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\title{Abstract Algebra}
\date{}
\author{V1.0}

\begin{document}
\maketitle

\section*{Groups:}

\begin{enumerate}
	\item A group $\langle G, * \rangle$ is a set $G$, closed under a binary operation $*$, such that the following axioms are satisfied.
	\begin{itemize}
		\item	For all $a, b, c \in G$, we have $(a*b)*c = a*(b*c)$. Associativity of $*$
		
		\item There is an element $e$ in $G$ such that for all $x \in G$, $e*x = x*e = x$. Identity $e$ for $*$
		
		\item Corresponding to each $a \in G$, there is an element $a'$ in G such that $a*a' = a'*a = e$. Inverse $a'$ of $a$
	\end{itemize}
	
	
	
	\item $\mathbb{Z}_n = {0, 1, ..., n-1}$ is a group under addition modulo $n$. 0=identity. $n-j$ = inverse of $j \in \mathbb{Z}_n$.
	
	
	
	\item General Linear Group of degree $n$ is the set of $n*n$ invertible matrices with matrix multiplication.



	\item $\mathbb{Z}_n'$ is a group under multiplication modulo $n$ if and only if the elements and $n$ are relatively prime. Identity=1. Inverse of $x$ = solution to $kx \ppmod{n} = 1$.



	\item $\{1, ..., p-1\}$ is a group under multiplication modulo $p$ if and only if $p$ is a prime. $\mathbb{Z}_p'$ has order $p-1$. Not every element in it is a generator. because $\mathbb{Z}_p'$ is cyclic and $p-1$ is not prime, and for every nontrivial divisor of $p-1$, $k$, there is a subgroup of order $k$, and the generator of that subgroup (which has order $k$) is definitely not the generator of $\mathbb{Z}_p'$. 
	
	
	
	\item The set $\mathbb{R}^n = \{(a_1, a_2, ..., a_n) | a_1, a_2, ..., a_n \in \mathbb{R}\}$ is a group under componentwise addition.\ (Direct product group)



	\item The number of elements of a group is its order = $|G|$.



	\item The order of an element $g$, which is denoted by $|g|$, in a group $G$ is the smallest positive integer $n$ such that $g^n = e$.



	\item $\{e\}$ is the trivial subgroup of $G$.



	\item Cyclic group $G = \{a^n | n \in \mathbb{Z}\} = \langle a \rangle$. $a$ is called a generator of $G$.



	\item The order of the generator is the same as the order of the group it generated.
	
	
	
	\item Let $G$ be a group, and let $a$ be any element of $G$. Then, $\langle a \rangle$ is a subgroup of $G$.
	
	
	
	\item $a^k = e$ implies that $|a|$ divides $k$.



	\item Let $G=\langle a \rangle$ be a cyclic group of order $n$. Then $G=\langle a^k \rangle$ if and only if $\gcd(n, k)=1$.



	\item An integer $k$ in $\mathbb{Z}_n$ is a generator of $\mathbb{Z}_n$ if and only if $\gcd(n, k)=1$.



	\item Every subgroup of a cyclic group is cyclic. Moreover, if $|\langle a \rangle|=n$, then the order of any subgroup of $\langle a \rangle$ is a divisor of $n$; and, for each positive divisor $k$ of $n$, the group $\langle a \rangle$ has exactly one subgroup of order $k$ - namely $\langle a^{(n/k)} \rangle$. This is not true for any group, but for cyclic groups (as well as for Abelian groups). For example, the group $A_4$ of order 12 has no subgroup of order 6.



	\item Every cyclic group is Abelian.



	\item For each positive divisor $k$ of $n$, the set $\langle n/k \rangle$ is the unique subgroup of $\mathbb{Z}_n$ of order $k$; moreover, these are the only subgroup of $\mathbb{Z}_n$.



	\item If two groups defined in different terms are really the same, we say that there is an isomorphism between the two groups.

	\textbf{Definition}: Let $\langle G, + \rangle$ and $\langle G', * \rangle$ be two groups. An isomorphism $\phi$ is a one-to-one onto function mapping $G$ to $G'$ such that: $\phi (g_1+g_2) = \phi (g_1)*\phi (g_2)$.
	
	
	\begin{description}
		\item[\small Step 1:] Mapping: Define a candidate for the isomorphism.
		\item[\small Step 2:] Isomorphism is ``1-1".
		\item[\small Step 3:] Isomorphism is ``onto".
		\item[\small Step 4:] Isomorphism is operation-preserving: show that $\phi (g_1+g_2) = \phi (g_1)*\phi (g_2)$ for all $g_1$ and $g_2$ in $G$.
	\end{description}

	Example:

	\begin{itemize}
		\item The real numbers under addition is isomorphic to positive real numbers under multiplication.
		\item infinite cyclic group is isomorphic to $\mathbb{Z}$. $\langle a \rangle$ of order $n$ is isomorphic to $\mathbb{Z}_n$ under addition modulo $n$ (other way of saying it is that cyclic group of order $n$ is isomorphic to $\mathbb{Z}_n$).
	\end{itemize}
	
	
	
	\item If $G$ is isomorphic to $G'$:
	
	\begin{itemize}
		\item $G$ is Abelian if and only if $G'$ is Abelian
		\item $G$ is cyclic if and only if $G'$ is cyclic
		\item If $K$ is a subgroup of $G$, then $\phi (K) = \{\phi (k) | k \in K\}$ is a subgroup of $G'$
	\end{itemize}

	
	\item Let $G$ be a group and $H$ a subset of $G$. For any $a \in G$, the set $\{ah | h \in H\}$ is denoted by $aH$. Analogously, $Ha = \{ha | h \in H\}$. When $H$ is a subgroup of $G$, the set $aH$ is called the left coset of $H$ in $G$ containing $a$, whereas $Ha$ is called the right coset of $H$ in $G$.
	
	
	Example:
	Let $H = \{0, 3, 6\}$ be subgroup in $G = \mathbb{Z}_9$ under addition. $a+H$:
	
	$0+H = \{0, 3, 6\} = 3+H = 6+h$
	
	$1+H = \{1, 4, 7\} = 4+H = 7+h$
	
	$2+H = \{2, 5, 8\} = 5+H = 8+h$
	
	
	
	\item Cosets are usually not subgroups. $aH$ may be the same as $bH$, even though $a$ is not the same as $b$.
	
	
	
	\item Let $H$ be a subgroup of $G$ ($G$ is finite).
	
	\begin{enumerate}[(1).]
	\item $a \in aH$
	\item $aH = H$ if and only if $a \in H$
	\item $aH$ is a subgroup of $G$ if and only if $a \in H$. (Hence, no coset of $H$ except $H$ itself is subgroup.)
	\item $aH = bH$ or $aH \cap bH = \emptyset$
	\item $|aH| = |bH|$
	\item $aH = bH$ iff $a^{-1}*b \in  H$
	\end{enumerate}
	
	
	
	\item \textbf{Lagrange's Theorem}: If $G$ is a finite group and $H$ is a subgroup of $G$, then $|H|$ divides $|G|$. The converse is not true. The number of distinct left (right) cosets of $H$ in $G$ is $|G|/|H|$ which is called the index of $H$.



