Reading Notes

MATH 402

Table of contents

Chapter 1: Arithmetic in $\mathbb{Z}$ Revisited
1. 1.2: Divisibility
2. 1.3: Primes and Unique Factorization
Chapter 2: Congruence in $\mathbb{Z}$ and Modular Arithmetic
Chapter 3: Rings
Chapter 4: Arithmetic in $F[x]$
Chapter 5: Congruence in $F[x]$ and Congruence-Class Arithmetic
Chapter 6: Ideals and Quotient Rings

Textbook: Abstract Algebra, an Introduction. Thomas Hungerford. Internet Archive link

Chapter 1: Arithmetic in $\mathbb{Z}$ Revisited

Algebra grows out of arithmetic and depends heavily on it

1.2: Divisibility

Let $a$ and $b$ be integers with $b \neq 0$. $b \vert a$ (divides $a$) if $a = bc$ for some integer $c$.

An important case of division happens when the remainder is zero
$a$ and $-a$ have the same divisors
$a$ and $b$ are not both 0; then, their GCD exists and is unique. It is also larger than or equal to one.

Theorem 1.2. Let $a$ and $b$ be integers, not both zero, and let $d$ be their greatest common divisor. Then there exist (not necessarily unique) integers $u$ and $v$ such that $d = au + bv$.

Proof:

Let $S$ be the set of all linear integral combinations of $a, b$: $S = \{ am + bn \vert m, n \in \mathbb{Z} \}$
Find the smallest positive element of $S$.
1. $a^2 + b^2 = aa + bb$, $aa + bb \in S$, $a^2 + b^2 \ge 0$.
2. $S$ must contain a smallest positive integer by the Well-Ordering Axiom.
3. Let $t = au + bv$ be the smallest positive element of $S$
Prove that $t = GCD(a, b)$
1. Prove $t \vert a$ and $t \vert b$
  1. By the Division Algorithm, $\exists q, r \in \mathbb{Z} : a = tq + r$, where $0 \le r < t$.
  2. Therefore, $r = a - tq$
  3. Substituting: $r = a - (au + bv)q$
  4. We get $a - aqu - bvq$ by expanding
  5. Rearranging into linear combination: $r = a(1 - qu) + b(-vq)$
  6. $r \in S$ because it is a linear combination of $a, b$
  7. $r < t$, the smallest positive element of $S$
  8. $r \ge 0$, so the only possibility is that $r = 0$
  9. Therefore, $t \vert a$. Repeat argument to show $t \vert b$.
  10. $t$ is a common divisor of $a$$ and $b$.
2. Prove $c \vert a$ and $c \vert b$ implies $c \vert t$
  1. Let $c$ be any common divisor of $a$ and $b$.
  2. Then $a = ck$ and $b = cs$, where $k, s \in \mathbb{Z}$
  3. Therefore, $t = au + bv = (ck)u + (cs)v = c(ku + sv)$
  4. We know from this that $c \vert t$
  5. $c \le \vert t \vert$: every divior of a nonzero integer $a$ is less than or equal to $\vert a \vert$
  6. However, $t$ is positive.
  7. Therefore, $c \le t$.
  8. Therefore, $t$ is the greatest common divisor.

Corollary 1.3. Let $a$ and $b$ be integers, not both 0, and let $d$ be a positive integer. Then $d$ is the greatest common divisor of $a$ and $b$ if and only if $d$ satisfies these conditions:

$d \vert a$ and $d \vert b$
$(c \vert a \wedge c \vert b)$ implies $c \vert d$

Theorem 1.4. If $a \vert bc$ and $(a, b) = 1$, then $a \vert c$.

Proof:

Since $(a, b) = 1$, $au + bv = 1$ for some $u, v \in \mathbb{Z}$
Multiply by $c$: $acu + bcv = c$
Since $a \vert bc$, $bc = ar$ for some $r \in \mathbb{Z}$
Therefore, $c = acu + bcv = acu + (ar) v = a(cu + rv)$
This shows $a \vert c$

1.3: Primes and Unique Factorization

Every nonzero integer has four distinct divisors.
Integers with only four divisors are important:

An integer $p$ is prime if $p \neq 0, \pm 1$ and the only divisors of $p$ are $\pm 1$ and $\pm p$.

Becuase an integer $p$ has the same divisors as $-p$, $p$ is prime iff $-p$ is prime.
If $p$ and $q$ are prime and $p \vert q$, then $p = \pm q$

Theorem 1.5. Let $p$ be an integer with $p \neq 0, \pm 1$. Then $p$ is prime iff whenever $p \vert ab$, $p \vert b$ or $p \vert c$.

Proof.

Suppose $p$ is prime and $p \vert bc$.
1. $(p, b)$ must be a positive divisor of the prime $p$.
2. If $(p, b) = \pm p$, $p \vert b$
3. If $(p, b) = \pm 1$, then $p \vert c$
Suppose $p \vert bc$ implies $p \vert b$ or $p \vert c$.
1. …

Corollary 1.6. If $p$ is prime and $p \vert a_1 a_2 ... a_n$, then $p$ divides at least one of $a_i$.

Theorem 1.7. Every integer $n$ except $0, \pm 1$ is the product of primes.

Proof: Let $S$ be the set of integers greater than 1 that are not a product of primes. Show that $S$ is the empty set. Since there are no integers in $S$, every integer greater than 1 is a product of primes.

Suppose $S$ is not empty.
By the Well-Ordering Principle, $S$ has a smallest element, call it $n$.
$n$ is not prime, so $n = ab$ for some integers $a, b$, where $1 < a, b < n$.
Since $n$ is the smallest element of $S$, $a$ and $b$ are not in $S$.
Therefore, $a$ and $b$ are products of primes.
Therefore, $n$ is a product of primes.
This contradicts the assumption that $n$ is not a product of primes.
Therefore, $S$ is empty.

Theorem 1.8 (The Fundamental Theorem of Arithmetic). Every integer $n$ except $0, \pm 1$ is a product of primes. The prime factorization is unique:

\[(n = p_1 p_2 ... p_n) \wedge (n = q_1 q_2 ... q_n) \implies p_1 = \pm q_1, p_2 = \pm q_2, ..., p_n \pm q_n\]

Proof:

Eery integer $n$ except $0, \pm 1$ is a product of primes by Theorem 1.7.
Suppose $n$ has two prime factorizations. Then,

\[n = p_1 p_2 ... p_n = q_1 q_2 ... q_n\]

We know that $p_1 \vert q_1 q_2 ... q_n$.
$p$ must divide one of the $q$.
After reordering, $p_1 \vert q_1$.
Since $p_1$ and $q_1$ are prime, $p_1 = \pm q_1$.
Therefore, replacing on the LHS: $\pm q_1 p_2 p_3 ... p_n = q_1 q_2 q_3 ... q_n$
Dividing both sides by $q_1$ gives $\pm p_2 p_3 ... p_n = q_2 q_3 ... q_n$
Continuing to eliminate one prime at each step, we get totaly equality.$$
The lengths of the prime factorizations must be equal. If they were not, then we would get something like $\pm p_{r+1} p_{r+2} ... p_n = 1$
This is a contradiction because $p_i = \pm 1$ but is also prime.

