Rings

The definition of a ring and first properties

Definition (Ring and commutative ring)

A ring is a set \(R\) together with two binary operations \(+, \cdot\) such that:

1. \((R, +)\) is an abelian group;
2. the operation \(\cdot\) is associative and has an identity element;
3. if \(a, b, c \in R\), then \(a \cdot (b + c) = a \cdot b + a \cdot c\) and \((a + b) \cdot c = a \cdot c + b \cdot c\).

If moreover \((R, +, \cdot)\) satisfies that for all \(a, b \in R\), \(a \cdot b = b \cdot a\), then \(R\) is a commutative ring.

Note (We do not assume that \(1 \neq 0\))

If \(0\) is the identity element of \(+\) and \(1\) is the identity element of \(\cdot\), we are not assuming that \(1 \neq 0\).

Proposition (Basic arithmetic rules in a ring)

Let \((R, +, \cdot)\) be a ring. Then:

1. for every \(a \in R\), \(a \cdot 0 = 0 \cdot a = 0\);
2. for all \(a, b \in R\), \((-a) \cdot b = -(a \cdot b) = a \cdot (-b)\);
3. for all \(a, b \in R\), \((-a) \cdot (-b) = a \cdot b\);
4. the identity element of \(\cdot\) is unique and \((-1) \cdot a = -a\).

Proof: Exercise.

Note (The trivial ring)

If \(0 = 1\), then

\[a = a \cdot 1 = a \cdot 0 = 0\]

for every \(a \in R\), whence \(R = \{0\}\) and the ring is the trivial ring with a single element.

Example (First examples of rings)

The following are basic examples:

1. \((\mathbb{Z}, +, \cdot)\) is a commutative ring.
2. Let \(n > 1\) and let \(R\) be a ring. Then \(M_{n}(R)\) is a ring with the matrix operations \(+, \cdot\). It is not a commutative ring.
3. \((\mathbb{R}, +, \cdot)\), \((\mathbb{Q}, +, \cdot)\) and \((\mathbb{C}, +, \cdot)\) are commutative rings.

Fields and subrings

Definition (Field)

A field \((F, +, \cdot)\) is a commutative ring such that \(1 \neq 0\) and for every \(a \in F\),

\[a \neq 0 \implies \exists b \in F\ (a \cdot b = 1).\]

Exercise (The usual number rings are fields)

Show that \((\mathbb{R}, +, \cdot)\), \((\mathbb{Q}, +, \cdot)\) and \((\mathbb{C}, +, \cdot)\) are fields.

Definition (Subring)

Let \((R, +, \cdot)\) be a ring and \(S \subseteq R\). We say that \(S\) is a subring of \(R\) if \((S, +\mid_{S \times S}, \cdot\mid_{S \times S})\) is a ring.

Note (Subring criterion)

\(S \subseteq R\) is a subring if and only if \(1 \in S\) and for all \(a, b \in S\) we have that \(a - b \in S\) and \(a \cdot b \in S\). (Exercise.)

Units, zero divisors and integral domains

Definition (Units and the group of units)

Let \((R, +, \cdot)\) be a ring. An element \(a \in R\) is a unit if there exists \(b \in R\) such that

\[a \cdot b = 1 = b \cdot a.\]

If \(a\) is a unit, then \(b\) is unique; in this case we write \(b = a^{-1}\). Moreover, we define the set

\[U_{R} = \{ a \in R :\ a \text{ is a unit} \}.\]

Sometimes one writes \(U_{R} = R^{\times}\).

Note (The units form a group)

\((U_{R}, \cdot)\) is a group.

Note (The units of a field)

If \((F, +, \cdot)\) is a field, then \(U_{F} = F \setminus \{0\}\).

Definition (Zero divisors, division rings and integral domains)

Let \((R, +, \cdot)\) be a ring.

1. \(a \in R\) is a zero divisor if \(a \neq 0\) and there exists \(b \in R\) such that \(b \neq 0\) and \(a \cdot b = 0\) or \(b \cdot a = 0\).
2. A ring without zero divisors is called a division ring.
3. A commutative ring without zero divisors is called an integral domain.

Note (Units are not zero divisors)

Every field is an integral domain. More generally, if \(a \in R\) is a unit, then \(a\) is not a zero divisor.

Lemma (Subrings of a field are integral domains)

Let \((F, +, \cdot)\) be a field and \(R \subseteq F\) a subring. Then \(R\) is an integral domain.

