21/finite-extensions.tex

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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}

part-4/chapters/chapter-21/properties-of-algebraic-extensions.tex

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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}