Corollary 1.9. Every integer $n > 1$ can be written in only one way in the form $n = p_1 p_2 p_3 ... p_n$, where the $p_i$ are primes and $p_1 \le p_2 \le p_3 \le ... \le p_n$.

Theorem 1.10. Let $n > 1$. If $n$ has no positive prime factor less than or equal to $\sqrt{n}$, then $n$ is prime.

Chapter 2: Congruence in $\mathbb{Z}$ and Modular Arithmetic

2.1: Congruence and Congruence Classes

Congruence is a generalization of the equality relation.
Two integers are equal if their difference is zero or a multiple of zero.
For $n \in \mathbb{Z}^+$, two integers are congruent modulo $n$ if their difference is a multiple of $n$.
Definition: Let $a, b, n$ be integers with $n > 0$. Then $a$ is congruent to $b$ modulo $n$, provided that $n$ divides $a-b$.
Congruence has many of the same properties as standard equality.
- Reflexivity: $a = a$ for every integer $a$
- Symmetric: $a = b \iff b = a$
- Transitive: $a = b \wedge b = c \implies a = c$

Theorem 2.1. Let $n$ be a positive integer. For all $a, b, c \in \mathbb{Z}$.

Reflexivity: $a \equiv_n a$
Symmetric: $a \equiv_n b \implies b \equiv_n a$
Transitive: $a \equiv_n b \wedge b \equiv_n c \implies a \equiv_n c$

Theorem 2.2. If $a \equiv_n b$ and $c \equiv_n d$, then

Distributivity: $a + c \equiv_n b + d$
Multiplicativity: $ac \equiv_n bd$

Definition. Let $a$ and $n$ be integers with $n > 0$. The congruence class of a modulo $n$ is the set of all integers that are congruent to $a$ modulo $n$. $[ a\ ] = \{ b \vert b \in \mathbb{Z} \text{ and } b \equiv_n a \}$ To say that $b \equiv_n a$ means that $b - a = kn$ for some integer $k$. Therefore we can rewrite it as $[a] = \{ a + kn \vert k \in \mathbb{Z} \}$

Theorem 2.3. $a \equiv_n c$ iff $[a] = [c]$

Corollary 2.4. Two congruence classes modulo $n$ are either disjoint or identical.

Corollary 2.5. Let $n > 1$ be an integer and consider congruence modulo $n$.

If $a$ is any integer and $r$ is the remainder when $a$ is divided by $n$, then $[a] = [r]$.
There are exactly $n$ distinct congruence classes, namely, $[0], [1], ..., [n-1]$.

The set of all congruence classes modulo $n$ is $\mathbb{Z}_n$. These elements are classes, not single integers.

2.2: Modular Arithmetic

The finite set $\mathbb{Z}_n$ is closely related to the infinite set $\mathbb{Z}$.
Can we define addition and multiplication on $\mathbb{Z}_n$?
The sum of classes $[a]$ and $[c]$ is $[a] \oplus [c] = [a + c]$
The product of classes $[a]$ and $[c]$ is $[a] \odot [c] = [ac]$
These operations do not depend on the choice of representatives from the various classes. (e.g. $[a] = [a + kn]$)

Theorem 2.6. If $[a] = [b]$ and $[c] = [d]$ in $\mathbb{Z}_n$, then $[a + c] = [b + d]$ and $[ac] = [bd]$

Theorem 2.7. For any classes $[a]$, $[b]$, $[c]$ in $\mathbb{Z}_n$,

Closure for addition: $[a] \oplus [b] \in \mathbb{Z}_n$
Associativity for addition: $[a] \oplus ([b] \oplus [c]) = ([a] \oplus [b]) \oplus [c]$
Commutativity for addition: $[a] \oplus [b] = [b] \oplus [a]$
Identity for addition: $[a] \oplus [0] = [a]$
Root existence: for each $[a]$ in $\mathbb{Z}_n$, the equation $[a] \oplus [x] = [0]$ has a solution in $\mathbb{Z}_n$
Closure for multiplication: $[a] \odot [b] \in \mathbb{Z}_n$
Associativity for multiplication: $[a] \odot ([b] \odot [c]) = ([a] \odot [b]) \odot [c]$
Distribution with addition: $[a] \odot ([b] \oplus [c]) = ([a] \odot [b]) \oplus ([a] \odot [c])$
Commutativity for multiplication: $[a] \odot [b] = [b] \odot [a]$
Identity for multiplication: $[a] \odot [1] = [a]$

For $k \in \mathbb{Z}^+$, $[a]^k$ denotes the product

\[[a]^k = [a] \odot [a] \odot ... \odot [a] , k\text{ times}\]

2.3: The Structure of $\mathbb{Z}_p$ and $\mathbb{Z}_n$

New notation: uses the same symbol to represent totally different entities.
The class notation $[a]$ will be replaced with $a$.

The structure of $\mathbb{Z}_p$ when $p$ is prime

When $a \neq 0$, the equation $ax = 1$ has a solution in $\mathbb{Z}$ iff $a = \pm 1$.
But $ax = 1$ has solutions when $p$ is prime for $\mathbb{Z}_p$

Theorem 2.8. If $p > 1$ is an integer, the following conditions are equivalent:

$p$ is prime
For any $a \neq 0$ in $\mathbb{Z}_p$, the equation $ax = 1$ has a solution in $\mathbb{Z}_p$
Whenever $bc = 0$ in $\mathbb{Z}_p$, then $b = 0$ or $c = 0$

Proof:

Show that 1 implies 2. Suppose $p$ is prime and $a \neq 0$ in $\mathbb{Z}_p$. Then $a \not\equiv_p 0$. Therefore, $p \nmid a$. Now, $(a, p)$ is a positive divisor of $p$ and is either $p$ or $1$. Since $(a, p)$ also divides $a$ and $p \nmid a$, $(a, p) = 1$. From theorem 1.2., $au + pv = 1$ for some integers $u, v$. Hence, $au - 1 = p(-v)$, so $au \equiv_p 1$. Therefore $[au] = [1]$ in $\mathbb{Z}_p$. Therefore $x = [u]$ is a solution to $ax = 1$ in $\mathbb{Z}_p$.
Show that 2 implies 3. Suppose $ab = 0$ in $\mathbb{Z}_p$. If $a = 0$, there is nothing to prove. If $a \neq 0$, then by 2 there exists $u \in \mathbb{Z}_p$ such that $au = 1$. Then $0 = u \cdot 0 = u(ab) = (ua)b = (au)b = 1 \cdot b = b$.
Show that 3 implies 1. Suppose $b, c \in \mathbb{Z}$, $p \vert bc$. Then $bc \equiv_p 0$. By 2.3, $[b][c] = [bc] = [0]$ in $\mathbb{Z}_p$. By 3, $[b] = 0$ or $[c] = [0]$. This means $b \equiv_p 0$ or $c \equiv_p 0$, or $p \vert b$ or $p \vert c$. This means $p$ is prime by theorem 1.5.

