§14 Rings

§14.1 Definition of a Ring

Definition 14.1 A monoid is a group without inverses. That is, a monoid is a set $M$ together with a function

\cdot:M\times M\to M

which has an identity element and is associative. We call a monoid commutative if

ab=ba

for all $a,b\in M$ .

Definition 14.2 An associative ring is a triple

(R,+,\cdot),

where $R$ is a set and

(1)

+:R\times R\to R

is a function making $(R,+)$ into an Abelian group. We call this operation addition, and its identity element is $0$ .

(2)

\cdot:R\times R\to R

is a function making $R$ into a monoid. We call the operation $\cdot$ multiplication, and its identity element is $1$ .

(3) Finally, we require multiplication to distribute over addition. This means that for all $a,b,c\in R$ ,

a(b+c)=ab+ac \quad\text{and}\quad (b+c)a=ba+ca,

where we write $a\cdot b$ as $ab$ .

We usually write simply $R$ for a ring, suppressing the operations $+$ and $\cdot$ .

Definition 14.3 If $(R,\cdot)$ is an Abelian monoid, we call $R$ a commutative ring.

When discussing groups, I quickly proved the cancellation law because it is useful to know. Here is another useful fact.

Proposition 14.1 Let $R$ be an associative ring, and let $0$ be the additive identity of $R$ . Then

0\cdot a=a\cdot0=0

for every $a\in R$ .

Proof:

\begin{align*} 0\cdot a &=(0+0)\cdot a \tag{since $0+0=0$}\\ &=0\cdot a+0\cdot a \tag{distributivity}. \end{align*}

Using the cancellation law in the Abelian group, we may subtract $0\cdot a$ from both sides of the equation. This leaves

0=0\cdot a.

Similarly,

a\cdot0=0.

~\tag*{$\square$}

Remark We will not explore the reason in detail, but rings behave very differently depending on whether or not they are commutative.

§14.2 Examples of Commutative Rings

Example 14.1 Consider the triple

(\mathbb{Z},+,\cdot).

This makes $(\mathbb{Z},+)$ an Abelian group, while $(\mathbb{Z},\cdot)$ is clearly a monoid: multiplication has an identity element called $1$ and is associative.

The distributive law is the familiar distributive law.

Example 14.2 With the usual addition and multiplication, the triple

(\mathbb{Q},+,\cdot)

is a ring.

The same is true of $\mathbb{R}$ and $\mathbb{C}$ with their usual multiplication.

These are special kinds of rings because every element of

R-\{0\}

has a multiplicative inverse.

Example 14.3 (Polynomial Rings)

Let

\mathbb{Z}[x]

denote the set of polynomials in $x$ with integer coefficients.

Thus an element is an expression

p(x) = a_0+a_1x+a_2x^2+\cdots+a_nx^n = \sum_{i=0}^{\infty}a_ix^i,

where

a_i=0

for all $i$ larger than some finite $n$ .

For example, the following are elements of $\mathbb{Z}[x]$ :

0,\quad5,\quad3+x-5x^4,\quad x.

In the third example,

a_0=3,\qquad a_1=1,\qquad a_2=0,\qquad a_3=0,\qquad a_4=-5.

Let $q(x)$ be a polynomial with coefficients $b_i$ .

Polynomial addition is defined by

p(x)+q(x) := \sum_{i=0}^{\infty}(a_i+b_i)x^i.

Notice that since $a_i=0$ for $i>n$ , and $b_i=0$ for every $i$ larger than some $m$ , the sum really is a polynomial because

a_i+b_i=0

for every

i>\max(m,n).

The product of two polynomials is defined in the usual way:

\begin{aligned} p(x)\cdot q(x) &= (a_0+a_1x+\cdots+a_nx^n) (b_0+b_1x+\cdots+b_mx^m)\\ &= a_0b_0 + (a_1b_0+a_0b_1)x + \cdots + a_nb_mx^{n+m}\\ &= \sum_{k\geqslant0} \left( \sum_{i+j=k}a_ib_j \right)x^k. \end{aligned}

Proposition 14.2 $\mathbb{Z}[x]$ is a commutative ring.

More generally, if $R$ is a commutative ring, then the set

R[x]

of polynomials with coefficients in $R$ is a commutative ring.

Proof:

$R[x]$ is a ring because:

(1) Addition is closed.

For two polynomials $p(x)$ and $q(x)$ in $R[x]$ , their sum

p(x)+q(x)

is a polynomial whose coefficients are the sums of the corresponding coefficients of $p(x)$ and $q(x)$ .

Since $R$ is a commutative ring and is closed under addition, this operation is well defined in $R[x]$ .

(2) Multiplication is closed.

For two polynomials $p(x)$ and $q(x)$ , their product can be written as

p(x)\cdot q(x) = \left( \sum_{i=0}^{n}a_ix^i \right) \left( \sum_{j=0}^{m}b_jx^j \right) = \sum_{k=0}^{n+m}c_kx^k,

where

c_k = \sum_{i+j=k}a_ib_j.