	\item In a finite group, the order of each element of the group divides the order of the group.



	\item A group of prime order is cyclic.



	\item For any $a \in G$, $a^{|G|} = e$.

	
	
	\item \textbf{Fermat's Little Theorem}: For every integer $a$ and every prime $p$. $a^p \ppmod{p} = a \ppmod{p}$. $\mathbb{Z}_p$ under multiplication mod $p = \{1, ..., p-1\}$, hence $a^p \ppmod{p} = a^{p-1}*a \ppmod{p} = e*a \ppmod{p}$.


	
	\item Definition: Let $G_1, G_2, \ldots, G_n$ be a finite collection of groups. The external direct product of $G_1, G_2, \ldots, G_n$, written $G_1*G_2*\ldots*G_n$ is the set of all $n$-tuples for which the $i$th component is an element of $G_i$ and the operation is componentwise:

	$G_1*G_2*\ldots*G_n = \{(g_1, g_2, \ldots, g_n) | g_i \in G_i\}\ \mathrm{where}\ (g_1, \ldots, g_n) * (g_1', \ldots, g_n') = (g_1 * g_1', \dots, g_n * g_n')$.



	\item The order of an element in a direct product of a finite number of finite groups is the least common multiple of the orders of the components of the element:
$|(g_1, g_2, \ldots, g_n)| = \mathrm{lcm}(|g_1|, \ldots, |g_n|)$



	\item Whenever you work with an arithmetic operation mod $q$, you are working in the group of $\mathbb{Z}_q$ or $\mathbb{Z}_q'$. For example: $k * 5 = 1 \ppmod{q}$, you are looking for the inverse of 5 in $\mathbb{Z}_q'$ under multiplication. XOR is just addition operation in $\mathbb{Z}_2*\dots*\mathbb{Z}_2$. 
	
	
	
	\item A subgroup $H$ of a group is called a normal subgroup of $G$ if $aH = Ha$ for all $a \in G$. We denote this by $H \triangleleft G$. It is wrong to assume that $H$ is normal in $G$ means $ah = ha$ for all $a \in G$ and $h \in H$. This is not what normality of $H$ means; rather, it means that if $a \in G$ and $h \in H$, then there exist elements $h'$ and $h''$ in $H$ such that $ah = h'a$ and $ha = ah''$. It is possible that $h' = h$ or $h'' = h$, but we may not assume this.
	
	Example:
	\begin{itemize}
	\item Every subgroup of an Abelian group is normal (in this case, $ah=ha$ for all $a \in G$ and $h \in H$).
	
	\item The center $Z(G)$ of a group is always normal (again, $ah=ha$ for any $a \in G$ and any $h \in Z(G)$).
	\end{itemize}
	
	
	
	\item We have yet to explain why normal subgroups are of special significance. The reason is simple. When the subgroup $H$ of $G$ is normal, then the set of left (or right) cosets of $H$ in $G$ is itself a group --- called the factor group of $G$ by $H$ (or the quotient group of $G$ by $H$). Quite often, one can obtain information about a group by studying one of its factor groups.

	\textbf{Definition}: Let $G$ be a group and $H$ a normal subgroup of $G$. The set $G/H = \{aH | a \in G\}$ is a group under the operation $(aH)(bH) = abH$.
	
	Example:

	Let $4\mathbb{Z} = \{0, \pm4, \pm8, \ldots\}$. To construct $\mathbb{Z}/4\mathbb{Z}$, we must first determine the left cosets of $4\mathbb{Z}$ in $\mathbb{Z}$. Consider the following four cosets:

	$0 + 4\mathbb{Z} = \{0, \pm4, \pm8, \ldots\}$

	$1 + 4\mathbb{Z} = \{1, 5, 9, \ldots, -3, -7, \ldots\}$
	
	$2 + 4\mathbb{Z} = \{2, 6, 10, \ldots\}$
	
	$3 + 4\mathbb{Z} = \{3, 7, 11, \ldots\}$

	We claim that there are no others. Since for any $k = 4q + r: k + 4\mathbb{Z} = r + 4q + 4\mathbb{Z} = r + 4\mathbb{Z}$. So $\mathbb{Z}/4\mathbb{Z} = \{0+4\mathbb{Z}, 1+4\mathbb{Z}, 2+4\mathbb{Z}, 3+4\mathbb{Z}\}$ and $\mathbb{Z}/4\mathbb{Z}$ is isomorphic to $\mathbb{Z}_4$ which is cyclic. Remember that in finite case, the number of cosets of $H$ is $|G|/|H|$.
	
	Example:
	\begin{itemize}
	\item	$\mathbb{Z}/n\mathbb{Z}$ is isomorphic to $\mathbb{Z}_n$.

	\item	 $G = \mathbb{Z}_{18}$ and $H = <6> = \{0, 6, 12\}$. Then $G/H = \{0+H, 1+H, 2+H, 3+H, 4+H,  5+H\}$. To illustrate how the group element are combined, consider $(5+H) + (4+H)$. This should be one of the six elements listed in the set $G/H$. $(5+H) + (4+H) = 5+4+H = 9+H =3+6+H = 3+H$, since $H$ absorbs all multiples of 6. 
	\end{itemize}
	
	
	
	\item \textbf{Homomorphism}: Let $\langle G, + \rangle$  and $\langle G', * \rangle$  be two groups. An homomorphism $\phi$ is a function mapping $G$ to $G'$ such that: $\phi(g_1+g_2) = \phi(g_1)*\phi(g_2)$, which is called operation preserving. (Remember that an isomorphism is 1-1 and onto.)



	\item The kernel of a homomorphism $\phi$ from a group $G$ to a group with identity $e$ is the set $Ker(\phi) = \{x \in G | \phi(x)=e\}$. 
	
	Example:
	
	The mapping $\phi$ from $R^*$ to $R^*$ defined by $\phi(x) = |x|$ is a homomorphism with $Ker(\phi) = \{1, -1\}$.

	
	
	\item If $G$ is homomorphic to $G'$
	
	\begin{enumerate}[(1).]
	\item $\phi$ carries the identity of $G$ to the identity of $G'$
	\item $\phi(g^n) = (\phi(g))^n$
	\item If $|g|$ is finite, then $|\phi(g)|$ divides $|g| : e = \phi(e) = \phi(g^n) = (\phi(g))^n$
	\item $Ker(\phi)$ is a subgroup of $G$.
	\item If $\phi(g) = g'$, then $\phi^{-1}(g') = gKer(\phi)$. and $gKer(\phi)$ is a coset of $Ker(\phi)$, so it implies that $\phi$ maps a set of elements in $G$ to an element in $G'$.
	\item $Ker(\phi)$ is normal, so $G/Ker(\phi)$ is the factor group of $G$ by $Ker(\phi)$ --- a group of cosets of $Ker(\phi)$.
	\end{enumerate}



	\item \textbf{First Isomorphism Theorem}: Let $\phi$ be a group homomorphism from $G$ to $G'$, Then the mapping from $G/Ker(\phi)$ to $\phi(G)$, given by $gKer(\phi) \rightarrow \phi(g)$, is an isomorphism.
	
