Created the Abstract Algebra theorems and definitions cheat sheet

part-3/chapters/chapter-17/chapter-17.tex

@@ -0,0 +1,4 @@
+\chapter{Factorization of Polynomials}
+\subimport{./}{reducibility-tests.tex}
+\subimport{./}{irreducibility-tests.tex}
+\subimport{./}{unique-factorization-in-zx.tex}

part-3/chapters/chapter-17/irreducibility-tests.tex

@@ -0,0 +1,29 @@
+\section{Irreducibility Tests}
+
+\begin{theorem}[Mod $\mathbf{p}$ Irreducibility Test]
+	Let $p$ be a prime and suppose that $f(x) \in \Z[x]$ with $\deg f(x) \geq 1$. Let $\overline{f}(x)$ be the polynomial in $\Z_p[x]$ obtained from $f(x)$ by reducing all the coefficients of $f(x)$ modulo $p$. If $\overline{f}(x)$ is irreducible over $\Z_p$ and $\deg \overline{f}(x) = \deg f(x)$, then $f(x)$ is irreducible over $\Q$.
+\end{theorem}
+
+\begin{theorem}[Eisenstein's Criterion (1850)]
+	Let
+	\[ f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_0 \in \Z[x] \]
+	\noindent If there is a prime $p$ such that $p \nmid a_n, p\ \vert\ a_{n-1}, \dots, p\ \vert\ a_0$ and $p^2 \nmid a_0$, then $f(x)$ is irreducible over $\Q$.
+\end{theorem}
+
+\begin{corollary}[Irreducibility of $\mathbf{p}$th Cyclotomic Polynomial]
+	For any prime $p$, the $p$th cyclotomic polynomial
+	\[ \Phi_p(x) = \frac{x^p - 1}{x-1} = x^{p-1} + x^{p-2} + \dots + x + 1 \]
+	\noindent is irreducible over $\Q$.
+\end{corollary}
+
+\begin{theorem}[$\mathbf{\lr{p(x)}}$ Is Maximal If and Only If $\mathbf{p(x)}$ Is Irreducible]
+	Let $\F$ be a field and let $p(x) \in \F[x]$. Then $\lr{p(x)}$ is a maximal ideal in $\F[x]$ if and only if $p(x)$ is irreducible over $\F$.
+\end{theorem}
+
+\begin{corollary}[$\mathbf{\F[x]/\lr{p(x)}}$ Is a Field]
+	Let $\F$ be a field and $p(x)$ be an irreducible polynomial over $\F$. Then $\F[x]/\lr{p(x)}$ is a field.
+\end{corollary}
+
+\begin{corollary}[$\mathbf{p(x)\ \vert\ a(x)b(x)}$ Implies $\mathbf{p(x)\ \vert\ a(x)}$ or $\mathbf{p(x)\ \vert\ b(x)}$]
+	Let $\F$ be a field and let $p(x), a(x), b(x) \in \F[x]$. If $p(x)$ is irreducible over $\F$ and $p(x)\ \vert\ a(x)b(x)$, then $p(x)\ \vert\ a(x)$ or $p(x)\ \vert\ b(x)$.
+\end{corollary}

part-3/chapters/chapter-17/reducibility-tests.tex

@@ -0,0 +1,21 @@
+\section{Reducibility Tests}
+
+\begin{definition}[Irreducible Polynomial, Reducible Polynomial]
+	Let $D$ be an integral domain. A polynomial $f(x)$ from $D[x]$ that is neither the zero polynomial nor a unit in $D[x]$ is said to be \textit{irreducible over $D$}, whenever $f(x)$ is expressed as a product $f(x) = g(x)h(x)$, with $g(x)$ and $h(x)$ from $D[x]$, then $g(x)$ or $h(x)$ is a unit in $D[x]$. A nonzero, nonunit element of $D[x]$ that is not irreducible over $D$ is called \textit{reducible over $D$}.
+\end{definition}
+
+\begin{theorem}[Reducibility Test for Degrees 2 and 3]
+	Let $\F$ be a field. If $f(x) \in \F[x]$ and $\deg f(x)$ is 2 or 3, then $f(x)$ is reducible over $\F$ if and only if $f(x)$ has a zero in $\F$.
+\end{theorem}
+
+\begin{definition}[Content of a Polynomial, Primitive Polynomial]
+	The \textit{content} of a nonzero polynomial $a_nx^n + a_{n-1}x^{n-1} + \dots + a_0$, where the $a$'a are integers, is the greatest common divisor of the integers $a_n,a_{n-1}, \dots, a_0$. A \textit{primitive polynomial} is an element of $\Z[x]$ with content 1.
+\end{definition}
+
+\begin{lemma}[Gauss's Lemma]
+	The product of two primitive polynomials is primitive.
+\end{lemma}
+
+\begin{theorem}[Reducibility over $\mathbf{\Q}$ Implies Reducibility over $\mathbf{\Z}$]
+	Let $f(x) \in \Z[x]$. If $f(x)$ is reducible over $\Q$, then it is reducible over $\Z$.
+\end{theorem}

part-3/chapters/chapter-17/unique-factorization-in-zx.tex

@@ -0,0 +1,7 @@
+\section{Unique Factorization In $\mathbf{\Z[x]}$}
+
+\begin{theorem}[Unique Factorization in $\mathbf{\Z[x]}$]
+	Every polynomial in $\Z[x]$ that is not the zero polynomial or a unit in $\Z[x]$ can be written in the form $b_1b_2\dots b_sp_1(x)p_2(x)\dots p_m(x)$, where the $b_i$'s are irreducible polynomials of degree 0 and the $p_i(x)$'s are irreducible polynomials of positive degree. Furthermore, if
+	\[ b_1b_2\dots b_sp_1(x)p_2(x) \dots p_m(x) = c_1c_2 \dots c_tq_1(x) q_2(x) \dots q_n(x) \]
+	\noindent where the $b_i$'s and the $c_i$'s are irreducible polynomials of degree 0 and the $p_i(x)$'s and $q_i(x)$'s are irreducible polynomials of positive degree, then $s=t, m=n$, and, after renumbering the $c$'s and $q(x)$'s, we have $b_i = \pm c_i$, for $i=1, \dots, s$, and $p_i(x)= \pm q_i(x)$, for $i = 1, \dots, m$.
+\end{theorem}