# Reading Notes

MATH 402

## Table of contents

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 \(S\).
- \(a^2 + b^2 = aa + bb\), \(aa + bb \in S\), \(a^2 + b^2 \ge 0\).
- \(S\) must contain a smallest positive integer by the Well-Ordering Axiom.
- Let \(t = au + bv\) be the smallest positive element of \(S\)

- Prove that \(t = GCD(a, b)\)
- Prove \(t \vert a\) and \(t \vert b\)
- By the Division Algorithm, \(\exists q, r \in \mathbb{Z} : a = tq + r\), where \(0 \le r < t\).
- Therefore, \(r = a - tq\)
- Substituting: \(r = a - (au + bv)q\)
- We get \(a - aqu - bvq\) by expanding
- Rearranging into linear combination: \(r = a(1 - qu) + b(-vq)\)
- \(r \in S\) because it is a linear combination of \(a, b\)
- \(r < t\), the smallest positive element of \(S\)
- \(r \ge 0\), so the only possibility is that \(r = 0\)
- Therefore, \(t \vert a\). Repeat argument to show \(t \vert b\).
- \(t\) is a common divisor of \(a\)$ and \(b\).

- Prove \(c \vert a\) and \(c \vert b\) implies \(c \vert t\)
- Let \(c\) be any common divisor of \(a\) and \(b\).
- Then \(a = ck\) and \(b = cs\), where \(k, s \in \mathbb{Z}\)
- Therefore, \(t = au + bv = (ck)u + (cs)v = c(ku + sv)\)
- We know from this that \(c \vert t\)
- \(c \le \vert t \vert\): every divior of a nonzero integer \(a\) is less than or equal to \(\vert a \vert\)
- However, \(t\) is positive.
- Therefore, \(c \le t\).
- Therefore, \(t\) is the greatest common divisor.

- Prove \(t \vert a\) and \(t \vert b\)

**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 \(p \vert bc\).
- \((p, b)\) must be a positive divisor of the prime \(p\).
- If \((p, b) = \pm p\), \(p \vert b\)
- If \((p, b) = \pm 1\), then \(p \vert c\)

- Suppose \(p \vert bc\) implies \(p \vert b\) or \(p \vert c\).
- …

**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:

Proof:

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

- 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

### 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

- 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 \(ac + I = bd + I\).

**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.