Rings

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updates: 31 May 2011

Rings and ring homomorphisms

A ring is a set $R$ with functions
$\begin{matrix} R \times R & \to & R \\ (r_{1}, r_{2}) & ⟼ & r_{1} + r_{2} \end{matrix}$ and $\begin{matrix} R \times R & \to & R \\ (r_{1}, r_{2}) & ⟼ & r_{1} r_{2} \end{matrix}$ ,
addition and multiplication, such that

(a) If $r_{1}, r_{2}, r_{3} \in R$ then $(r_{1} + r_{2}) + r_{3} = (r_{1} + (r_{2} + r_{3})$ ,

(b) If $r_{1}, r_{2} \in R$ then $r_{1} + r_{2} = r_{2} + r_{1}$ ,

(c) There exists a zero (sometimes called the additive identity, $0 \in R$ , such that if $r \in R$ then $0 + r = r + 0 = r$ ,

(d) If $r \in R$ then there exists an additive inverse to $r$ , $- r \in R$ , such that $(- r) + r = r + (- r) = 0$ ,

(e) If $r_{1}, r_{2}, r_{3} \in R$ then $r_{1} (r_{2} r_{3}) = (r_{1} r_{2}) r_{3}$ ,

(f) There exists an identity (sometimes called the multiplicative identity), $1 \in R$ , such that if $r \in R$ then $1 r = r 1 = r$ ,

(g) Distributive law. If $r, s, t \in R$ then $r (s + t) = r s + r t and (s + t) r = s r + t r .$

Note that (a), (b), (c) and (d) in the definition of a ring $R$ mean that $R$ is an abelian group under addition. The definition of a ring is motivated by the properties of the integers.

HW: Show that the additive identity $0 \in R$ is unique.
HW: Show that if $r \in R$ then its additive inverse $- r \in R$ is unique.
HW: Show that the identity $1 \in R$ is unique.
HW: Show that if $r \in R$ then $0 \cdot r = 0$ .
HW: Show that if $r \in R$ and $1 \in R$ is the identity in $R$ then $(- 1) \cdot r = r \cdot (- 1) = - r$ .

Important examples of rings are:

(a) The integers,

ℤ

(b) The

n \times n

matrices,

M_{n} (R)

R [x]

Ring homomorphisms are for comparing rings.

Let $R$ and $S$ be rings with identities $1_{R}$ and $1_{S}$ , respectively.

A ring homomorphism from $R$ to $S$ is a function $f : R \to S$ such that

(a) If $r, s \in R$ then $f (r + s) = f (r) + f (s)$ ,

(b) If $r, s \in R$ then $f (r s) = f (r) f (s)$ ,

(c) $f (1_{R}) = 1_{S}$ .
A ring isomorphism is a bijective ring homomorphism.
Two rings $R$ and $S$ are isomorphic, $R ≃ S$ , if there exists a ring isomorphism $f : R \to S$ between them.

Two rings are isomorphic if the elements of the rings and their operations match up exactly. Think of rings that are isomorphic as being "the same".
HW: Give an example of two rings $R$ and $S$ that are isomorphic as groups but not as rings.
In the case of group, condition (b) in the definition of ring homomorphism forced condition (c). (See Proposition 1.1.11). This does not happen here since rings don't necessarily have multiplicative inverses.

Let $f : R \to S$ be a ring homomorphism. Let $0_{R}$ and $0_{S}$ be the zeros for $R$ and $S$ respectively. Then

(a)

f (0_{R}) = 0_{S}

(b) If

r \in R

then

f (- r) = - f (r)

A subring $S$ of a ring $R$ is a subset $S \subseteq R$ such that

(a) If $s_{1}, s_{2} \in S$ then $s_{1} + s_{2} \in S$ ,

(b) $0 \in S$ ,

(c) If $s \in S$ then $- s \in S$ ,

(d) If $s_{1}, s_{2} \in S$ then $s_{1} s_{2} \in S$ ,

(e) $1 \in S$ ,
The zero ring, $(0)$ , is the set containing only $0$ with operations $0 + 0 = 0$ and $0 \cdot 0 = 0$ .

Cosets

An additive subgroup of a ring $R$ is a subset $I \subseteq R$ of $R$ such that

(a) If $h_{1}, h_{2} \in I$ then $h_{1} + h_{2} \in I$ ,

(b) $0 \in I$ ,

(c) If $h \in I$ then $- h \in I$ ,

Let $R$ be a ring and let $I$ be an additive subgroup of $R$ . We will use the subgroup $I$ to divide up the ring $R$ .

A coset of $I$ in $R$ is a set $r + I = {r + h | h \in I}$ , where $r \in R$ .
$R / I$ (pronounced " $R$ mod $I$ ") is the set of cosets of $I$ in $R$ .

Let $R$ be a ring and let $I$ be an additive subgroup of $R$ . Then the cosets of $I$ in $R$ partition $R$ .