	
	
	\item Every normal subgroup of a group $G$ is the kernel of a homomorphism of $G$. In particular, a normal subgroup $N$ is the kernel of the mapping $g \rightarrow gN$ from $G$ to $G/N$.



	\item A group homomorphism $\phi$ is a one-to-one map if and only if $Ker(\phi) = \{e\}$.

 
 
	\item To show $\phi: G  \rightarrow G'$ is an isomorphism:
	
	\begin{itemize}
	\item 	Show $\phi$ is a homomorphism
	
	\item	Show $Ker(\phi) = \{e\}$
	
	\item	Show $\phi$ maps $G$ onto $G'$.
	\end{itemize}
	
	
	
	\item \textbf{Fundamental Theorem of Finite Abelian Groups}: Every finite Abelian group is a direct product of cyclic groups of prime-power order. Moreover, the number of terms in the product and the orders of the cyclic groups are uniquely determined by the group.

	This theorem shows that every finite Abelian group G is isomorphic to a group of the form
\[\mathbb{Z}_{{p_1}^{n_1}} + \mathbb{Z}_{{p_2}^{n_2}} + \ldots + \mathbb{Z}_{{p_k}^{n_k}}\]
where the $p_i$'s are not necessarily distinct primes and the prime powers $p_1^{n_1}, p_2^{n_2}, \ldots, p_k^{n_k}$ are uniquely determined by $G$. Writing a group in this form is called determining the isomorphism class of $G$.

	This is a powerful theorem which allows you to construct all Abelian groups of any order.

	Example:
	
	Construct Abelian groups of order 1176:
	
	$1176 = 2^3*3*7^2$. Then the complete list of the distinct isomorphism classes of Abelian groups of order 1176 is:
	
	$\mathbb{Z}_8+\mathbb{Z}_3+\mathbb{Z}_{49}$
	
	$\mathbb{Z}_4+\mathbb{Z}_2+\mathbb{Z}_3+\mathbb{Z}_{49}$
	
	$\mathbb{Z}_2+\mathbb{Z}_2+\mathbb{Z}_2+\mathbb{Z}_3+\mathbb{Z}_{49}$
	
	$\mathbb{Z}_8+\mathbb{Z}_3+\mathbb{Z}_7+\mathbb{Z}_7$
	
	$\mathbb{Z}_4+\mathbb{Z}_2+\mathbb{Z}_3+\mathbb{Z}_7+\mathbb{Z}_7$
	
	$\mathbb{Z}_2+\mathbb{Z}_2+\mathbb{Z}_2+\mathbb{Z}_3+\mathbb{Z}_7+\mathbb{Z}_7$

	Given any particular Abelian group $G$ of order 1176, to find which of the preceding six isomorphism classes represents the structure of $G$ we only need to compare the orders of the elements of $G$ with the orders of the elements in the six direct products, since two finite Abelian groups are isomorphic if and only if they have the same number of elements of each order. For instance, we could determine whether $G$ has any elements of order 8. If so, then $G$ must be isomorphic to the first or fourth group above, since these are the only ones with elements of order 8. To narrow $G$ down to a single choice, we now only need to check whether or not $G$ has an element of order 49, since the first product above has such an element, whereas the fourth one does not.
	
	 Corollary: If $m$ divides the order of a finite Abelian group $G$, then $G$ has a subgroup of order $m$.	
	
\end{enumerate}

\newpage

%************************************************************************************************************************************************************************%

\section*{Rings:}

\begin{enumerate}

	\item A ring $\langle R, +, * \rangle$ is a non-empty set with two binary operations, addition and multiplication (as oppose to Group which only has one operation), such that for all $a, b, c \in R$:
	
	\begin{enumerate}[(1).]
		\item $a+b = b+a$
		\item $(a+b)+c = a+(b+c)$
		\item additive identity 0
		\item additive inverse
		\item $a*(b*c) = (a*b)*c$
		\item $a*(b+c) = a*b+a*c$  and   $(b+c)*a = b*a + c*a$
	\end{enumerate}

	or we can use group to define ring:
	
	\begin{itemize}
		\item $\langle R, + \rangle$ is an abelian group
		\item multiplication is associative
		\item for all $a, b, c \in R$, the left and right distributive law holds.
	\end{itemize}
	
	Facts:
	\begin{itemize}
		\item multiplication need not be commutative, when it is we say that the ring is commutative.
	
		\item a ring need not have an identity under multiplication. A unity in a ring is a nonzero element that is an identity under multiplication.
		
		\item a nonzero element of a ring with unity need not have a multiplicative inverse. When it does, we say that it is a unit of the ring.

		\item $\mathbb{Z}$ under ordinary addition and multiplication is a commutative ring with unity 1.

		\item  $\mathbb{Z}_n = \{0, 1, ..., n-1\}$ under addition and multiplication modulo $n$ is a commutative ring with unity 1. The set of units is the group $\mathbb{Z}_n'$ under multiplication.

		\item $M_n(\mathbb{R}), M_n(\mathbb{Q}), M_n(\mathbb{Z})$ are rings.

		\item  $\mathbb{Z}[x]$ of all polynomials in the variable $x$ with integer coefficients under ordinary addition and multiplication is a commutative ring with unity $f(x) = 1$.

		\item Let $R_1, R_2, \ldots, R_n$ be rings. The direct product of these rings is also a ring and is defined by
	
		$R_1*R_2*\ldots*R_n = \{(a_1, a_2, \ldots, a_n) | a_i \in R_i \}$
	
		addition: $(a_1, a_2, \ldots, a_n) + (b_1, b_2, \ldots, b_n) = (a_1+b_1, \ldots, a_n+b_n)$
	
		multiplication: $(a_1, \ldots, a_n)*(b_1, \ldots, b_n) = (a_1*b_1, \ldots, a_n*b_n)$
	\end{itemize}
	
	
	
	\item If a ring has a unity, it is unique. If a ring element has a multiplicative inverse, it is unique.

	
	
	\item A nonempty subset $S$ of a ring $R$ is a subring if $S$ is closed under subtraction and multiplication.
	
	Example:
	 $\{0, 2, 4\}$ is a subring of the ring $\mathbb{Z}_6$. Note that although 1 is the unity in $\mathbb{Z}_6$, 4 is the unity in $\{0, 2, 4\}$.

