notes-archive/docs/math/abstract-algebra/16-rings.md

Chapter 16 - Rings

Section 16.1 - Rings

Definition. A nonempty set S is a ring if, with two binary operations called addition and multipllication, the following are satisfied:

1. Addition is commutative. a + b = b + a for a, b \in R
2. Addition is associative. (a + b) + c = a + (b + c) for a, b, c \in R
3. There exists a zero-element 0_R in R such that a + 0 = a for all a \in
4. Every element a has an additive inverse -a \in R such that a + (-a) = 0_R
5. Multiplication is associative. That is, a(bc) = (ab)c for a, b, c \in R
6. The Distributive Property holds. That is, \forall a, b, c \in R,

a(b+c) = ab+bc \\
(a+b)c = ac + bc

Definition. If there exists some element 1_R \in R such that 1a = a1 = a for all a \in R, we say that R is a ring with unity or identity.

Note that some books impose the condition that 1 \neq 0. If 1 = 0, we can show the ring only has one element.

Definition. If ab = ba for all a, b \in R, the ring is said to be a commutative ring.

Definition. If a ring R is commutative, R is an integral domain if and only if for every a, b \in R, ab = 0 implies that either a = 0 or b = 0.

Definition. An element a \in R is called a unit if there exists some a^{-1} such that a a^{-1} = a^{-1} a = 1.

Definition. A ring R with identity is called a division ring if every nonzero element in R is a unit.

Definition. A commutative division ring is called a field. That is, in a field, every element has an inverse.

Section 16.2 - Integral Domains and Fields

Definition. If R is a commutative ring and r \in R, then r is said to be a zero divisor if there is some nonzero s \in R such that rs = 0.

Definition. A commutative ring with no zero divisors is called an integral domain.

Example. Consider the set \mathbb{Z}[i] = \{m + ni | m, n \in \mathbb{Z}\}. This ring is called the Gaussian integers. Prove that the Gaussian integers are not a field, and are an integral domain.

Example. Proposition 16.15: Cancellation law. Let D be a commutative ring with identity. Then, D is an integral domain if and only if for every nonzero a \in R, ab = ac implies b = c.

Theorem. 16.16: Every finite integral domain is a field.

Definition. For any non-negative integer n \in \mathbb{N} and r \in R, we say that nr = r + \ldots + r \text{(n times)}.

Definition. The charactaristic of a ring is the leat possible n \in \mathbb{N} such that nr = 0 for all r \in R.

Example. For every prime number p, \mathbb{N}_p is a field of charactaristic p.

Lemma. 16.18: Given R is a ring with identity, the charactaristic of 1 is the charactartistic of the field.

Theorem. 16.19: The charactaristic of an integral domain is prime or zero.

Section 16.3 - Ring Homomorphisms and Ideals

Definition Given rins R and S, and a mapping \phi: R \rightarrow S, we say that \phi is a ring homomorphism if the following are satisfied for all elements of R:

\begin{align}
    \phi(a + b) &= \phi(a) + \phi(b) \\
    \phi(ab) &= \phi(a) \phi(b)
\end{align}

Definition. If \phi is one-to-one and onto, it is an isomorphism.

Definition. For any ring homomorphism \phi, the kernel of \phi is the set

\ker \phi = \{ r \in R | \phi(r) = 0 \}

Definition. Proposition 16.22: Let \phi: R \rightarrow S be a ring homomorphism. Then,

1. If R is a commutative ring, then \phi(R) \subseteq S is a commutative ring.
2. \phi(0_R) = 0_S
3. Let 1_R and 1_S be the identities in R and S. If \phi is onto, then \phi(1_R) = 1_S
4. If R is a field an \phi(R) \neq \{0\}, then \phi(R) \subseteq S is a field.

Definition. A subring I \subseteq R is asn ideal of R if, when given a \in I, r \in R, then ar and ra are both in I. That is, rI \subseteq I and Ir \subseteq I.

Definition. Given a commutative ring R with identity, and r \in R, the set

\langle a \rangle = (r)R = \{ ar : r \in R \}

is an ideal in R. Specifically, \langle a \rangle is a principal ideal.

Example. Theorem 16.25. Every ideal in \mathbb{Z} is a principal ideal.

Examplee. With \phi: R \rightarrow S, \ker \phi is an ideal of R.

Remark. 16.28: We are working with two-sided ideals. If rings are not commutative, we may deal with left ideals and right ideals.

Theorem. 16.29: Let I be an ideal of R. Then, the factor/quotient ring R/I is a ring with multiplication defined by

(r + I)(s + I) = rs + I

Theorem. 16.30: Let I be an ideal of R. Then, the map \phi: R \rightarrow R/I defined by \phi(r) = r + I is a ring homomorphism of R onto R/I with \ker \phi = I.

Theorem. 16.31, First Isomorphism Theorem. Let \psi: R \rightarrow S. Then, \ker \psi is an ideal of R. Consider the isomorphism \phi: R \rightarrow R/\ker \psi. There exists an isomorphism \eta: R / \ker \psi \rightarrow \psi(R) such that \psi = \eta \phi.

Theorem. 16.32, Second Isomorphism Theorem. Let I be a subring of R and J be an ideal of R. Then, I \cap J is an ideal of I and

I/I \cap J \cong (I + J) / J

Theorem. 16.33, Third Isomorphism Theorem. Let R be a ring and I, J be ideals of J. If J \subsetneq I, then

R/I \cong \frac{R/J}{I/J}

Theorem. 16.34, Correspondence Theorem. Let I be an ideal of R. Then, S \mapsto S/I is a one-to-one correspeondence between the set of subrings S containing I (that is, I \in S) and the set of subrings of R/I. Furthermore, the ideals of R containing I correspond to the ideals of R/I.

Section 16.4 - Maximal and Prime Ideals

Definition. Consider ring R and proper ideal M \subseteq R. Then, M is a maximal ideal of R if the ideal M is not a subset of any ideal except R itself. That is, given any ideal I properly containing M, I = R.

Theorem. 16.35: Given a commutative ring with identity R, M is a maximal ideal if and only if R/M is a field.

Definition. Consider ring R and proper ideal P \subseteq R. Then, P is a prime ideal if given ab \in P, either a \in P or b \in P.

Theorem. 16.38: Let R be a commutative ring with identity 1. Then, P \subseteq R is a prime ideal of R if and only if R/P is a field.

Let us assume that P is an ideal in R and R/P is an integral domain. Take two elements ab \in P. Now, consider a + P and b + P in R/P such that (a+P)(b+P) = 0+P = P. As R/P is a field, either a + P = 0 + P = P or b + P = 0 + P = P, meaning either a \in P or b \in P. Thus, P is as prime ideal.

Now, assume the opposite. Let P be prime. Now, we want to show that R/P is an integral domain.

Consider two elements a + P, b + P in R/P. We know that

(a + P)(b + P) = ab + P = 0 + P = P

Thus, ab \in P. By symnetry, assume a \notin P. Thus, b \in P by the devinition of a prime ideal, so b + P = 0 + P, meaning R/P is an integral domain.

Theorem. 16.40: In a commutative ring with identity, every maximal ideal is also a prime ideal.