Rings

8.1 Definition of a Ring

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.

8.2 Examples

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.

8.3 Subrings

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. $ab \in S$.

8.4 Integral Domains

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, $b - c = 0$. $\blacksquare$

8.5 Fields: Further Examples

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. $\blacksquare$

8.6 Ring Homomorphisms

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 $\{f \in \mathbb{R}[x] : f(a) = 0\} = (x - a)$.

8.7 Worked Examples: Ring Homomorphisms

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.