notes-archive/docs/math/abstract-algebra/18-integral-domains.md

Chapter 18 - Integral Domains

Section 18.1 - Fields of Fractions

Definition. Given an integral domain D, we can construct a field F containing D by stating that any p/q \in F, annd that any two elements a/b = c/d if and only if ad = bc. We can consider this akin o a set of ordered pairs

S = \{(a, b) : a, b \in D \text{ and } b \neq 0 \}

Lemma. 18.1: The relation (a, b) ~ (c, d) \text{ if } ad = bc is an equivalence relation.

Lemma. 18.2: The operations of addition and multiplication on F are well-defined.

Lemma. 18.3: The set of equivalence classes of S, F under ~ form a field.

Theorem. 18.4: Let D be an integral domain. Then, D can be embedded in a field of fractions F_D where any element in F_D can be expressed as the quotient of two elements in D.

Additionally, F_D is unique. That is, given field E such that E \supset D, there exists a map \psi: F_D \rightarrow D giving an isomorphism such that \psi(a) = a for all a \in D.

Collary. 18.6: Let F be a field of charactaristic 0. Then, F contains a subfield isomorphic to \mathbb{Q}.

Collary. 18.6: Let F be a field of charactaristic p. Then, F contains a subfield isomorphic to \mathbb{Z}_p.

Section 18.2 - Factorization in Integral Domains

Definition. Let R be a commutative ring with identity, and a, b \in R. We say that a divideds b, that is, a | b, if there exists some c \in R such that b = ac.

Definition. A unit element is any element that has a multiplicative inverse.

Definition. Two elements a, b \in R are said to be associates if there exists some unit u \in R such that a = ub.

Definition. Let D be an integral domain. A nonzero element p \in D is said to be irreducible if when given p = ab, either a or b is a unit.

Definition. Let D be an integral domain. A nonzero element p is prime if when given p = ab, either p | a or p | b.

Definition. Given integral domain D, we say that D is a Unique Factorization Domain (UFD) if it satisfies the following criteria:

1. Given a \in D, a \neq 0, and a is not a unit, a can be written as a product of irreducible elements in D.
2. Let a = p_1 \ldots p_r = q_1 \ldots q_s, where p_i and q_i are all irreducible. Then, r = s, and there exists some fuction \pi \in S_r such that p_i and q_{\pi(j)} are associates for j = 1, \ldots, r.

Definition. A ring R is a principal ideal domain (PID) if every ideal of R is principal.

Lemma. 18.11: Let D be an integral domain and a, b \in D. Then,

1. a | b if and only if \langle b \rangle \subseteq \langle a \rangle
2. a and b are associates if and only if \langle b \rangle = \langle a \rangle
3. a is a unit in D if and only if \langle a \rangle = D.

Theorem. 18.12: Let D be a PID, and let \langle p \rangle be a nonzero ideal in D. Thus, \langle p \rangle is a maximal ideal if and only if p is irreducible.

Collary. 18.13: Let D be a PID. For any p \in D, if p is irreducible, then p is prime.

Lemma. 18.14: Let D be a PID. Let I_1 \subseteq I_2 \subseteq \ldots. Then, there exists some integer N such that I_n = I_N for all n > N. That is, any chain of ideals converges.

Definition. Any commutative ring that satisfies the above condition (the ascending chain condition), even if it's not a PID, is called a Noetherien ring.

Theorem. 18.15: Every PID is a UFD. Note that the converse is not true.

Collary 18.16: Let F be a field. Then, F[x] is a UFD.

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Definition. Any integral domain D is a euclidian domain with a euclidian function nu: D \\ \{0\} \rightarrow \mathbb{N} that satisfies the following:

1. Given a, b \neq 0, then \nu(a) \leq \nu(ab).
2. Given, a, b \in D and b \neq 0, there exists some q, r \in D such that a = bq + r and either r = 0 or \nu(r) < \nu(b).

Example. Absolute value on \mathbb{Z} is a Euclidian validation.

Example. Degree on F[x] is a Euclidian validation.

Example. \nu(a + bi) = a^2 + b^2 is a Euclidian validation over \mathbb{Z}[i].

Theorem. 18.21: Every Euclidian domain is a PID.

Collary. Every Euclidian domain is a UFD.

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Definition. Given a polynomial p(x) \in D, with D bein an integer domain, we say that the content of p(x) is the greatest common divisor of its coefficients. Additionally, if the content is 1, we say that p(x) is primitive.

Theorem. 18.24: Let D be a UFD, and f(x), g(x) \in D[x] be primitive. Then, f(x)g(x) is primitive.

Lemma. 18.25: Given D is a UFD, and p(x), q(x) \in D[x], the content of p(x)q(x) is equal to the product of the contents of the individual polynomials

Lemma. 18.26: Let D be a UFD and F = F_D be its field of fractions. Given p(x) \in D[x], and p(x) = f(x)g(x) with f(x), g(x) \in F_D, we can say that p(x) = f_1(x)g_1(x) with f_1(x), g_1(x) \in D. Additionally, \deg f_1(x) = \deg f(x) and \deg g_1(x) = \deg g(x).

As a direct consequence, we see the following.

Collary. Let D be a UFD, and F = F_D. Then, a primitive polynomial p(x) \in D[x] is irreducible in D[x] if and only if it is irreducible in F[x].

Collary. Let D be a UDF, and F = F_D. Then, if a monic polynomial p(x) \ in D[x] can be written as p(x) = f(x)g(x) with f(x), g(x) \in F_D[x], then p(x) can be written as p(x) = f_1(x)g_1(x), where f_1(x), g_1(x) \in D[x].

Theorem. If D is as UFD, then D[x] is a UFD.

Collary. This theorem has several collaries:

1. Given a field F, since F is a PID, it is also a UFD. Thus, F[x] is a UFD.
2. The ring of polynomials over integers, \mathbb{Z}[x] is a UFD.
3. Given D is a UFD, D[x] is a UFD. Thus, D[x_1, x_2] is a UFD, and by induction, D[x_1, \ldots, x_n] is a UFD.