	
	
	\item A zero-divisor is a nonzero element $a$ of a commutative ring $R$ such that there is a nonzero element $b \in R$ with $a*b=0$.
	
	

	\item To a certain degree, the notion of a ring was invented in an attempt to put the algebraic properties of the integers into an abstract setting. A ring is not the appropriate abstraction of the integers, however, for too much is lost in the process: commutativity, existence of unity, and cancellation. Hence we present integral domains which have all three properties.

	\textbf{Definition}: An integral domain is a commutative ring with unity and no zero-divisor.
	
	Facts:
	\begin{itemize}
		\item $\mathbb{Z}$ is an integral domain.
		\item $\mathbb{Z}[x]$ is an integral domain. Unity: $f(x)=1$.
		\item $\mathbb{Z}_p$ where $p$ is a prime is an integral domain.
		\item $\mathbb{Z}_n$ of integers is not an integral domain when $n$ is not prime. (because it has zero-divisors)
		\item $M_n(\mathbb{Z})$ is not an integral domain. (because it has zero-divisors)
	\end{itemize}
	
	
	
	\item A field is a commutative ring with unity (or integral domain) in which every nonzero element is a unit. A field is an integral domain.
		
	Facts:
	\begin{itemize}
		\item A field can be thought of as simply an algebraic system that is closed under addition, subtraction, multiplication, and division (except by 0).
		\item $\mathbb{Z}$ is not a field.
		\item $\mathbb{Q}$ is a field.
		\item Even if $F$ is a field, $F[x]$ is not a field.
	\end{itemize}
	
	
	
	\item A finite integral domain is a field.
	
	Facts:
	\begin{itemize}
		\item  So, in the finite case, fields and integral domains are the same.
		\item $\mathbb{Z}_p$ where $p$ is a prime is a field.
		\item $\mathbb{Q}[\sqrt{2}] = \{a + b\sqrt{2} | a, b \in \mathbb{Q}\}$ is a field, because the multiplicative inverse $1/(a+b\sqrt{2})$ can be expressed in $c + d\sqrt{2}$.
	\end{itemize}

	
	
	\item The characteristic of a ring $R$, $char(R)$, is the least positive integer $n$ such that $nx = 0$ for all $x \in R$. If no such integer exists, we say that $R$ has characteristic 0.
	
	Facts:
	\begin{itemize}
		\item $char(\mathbb{Z}) = 0$.
		\item an infinite ring can have a nonzero characteristic: the ring $\mathbb{Z}_2[X]$ has characteristic 2.
	\end{itemize}
	
	
	
	\item Let $R$ be a ring with unity 1. If 1 has infinite order under addition, then the characteristic of $R$ is 0. If 1 has order $n$ under addition, then the characteristic of $R$ is $n$.

	
	
	\item The characteristic of an integral domain is 0 or prime. 
	
	
	
	\item A subring $A$ of a ring $R$ is called a (two-sided) ideal of $R$ if for every $r \in R$ and every $a \in A$ both $ra$ and $ar$ are in $A$.

	So a subring $A$ of a ring $R$ is an ideal of $R$ if $A$ ``absorbs" elements from $R$ --- that is, if $rA = \{ra | a \in A\} \subseteq A\ \mathrm{for\ all}\ r \in R$. An ideal $A$ of $R$ is called a proper ideal of $R$ if $A$ is a proper subset of $R$.

	Let $R$ be a commutative ring with unity and let $a \in R$. The set $\langle a \rangle = \{ra | r \in R\}$ is an ideal of $R$ called the principal ideal generated by $a$.

	Ideal test: A nonempty subset $A$ of a ring $R$ is an ideal of $R$ if
	
	\begin{itemize}
		\item $a-b \in A$ whenever (for all) $a, b \in A$.
		\item $ra$ and $ar$ are in $A$ whenever (for all) $a \in A$ and $r \in R$.
	\end{itemize}
	
	Facts:
	
	\begin{itemize}
		\item $\{0\}$ is the trivial ideal of $R$.
		\item $R$ is an ideal of $R$.
		\item $n\mathbb{Z}$ is an ideal of $\mathbb{Z}$.
		\item Let $\mathbb{R}[x]$ denote the set of all polynomials with real coefficients and let $A$ denote the subset of all polynomials with constant term 0. Then $A$ is an ideal of $\mathbb{R}[x]$ and $A = \langle x \rangle$.
		\item Let $I$ be the subset of $\mathbb{Z}[x]$ of all polynomials with even constant term. Then $I$ is an ideal of $\mathbb{Z}[x]$ and $I = \langle x, 2 \rangle$. ($R$ be a commutative ring with unity and let $a_1, \ldots, a_n$ belong to $R$. Then $I =\langle a_1, a_2, \ldots, a_n \rangle = \{r_1*a_1 + \ldots + r_n*a_n\}$ is an ideal of R.) 
	\end{itemize}
	
	
	
	\item If $A$ is an ideal of a ring $R$ and $1$ belongs to $A$, then $A = R$.


	
	\item A ring $R$ is a abelian group under addition, an ideal $A$ is a normal subgroup of $R$.
	
	
	
	\item \textbf{Factor rings}: Let $R$ be a ring and let $A$ be an ideal of $R$. Since $R$ is a abelian group under addition and $A$ is a normal subgroup of $R$, we may form the factor group $R/A = \{r + A | r \in R\}$. The natural question at this point is: How may we form a ring of this group of cosets? The addtion is already taken care of, and, by analogy with groups of cosets, we define the product of two cosets of $s + A$ and $t + A$ as $st + A$. This works as long as $A$ is an ideal, and not just a subring of $R$. 

	More rigorous definition:
	Let $R$ be a ring and let $A$ be a subring of $R$. The set of cosets $\{r + A | r \in R\}$ is a ring under the operations $(s + A) + (t + A) = s + t + A$ and $(s + A)(t + A) = st + A$ if and only if $A$ is an ideal of $R$.
	
	Examples:
	
	\begin{itemize}
		\item $\mathbb{Z}/4\mathbb{Z} = \{0 + 4\mathbb{Z}, 1 + 4\mathbb{Z}, 2 + 4\mathbb{Z}, 3 + 4\mathbb{Z}\}$.
		\item $(2 + 4\mathbb{Z}) + (3 + 4\mathbb{Z}) = 1 + 4 + 4\mathbb{Z} = 1 + 4\mathbb{Z}$; $(2 + 4\mathbb{Z})(3 + 4\mathbb{Z}) = 2 + 4 + 4\mathbb{Z} = 2 + 4\mathbb{Z}$. The two operations are essentially modulo 4 arithmetic. (is $\mathbb{Z}/4\mathbb{Z}$ ring isomorphic to $\mathbb{Z}_4$?)

		\item $2\mathbb{Z}/6\mathbb{Z} = \{0 + 6\mathbb{Z}, 2 + 6\mathbb{Z}, 4 + 6\mathbb{Z}\}$. Here the operations are essentially modulo 6 arithmetic.
	