The Structure of $\mathbb{Z}_n$

When $n$ is not prime, $ax = 1$ does not necessarily have a solution in $\mathbb{Z}_n$.

Theorem 2.9. Let $a$ and $n$ be integers with $n > 1$. Then the equation $[a]x = [1]$ has a solution in $\mathbb{Z}_n$ iff $(a, n) = 1$ in $\mathbb{Z}$.

Units and Zero Divisors

An element $a$ in $\mathbb{Z}_n$ is a unit if the equation $ax = 1$ has a solution

Theorem 2.10. Let $a, n$ be integers with $n > 1$. Then $[a]$ is a unit in $\mathbb{Z}_n$ iff $(a, n) = 1$ in $\mathbb{Z}$.

A nonzero element $a$ of $\mathbb{Z}_n$ is called a zero divisor if the equation $ax = 0$ has a noznero solution
From 2.8. part 3, when $p$ is prime, there are no zero divisors in $\mathbb{Z}_p$

Chapter 3: Rings

High-school algebra: arithmetic of polynomials
What are common features of arithmetic across different systems?
Abstraction allows us to understand the real reasons for why a particular statement is true.
Rings – systems which share a minimal number of fundamental properties of $\mathbb{Z}$ and $\mathbb{Z}_n$.

3.1: Definition and Examples

A ring is a nonempty set $R$ equipped with two operations, addition and multiplication, which satisfy the following axioms. For all $a, b, c \in R$:

Closure for addition: $a + b \in R$
Associative addition: $(a + b) + c = a + (b + c)$
Commutative addition: $a + b = b + a$
Additive identity or zero element: there is an element $O_R$ in $R$ such that $a + O_R = a$ for all $a \in R$
Additive identity or zero element: for each $a \in R$, the equation $a + x = O_R$ has a solution in $R$
Closure for multiplication: $ab \in R$
Associative multiplication: $(ab)c = a(bc)$
Distributive laws. $a(b + c) = ab + ac$ and $(b + c)a = ba + ca$

A commutative ring is a ring which additionally satisfies the axiom that for all $a, b \in R$, $ab = ba$.

A ring with identity is a ring which additionally satisfies the axiom that there is an element $1_R$ in $R$ such that $1_R a = a = a 1_R$ for all $a \in R$ (multiplicative identity).

Examples

$\mathbb{Z}$ is a commutative ring with identity.
$\mathbb{Z}_n$ is a commutative ring with identity.
$\mathbb{R}$ is a commutative ring with identity.
The set of odd integers with integer addition and multiplication is not a ring. The sum of two odd integers is not odd.
The set of even integers is a commutative ring.
$M(\mathbb{R})$ is the set of all $2 \times 2$ matrices over the real numbers. It is a ring with identity.
$T$ is the set of all functions from $\mathbb{R}\to\mathbb{R}$. $f + g$ and $fg$ are defined by $(f + g)(x) = f(x) + g(x)$ and $(fg)(x) = f(x)g(x)$. It is a commutative ring with identity.

An integral domain is a commutative ring $R$ with identity $1_R \neq 0_R$ that satisfies this axiom: whenever $a, b \in \mathbb{R}$ and $ab = 0_R$, then $a = 0_R$ or $b = 0_R$ (Note: $1_R \neq 0_R$ excludes the zero ring from integral domains.) Contrapositive: whenever $a, b \in \mathbb{R}$ and $a \neq 0_R$ and $b \neq 0_R$, then $ab \neq 0_R$

The ring $\mathbb{Z}$ of integers is an integral domain.
The ring $\mathbb{Z}_p$ is an integral domain.
The ring $\mathbb{Z}_n$ is not an integral domain when $n$ is not prime. For example, $[2] \odot [3] = [0]$ in $\mathbb{Z}_6$.
The ring $\mathbb{Q}$ is an integral domain.

A field is a commutative ring $R$ with identity $1_R \neq 0_R$ which satisfies this axiom: for each $a \neq 0_R$ in $R$, the equation $ax = 1_R$ has a solution in $R$.

$\mathbb{R}$ is a field
$\mathbb{Z}_p$ is a field
$\mathbb{C}$ is a field
The set of all matrices of the form $\begin{pmatrix} a & b \\ -b & a \end{pmatrix}$ is a field.

Theorem 3.1. Let $R$ and $S$ be rings. Define addition and multiplication on the Cartesian product $R \times S$ by $(r, s) + (r', s') = (r + r', s + s')$ $(r, s) \cdot (r', s') = (rr', ss')$ Then $R \times S$ is a ring. If $R$ and $S$ are both commutative, then $R \times S$ is commutative. If both $R$ and $S$ have an identity, then so does $R \times S$.

When a subset $S$ of a ring $R$ is itself a ring under the addition and multiplication in $R$, $S$ is a subring of $R$.

$\mathbb{Z}$ is a subring of the ring $\mathbb{Q}$
$\mathbb{Q}$ is a subring of the ring $\mathbb{R}$
$\mathbb{Q}$ is a subfield of $\mathbb{R}$
$\mathbb{R}$ is a subfield of $\mathbb{C}$
$M(\mathbb{Z})$ is a subring of $M(\mathbb{R})$
The set of all continuous functions $\mathbb{R} \to \mathbb{R}$ is a subring of the ring of all functions from $\mathbb{R} \to \mathbb{R}$

Often, proving $S$ is a subring is easier than proving that $S$ is a ring.

Theorem 3.2. Suppose $R$ is a ring and that $S$ is a subset of $R$ such that

$S$ is closed under addition
$S$ is closed under multiplication
Zero element: $0_R \in S$
If $a \in S$, then the solution to the equation $a + x = 0_R$ is in $S$ Then $S$ is a subring of $R$.

$\{0, 3\}$ is a subring of $\mathbb{Z}_6$.
The set of all matrices with form $\begin{pmatrix} a & 0 \\ b & c \end{pmatrix}$ is a subring of $M(\mathbb{R})$.
The set $\{ a + b \sqrt{2} \vert a, b \in \mathbb{Z} \}$ is a subring of $\mathbb{R}$.

3.2: Basic Properties of Rings

Arithmetic in Rings

We cannot assume subtraction exists in an arbitrary ring

Theorem 3.3. For any element $a$ in a ring $R$, the equation $a + x = 0_R$ has a unique solution.

You can define negatives and subtraction: $-a$ is the solution to $a + x = 0_R$
Since addition is commutative $-a$ is the unique element of $R$ such that $a + (-a) = 0_R = (-a) + a$
Subtraction in a ring $b-a$ is defined as $b+(-a)$.

Theorem 3.4. If $a + b = a + c$ in a ring $R$, then $b = c$.

