MATH 402

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:

1. Let $$S$$ be the set of all linear integral combinations of $$a, b$$: $$S = \{ am + bn \vert m, n \in \mathbb{Z} \}$$
2. 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 $$S$$
3. 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: 1. Since $$(a, b) = 1$$, $$au + bv = 1$$ for some $$u, v \in \mathbb{Z}$$ 2. Multiply by $$c$$: $$acu + bcv = c$$ 3. Since $$a \vert bc$$, $$bc = ar$$ for some $$r \in \mathbb{Z}$$ 4. Therefore, $$c = acu + bcv = acu + (ar) v = a(cu + rv)$$ 5. 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. 1. 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 $$p \vert c$$ 2. 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. 1. Suppose $$S$$ is not empty. 2. By the Well-Ordering Principle, $$S$$ has a smallest element, call it $$n$$. 3. $$n$$ is not prime, so $$n = ab$$ for some integers $$a, b$$, where $$1 < a, b < n$$. 4. Since $$n$$ is the smallest element of $$S$$, $$a$$ and $$b$$ are not in $$S$$. 5. Therefore, $$a$$ and $$b$$ are products of primes. 6. Therefore, $$n$$ is a product of primes. 7. This contradicts the assumption that $$n$$ is not a product of primes. 8. 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: 1. Eery integer $$n$$ except $$0, \pm 1$$ is a product of primes by Theorem 1.7. 2. Suppose $$n$$ has two prime factorizations. Then, $n = p_1 p_2 ... p_n = q_1 q_2 ... q_n$ 1. We know that $$p_1 \vert q_1 q_2 ... q_n$$. 2. $$p$$ must divide one of the $$q$$. 3. After reordering, $$p_1 \vert q_1$$. 4. Since $$p_1$$ and $$q_1$$ are prime, $$p_1 = \pm q_1$$. 5. Therefore, replacing on the LHS: $$\pm q_1 p_2 p_3 ... p_n = q_1 q_2 q_3 ... q_n$$ 6. Dividing both sides by $$q_1$$ gives $$\pm p_2 p_3 ... p_n = q_2 q_3 ... q_n$$ 7. Continuing to eliminate one prime at each step, we get totaly equality.$$8. The lengths of the prime factorizations must be equal. If they were not, then we would get something like $$\pm p_{r+1} p_{r+2} ... p_n = 1$$ 9. This is a contradiction because $$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}$$. 1. Reflexivity: $$a \equiv_n a$$ 2. Symmetric: $$a \equiv_n b \implies b \equiv_n a$$ 3. Transitive: $$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$$, then 1. Distributivity: $$a + c \equiv_n b + d$$ 2. Multiplicativity: $$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$$. 1. If $$a$$ is any integer and $$r$$ is the remainder when $$a$$ is divided by $$n$$, then $$[a] = [r]$$. 2. There are exactly $$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$$, 1. Closure for addition: $$[a] \oplus [b] \in \mathbb{Z}_n$$ 2. Associativity for addition: $$[a] \oplus ([b] \oplus [c]) = ([a] \oplus [b]) \oplus [c]$$ 3. Commutativity for addition: $$[a] \oplus [b] = [b] \oplus [a]$$ 4. Identity for addition: $$[a] \oplus [0] = [a]$$ 5. Root existence: for each $$[a]$$ in $$\mathbb{Z}_n$$, the equation $$[a] \oplus [x] = [0]$$ has a solution in $$\mathbb{Z}_n$$ 6. Closure for multiplication: $$[a] \odot [b] \in \mathbb{Z}_n$$ 7. Associativity for multiplication: $$[a] \odot ([b] \odot [c]) = ([a] \odot [b]) \odot [c]$$ 8. Distribution with addition: $$[a] \odot ([b] \oplus [c]) = ([a] \odot [b]) \oplus ([a] \odot [c])$$ 9. Commutativity for multiplication: $$[a] \odot [b] = [b] \odot [a]$$ 10. Identity for multiplication: $$[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$$ is an integer, the following conditions are equivalent: 1. $$p$$ is prime 2. For any $$a \neq 0$$ in $$\mathbb{Z}_p$$, the equation $$ax = 1$$ has a solution in $$\mathbb{Z}_p$$ 3. Whenever $$bc = 0$$ in $$\mathbb{Z}_p$$, then $$b = 0$$ or $$c = 0$$ Proof: 1. Show that 1 implies 2. Suppose $$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$$. 2. Show that 2 implies 3. Suppose $$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$$. 