		\item $R = \{M_2(\mathbb{Z})\}$ and let $I$ be the subset of $R$ consisting of matrices with even entries. It is easy to show that $I$ is indeed an ideal of $R$. The factor ring $R/I$ has size 16.

		\item Let $\langle x^2 + 1 \rangle$ denote the principal ideal of $\mathbb{R}[x]$ generated by $x^2 + 1$. $\langle x^2 + 1 \rangle = \{f(x)(x^2 + 1) | f(x) \in \mathbb{R}[x]\}$.

		Then $\mathbb{R}[x]/\langle x^2 + 1 \rangle = \{g(x) + \langle x^2 + 1\rangle | g(x) \in \mathbb{R}[x]\}$ = $\{ax + b + \langle x^2 + 1 \rangle | a, b \in \mathbb{R}\}$.

		(Why is $\langle x^2 + 1 \rangle$ a principle ideal? Proof: For any $f(x) \in \langle x^2 + 1 \rangle$, then $f(x) = g(x)(x^2 + 1)$ for some $g(x) \in \mathbb{R}[x]$. For any $w(x) \in \mathbb{R}[x], w(x)f(x) = w(x)g(x)(x^2 + 1) = u(x)(x^2 + 1) \in \langle x^2 + 1 \rangle$, for some $u(x) = w(x)g(x)$.)

		To see this last equality, note that if $g(x)$ is any member of $\mathbb{R}[x]$, then we may write $g(x)$ in the form of $q(x)(x^2 + 1) + r(x)$, where $q(x)$ is the quotient and $r(x)$ is the remainder upon dividing $g(x)$ by $x^2 + 1$. In particular, $r(x) = 0$ or the degree of $r(x)$ is less than 2, so that $r(x) = ax + b$ for some $a$ and $b$ in $\mathbb{R}$. Thus $g(x) + \langle x^2 + 1 \rangle = q(x)(x^2 + 1) + r(x) + \langle x^2 + 1 \rangle = r(x) + \langle x^2 + 1 \rangle$.

		How is multiplication done? Since $x^2 + 1 + \langle x^2 + 1 \rangle = 0 + \langle x^2 + 1 \rangle$, one should think of $x^2 + 1$ as 0 or, equivalently, as $x^2 = -1$. So, for example, $(x + 3 + \langle x^2 + 1 \rangle)*(2x + 5 + \langle x^2 + 1\rangle) = 2x^2 + 11x + 15 + \langle x^2 + 1 \rangle = 11x + 13 + \langle x^2 + 1 \rangle$.
	\end{itemize}
	

		
	\item A prime ideal $A$ of a commutative ring $R$ is a proper ideal of $R$ such that $a, b \in R$ and $ab \in A$ implies $a \in A\ \mathrm{or}\ b \in A$. The motivation for the definition of a prime ideal comes from the integers.

	Examples:
	\begin{itemize}
	\item Let $n$ be an integer greater than 1. Then, in the ring of integers, the ideal $n\mathbb{Z}$ is prime if and only if $n$ is prime. (Hence $4\mathbb{Z}$ is not a prime ideal, since $12 = 2*6$, and $2, 6 \notin 4\mathbb{Z}$.)
	
	\item The ideal $\langle x^2+1\rangle$ is not prime in $\mathbb{Z}_2[x]$, since it contains $(x+1)^2 = x^2 + 2x + 1 = x^2 + 1$ but does not contain $x+1$.
	\end{itemize}
	
	
	
	\item A maximum ideal $A$ of $R$ is a proper ideal of $R$ if, whenever $B$ is an ideal of $R$ and $A \subseteq B \subseteq R$, then $B = A$ or $B = R$. So, the only ideal that properly contains a maximal ideal is the entire ring.

	Example:
	
	 The ideal $\langle x^2+1\rangle$ is maximal in $\mathbb{R}[x]$.



	\item Let $R$ be a commutative ring with unity and let $A$ be an ideal of $R$. Then $R/A$ is an integral domain if and only if $A$ is prime.
	\label{integral}



	\item Let $R$ be a commutative ring with unity and let $A$ be an ideal of $R$. Then $R/A$ is a field if and only if $A$ is maximal.
	\label{field}



	\item Since a field is a special kind of integral domain, a maximal ideal is a prime ideal (follows from \ref{integral}. and \ref{field}.)\@. However, a prime ideal need not be maximal.
	
	
	
	\item  A ring homomorphism $\phi$ from a ring $R$ to a ring $S$ is a mapping from $R$ to $S$ that preserves the two ring operations; that is, for all $a$ and $b$ in $R$:
$\phi(a+b) = \phi(a)+\phi(b)$  and  $\phi(a*b) = \phi(a)*\phi(b)$.

	A ring homomorphism that is both $1-1$ and onto is called a ring isomorphism.

	Examples:
	\begin{itemize}
	\item The mapping $f(x) \rightarrow f(1)$ is a ring homomorphism from $\mathbb{R}[X]$ onto $\mathbb{R}$.
	
	\item Although $2\mathbb{Z}$, the group of even integers under addition, is group-isomorphic to the group $\mathbb{Z}$ under addition, the ring $2\mathbb{Z}$ is not ring-isomorphic to the ring $\mathbb{Z}$, because $2\mathbb{Z}$ doesn't have any unity hence structurally different from $\mathbb{Z}$ which has unity 1.
	\end{itemize}
	


	\item Let $\phi$ be a ring homomorphism from a ring $R$ to a ring $S$. Let $A$ be a subring of $R$ and $B$ an ideal of $S$.
	\begin{enumerate}[(1).]
	\item For any $r \in R$ and any positive integer $n$, $\phi(nr) = n\phi(r)$ and $\phi(r^n) = (\phi(r))^n$
	
	\item $\phi(A) = \{\phi(a) | a \in A\}$ is a subring of $S$.
	
	\item If $A$ is an ideal and $\phi$ is onto $S$, then $\phi(A)$ is an ideal.
	
	\item $\phi^{-1}(B) = \{r \in R | \phi(r) \in B\}$ is an ideal of $R$.
	
	\item If $R$ is commutative, the $\phi(R)$ is commutative.
	
	\item $\phi$ is an isomorphism if and only if $\phi$ is onto and $Ker(\phi) = \{r \in R | \phi(r) = 0\} = 0$.
	\end{enumerate}
	
	
	
	
	\item  Let $\phi$ be a homomorphism from a ring $R$ to a ring $S$. Then $Ker(\phi) = \{r \in R | \phi(r) = 0\}$ is an ideal of $R$.



	\item \textbf{First isomorphism theorem for rings}: Let $\phi$ be a ring homomorphism from $R$ to $S$. Then the mapping from $R/Ker(\phi)$ to $\phi(R)$, given by $r + Ker(\phi) \rightarrow \phi(r)$, is an isomorphism. In symbol, $R/Ker(\phi) \sim \phi(R)$.