Theorem 3.5. For any elements $a, b$ of a ring $R$,

Multiplication of the zero element: $a \cdot 0_R = 0_R = 0_R \cdot a$
Negative of a product: $(-a)b = -(ab) = a(-b)$
Double negation: $-(-a) = a$
Distribution of negative: $-(a + b) = (-a) + (-b)$
Distribution of negative: $-(a - b) = (-a) + b$
Double negation: $(-a)(-b) = ab$
(If $R$ has an identity) $(-1_R)a = -a$

Exponents: $a^n = aaa \cdot a$
We can verify that $a^ma^n = a^{m+n}$ and $(a^m)^n = a^{mn}$
$na = a + a + a + ... + a$, $-na = (-a) + (-a) + (-a) + ... + (-a)$
We can give a product of an integer $n$ and a ring element $a$

Theorem 3.6. Let $S$ be a nonempty subset of a ring $R$ such that

$S$ is closed under subtraction (if $a, b \in S$, then $a - b \in S$)
$S$ is closed under multiplication (if $a, b \in S$, then $ab \in S$) Then $S$ is a subring of $R$.

Units and Zero Divisors

An element $a$ in a ring $R$ is a unit if the equation $au = 1_R = ua$ has a solution in $R$.
The element $u$ is the multiplicative inverse of $a$ and denoted $a^{-1}$
The only units in $\mathbb{Z}$ are $\pm 1$
Every nonzero element of a field is a unit.
Invertible matrices are units in a matrix ring.
An element $a$ in a ring $R$ is a zero divisor provided that $a \neq 0_R$ and there exists a nonzero element $c$ in $R$ such that $ac = 0_R$ or $ca = 0_R$.
An integral domain contains no zero divisors.

Theorem 3.7. Cancellation is valid in any integral domain $R$. If $a \neq 0_R$ and $ab = ac$ in $R$, then $b = c$.

Theorem 3.8. Every field $F$ is an integral domain.

Theorem 3.9. Every finite integral domain $R$ is a field.

3.3: Isomorphisms and Homomorphisms

Consider the subset $S = \{0, 2, 4, 6, 8\}$ of $\mathbb{Z}_{10}$. $S$ is essentially the same as the field $\mathbb{Z}_5$, except for the labels on the elements.
That is, $S$ is isomorphic to $\mathbb{Z}_5$.
Isomorphic rings are rings with the same structure, in that addition and multiplication tables are equivalent when relabelled.
Relabing: every element of $R$ is paired with a unique element of $S$.
There is a function $f: R \to S$ which assigns each $r$ to a new label $f(r) \in S$.
Properties of $f$:
- Distinct elements of $R$ must get distinct new labels: if $r \neq r'$ in $R$, then $f(r) \neq f(r')$ in $S$. (Injective)
- Every element of $S$ must be the label of some element in $R$: for each $s \in S$, there is an $r \in R$ such that $f(r) = s$. (Surjective)
The function $f$ is a bijection.
If $a + b = c$ in $R$, then $f(a) + f(b) = f(c)$ in $S$. And $f(a + b) = f(c)$. Therefore, $f(a + b) + f(a) + f(b)$.

Definition. A ring $R$ is isomorphic to a ring $S$ is there is a function $f: R \to S$ such that

$f$ is injective,
$f$ is surjective,
$f(a + b) = f(a) + f(b)$ and $f(ab) = f(a) f(b)$ for all $a, b \in R$,

Example. The field $K$ of all matrices of the form $\begin{pmatrix} a & b \\ -b & a \end{pmatrix}$ is ismomorphic to the field $\mathbb{C}$. Let $f: K \to \mathbb{C}$ be defined as follows: $f(\begin{pmatrix} a & b \\ -b & a \end{pmatrix}) = a + bi$ We will first show that $f$ is injective. Suppose $f(\begin{pmatrix} a & b \\ -b & a \end{pmatrix}) = f(\begin{pmatrix} c & d \\ -d & c \end{pmatrix})$. Then $a + bi = c + di$. Therefore, $a = c$ and $b = d$. Therefore, $\begin{pmatrix} a & b \\ -b & a \end{pmatrix} = \begin{pmatrix} c & d \\ -d & c \end{pmatrix}$. Therefore, $f$ is injective. We will next show that $f$ is surjective. Let $a + bi$ be any element of $\mathbb{C}$. Then $f(\begin{pmatrix} a & b \\ -b & a \end{pmatrix}) = a + bi$. Therefore, $f$ is surjective. Finally, we will show that $f$ preserves addition and multiplication. Let $\begin{pmatrix} a & b \\ -b & a \end{pmatrix}, \begin{pmatrix} c & d \\ -d & c \end{pmatrix}$ be any elements of $K$. Then $f(\begin{pmatrix} a & b \\ -b & a \end{pmatrix} + \begin{pmatrix} c & d \\ -d & c \end{pmatrix}) = f(\begin{pmatrix} a + c & b + d \\ -(b + d) & a + c \end{pmatrix}) = a + c + (b + d)i = (a + bi) + (c + di) = f(\begin{pmatrix} a & b \\ -b & a \end{pmatrix}) + f(\begin{pmatrix} c & d \\ -d & c \end{pmatrix})$. Therefore, $f$ preserves addition. Similarly, $f$ preserves multiplication. Therefore, $f$ is an isomorphism.

You can even relabel the elements of a ring such that the ring is ismorphic to itself. For example, $f : \mathbb{C} \to \mathbb{C}$ where $f(a + bi) = a - bi$.

Isomorphisms are intuitively symmetric. But the definition of an isomorphism is not necessarily symmetric. Requires a function from $R$ onto $S$ but not vice versa. Yet, $f$ is a bijective function of sets, which means the inverse exists. Therefore ismorphism is symmetric.

Definition. Let $R, S$ be rings. A function $f : R \to S$ is a homomorphism if $f(a + b) = f(a) + f(b)$ and $f(ab) = f(a) f(b)$ for all $a, b \in R$.

Every isomorphism is a homomorphism.
However a homomorphism may fail to be injective or surjective.

Example. The zero map $z : R \to S$, $z(r) = 0_S$ is a homomorphism. The function $f: \mathbb{Z} \to \mathbb{Z}_6$, $f(a) = [a]$ is a homomorphism.

Theorem 3.10. Let $f : R \to S$ be a homomorphism of rings. Then

Identity preservation: $f(0_R) = 0_S$
Negation preservation: $f(-a) = -f(a)$
Subtrative cancellation: $f(a - b) = f(a) - f(b)$ If $R$ is a ring with identity and $f$ is surjective, then
$S$ is a ring with identity $f(1_R)$.
Whenever $u$ is a unit in $R$, then $f(u)$ is a unit in $S$ and $f(u)^{-1} = f(u^{-1})$.

Corollary 3.11. If $f : R \to S$ is a homomorphism of rings, then the image of $f$ is a subring of $S$.

How to show that there is no possible function from one ring to another? Proceed indirectly. Assume that there is a contradiction which exists, and show a contradiction. Alternatively, isomorphisms should preserve features such as zero elements, units, or identities. For instance, we know that $\mathbb{Q}, \mathbb{R}$ are not isomorphic to $\mathbb{Z}$ because every nonzero element in $\mathbb{Q}, \mathbb{R}$ is a unit, but not in $\mathbb{Z}$.

No commutative ring can be isomorphic to a noncommutative ring.