Notice that the proofs of Proposition (rgptn) and Proposition (gpptn) are essentially the same.
HW: Write a very short proof of Proposition (rgptn) by using (gpptn).

Quotient rings $\leftrightarrow$ Ideals

Let $R$ be a ring and let $I$ be an additive subgroup of $R$ . We can try to make the set $R / I$ of cosets of $I$ in $R$ into a ring by defining an addition operation and a multiplication operation on cosets. The only problem is that this doesn't work for the cosets of just any additive subgroup, the subgroup has to have special properties.

HW: Let $R$ be a ring and let $I$ be an additive subgroup of $R$ . Show that $I$ is a normal subgroup of $R$ .

An ideal $I$ is a subset of a ring $R$ such that

(a) If $a, b \in I$ then $a + b \in I$ ,

(b) If $i \in I$ and $r \in R$ then $i r \in R$ and $r i \in R$ .
The zero ideal $(0)$ of $R$ is the ideal containing only the zero element of $R$ .

HW: Show that if $I$ is an ideal of a ring $R$ then $0 \in I$ and if $a \in I$ then $- a \in I$ .
HW: Show that an ideal $I$ of a ring $R$ is an additive subgroup of $R$ .

Let $I$ be an additive subgroup of a ring $R$ . Then $I$ is an ideal of $R$ if and only if $R / I$ with operations given by $(r_{1} + I) + (r_{2} + I) = (r_{1} + r_{2}) + I and r_{1} (r_{2} + I) = r_{1} r_{2} + I$ is a ring.

Notice that the proofs of Proposition 2.1.6 and proposition 1.1.8 are essentially the same.

HW: Write a shorter proof of Proposition 2.1.6 by using Proposition 1.1.8.

The quotient ring $R / I$ is the ring of cosets of an ideal $I$ of a ring $R$ with operations given by $(r_{1} + I) + (r_{2} + I) = (r_{1} + r_{2}) + I$ and $(r_{1} + I) (r_{2} + I) = r_{1} r_{2} + I$ .

So we have successfully made $R / I$ into a ring when $I$ is an ideal of $R$ .

Kernel and image of a homomorphism

The kernel of a ring homomorphism $f : R \to S$ is the set $\ker f = {r \in R | f (r) = 0_{S}},$ where $0_{S}$ is the zero element of $S$ .
The image of a ring homomorphism $f : R \to S$ is the set $im f = {f (r) | r \in R} .$

Note that $\ker f = {r \in R | f (r) = 0_{S}}$ not ${r \in R | f (r) = 1_{S}}$ . If $\ker f$ was ${r \in R | f (r) = 1_{S}}$ then $\ker f$ would not necessarily be a subgroup of $R$ (not to mention an ideal) and we couldn't even hope to get homomorphism theorems like we did for groups.

Let $f : R \to S$ be a ring homomorphism. Then

(a)

\ker f

is an ideal of

R

(b)

im f

is a subring of

S

Let $f : R \to S$ be a ring homomorphism. Let $0_{R}$ be the zero element of $R$ . Then

(a)

\ker f = {0_{R}}

if and only if

f

is injective.

(b)

im f = S

if and only if

f

is surjective.

Notice that the proof of Proposition (rginjsur)(b) does not use the fact that $f : R \to S$ is a homomorphism, only the fact that $f : R \to S$ is a function.

(a) Let

f : R \to S

be a ring homomorphism and let

K = \ker f

. Define

\begin{matrix} \hat{f} : & R / \ker f & ⟶ & S \\ r + K & ⟼ & f (r) \end{matrix}

Then

\hat{f}

is a well defined injective ring homomorphism.

(b) Let

f : R \to S

be a ring homomorphism and define

\begin{matrix} f' : & R & ⟶ & im f \\ r & ⟼ & f (r) \end{matrix}

Then

f'

is a well defined surjective ring homomorphism.

f : R \to S

is a ring homomorphism then

R / \ker f ≃ im f,

where the isomorphism is a ring isomorphism.

Direct sums

Suppose $S$ and $T$ are rings. The idea is to make $S \times T$ into a ring.