	\item Every ideal of a ring $R$ is the kernel of a ring homomorphism of $R$. In particular, an ideal $A$ is the kernel of the mapping $r \rightarrow r+A$ from $R$ to $R/A$, and this is called the natural homomorphism from $R$ to $R/A$.

	Example:
	
	Since the mapping $\phi$ from $\mathbb{Z}[x]$ onto $\mathbb{Z}$ given by $\phi(f(x)) = f(0)$ is a ring homomorphism with $Ker(\phi) = \langle x \rangle$, we have $\mathbb{Z}[x]/\langle x \rangle \sim \mathbb{Z}$. And because $\mathbb{Z}$ is an integral domain but not a field, we know that the ideal $\langle x \rangle$ is prime but not maximal in $\mathbb{Z}[X]$.



	\item If $R$ is a ring with unity and the characteristic of R is $n > 0$, then $R$ contains a subring isomorphic to $\mathbb{Z}_n$. If the characteristic of $R$ is 0, then $R$ contains a subring isomorphic to $\mathbb{Z}$.



	\item If $F$ is a field of characteristic $p$, then $F$ contains a subfield isomorphic to $\mathbb{Z}_p$. If $F$ is a field of characteristic 0, then $F$ contains a subfield isomorphic to the rational numbers $\mathbb{Q}$.



	\item Although the integral domain $\mathbb{Z}$ is not a field, it is at least contained in a field --- the field of rational numbers. And notice that the field of rational numbers is nothing more than  quotients of integers. Can we mimic the construction of the rationals from the integers for other integral domains?
	
	
	
	\item Let $D$ be an integral domain. Then there exists a field $F$ (called the field of quotients of $D$) that contains a subring isomorphic to $D$.

	Examples:
	\begin{itemize}
	\item Let $D = \mathbb{Z}[x]$. Then the field of quotients of $D$ is $\{f(x)/g(x) | f(x), g(x) \in D,\ \mathrm{where}\ g(x)$ is not the zero polynomial$\}$.
	
	\item Let $p$ be a prime. Then $\mathbb{Z}_p(x) = \{f(x)/g(x) | f(x), g(x) \in \mathbb{Z}_p[x], g(x) \neq 0\}$ is an infinite field of characteristic $p$.
	\end{itemize}	
	
	
	
	\item Let $R$ be a commutative ring. The set of formal symbols $R[x] = \{a_nx^n + a_{n-1}x^{n-1} + \ldots + a_0 | a_i \in R\}$ is called the ring of polynomials over $R$ in the indeterminate $x$.
	
	In this definition, the symbols $x, x^2, ..., x^n$ do not represent ``unknown" elements or variables from the ring $R$. Rather their purpose is to serve as convenient placeholders that separate the ring elements $a_n, a_{n-1}, \ldots, a_0$. We could have avoided the $x$'s by defining a polynomial as an infinite sequence $a_0, a_1, a_2, \ldots, a_n, 0, 0, \ldots$

	
	
	\item If $D$ is an integral domain, then $D[x]$ is an integral domain.

	
	
	\item Let $F$ be a field and let $f(x)$ and $g(x) \in F[x]$ with $g(x) \neq 0$. Then there exist unique polynomials $q(x)$ and $r(x)$ in $F[x]$ such that $f(x) = g(x)q(x) + r(x)$ and either $r(x) = 0$ or $deg(r(x)) < deg(g(x))$. 
	
	
	
	\item Let $D$ be an integral domain. If $f(x)$ and $g(x) \in D[x]$, we say that $g(x)$ divides $f(x) \in D[x]$ if there exists an $h(x) \in D[x]$ such that $f(x) = g(x)h(x)$. In this case, we also call $g(x)$ a factor of $f(x)$. An element $a$ is a zero (or a root) of a polynomial $f(x)$ if $f(a) = 0$. When $F$ is a field, $a \in F$, and $f(x) \in F[x]$, we say that $a$ is a zero of multiplicity $k$ if $(x - a)^k$ is a factor of $f(x)$ but $(x - a)^{k+1}$ is not a factor of $f(x)$. 

	
	
	\item Let $F$ be a field, $a \in F$, and $f(x) \in F[x]$. Then $f(a)$ is the remainder in the division of $f(x)$ by $x-a$.

	
	
	\item Let $F$ be a field, $a \in F$, and $f(x) \in F[x]$. Then $a$ is a zero of $f(x)$ if and only if $x-a$ is a factor of $f(x)$.

	
	
	\item A polynomial of degree $n$ over a field has at most $n$ zeros counting multiplicity.

	
	
	\item A principal ideal domain (PID) is an integral domain $R$ in which every ideal has the form $\langle a \rangle = \{ra | r \in R\}$ for some $a \in R$.

	
	
	\item Let $F$ be a field. Then $F[x]$ is a principal ideal domain.
	\label{PID}

	
	
	\item Criterion for $I = \langle g(x) \rangle$: Let $F$ be a field, $I$ a nonzero ideal in $F[x]$, and $g(x)$ an element of $F[x]$. Then, $I = \langle g(x) \rangle$ if and only if $g(x)$ is a nonzero polynomial of minimum degree in $I$. 
	
	
	
	\item In mathematics, irreducible means that an object cannot be expressed as a product of at least two non-trivial factors in a given set.
For any field $F$ (actually it can be just integral domain), a polynomial $p(x) \in F[x]$ is called irreducible over $F$ if it is non-constant and cannot be represented as the product of two or more non-constant polynomials from $F[x]$. Here is a formal definition.

	\textbf{Definition}: 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 irreducible over $D$ if, 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, non-unit element of $D[x]$ that is not irreducible over $D$ is called reducible over $D$.
	
	Examples:
	\begin{itemize}
	\item The polynomial $f(x) = 2x^2 + 4$ is irreducible over $\mathbb{Q}$ but reducible over $\mathbb{Z}$, since $2x^2 + 4 = 2(x^2+2)$ and neither 2 nor $x^2 + 2$ is a unit in $\mathbb{Z}[x]$.
	
	\item $x^2 -2$ is irreducible over $\mathbb{Q}$ but reducible over $\mathbb{R}: (x + \sqrt{2})(x - \sqrt{2})$.
	
	\item The polynomial $x^2+1$ is irreducible over $\mathbb{Z}_3$ but reducible over $\mathbb{Z}_5: (2x+1)(3x+1)$.
	\end{itemize}
	
	As we have mentioned above, in the case that an integral domain is a field $F$, it is equivalent and more convenient to define a nonconstant $f(x) \in F[x]$ to be irreducible if $f(x)$ cannot be expressed as a product of two polynomials of lower degree.

	It is helpful to compare irreducible polynomials to prime numbers: prime numbers are irreducible integers. They exhibit many of the general properties of the concept ``irreducibility" that equally apply to irreducible polynomials, such as the essentially unique factorization into prime or irreducible factors. Every polynomial $p(x) \in F[x]$ can be factorized into polynomials that are irreducible over $F$. This factorization is unique. 
	
