7.1 Basic Definitions and Examples

Definition. A ring $R$ is an abelian group $(R, +)$ along with a multiplication operation $\times$ that is associatively and distributively compatible with the addition operation $+$. If multiplication is commutative, then we say that $R$ is commutative. If $R$ has a multiplicative identity, then we say $R$ is a ring with identity or a unital ring.

Examples

1. The trivial ring ${0}$ with the forced choice of addition and multiplication. This ring is also commutative and unital.
2. The integers $\mathbb{Z}$ form a commutative unital ring under usual addition and multiplication.
3. The quotient group $\mathbb{Z}/n\mathbb{Z}$ is a commutative unital ring with the canonical choice of multiplication.
4. The rational numbers $\mathbb{Q}$, real numbers $\mathbb{R}$, and complex numbers $\mathbb{C}$ are all commutative unital rings.
5. If $X$ is any nonempty set and $R$ is a ring, then the set of all functions $f:X \to R$ forms a ring under pointwise addition and multiplication. Notice that commutativity and existence of identity both are inherited from $R$.
6. The ring $2\mathbb{Z}$ is unital but has no identity.
7. The ring of functions $f:\mathbb{R} \to \mathbb{R}$ of compact support (pointwise addition and multiplication) form a nonunital ring.
8. The $n\times n$ matrices $M_n(R)$ with entries in the ring $R$ form a (usually) noncommutative unital ring under entrywise additiona and matrix multiplication.
9. If $A$ is an abelian group, then the endomorphism form a ring under pointwise addition and composition as multiplication.

The latter two are common ways of constructing noncommutative rings.

Definition. A unital ring $R$ with $1\neq 0$ is called a division ring if every nonzero element has a multiplicative inverse. If $R$ is also commutative, it is called a field.

Examples

1. $\mathbb{Q}$, $\mathbb{R}$, and $\mathbb{C}$ are all fields.
2. The Hamilton Quaternions $\mathbb{H}$ generated by formal real linear combinations of $1$, $i$, $j$, and $k$ subject to the relations $i^2 = j^2 = k^2 = -1$, $ij = -ji = k$, $jk = -kj = i$, $ki = -ik = j$, form a noncommutative division ring. The multiplicative inverses are given by $(a + bi + cj + df)^{-1} = \dfrac{a - bi - cj - dk}{a^2 + b^2 + c^2 + d^2}$.

The very general structure of rings allow for two arbitrarily chosen rings to have vastly different structures. That being said, though, there are some properties that all rings share in common.

Proposition. Let $R$ be a ring. Then $0a = a0 = 0$ for all $a \in R$.

Proof. $a0 = a(0 + 0) = a0 + a0$. Thus, $0 = a0$. The proof for $0a = 0$ is the same argument.

Proposition. Let $R$ be a ring. Then $(-a)b = a(-b) = -(ab)$ for all $a, b \in R$.

Proof. $(-a)b - (-(ab)) = (-a)b + ab = (-a + a)b = 0$. The proof for $a(-b) = -(ab)$ is the same argument.

Proposition. Let $R$ be a ring. Then $(-a)(-b) = ab$ for all $a, b \in R$.

Proof. $(-a)(-b) - (ab) = (-a)(-b) - (-a)b = (-a + a)(-b) = 0$.

Proposition. If $R$ has an identity $1$, then the identity is unique.

Proof. Suppose $q$ is an identity of $R$. Then $q = 1q = 1$.

Proposition. If $R$ has an identity $1$, then $-a = (-1)a$.

Proof. $(-1)a + a = (-1 + 1)a = 0$.

One way in which rings can differ significantly from the integers is when we look at the rings $\mathbb{Z}/n\mathbb{Z}$ for some positive integer $n$ that is not a prime number. Then we can factor $n = ab$ nontrivially and the image of $a$ and $b$ under the quotient map have product $0$.

Definition. Let $R$ be a ring. We say that the two factors of a nontrivial factoring (that is, neither is $0$) of $0$ is a zero divisor. An element is a unit if has a multiplicative inverse.

Proposition. A zero divisor is never a unit and vice versa.