Since $R$ is closed under multiplication, each coefficient $c_k$ is an element of $R$ . Thus the product is a polynomial in $R[x]$ .

(3) Addition is associative and commutative.

Polynomial addition in $R[x]$ is associative and commutative because addition in $R$ is associative and commutative, and the operations on individual coefficients obey the corresponding properties in $R$ .

(4) Multiplication is associative.

Polynomial multiplication in $R[x]$ is associative because the distributive laws hold and multiplication in $R$ is associative.

Thus, for polynomials $p(x),q(x),r(x)\in R[x]$ ,

(p(x)\cdot q(x))\cdot r(x) = p(x)\cdot(q(x)\cdot r(x)).

(5) Distributivity.

Multiplication distributes over addition because, for $p(x),q(x),r(x)\in R[x]$ ,

p(x)\cdot(q(x)+r(x)) = p(x)\cdot q(x) + p(x)\cdot r(x).

This follows from the distributive law for the individual coefficients in $R$ .

Commutativity of multiplication.

For any polynomials

p(x),q(x)\in R[x],

we have

p(x)\cdot q(x) = q(x)\cdot p(x).

This is because multiplication in $R$ is commutative.

In particular, if

p(x) = \sum_{i=0}^{n}a_ix^i

and

q(x) = \sum_{j=0}^{m}b_jx^j,

then

p(x)\cdot q(x) = \sum_{k=0}^{n+m} \sum_{i+j=k} a_ib_jx^k.

Since multiplication in $R$ is commutative,

a_ib_j=b_ja_i.

Therefore,

p(x)\cdot q(x) = q(x)\cdot p(x).

~\tag*{$\square$}

Example 14.4 (Smooth Functions)

Here is an example in which it is more difficult to explicitly list all the elements of the set.

Let

C^{\infty}(\mathbb{R}^{n})

denote the set of all infinitely differentiable functions from $\mathbb{R}^{n}$ to $\mathbb{R}$ .

Given two functions $f$ and $g$ , define their sum $f+g$ to be the function sending

x\in\mathbb{R}^{n}

f(x)+g(x),

where the addition takes place in $\mathbb{R}$ .

Their product $fg$ is the function sending

x\in\mathbb{R}^{n}

f(x)\cdot g(x)\in\mathbb{R}.

This is also a ring.

§14.3 The Ring $\mathbb{Z}/n\mathbb{Z}$

We have already seen three examples of familiar rings.

They are all infinite.

Now let us look at some finite examples.

Lemma 14.3 Let

\mathbb{Z}/n\mathbb{Z}

be the set of integers modulo $n$ .

The function

\cdot: \mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}, \qquad \overline{a}\cdot\overline{b} := \overline{a\cdot b}

is well defined.

Remark We use $\overline{a}$ to denote the equivalence class associated with $a\in\mathbb{Z}$ .

Here, $a\cdot b$ is ordinary multiplication of integers, while $\overline{a\cdot b}$ denotes the corresponding equivalence class modulo $n$ .

Proof: We need to prove that if

\overline{a} = \overline{a}^{\prime}

and

\overline{b} = \overline{b}^{\prime},

then

\overline{a\cdot b} = \overline{a^{\prime}\cdot b^{\prime}}.

Since $a=a^{\prime}$ modulo $n$ if and only if

a=a^{\prime}+An

for some integer $A$ .

Similarly,

b=b^{\prime}+Bn

for some integer $B$ .

Therefore,

ab = (a^{\prime}+An)(b^{\prime}+Bn) = a^{\prime}b^{\prime} + (a^{\prime}B+Ab^{\prime}+AB)n.

Hence

ab

is equal to

a^{\prime}b^{\prime}

modulo $n$ .

~\tag*{$\square$}

Corollary 14.4 Let $+$ be the usual addition on $\mathbb{Z}/n\mathbb{Z}$ , and let $\cdot$ be the operation above.

Then:

(1)

(\mathbb{Z}/n\mathbb{Z},+)

is an Abelian group with identity $\overline{0}$ .

(2)

(\mathbb{Z}/n\mathbb{Z},\cdot)

is an Abelian monoid with identity $\overline{1}$ .

(3) The operation $\cdot$ distributes over $+$ .

Proof:

(1) This is a result we have already proved.

(2) To prove associativity, note that

\begin{aligned} (\overline{a}\cdot\overline{b})\cdot\overline{c} &= \overline{a\cdot b}\cdot\overline{c}\\ &= \overline{(a\cdot b)\cdot c}\\ &= \overline{a\cdot(b\cdot c)}\\ &= \overline{a}\cdot\overline{b\cdot c}\\ &= \overline{a}\cdot (\overline{b}\cdot\overline{c}). \tag{14.1} \end{aligned}

Every line except (14.1) follows from the definition of $\cdot$ , while (14.1) uses associativity of integer multiplication.

Commutativity holds because

\overline{a}\cdot\overline{b} = \overline{a\cdot b} = \overline{b\cdot a} = \overline{b}\cdot\overline{a}.

The middle equality is simply commutativity of integer multiplication.

