Rings | Mathematics - Wyatt's Notes

A ring $(R, +, \cdot)$ is a set $R$ with two binary operations satisfying:

1. $(R, +)$ is an abelian group.
2. Multiplication is associative: $(ab)c = a(bc)$.
3. Distributive laws: $a(b + c) = ab + ac$ and $(a + b)c = ac + bc$.

A ring is commutative if $ab = ba$ for all $a, b \in R$. A ring with a multiplicative identity $1$ is A ring with unity. A field is a commutative ring with unity in which every non-zero element has A multiplicative inverse.

Example 1. $\mathbb{Z}$ is a commutative ring with unity, but not a field.

Example 2. $\mathbb{Z}/n\mathbb{Z}$ is a commutative ring with unity. It is a field if and only if $n$ is prime.

Example 3. $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$ are fields.

Example 4. The set $M_n(\mathbb{R})$ of $n \times n$ real matrices is a non-commutative ring with unity.

Example 5. $\mathbb{Z}[i] = \{a + bi : a, b \in \mathbb{Z}\}$ (Gaussian integers) is a commutative ring with unity.

A subring $S \subseteq R$ is a subset that is itself a ring under the operations of $R$.

Proposition 8.1 (Subring Criterion). A non-empty subset $S \subseteq R$ is a subring if and only if For all $a, b \in S$:

1. $a - b \in S$.
2. $ab \in S$.

An integral domain is a commutative ring $R$ with unity $1 \neq 0$ in which there are no zero divisors: if $ab = 0$ for $a, b \in R$ Then $a = 0$ or $b = 0$.

Proposition 8.2. Every field is an integral domain.

Proof. Let $F$ be a field and suppose $ab = 0$ with $a \neq 0$. Then $b = a^{-1}(ab) = a^{-1} \cdot 0 = 0$. $\blacksquare$

Proposition 8.3. $\mathbb{Z}/n\mathbb{Z}$ is an integral domain if and only if $n$ is prime.

Proof. If $n = p$ is prime, then $\mathbb{Z}/p\mathbb{Z}$ is a field, hence an integral domain. If $n = ab$ with $1 \lt a, b \lt n$ Then $[a][b] = [ab] = [n] = [0]$ in $\mathbb{Z}/n\mathbb{Z}$ But $[a] \neq [0]$ and $[b] \neq [0]$ So $\mathbb{Z}/n\mathbb{Z}$ has zero divisors. $\blacksquare$

Proposition 8.4 (Cancellation Law for Integral Domains). In an integral domain, if $ab = ac$ and $a \neq 0$ Then $b = c$.

Proof. $ab = ac$ implies $a(b - c) = 0$. Since $a \neq 0$ and there are no zero divisors, $b - c = 0$. $\blacksquare$

Example. $\mathbb{Q}(\sqrt{2}) = \{a + b\sqrt{2} : a, b \in \mathbb{Q}\}$ is a field. The inverse of $a + b\sqrt{2}$ (with $a, b$ not both zero) is $\frac{a - b\sqrt{2}}{a^2 - 2b^2}$.

Example. For any prime $p$, $\mathbb{Z}/p\mathbb{Z}$ is a field with $p$ elements, denoted $\mathbb{F}_p$.

Proposition 8.5. In a finite integral domain $R$Every non-zero element is a unit. Hence every finite Integral domain is a field.

Proof. Let $a \in R$ with $a \neq 0$. The map $\phi : R \to R$ given by $\phi(x) = ax$ is injective (since $ax = ay$ implies $a(x-y) = 0$ implies $x = y$ by the cancellation law). Since $R$ is finite, $\phi$ is also surjective, so there exists $b \in R$ with $ab = 1$. Thus $a$ is a unit. $\blacksquare$

A ring homomorphism $\phi : R \to S$ is a function satisfying:

1. $\phi(a + b) = \phi(a) + \phi(b)$ for all $a, b \in R$.
2. $\phi(ab) = \phi(a)\phi(b)$ for all $a, b \in R$.
3. $\phi(1_R) = 1_S$ (for rings with unity).

A ring homomorphism that is bijective is a ring isomorphism.

Proposition 8.6. If $\phi : R \to S$ is a ring homomorphism, then:

1. $\phi(0_R) = 0_S$.
2. $\phi(-a) = -\phi(a)$ for all $a \in R$.
3. $\ker(\phi) = \{r \in R : \phi(r) = 0_S\}$ is an ideal of $R$.