	
	
	\item Let $F$ be a field. If $f(x) \in F[x]$ and $deg(f(x)) = 2$ or 3, then $f(x)$ is reducible over $F$ if and only if $f(x)$ has a zero in $F$.

	
	
	\item Let $F$ be a field and let $p(x) \in F[x]$. Then $\langle p(x) \rangle$ is a maximal ideal in $F[x]$ if and only if $p(x)$ is irreducible over $F$.
	\label{irreducible}
	
	
	
	\item Let $F$ be a field and $p(x)$ and irreducible polynomial over $F$. Then $F[x]/\langle p(x) \rangle$ is a field. (directly from \ref{field} and \ref{irreducible}).
	
	Example:
	
	We construct a field with eight elements. $x^3 + x + 1$ has no zero in $\mathbb{Z}_2$, hence it is irreducible over $\mathbb{Z}_2[x]$. Thus, $\mathbb{Z}_2[x]/\langle x^3 + x + 1 \rangle = \{ax^2 + bx + c + \langle x^3 + x + 1\rangle | a, b, c \in \mathbb{Z}_2\}$ is a field with eight elements. $a, b, c \in \{0, 1\}$.
	
	Addition: $(x^2 + x + 1 + \langle x^3 + x + 1 \rangle) + (x^2 + 1 + \langle x^3 + x + 1 \rangle) = x + \langle x^3 + x + 1 \rangle$.
	
	Multiplication: $(x^2 + x + 1 + \langle x^3 + x + 1\rangle)(x^2 + 1 + \langle x^3 + x + 1\rangle) = x^4 + x^3 + x + 1 + \langle x^3 + x + 1\rangle = x^4 + \langle x^3 + x + 1 \rangle$. Then we long divide $x^4$ by $x^3 + x + 1$. We obtain $x^2 + x + \langle x^3 + x + 1\rangle$.

	
	
	\item Let $F$ be a field, then every nonconstant polynomial $f(x) \in F[x]$ can be factored in $F[x]$ into a unique product of irreducible polynomials.

	
	
	\item An integral domain $D$ is a unique factorization domain (UFD) if:
	\begin{itemize}
	\item every nonzero element of $D$ that is not a unit can be written as a product of irreducibles of $D$.
	
	\item the factorization into irreducibles is unique.
	\end{itemize}
	
	
	
	\item Every principal ideal domain is a unique factorization domain.
	\label{UFD}

	
	
	\item Let $F$ be a field. Then $F[x]$ is a unique factorization domain. (Directly from \ref{PID} and \ref{UFD}).
\end{enumerate}



\newpage

%************************************************************************************************************************************************************************%

\section*{Fields:}

\begin{enumerate}

	\item Vector Space: A set $V$ is said to be a 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}[(1).]
	\item $a(v + u) = av + au$
	
	\item $(a + b) v = av + bv$
	
	\item $a(bv) = (ab)v$
	
	\item $1v = v$
	\end{enumerate}

	The members of a vector space are called vectors. The members of the field are called scalars.
	
	Examples:
	\begin{itemize}
	\item The set $\mathbb{R}^n = \{(a_1, \ldots, a_n) | a_i \in \mathbb{R}\}$ is a vector space over $\mathbb{R}: (a_1, \ldots, a_n) + (b_1, \ldots, b_n) = (a_1+b_1, \ldots, a_n+b_n)$; $b(a_1, \ldots, a_n) = (ba_1, \ldots, ba_n)$.
	
	\item The set $M_2(\mathbb{Q})$ is a vector space over $\mathbb{Q}$.
	
	\item The set $\mathbb{Z}_p[x]$ of polynomials with coefficients from $\mathbb{Z}_p$ is a vector space over $\mathbb{Z}_p$, where $p$ is a prime.
	
	\item Let $E$ be a field and let $F$ be a subfield of $E$. Then $E$ is a vector space over $F$. The operations are the operations of $E$.
	\end{itemize}
	
	
	
	\item Let $V$ be a vector space over a field $F$ and let $U$ be a subset of $V$. We say that $U$ is a subspace of $V$ if $U$ is also a vector space over $F$ under the operations of $V$.
	
	Example:
	
	The set $\{a_2x^2 + a_1x + a_0 | a_0, a_1, a_2 \in R\}$ is a subspace of the vector space of all polynomials with real coefficients over $R$. 



	\item Let $V$ be a vector space over $F$ and let $v_1, \ldots, v_n$ be elements of $V$. Then the subset $\langle v_1, \ldots, v_n \rangle = \{a_1v_1 + \ldots + a_nv_n | a_1, \ldots, a_n \in F\}$ is called the subspace of $V$ spanned by $v_1, \ldots, v_n$.  Any summand of the form $a_1v_1 + \ldots + a_nv_n$ is called a linear combination of $v_1, \ldots, v_n$. If $\langle v_1, \ldots, v_n \rangle = V$, we say that $\{v_1, \ldots, v_n\}$ spans $V$.

	
	
	\item A set of vectors is said to be linearly dependent over the field $F$ if there are vectors $v_1, \ldots, v_n$ from $S$ and elements $a_1, \ldots, a_n$ from $F$, not all zero, such that $a_1v_1 + \ldots + a_nv_n = 0$. A set of vectors that not linearly dependent over $F$ is called linearly independent over $F$.
	
	

	\item Let $V$ be a vector space over $F$. A subset $B$ of $V$ is called a basis for $V$ if $B$ is linearly independent over $F$ and every element of $V$ is a linear combination of elements of $B$.
	
	
	
	\item All bases of the same vector space have the same size.

	
	
	\item A vector space that has a basis consisting of $n$ elements is said to have dimension $n$. For completeness, the trivial vector space $\{0\}$ is said to be spanned by the empty set and to have dimension 0. Every vector space has a basis. A vector space that has a finite basis is called finite dimensional; otherwise, it is called infinite dimensional.

	
	
	\item A field $E$ is an extension field of a field $F$ if $F \subseteq E$ and the operations of $F$ are those of $E$ restricted to $F$.



	\item \textbf{Fundamental Theorem of Field Theory}: 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.