3. Show that 3 implies 1. Suppose $$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$$: 1. Closure for addition: $$a + b \in R$$ 2. Associative addition: $$(a + b) + c = a + (b + c)$$ 3. Commutative addition: $$a + b = b + a$$ 4. Additive identity or zero element: there is an element $$O_R$$ in $$R$$ such that $$a + O_R = a$$ for all $$a \in R$$ 5. Additive identity or zero element: for each $$a \in R$$, the equation $$a + x = O_R$$ has a solution in $$R$$ 6. Closure for multiplication: $$ab \in R$$ 7. Associative multiplication: $$(ab)c = a(bc)$$ 8. Distributive laws. $$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$$ such that 1. $$S$$ is closed under addition 2. $$S$$ is closed under multiplication 3. Zero element: $$0_R \in S$$ 4. If $$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$$, 1. Multiplication of the zero element: $$a \cdot 0_R = 0_R = 0_R \cdot a$$ 2. Negative of a product: $$(-a)b = -(ab) = a(-b)$$ 3. Double negation: $$-(-a) = a$$ 4. Distribution of negative: $$-(a + b) = (-a) + (-b)$$ 5. Distribution of negative: $$-(a - b) = (-a) + b$$ 6. Double negation: $$(-a)(-b) = ab$$ 7. (If $$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$$ such that 1. $$S$$ is closed under subtraction (if $$a, b \in S$$, then $$a - b \in S$$) 2. $$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$$ such that 1. $$f$$ is injective, 2. $$f$$ is surjective, 3. $$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$$ be a homomorphism of rings. Then 1. Identity preservation: $$f(0_R) = 0_S$$ 2. Negation preservation: $$f(-a) = -f(a)$$ 3. Subtrative cancellation: $$f(a - b) = f(a) - f(b)$$ If $$R$$ is a ring with identity and $$f$$ is surjective, then 4. $$S$$ is a ring with identity $$f(1_R)$$. 5. Whenever $$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$$ and has these properties: 1. $$R$$ is a subring of $$T$$ 2. $$xa = ax$$ for every $$a \in R$$ 3. The set $$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$$. 4. The representation of elements is unique. 5. $$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. 1. If $$b(x) \vert a(x)$$, then $$cb(x) \vert a(x)$$ for each nonzero $$c \in F$$ 2. 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 1. $$d(x) \vert a(x)$$ and $$d(x) \vert b(x)$$ 2. 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]$$. Then the following conditions are equivalent. 1. $$p(x)$$ is irreducible 2. If $$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)$$ 3. If $$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)$$ is 1. reflexive: $$f(x) \equiv_{p(x)} f(x), \forall f(x) \in F[x]$$ 2. symmetric: if $$f(x) \equiv_{p(x)} g(x)$$, then $$g(x) \equiv_{p(x)} f(x)$$ 3. transitive: if $$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)$$, then 1. $f(x) + h(x) \equiv_{p(x)} g(x) + k(x)$ 2. $f(x) h(x) \equiv_{p(x)} g(x) k(x)$ Definition. The congruence class / residue class of $$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 $$p(x)$$. 1. 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)]$$. 2. Let $$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:

1. Use previous proof of commutative ring with identity for congruence classes.
2. Let $$F^*$$ be the subset of $$F[x] / p(x)$$ consisting of the congruence classes of all constant polynomials: $$F^* = \{ [a] \vert a \in F \}$$.
3. Define a map $$\phi : F \to F^*$$ by $$\phi(a) = [a]$$. $$\phi$$ is surjective. It is easy to prove that it is a homomorphism.
4. 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:

1. $$p(x)$$ is irreducible in $$F[x]$$
2. $$F[x] / p(x)$$ is a field
3. $$F[x] / p(x)$$ 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

1. reflexive: $$a \equiv_I a$$, $$\forall a \in R$$
2. symmetric: if $$a \equiv_I b$$, then $$b \equiv_I a$$
3. 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

1. $a + c \equiv_I b + d$
2. $ac \equiv_I bd$
• 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

1. $$R/I$$ is a ring, with addition and multiplication of cosets as defined previously
2. If $$R$$ is commutative, $$R / I$$ is a commutative ring
3. 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.