Chapter 4: Arithmetic in $F[x]$

4.1: Polynomial Arithmetic and the Division Algorithm

A polynomial with coefficients in $R$ is an expression of the form $a_0 + a_1 x + a_2 x^2 + ... + a_n x^n$, where $n$ is a nonnegative integer and the $a_i$ are elements of $R$.
What is $x$? What does it mean to multiply $x$ by a ring element? Is $x \in R$?
Polynomials are coefficients in a ring $R$, which are elements of a larger ring containing both $R$ and the special element $x$ if not in $R$

Theorem 4.1. If $R$ is a ring, then there exists a ring $T$ containing an element $x$ that is not in $R$ and has these properties:

$R$ is a subring of $T$
$xa = ax$ for every $a \in R$
The set $R[x]$ of all elements of $T$ of the form $a_0 + a_1 x + a_2 x^2 + ... + a_n x^n$ for $n \ge 0$ and $a_i \in R$ is a subring of $T$ that contains $R$.
The representation of elements is unique.
$a_0 + a_1 x + a_2 x^2 + ... + a_n x^n = 0_R$ iff $\forall i : a_i = 0_R$.

Elements of $R[x]$ are polynomials with coefficients in $R$ and the elements $a_i$ are coefficients
$x$ is an indeterminate
The ring $T$ is not necessarily commutative, but rather that $x$ is commutative with every element in $R$
Examples
- $\mathbb{Z}[x], \mathbb{Q}[x], \mathbb{R}[x]$ are all familiar rings
- $4 - 6x + 4x^3 \in E[x]$, $E$ ring of even integers. The polynomial $x$ is not in $E[x]$ becuase it cannot be written with even coefficients.

Polynomial Arithmetic

Rules for adding and multiplying polynomials follow from the fact that $R[x]$ is ar ing.
Polynomial addition and multiplication

\[\sum_{i=0}^n a_i x^i + \sum_{i=0}^m b_i x^i = \sum_{i=0}^k (a_i + b_i) x^i\] \[\sum_{i=0}^n a_i x^i \cdot \sum_{i=0}^m b_i x^i = \sum_{i=0}^{n+m} \left( \sum_{j=0}^i a_{j} b_{i-j} \right)x^i\]

If $R$ is commutative, then so is $R[x]$
If $1_R \in R$, then $1_R$ is the multiplicative identity of $R[x]$.
$a_n$ is the leading coefficient of $f(x)$
Degree is the largest exponent of $x$ which appears as a nonzero coeficient
The constant polynomial does not have a degree

Theorem 4.2. If $R$ is an integral domain and $f(x), g(x)$ are nonzero polynomials in $R[x]$, then $\deg(f(x)g(x)) = \deg(f(x)) + \deg(g(x))$.

Corollary 4.3. If $R$ is an integral domain, then so is $R[x]$.

Corollary 4.4. Let $R$ be a ring. If $f(x), g(x), f(x)g(x)$ are nonzero in $R[x]$, then $\deg[ f(x) g(x) ] \le \deg f(x) + \deg g(x)$.

Corollary 4.5. Let $R$ be an integral domain and $f(x) \in R[x]$. Then $f(x)$ is a unit in $R[x]$ iff $f(x)$ is a constant polynomial that is a unit in $R$. In particular, if $F$ is a field, the units in $F[x]$ are the nonzero constants in $F$.

The Division Algortihm in $F[x]$

What about polynomials with coefficients in a field $F$?

Theorem 4.6: The Division Algorithm in $F[x]$. Let $F$ be a field and $f(x), g(x) \in F[x]$ with $g(x) \neq 0_F$. Then there exist unique polynomials $q(x)$ and $r(x)$ such that $f(x) = g(x) q(x) + r(x)$and either$r(x) = 0_F$or$\deg r(x) < \deg g(x)$$.

4.2: Divisibility in $F[x]$

Let $F$ be a field and $a(x), b(x) \in F[x]$ with $b(x)$ nonzero.
$b(x)$ divides $a(x)$ if $a(x) = b(x) h(x)$ for some $h(x) \in F[x]$.

Theorem 4.7. Let $F$ be a field and $a(x), b(x) \in F[x]$ with $b(x)$ nonzero.

If $b(x) \vert a(x)$, then $cb(x) \vert a(x)$ for each nonzero $c \in F$
Every divisor of $a(x)$ has degree less than or equal to $\deg a(x)$.

A polynomial is monic if its leading coefficient is $1_F$

Definition. The GCD of $a(x)$ and $b(x)$ is the monic polynomial of highest degree that divides both $a(x)$ and $b(x)$. $d(x)$ is the gcd of $a(x)$ and $b(x)$, provided that $d(x)$ is monic and

$d(x) \vert a(x)$ and $d(x) \vert b(x)$
If $c(x) \vert a(x)$ and $c(x) \vert b(x)$, then $\deg c(x) \le \deg d(x)$.

Theorem 4.8. Let $F$ be a field and $a(x), b(x) \in F[x]$, both nonzero. Then there is a unique GCD $d(x)$ of $a(x)$ and $b(x)$. There are polynomials $u(x), v(x) : d(x) = a(x) u(x) + b(x) v(x)$.

Theorem 4.10. Let $F$ be a field and $a(x), b(x), c(x) \in F[x]$. If $a(x) \vert b(x) c(x)$ and $a(x)$ and $b(x)$ are relatively prime, then $a(x) \vert c(x)$.

4.3: Irreducibles and Unique Factorization

Here, $F$ always denotes a field
In $\mathbb{Z}$ there are only two units, $\pm 1$ - polynomial rings can have more units
Elements $a$ in commutative rings with identity $R$ are associates of an element $b \in R$ if $a = bu$ for some unit $u$
- In $\mathbb{Z}$, $a$ and $b$ are associates iff $a = \pm b$

$f(x)$ is an associate of $g(x)$ in $F[x]$ iff $f(x) = cg(x)$ for some nonzero $c \in F$

A nonzero integer $p$ is prime in $\mathbb{Z}44 if it is not$ \pm 1$and its only divisors are$\pm 1$and$\pm p$, the associates of$p$$

Definition. Let $F$ be a field. A nonconstant polynomial $p(x) \in F[x]$ is irreducible if its only divisor are its associates and units (nonzero constant polynomials). A nonconstant polynomial that is not irreducible is reducible.

Every polynomial of degree 1 in $F[x]$ is irreducible in $F[x]$.

Theorem 4.11. Let $F$ be a field. A nonzero polynomial $f(x)$ is reducible in $F[x]$ iff $f(x)$ can be written as the product of two polynomials of lower degree.

Irreducibles in $F[x]$ have the same divisibility properties as primes in $\mathbb{Z}$

Theorem 4.12. Let $F$ be a field and $p(x)$ be a nonconstant polynomial in $F[x]$. Then the following conditions are equivalent.

$p(x)$ is irreducible
If $b(x)$ and $c(x)$ are polynomials such that $p(x) \vert b(x) c(x)$, then $p(x) \vert b(x)$ or $p(x) \vert c(x)$
If $r(x)$ and $s(x)$ are any polynomials such that $p(x) = r(x) s(x)$, then $r(x)$ or $s(x)$ is a nonzero constant polynomial.

Corollary 4.13. Let $F$ be a field and $p(x)$ an irreducible polynomial in $F[x]$. If $p(x) \vert a_1(x) a_2(x) ... a_n(x)$, then $p(x)$ divides at least one of the $a_i(x)$.