Proof. (1) $\phi(0) = \phi(0 + 0) = \phi(0) + \phi(0)$ So $\phi(0) = 0$ by cancellation in $(S, +)$. (2) $\phi(a) + \phi(-a) = \phi(a + (-a)) = \phi(0) = 0$ So $\phi(-a) = -\phi(a)$. (3) $\ker(\phi)$ Is an ideal: it is a subgroup of $(R, +)$ by the group homomorphism property, and for any $r \in R$ and $a \in \ker(\phi)$, $\phi(ra) = \phi(r)\phi(a) = \phi(r) \cdot 0 = 0$ and $\phi(ar) = \phi(a)\phi(r) = 0$ So $ra, ar \in \ker(\phi)$. $\blacksquare$

Example. The map $\phi : \mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$ given by $\phi(k) = [k]$ is a surjective Ring homomorphism with kernel $n\mathbb{Z}$.

Example. The evaluation map $\phi : \mathbb{R}[x] \to \mathbb{R}$ given by $\phi(f) = f(a)$ for a fixed $a \in \mathbb{R}$ is a surjective ring homomorphism with kernel $\{f \in \mathbb{R}[x] : f(a) = 0\} = (x - a)$.

A ring is the algebraic structure that captures what it means to do arithmetic. Where a group captures symmetry (one operation), a ring captures the idea of having both addition and multiplication that interact through distributivity. The integers $\mathbb{Z}$ are the prototypical ring: you can add, subtract, and multiply, but you cannot always divide.

The key distinction within rings is whether division is possible. A field allows division by every non-zero element, like the rational or real numbers. An integral domain prevents “zero divisors,” meaning $ab = 0$ implies $a = 0$ or $b = 0$ — a reasonable expectation inherited from ordinary arithmetic. Non-commutative rings, like matrix rings, reveal that multiplication need not be commutative while still satisfying all other ring axioms. Ring homomorphisms preserve both addition and multiplication, which is stronger than group homomorphisms that only preserve one operation. The kernel of a ring homomorphism is always an ideal, which is the ring-theoretic analogue of a normal subgroup, enabling quotient constructions that build new rings from old ones.

Problem. Let $\phi : \mathbb{Z}[i] \to \mathbb{Z}/5\mathbb{Z}$ be defined by $\phi(a + bi) = (a + 2b) + 5\mathbb{Z}$. Show that $\phi$ is a surjective ring homomorphism and find its kernel.

Problem. Let $R = \mathbb{Z}[x]/(x^2 - x)$. Describe the elements of $R$ and show that $R$ has zero divisors.

Mistake 1: Confusing ideals with subrings An ideal $I$ of $R$ is a subring that is also closed under multiplication by arbitrary ring elements: $rI \subseteq I$ and $Ir \subseteq I$ for all $r \in R$. Not every subring is an ideal. For example, $2\mathbb{Z}$ is a subring of $\mathbb{Z}$ but not an ideal of $\mathbb{Z}[x]$ because $x \cdot 2 = 2x \notin 2\mathbb{Z}$ as a polynomial. Always verify the absorption property.

Mistake 2: Confusing zero divisors with units A zero divisor $a$ satisfies $ab = 0$ for some nonzero $b$, while a unit $a$ satisfies $ab = 1$ for some $b$. In $\mathbb{Z}/6\mathbb{Z}$, $[2]$ is a zero divisor (since $[2][3] = [0]$) but not a unit. In $\mathbb{Z}/5\mathbb{Z}$, every nonzero element is a unit and there are no zero divisors. These are opposite extremes of multiplicative behaviour.

Mistake 3: Assuming all rings have a multiplicative identity The definition of a ring does not require a multiplicative identity $1$. Some authors include $1$ in the definition (called “rings with unity”), while others do not. When working with rings without unity, statements about units and the cancellation law may fail. Always check whether the ring is assumed to have unity.

flowchart TD
    A[8_Rings] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

• Ideals and Quotient Rings: Ideals are the ring-theoretic analogue of normal subgroups and enable quotient ring constructions.

• Homomorphisms and Isomorphism Theorems: Ring homomorphisms preserve both addition and multiplication, generalising group homomorphisms.

• Field Theory: Fields are commutative rings in which every nonzero element has a multiplicative inverse.

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• Classical Mechanics

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