The identity is $1$ because

\overline{a}\cdot\overline{1} = \overline{a\cdot1} = \overline{a}.

(3) Distributivity holds because

\begin{aligned} \overline{a}\cdot (\overline{b}+\overline{c}) &= \overline{a}\cdot\overline{b+c}\\ &= \overline{a\cdot(b+c)}\\ &= \overline{ab+ac}\\ &= \overline{ab}+\overline{ac}\\ &= \overline{a}\cdot\overline{b} + \overline{a}\cdot\overline{c}. \tag{14.2} \end{aligned}

Except for (14.2), each line follows from the definitions, while (14.2) uses the ordinary distributive law for integers.

~\tag*{$\square$}

§14.4 Motivation for Commutative Rings

There is a philosophy in modern mathematics that the properties of a space can be inferred by studying the properties of collections of functions on that space.

For example, by studying the collection of polynomial functions on a space $X$ , one can infer certain properties of the space $X$ itself.

In fact, collections of functions naturally form commutative rings.

Properties of these rings determine certain features of the space $X$ .

This is far from obvious.

Some of the most important developments related to this idea did not appear until the 1880s—almost two hundred years after Descartes first observed that algebraic equations could describe concrete geometry.

So if you consider that algebra, beginning in the Islamic Golden Age around the 800s, and geometry, originating with the Greeks, developed for almost a thousand years before Descartes brought them together, and that it then took another two hundred years before we systematically understood rings as powerful tools for studying geometry, you may begin to appreciate that these are very deep ideas.

We will not be able to explore the theory of using rings to study geometry in detail.

But if you are interested, you can look into commutative algebra and algebraic geometry.

§14.5 Examples of Noncommutative Rings

Example 14.5 (Matrix Rings)

Fix an integer

n\geqslant0,

and consider the set

M_{n\times n}(\mathbb{R})

of all $n\times n$ matrices with entries in $\mathbb{R}$ .

You can add and multiply matrices, and matrix multiplication distributes over addition.

Therefore,

M_{n\times n}(\mathbb{R})

is a ring.

To make distributivity explicit, consider three matrices $A,B,C$ with entries

a_{ij}, \qquad b_{ij}, \qquad c_{ij}.

The $ij$ entry of

A(B+C)

\sum_{k=1}^{n} a_{ik}(b_{kj}+c_{kj}) = \sum_{k=1}^{n}a_{ik}b_{kj} + \sum_{k=1}^{n}a_{ik}c_{kj}.

But the expression on the right is exactly the $ij$ entry of

AB+AC.

Similarly, one can prove

(B+C)A = BA+CA.

Example 14.6 (Group Rings)

Let $G$ be a finite group and let $R$ be a commutative ring.

As a set,

R[G]

is the set of all functions from $G$ to $R$ .

Thus each element

g\in G

corresponds to an element

r_g\in R.

We write such a function using the notation

\sum_{g\in G}r_gg.

For example, the following is an element of

\mathbb{Z}[S_3]:

5(~)+3(12)-8(123).

Addition is straightforward: we simply add the corresponding terms:

\left( \sum r_gg \right) + \sum(s_gg) = \sum_{g\in G}(r_g+s_g)g.

That is, this is simply addition of functions.

Multiplication is not just multiplication of functions.

The coefficient of $g$ in the product of

\sum r_gg

and

\sum s_gg

is given by

\sum_{(g_1,g_2)~\text{s.t.}~g_1g_2=g} r_{g_1}s_{g_2}.

In other words,

\left( \sum_{g\in G}r_gg \right) \left( \sum_{h\in G}s_hh \right) = \sum_{g\in G} \sum_{h\in G} r_gs_h(gh) = \sum_{k\in G} \left( \sum_{(g_1,g_2)~\text{s.t.}~g_1g_2=g} r_gs_h \right)k.

Notice that this multiplication is not commutative.

§14.6 Ring Homomorphisms

Definition 14.4 Let $R$ and $S$ be rings, and let

f:R\to S

be a function.

We call $f$ a ring homomorphism if

(1) $f$ is a group homomorphism with respect to addition;

(2)

f(1)=1,

i.e. $f$ sends the multiplicative identity of $R$ to the multiplicative identity of $S$ ;

(3) for all $a,b\in R$ ,

f(ab)=f(a)f(b).

Definition 14.5 If $f$ is a bijection, we call $f$ an isomorphism.

Now I would like to explain further why

\mathbb{Z}/n\mathbb{Z}

is a ring.

How did we prove that it is a group?

By applying a general principle:

H\triangleleft G,

then

G/H

is a group.

I would like to do the same thing for rings.

But in this context, whenever we say “ring”, we will mean a commutative ring.

Contents

§14 Rings

§14.1 Definition of a Ring

§14.2 Examples of Commutative Rings

§14.3 The Ring Z/nZ\mathbb{Z}/n\mathbb{Z}Z/nZ

§14.4 Motivation for Commutative Rings

§14.5 Examples of Noncommutative Rings

§14.6 Ring Homomorphisms

§14.3 The Ring $\mathbb{Z}/n\mathbb{Z}$