Theorem 4.14. Let $F$ be a field. Every nonconstant polynomial $f(x)$ in $F[x]$ is a product of irreducible polynomials in $F[x]$. This factorization is unique, such that when ordered and relabelled, every element is an associate in the parallel factorization.

4.4: Polynomial Functions, Roots, and Reducibility

What are criteria for the irreducibility of polynomials (e.g. primality testing)
Every polynomial induces a function from $F$ to $F$
$R$ is a commutative ring
Polynomial function $f : R \to R$, for each $r \in R$, $f(r) = a_n r^n + ... + a_2 r^2 + a_1r + a_0$
$x$ can be treated as indeterminate or determinate, and yielding different results

Definition. Let $R$ be a commutative ring and $f(x) \in R[x]$. An element $a$ of $R$ is said to be a root (or zero) of the polynomial $f(x)$ if $f(a) = 0_R$, that is, if the induced function $f : R \to R$ maps $a$ to $0_R$.

Theorem 4.15. The Remainder Theorem. Let $F$ be a field, $f(x) \in F[x]$, and $a \in F$. The remainder when $f(x)$ is divided by the polynomial $x - a$ is $f(a)$.

Theorem 4.16. The Factor Theorem. Let $F$ be a field, $f(x) \in F[x]$, and $a \in F$. Then $a$ is a root of the polynomial $f(x)$ iff $x - a$ is a factor of $f(x)$ in $F[x]$.

Corollary 4.17. Let $F$ be a field and $f(x)$ a nonzero polynomial of degree $n$ in $F[x]$. Then $f(x)$ has at most $N$ roots in $F$.

Corollary 4.18. Let $F$ be a field and $f(x) \in F[x]$, with $\deg f(x) \ge 2$. If $f(x)$ is irreducible in $F[x]$, then $f(x)$ has no roots in $F$.

Corollary 4.19. Let $F$ be a field and let $f(x) \in F[x]$ be a polynomial of degree $2$ or $3$. Then $f(x)$ is irreducible in $F[x]$ iff $f(x)$ has no roots in $F$.

Corollary 4.20. Let $F$ be an infinite field and $f(x), g(x) \in F[x]$. Then $f(x)$ and $g(x)$ induce the same function from $F$ to $F$ iff $f(x) = g(x)$ in $F[x]$.

4.5: Irreducibility in $\mathbb{Q}[x]$

Factoring in $\mathbb{Q}[x]$ can be reduced to factoring in $\mathbb{Z}[x]$.
If $f(x) \in \mathbb{Q}[x]$, then $cf(x)$ has integer coefficients for some nonzero integer $c$.
Factor theorem: finding first-degree factors of a polynomial in $\mathbb{Q}[x]$ is equivlaent to finding the roots of $g(x) \in \mathbb{Q}$.

Theorem 4.21. The Rational Root Test. Let $f(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0$ be a polynomial with integer coefficients. If $r \neq 0$ and the rational number $r/s$ in lowest terms is a root of $f(x)$, then $r \vert a_0$ and $s \vert a_n$.

If $f(x) \in \mathbb{Q}[x]$, then $cf(x)$ has integer coefficients for some nonzero integer $c$.

Lemma 4.22. Let $f(x), g(x), h(x) \in \mathbb{Z}[x]$ with $f(x) = g(x) h(x)$. If $p$ is a prime that divides every coefficient of $h(x)$, then either $p$ divides every coefficient of $g(x)$ or $p$ divides every coefficient of $h(x)$.

Theorem 4.23. Let $f(x)$ be a polynomial with integer coefficients. Then $f(x)$ factors as a product of polynomials of degrees $m$ and $n$ in $\mathbb{Q}[x]$ iff $f(x)$ factors as a product of polynomials of degrees $m$ and $n$ in $\mathbb{Z}[x]$.

Theorem 4.24. Eisenstein’s Criterion. Let $f(x) = a_n x^n + ... + a_1 x + a_0$ be a nonconstant polynomial with integer coefficients. If there is a prime $p$ such that $p$ divides each of $a-0, a_1, ..., a_{n-1}$ but $p$ does not divide $a_n$ and $p^2$ does not divide $a_0$, then $f(x)$ is irreducible in $\mathbb{Q}[x]$.

The polynomial $x%n + 5$ is reducible in $\mathbb{Q}[x]$ for each $n \ge 1$. Therefore, there are irreducible polynomials of every degree in $\mathbb{Q}[x]$.

If $f(x) = a_k x^k + ... + a_1 x + a_0$, $\bar{f}(x) = [a_k] x^k + ... + [a_1] x + [a_0] \in \mathbb{Z}_p[x]$. These two polynomials will have the same degree when the leading coefficient of $f(x)$ is not divisible by $p$.

Theorem 4.25. Let $f(x) = a_k x^k + ... + a_1 x + a_0$ be a polynomial with integer coefficients, and let $p$ be a positiv eprime that does not divide $a_k$. If $\bar{f}(x)$ is irreducible in $\mathbb{Z}_p[x]$, then $f(x)$ is irreducible in $\mathbb{Q}[x]$.

For every non-negative integer $k$, there are only finitely many polynomials of degree $k$ in $\mathbb{Z}_p[x]$, so in fact you can determine whether any given polynomial in $\mathbb{Z}_p$ is irreducible by checking a finite number of factors.

Chapter 5: Congruence in $F[x]$ and Congruence-Class Arithmetic

Concepts of congruence and congruence-class arithmetic carry over from $\mathbb{Z}$ to $F[x]$ with basically no changes.
New congruence-class rings have a much richer structure than rings $\mathbb{Z}_n$
Given any polynomial over any field, we can find a root of that polynomial in some larger field.

5.1: Congruence in $F[x]$ and Congruence Classes

Basic facts about divisibility in $\mathbb{Z}$ support the concept of congruence in the integers.
If $F$ is a field, the polynomial ring $F[x]$ has the same divisibility properties as $\mathbb{Z}$.

Definition. Let $F$ be a field and $f(x), g(x), p(x) \in F[x]$ with $p(x)$ nonzero. Then $f(x)$ is congruent to $g(x)$ mod $p(x)$, $f(x) \equiv_{p(x)} g(x)$, given $p(x) \vert f(x) - g(x)$.

Theorem 5.1. Let $F$ be a field and $p(x)$ a nonzero polynomial in $F[x]$. Then the relation of congruence modulo $p(x)$ is

reflexive: $f(x) \equiv_{p(x)} f(x), \forall f(x) \in F[x]$
symmetric: if $f(x) \equiv_{p(x)} g(x)$, then $g(x) \equiv_{p(x)} f(x)$
transitive: if $f(x) \equiv_{p(x)} g(x)$ and $g(x) \equiv_{p(x)} h(x)$, then $f(x) \equiv_{p(x)} h(x)$

Theorem 5.2. Let $F$ be a field and $p(x)$ a nonzero polynomial in $F[x]$. If $f(x) \equiv_{p(x)} g(x)$ and $h(x) \equiv_{p(x)} k(x)$, then

\[f(x) + h(x) \equiv_{p(x)} g(x) + k(x)\]
\[f(x) h(x) \equiv_{p(x)} g(x) k(x)\]

Definition. The congruence class / residue class of $f(x)$ modulo $p(x)$ is $[f(x)]$ and consists of all polynomials in $F[x]$ which are congruent to $f(x)$ modulo $p(x)$.

$$[f(x)] = { g(x) \vert g(x) \in F[x] \wedge g(x) \equiv_{p(x)} f(x) }

Some algebra shows that

\[[f(x)] = \{ f(x) + k(x) p(x) \vert k(x) \in F[x] \}\]

Theorem 5.3. $f(x) \equiv_{p(x)} g(x)$ iff $[f(x)] = [g(x)]$.