Proof: First, \(R\) is commutative, since the operation \(\cdot\) of \(R\) is the restriction of that of \(F\), which is commutative. Now, suppose that there exists a zero divisor \(a \in R\). Then \(a \neq 0\) and there exists \(b \in R\) with \(b \neq 0\) such that \(a \cdot b = 0\) or \(b \cdot a = 0\). Since \(R \subseteq F\), we have that \(a, b \in F\), and so \(a\) is a zero divisor in \(F\). But \(F\) is a field and therefore an integral domain, so \(F\) has no zero divisors, a contradiction. We conclude that \(R\) has no zero divisors, that is, \(R\) is an integral domain.

Proposition (Cancellation law)

Let \((R, +, \cdot)\) be a ring and \(a \in R\) such that \(a \neq 0\) and \(a\) is not a zero divisor. If \(a \cdot x = a \cdot y\), then \(x = y\).

Proof: Suppose that \(ax = ay\). Then, subtracting and using distributivity,

\[ax = ay \implies ax - ay = 0 \implies a(x - y) = 0.\]

If we had \(x - y \neq 0\), the equality \(a(x - y) = 0\) would exhibit \(a\) as a zero divisor, since \(a \neq 0\). As \(a\) is not a zero divisor, necessarily \(x - y = 0\), that is, \(x = y\).

Lemma (Every finite integral domain is a field)

Let \(R \neq \{0\}\) be a finite integral domain. Then \(R\) is a field.

Proof: Note first that, since \(R \neq \{0\}\), we have that \(1 \neq 0\). Let \(a \in R\) with \(a \neq 0\), and consider the function \(\mu : R \to R\) given by \(\mu(x) = a \cdot x\). Note that \(\mu\) is injective: if \(\mu(x) = \mu(y)\), then \(a \cdot x = a \cdot y\), and since \(a \neq 0\) is not a zero divisor (as \(R\) is an integral domain), the cancellation law implies that \(x = y\). Moreover, since \(R\) is finite and \(\mu : R \to R\) is injective, \(\mu\) is surjective. In particular, there exists \(b \in R\) such that

\[\mu(b) = a \cdot b = 1.\]

Note that \(b \neq 0\), since if \(b = 0\) we would have \(1 = a \cdot 0 = 0\), and we know that \(0 \neq 1\) in \(R\). We conclude that \(a\) is invertible and, since \(a \neq 0\) was arbitrary and \(R\) is commutative with \(1 \neq 0\), \(R\) is a field.

Example (The integers and the integers modulo \(n\))

The following examples will recur:

1. \((\mathbb{Z}, +, \cdot)\) is a ring, \(\mathbb{Z}^{\times} = \{1, -1\}\) and \(\mathbb{Z}\) is an integral domain.

2. \((\mathbb{Z}_{n}, +, \cdot)\), with \(n > 1\), is a ring. It is a field if and only if \(n\) is prime, and

   \[\mathbb{Z}_{n}^{\times} = \{ [k] \in \mathbb{Z}_{n} :\ \operatorname{gcd}(k, n) = 1 \ \text{and}\ 0 \leq k \leq n - 1 \}.\]

   If \([x] \in \mathbb{Z}_{n}\), there are exactly two possibilities: \([x] \in \mathbb{Z}_{n}^{\times}\) or \([x]\) is a zero divisor (exercise).

Example (The endomorphism ring of an abelian group)

Let \(G\) be an abelian group and let

\[\operatorname{End}(G) := \{ f : G \to G\ \text{homomorphism} \}.\]

Define \(+, \cdot\) on \(\operatorname{End}(G)\) as follows: if \(f, g \in \operatorname{End}(G)\), then \(f + g : G \to G\) is such that \((f + g)(x) = f(x) + g(x)\), and \(f \cdot g : G \to G\) is such that \((f \cdot g)(x) = f(g(x))\). Exercise: \(\operatorname{End}(G)\) is a non-commutative ring.

Example (The ring of functions with values in a ring)

Let \(R \neq \{0\}\) be a ring and \(X\) a non-empty set. Define

\[\mathcal{F}(X, R) := \{ f : X \to R\ \text{functions} \},\]

with the operations \(+, \cdot\) given pointwise: if \(f, g \in \mathcal{F}(X, R)\), then

\[f + g : X \to R, \quad x \mapsto f(x) +_{R} g(x), \qquad f \cdot g : X \to R, \quad x \mapsto f(x) \cdot_{R} g(x).\]

Exercise: \(\mathcal{F}(X, R)\) is a ring, and it is commutative if and only if \(R\) is commutative.

Example (The Gaussian integers)

Let \(\mathbb{Z}[i] := \{ a + bi :\ a, b \in \mathbb{Z} \} \subseteq \mathbb{C}\). This set is called the set of Gaussian integers. Since \(\mathbb{Z}[i]\) is a subring of the field \(\mathbb{C}\), we can see that it is an integral domain.