	\textbf{Proof} (so far the only proof in this write-up):
	
	Since $F[x]$ is a unique factorization domain, $f(x)$ has an irreducible factor, say, $p(x)$. It is clearly sufficient to find an extension field $E$ of $F$ containing an element $z$ such that $p(z) = 0$.
By (Rings: \ref{irreducible}), $\langle p(x) \rangle$ is a maximal ideal in $F[x]$, so $E = F[x]/\langle p(x) \rangle$ is a field. We claim that $F$ can be identified as a subfield of $F[x]/\langle p(x)\rangle$ in a natural way by use of the map $\phi: F \rightarrow F[x]/\langle p(x) \rangle$ given by $\phi(a) = a + \langle p(x) \rangle$ for $a \in F$. This map is one to one, for if $\phi(a) = \phi(b)$, that is, if $a+\langle p(x) \rangle = b+\langle p(x) \rangle$ for some $a, b \in F$, then $(a-b) \in \langle p(x) \rangle$, so $a-b$ must be a multiple of the polynomial $p(x)$, which is of degree $ \geq 1$. Now $a, b \in F$ implies that $a-b$ is in $F$. Thus we must have $a-b=0$, so $a=b$. Hence $\phi: F \rightarrow E$ given by $\phi(a) = a + \langle p(x) \rangle$ is one-to-one and preserves both operations, $E$ has a subfield isomorphic to $F$. We may think of $E$ as containing $F$ if we simply identify the coset $a+\langle p(x) \rangle$ with its unique coset representative $a$ that belongs to $F$. Finally, to show that $p(x)$ has a zero in $E$, write \[p(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_0.\]
Then, in $E$, $x+\langle p(x)\rangle$ is a zero of $p(x)$, because
\begin{eqnarray*}
p(x + \langle p(x) \rangle) & = & a_n(x + \langle p(x) \rangle)^n + a_{n-1}(x + \langle p(x) \rangle)^{n-1} + \ldots + a_0 \\
& = & a_n(x^n + \langle p(x) \rangle) + a_{n-1}(x^{n-1} + \langle p(x) \rangle) + \ldots + a_0 \\
& = & a_nx^n + a_{n-1}x^{n-1} + \ldots + a_0 + \langle p(x)\rangle \\
& = & p(x) + \langle p(x) \rangle = 0 + \langle p(x) \rangle                                  \\
\end{eqnarray*}

	Example:
	\begin{itemize}
	\item Let $f(x) = x^2 + 1 \in \mathbb{Q}[x]$. Then in $E = \mathbb{Q}[x]/\langle x^2 + 1 \rangle$, we have
\[f(x + \langle x^2+1\rangle) = (x + \langle x^2+1\rangle)^2 + 1 = x^2 + \langle x^2 + 1\rangle  + 1 = x^2 + 1 + \langle x^2 + 1\rangle  = 0 + \langle x^2 + 1\rangle\]

	\item Let $f(x) = x^5 + 2x^2 + 2x + 2 \in \mathbb{Z}_3[x]$. Then, the irreducible factorization of $f(x)$ over $\mathbb{Z}_3$ is $(x^2+1)(x^3+2x+2)$. So, to find an extension $E$ of $\mathbb{Z}_3$ in which $f(x)$ has a zero, we may take $E = \mathbb{Z}_3[x]/\langle x^2 + 1\rangle$, a field with nine elements, (or $E=\mathbb{Z}_3[x]/\langle x^3+2x+2\rangle$ a field with 27 elements). we have $f(x + \langle x^2+1\rangle) = 0 + \langle x^2 + 1\rangle$. 
	\end{itemize}



	\item Splitting Field: Let $E$ be an extension field of $F$ and let $f(x) \in F[x]$. We say that $f(x)$ splits in $E$ if $f(x)$ can be factored as a product of linear factors in $E[x]$. We call $E$ a splitting field for $f(x)$ over $F$ if $f(x)$ splits in $E$ but in no proper subfield of $E$. So, a splitting field of $f(x)$ over $F$ is just a smallest extension field of $F$ in which $f(x)$ splits.
	
	There is a useful analogy between the definition of a splitting field and the definition of an irreducible polynomial. just as it makes no sense to say``$f(x)$ is irreducible", it makes no sense to say ``$E$ is a splitting field for $f(x)$". In each case, the underlying field must be specified; that is, one must say ``$f(x)$ is irreducible over $F$" and ``$E$ is a splitting field for $f(x)$ over $F$".

	
	
	\item 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 $F$.

	
	
	\item Types of Extensions: Let $E$ be an extension field of a field $F$ and let $a \in E$. We call $a$ 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 transcendental over $F$. An extension $E$ of $F$ is called an 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 transcendental extension of $F$. An extension of $F$ of the form $F(a)$ is called a simple extension of $F$.
	
	Leonhard Euler used the term transcendental for numbers that are not algebraic because ``they transcended the power of algebraic methods". $e$ and $\pi$ are transcendental numbers over $Q$.



	\item Characterization of Extensions:
	Let $E$ be an extension field of the field $F$ and let $a \in E$. If $a$ is transcendental over $F$, the $F(a)$ is isomorphic to $F(x)$. If $a$ is algebraic over $F$, then $F(a)$ is isomorphic to $F[x]/\langle p(x) \rangle$, where $p(x)$ is a polynomial in $F[x]$ of minimum degree such that $p(a) = 0$. Moreover, $p(x)$ is irreducible over $F$.
	
	Facts:
	\begin{itemize}
	\item If $a$ is algebraic over a field $F$, then there is a unique monic irreducible polynomial $p(x) \in F[x]$ such that $p(a) = 0$.
	
	\item 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{itemize}

	
	
	\item Degree of an Extension:
	Let $E$ be an extension field of a field $F$. We say that $E$ 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 finite extension of $F$; otherwise, we say that $E$ is an infinite extension of $F$.
	
	Example:
	
	The field of complex numbers has degree 2 over the reals, since $\{1, i\}$ is a basis. The field of complex numbers is an infinite extension of the rationals.



	\item If $E$ is a finite extension of $F$, then $E$ is an algebraic extension of $F$.

	
	
	\item If $K$ is an algebraic extension of $E$ and $E$ is an algebraic extension of $F$, then $K$ is an algebraic extension of $F$.
 
	
	
	\item For each prime $p$ and each positive integer $n$, there is, up to isomorphism, a unique field of order $p^n$. 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 Galois field of order $p^n$.

	
	
	\item Elements of $GF(p^n)$ are distinct roots of $x^{p^n} - x$ in $\mathbb{Z}_p[x]$.

	
	
	\item \textbf{Structure of Finite Field}:
	As a group under addition, $GF(p^n)$ is isomorphic to $\mathbb{Z}_p + \ldots + \mathbb{Z}_p$ ($n$ factors). Under multiplication, the set of nonzero elements of $GF(p^n)$ forms a cyclic group.

	Suppose that $F$ is a finite field and that $|F| = p^n$ where $p$ is a prime and $n$ is a positive integer. Then $F$ is isomorphic to $\mathbb{Z}_p[x]/\langle p(x) \rangle$, where $p(x)$ is an irreducible polynomial in $\mathbb{Z}_p[x]$, and $deg(f) = n$.

	
	
	\item $[GF(p^n):GF(p)] = n$

	
	
	\item A simple construction of finite field with order $p^n$: First find an irreducible polynomial $p(x)$ in $\mathbb{Z}_p[x]$ with degree $n$, and then $\mathbb{Z}_p/\langle p(x) \rangle$ gives you the finite field of order $p^n$. 
\end{enumerate}











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