Corollary 5.4. Two congruence classes modulo $p(x)$ are either disjoint or identical.

Corollary 5.5. Let $F$ be a field and $p(x)$ a polynomial of degree $n$ in $F[x]$, and consider congruence modulo $p(x)$.

If $f(x) \in F[x]$ and $r(x)$ is the remainder when $f(x)$ is divided by $p(x)$, then $[f(x)] = [r(x)]$$.
Let $S$ be the set consisting of the zero polynomial and all the polynomials of degree less than $n$ in $F[x]$. Then every congruence class modulo $p(x)$ is the class of some polynomial in $S$, and the congruence calsses of different polynomials in $S$ are distinct.

5.2: Congruence-Class Arithmetic

Congruence in the integers lead to rings $\mathbb{Z}_n$
Congruence in $F[x]$ can also produce new rings and fields

Theorem 5.6. Let $F$ be a field and $p(x)$ a nonconstant polynomial in $F[x]$. If $[f(x)] = [g(x)]$ and $[h(x)] = [k(x)]$ are both in $F[x] / p(x)$, then $[f(x) + h(x)] = [g(x) + k(x)]$ and $$[f(x) h(x)] = [g(x) k(x)]

Theorem 5.7. Let $F$ be a field and $p(x)$ a nonconstant polynomial in $F[x]$. Then the set $F[x] / p(x)$ of congruence classes modulo $p(x)$ is a commutative ring with identity. Furthermore, $F[x] / p(x)$ contains a subring $F^*$ that is isomorphic to $F$.

Proof:

Use previous proof of commutative ring with identity for congruence classes.
Let $F^*$ be the subset of $F[x] / p(x)$ consisting of the congruence classes of all constant polynomials: $F^* = \{ [a] \vert a \in F \}$.
Define a map $\phi : F \to F^*$ by $\phi(a) = [a]$. $\phi$ is surjective. It is easy to prove that it is a homomorphism.
Suppose $\phi(a) = \phi(b)$. Then $[a] = [b]$, so $a \equiv_{p(x)} b$, and $a - b \vert p(x)$. Because $\deg p(x) \ge 1$ and $a - b \in F$, this is only possible if $a = b$. Therefore, $\phi$ is injective. So $\phi$ is an isomorphism.

We have constructed a ring $F[x] / p(x)$ that contains an isomorphic copy of $F$. How can we get a ring which contains the field $F$ itself?
In general, given a field $F$ and a polynomial $p(x)$ in $F[x]$, we can construct a ring containing $F$ as a subset. Identify $F$ with its isomorphic copy $F^*$ inside $F[x] / p(x)$ and consider $F$ to be a subset of $F[x] / p(x)$.

Theorem 5.8. Let $F$ be a field and $p(x)$ a nonconstant polynomial in $F[x]$. Then $F[x] / p(x)$ is a commutative ring with identity that contains $F$.

Theorem 5.9. Let $F$ be a field and $p(x)$ a nonconstant polynomial in $F[x]$. If $f(x) \in F[x]$ and $f(x)$ is relatively prime to $p(x)$, then $[f(x)]$ is a unit in $F[x] / p(x)$.

5.3: The Structure of $F[x]/(p(x))$ when $p(x)$ is Irreducible

When $p$ is a prime integer, $\mathbb{Z}_p$ is a field (+ integral domain)

Theorem 5.10. Let $F$ be a field and $p(x)$ a nonconstant polynomial in $F[x]$. Then the following statements are equivalent:

$p(x)$ is irreducible in $F[x]$
$F[x] / p(x)$ is a field
$F[x] / p(x)$ is an integral domain

You can construct finite fields with Theorem 5.10
If $p$ is prime and $f(x)$ is irreducible in $\mathbb{Z}_p[x]$ of degree $k$, then $\mathbb{Z}_p[x] / f(x)$ is a field; this field has $p^k$ elements.
Let $F$ be a field and $p(x)$ an irreducible polynomial in $F[x]$. Let $K$ be the field of congruence classes $F[x] / p(x)$. $F$ is a subfield of the field $K$; $K$ is an extension field of $F$.
What can be said about roots of polynomials in $K$?

Theorem 5.11. Let $F$ be a field and $p(x)$ an irreducible polynomial in $F[x]$. Then $F[x] / p(x)$ is an extension field of $F$ that contains a root of $p(x)$.

Corollary 5.12. Let $F$ be a field and $f(x)$ a nonconstant polynomial in $F[x]$. Then there is an extension field $K$ of $F$ that contains a root of $f(x)$.

The implications of 5.11 run very deep
The passage from known number systems to new larger systems have been greeted with mistrust
Negative numbers, complex numbers, etc.
Abstract algebra provides a framework to view the situation.
Turn from “is there a number whose square is -1” to “is there a field containing $\mathbb{R}$ in which the polynomial $x^2 + 1$ has a root? Since $x^2 + 1$ is irreducible in $\mathbb{R}[x]$, the answer is yes: $K = \mathbb{R}[x]/(x^2 + 1)$ is an extension field of $\mathbb{R}$ that contains a root of $x^2 + 1$, $\alpha = [x]$
You can write every element of $K$ as $[ax + b]$
$\alpha = i$.
The field $K$ is isomorphic to the field $\mathbb{C}$, with the isomorphism $f$ being given by $f(a + b \alpha) = a + bi$.

Chapter 6: Ideals and Quotient Rings

Congruence in the integers led us to finite arithmetics $\mathbb{Z}_n$
Polynomial ring $F[x]$ allows us to understand $F[x] / p(x)$, and to construct extension fields of $F$ which contain the roots of the polynomial $p(x)$
But we can extend the concept of congruence to arbitrary rings

6.1: Ideals and Congruence

Develop notion of congruence in arbitrary rings
Whenever $k \in \mathbb{Z}$ and $i \in I$, then $ki \in I$
Congruence in $R$ can be defined in terms of certain subrings
$I$ “absorbs” products: multiplying an element of $I$ by any element of the ring results in a product which is an element of $I$

Definition. A subring $I$ of a ring $R$ is an ideal when whenever $r \in R$ and $a \in I$, then $ra \in I$ and $ar \in I$.