Proof. If $a$ is a zero divisor, then there is some nonzero $b \in R$ such that $ab = 0$. If $a^{-1}$ exists, then $b = a^{-1}ab = 0$, contradiction. The same argument works to prove the other direction.

Remark. Notice that this essentially proves that the set of units $R^\times$ in $R$ forms a multiplicative group.

Definition. A commutative ring with identity $1\neq 0$ is called an integral domain if it has no zero divisors.

The lack of zero divisors gives rise to a cancellation property:

Proposition. Let $a, b, c$ with $a$ not a zero divisor. Then $ab = ac$ implies that $a = 0$ or $b = c$.

Proof. $ab = ac$ implies $a(b - c) = 0$. Since $a$ is not a zero divisor, either $a = 0$ or $b - c = 0$.

Theorem. Let $R$ be a finite integral domain. Then $R$ is a field.

Proof. List out the elements of $R$ as

Pick some nonzero $r_j \in R$. Then multiply each element by $r_j$:

Since $R$ is an integral domain, the cancellation law implies that $r_jr_k = r_j r_m$ implies $r_k = r_m$. Thus, by the Pigeonhole Principle, there must be some $r_i$ such that $r_j r_i = 1$. Thus, $r_j$ is a unit.

Exercises

Let $R$ be a unital ring.

Problem 11. Prove that if $R$ is an integral domain and $x^2 = 1$ for some $x \in R$, then $x = \pm 1$.

Proof. $x^2 = 1$ implies that $(x + 1)(x - 1) = 0$. Thus, $x = \pm 1$.

Problem 14. Let $x$ be a nilpotent of the commutative ring $R$.

• (a) Prove that $x$ is either zero or a zero divisor.
• (b) Prove that $rx$ is nilpotent for all $r \in R$.
• (c) Prove that $1 + x$ is a unit in $R$.
• (d) Deduce that the sum of a nilpotent element and a unit is a unit.

Proof. (a) If $x = 0$, we are done (clearly $0$ is nilpotent). If $x$ is not zero, then there is some minimal positive integer $m > 1$ such that $x^m = 0$ by the well-ordering principle. Then $x\cdot x^{m-1} = 0$. The minimality of $m$ implies that $x^{m-1} \neq 0$ so $x$ is a zero divisor.

(b) Since $R$ is commutative, $(rx)^m = r^m x^m$ and the claim follows.

(c) A straightforward computation shows that $(1 + x)^{-1} = 1 - x + x^2 - x^3 + \cdots + (-1)^{m-1}x^{m-1}$ where we choose $m > 0$ minimally such that $x^m = 0$. (For intuition: note that $1/(1 + x)$ is formally a geometric power series then nilpotence kills off everything past $x^{m-1}$.)

(d) Since $u$ is invertible, we can factor $u + x = u(1 + u^{-1}x)$. Part (b) and (c) imply that $1 + u^{-1}x$ is a unit. Thus, $(u + x)^{-1} = u^{-1}(1 + u^{-1}x)^{-1}$ (I’m lazy because commutativity).

Problem 25. Let $I$ be the ring of integral Hamilton Quaternions and define

(a) Prove that $N(\alpha) = \alpha \overline{\alpha}$ for all $\alpha \in I$. (b) Prove that $N(\alpha\beta) = N(\alpha)N(\beta)$ for all $\alpha, \beta \in I$. (c) Prove that an element of $I$ is a unit if and only if it has norm $+1$. Show that $I^\times$ is isomorphic to the quaternion group of order $8$.

Proof. (a) and (b) are straightforward computations so we skip them. For (c), we begin by noting that the inverse of $\alpha \neq 0$ in the ring of rational quaternions is $\overline{\alpha}/N(\alpha)$. As the only units in $\mathbb{Z}$ are $\pm 1$, it follows that $\alpha$ is a unit if and only if it has $+1$ norm. The only solutions to $a^2 + b^2 + c^2 + d^2$ are $\pm 1$, $\pm i$, $\pm j$, and $\pm k$. Thus, it is clear that $I^\times$ is isomorphic to $Q_8$.