Every ring has a zero ideal
The entire ring $R$ is also an ideal
Left ideals

Theorem 6.1. A nonempty subset $I$ of a ring $R$ is an ideal iff it has these properties:

If $a, b \in I$, then $a - b \in I$
If $r \in R$ and $a \in I$, then $ra \in I$ and $ar \in I$

Finitely Generated Ideals

Theorem 6.2. Let $R$ be a commutative ring with identity, $c \in R$, and $I$ the set of all multiples of $c$ in $R$, that is, $I = \{ rc \vert r \in R \}$. Then $I$ is an ideal.

The ideal $I$ in 6.2. is the principal ideal generated by $c$ and is denoted $(c)$.
Every ideal in $\mathbb{Z}$ is a principal ideal
There are ideals in other rings which are not principal.

Theorem 6.3. Let $R$ be a commutative ring with identity and $c_1, c_2, ..., c_n \in R$. Then the set $I = \{ r_1 c_1 + r_2 c_2 + ... + r_n c_n \vert r_1, r_2, ..., r_n \in R \}$ is an ideal in $R$.

Ideals generated in 6.3. are finitely generated

Congruence

Definition. Let $I$ be an ideal in a ring $R$ and let $a, b \in R$. Then $a \equiv_I b$ iff $a - b \in I$.

Theorem 6.4. Let $I$ be an ideal in a ring $R$. Then the relation of congruence modulo $I$ is

reflexive: $a \equiv_I a$, $\forall a \in R$
symmetric: if $a \equiv_I b$, then $b \equiv_I a$
transitive: if $a \equiv_I b$ and $b \equiv_I c$, then $a \equiv_I c$

Theorem 6.5. Let $I$ be an ideal in a ring $R$. If $a \equiv_I b$ and $c \equiv_I d$, then

\[a + c \equiv_I b + d\]
\[ac \equiv_I bd\]

If $I$ is an ideal in a ring $R$ and $a \in R$, the congruence class of $a \mod I$ is the set of all elements in $R$ congruent to $a$ modulo $I$
Denote the congurence class of $a$ modulo $I$ by $a + I$

Theorem 6.6. Let $I$ be an ideal in a ring $R$ and let $a, c \in R$. Then $a \equiv_I c$ iff $a + I = c + I$.

Corollary 6.7. Let $I$ be an ideal in a ring $R$. Then two cosets of $I$ are either disjoint or identical.

If $I$ is an ideal in a ring $R$, then the set of all cosets of $I$ (congruence classes modulo $I$) is denoted $R / I$.

6.2: Quotient Rings and Homomorphisms

The set of congruence classes modulo an ideal is itself a ring
Let $I$ be an ideal in $R$.
Then the elements of $R/I$ are cosets of $I$ (congruence classes modulo $I$), i.e. $a + I = \{ a + i \vert i \in I\}$.

Theorem 6.8. Let $I$ be an ideal in a ring $R$. If $a + I = b + I$ and $c + I = d + I$ in $R / I$, then $(a + c) + I = (b + d) + I$ and $ac + I = bd + I$.

\[(a + I) + (c + I) = (a + c) + I\]

Theorem 6.9. Let $I$ be an ideal in a ring $R$. Then

$R/I$ is a ring, with addition and multiplication of cosets as defined previously
If $R$ is commutative, $R / I$ is a commutative ring
If $R$ has an identity, then so does the ring $R / I$

$R / I$ is the quotient ring or factor ring of $R$ by $I$.

Homomorphisms

Quotients generalize congruence-class arithmetic in $\mathbb{Z}$ and $F[x]$.

Definition. Let $f : R \to S$ be a homomporhism of rings. Then the kernel of $f$ is the set $K = \{ r \in R \vert f(r) = 0_S \}$.

Note that $0_R$ is in the kernel since $f(0_R) = 0_S$

Theorem 6.10. Let $f : R \to S$ be a homomorphism of rings. Then the kernel $K$ of $f$ is an ideal in the ring $R$. Every kernel is an ideal.

Theorem 6.11. Let $f : R \to S$ be a homomorphism of rings with kernel $K$. Then $K = (0_R)$ iff $f$ is injective.

Theorem 6.12. Let $I$ be an ideal in a ring $R$. Then the map $\pi: R \to R / I$ given by $\pi(r) = r + I$ is a surjective homomorphism with kernel $I$. The map $\pi$ is the natural homomorphism from $R$ to $R / I$. Every ideal is the kernel of a homomorphism.

If $f : R \to S$ is a surjective homomorphism of rings, $S$ is a homomorphic image of $R$.

Theorem 6.13. (The First Isomorphism Theorem.) Let $f: R \to S$ be a surjective homomorphism of rings with kernel $K$. Then the quotient ring $R / K$ is isomorphic to $S$.

6.3: The Structure of $R/I$ When $I$ is Prime or Maximal

Quotient rings are natural generalizations of $\mathbb{Z}_p$, $F[x] / (p(x))$
When $p$ is prime and $p(x)$ is irreducible, then we get fields
An ideal $P$ in a commutative ring $R$ is prime if $P \neq R$ and whenever $bc \in P$, then $b \in P$ or $c \in P$.
If $F$ is a field and $p(x)$ is irreducible in $F[x]$, then the principal ideal $(p(x))$ is prime in $F[x]$.
$R/P$ may not always be a field when $P$ is prime.

Theorem 6.14. Let $P$ be an ideal in a commutative ring $R$ with identity. Then $P$ is a prime ideal iff the quotient ring $R/P$ is an integral domain.

Definition. An ideal $M$ in a ring $r$ is maximal if $M \neq R$ and whenever $J$ is an ideal such that $M \subseteq J \subseteq R$, then $M = J$ or $J = R$.

Theorem 6.15. Let $M$ be an ideal in a commutative ring $R$ with identity. Then $M$ is a maximal ideal iff the quotient ring $R / M$ is a field.

Corollary 6.16. In a commutative ring $R$ with identity, every maximal ideal is prime.

Reading Notes

Chapter 1: Arithmetic in \(\mathbb{Z}\) Revisited

1.2: Divisibility

1.3: Primes and Unique Factorization

Chapter 2: Congruence in \(\mathbb{Z}\) and Modular Arithmetic

2.1: Congruence and Congruence Classes

2.2: Modular Arithmetic

2.3: The Structure of \(\mathbb{Z}_p\) and \(\mathbb{Z}_n\)

Chapter 3: Rings

3.1: Definition and Examples

3.2: Basic Properties of Rings

3.3: Isomorphisms and Homomorphisms

Chapter 4: Arithmetic in \(F[x]\)

4.1: Polynomial Arithmetic and the Division Algorithm

4.2: Divisibility in \(F[x]\)

4.3: Irreducibles and Unique Factorization

4.4: Polynomial Functions, Roots, and Reducibility

4.5: Irreducibility in \(\mathbb{Q}[x]\)

Chapter 5: Congruence in \(F[x]\) and Congruence-Class Arithmetic

5.1: Congruence in \(F[x]\) and Congruence Classes

5.2: Congruence-Class Arithmetic

5.3: The Structure of \(F[x]/(p(x))\) when \(p(x)\) is Irreducible

Chapter 6: Ideals and Quotient Rings

6.1: Ideals and Congruence

6.2: Quotient Rings and Homomorphisms

6.3: The Structure of \(R/I\) When \(I\) is Prime or Maximal