The direct sum $S \oplus T$ of two rings $S$ and $T$ is the set $S \times T$ with operations given by $(s_{1}, t_{1}) + (s_{2}, t_{2}) = (s_{1} + s_{2}, t_{1} + t_{2}) and (s_{1}, t_{1}) (s_{2}, t_{2}) = (s_{1} s_{2}, t_{1} t_{2}),$ for $s_{1}, s_{2} \in S$ , $t_{1}, t_{2} \in T$ .
More generally, given rings $R_{1}, R_{2}, \dots, R_{n}$ , the direct sum $R_{1} \oplus R_{2} \oplus \dots \oplus R_{n}$ is the set $R_{1} \times R_{2} \times \dots \times R_{n}$ with the operations given by $(s_{1}, \dots, s_{i}, \dots, s_{n}) + (t_{1}, \dots, t_{i}, \dots, t_{n}) = (s_{1} + t_{1}, \dots, s_{i} + t_{i}, \dots, s_{n} + t_{n})$ $and (s_{1}, \dots, s_{i}, \dots, s_{n}) (t_{1}, \dots, t_{i}, \dots, t_{n}) = (s_{1} t_{1}, \dots, s_{i} t_{i}, \dots, s_{n} t_{n})$ where $s_{i}, t_{i} \in R_{i}$ , and $s_{i} + t_{i}$ and $s_{i} t_{i}$ are given by the operations for the ring $R_{i}$ . The operations in the direct sum are just the operations from the original rings acting componentwise.

HW: Show that these are good definitions, i.e. that, as defined above, $S \oplus T$ and $R_{1} \oplus R_{2} \oplus \dots \oplus R_{n}$ are rings with zeros given by $(0_{S}, 0_{T}$ and $(0_{R_{1}}, \dots, 0_{R_{n}}$ , respectively, and identities given by $(1_{S}, 1_{T}$ and $(1_{R_{1}}, \dots, 1_{R_{n}}$ , respectively.

Further definitions

There are many things which help to characterize a ring; some of these will be studied in depth in later chapters. Some definitions are given here for reference.

A commutative ring is a ring $R$ such that if $a, b \in R$ then $a b = b a$ .
The center of a ring $R$ is the set $Z (R) = {z \in R | if r \in R then z r = r z} .$

HW: Give an example of a noncommutative ring.
HW:Prove that $Z (R)$ is a subring of $R$ .
HW: Give an example to show that $Z (R)$ is not necessarily an ideal of $R$ .
HW: What two elements are always in the center of $R$ ?

The characteristic, $char (R)$ , of a ring $R$ is the smallest positive integer $n$ such that $1 + 1 + \dots + 1$ ( $n$ times) is 0. If such an integer does not exist, $char (R)$ is 0.

Let $R$ be a ring. Let $0_{R}$ and $1_{R}$ be the zero and the identity in $R$ , respectively.

(a) The function

ϕ : ℤ \to R

given by

ϕ (0) = 0_{R}, ϕ (m) = \underset{\underset{m times}{⏟}}{1_{R} + \dots + 1_{R}}, and ϕ (- m) = - ϕ (m), for m \in ℤ_{> 0},

is the unique ring homomorphism from

ℤ

R

(b)

\ker ϕ = n ℤ = {n k | k \in ℤ}

, where

n = char (R)

is the characteristic of the ring

R

HW: Show that if $char (R) = 2$ then $1 = - 1$ in $R$ .

A left inverse of an element $b$ of a ring $R$ is an element $c \in R$ such that $c b = 1$ .
A right inverse of an element $b$ of a ring $R$ is an element $c \in R$ such that $b c = 1$ .
An inverse of an element $b$ of a ring $R$ is an element $c \in R$ such that $c b = b c = 1$ .
A unit is an element of a ring that has an inverse.
If $R$ is a ring, $R^{\times}$ is the set of units of $R$ .

HW: Show that if $b \in R$ has both a left inverse and a right inverse then they must be equal.
HW: Give an example of a ring $R$ and an element of $R$ that has a left inverse but not a right inverse.
HW: What element of a ring is always a unit?
HW: Give an example of a ring such that $R^{\times} = R - {0}$ .

Let $R$ be a ring and let $S$ be a subset of $R$ . The ideal generated by $S$ , $(S)$ , is the ideal of $R$ such that

(a) $S \subseteq (S)$ ,

(b) If $J$ is an ideal of $R$ and $S \subseteq J$ then $(S) \subseteq J$ .
An ideal is principal if it is generated by one element.

The ideal $(S)$ is the smallest ideal of $R$ containing $S$ . Think of $(S)$ as gotten by adding to $S$ exactly those elements of $R$ that are needed to make an ideal.

A proper ideal of a ring $R$ is an ideal that is not equal to $R$ .
A maximal ideal of a ring $R$ is a proper ideal of $R$ that is not contained in any other proper ideal of $R$ .

HW: Show that a proper ideal does not contain any units.

Every proper ideal $I$ of a ring $R$ is contained in a maximal ideal of $R$ .

A local ring is a commutative ring with only one maximal ideal.
A simple ring is a ring with no ideals except $(0)$ and $R$ .
A ring $R$ is a division ring if every nonzero element of $R$ has an inverse in $R$ .
A field is a commutative ring $𝔽$ such that every nonzero element of $𝔽$ has an inverse in $𝔽$ .

Notes and References

These notes are written to highlight the analogy between groups and group actions, rings and modules, and fields and vector spaces.

References

[Ram] A. Ram, Notes in abstract algebra, University of Wisconsin, Madison 1993-1994.

[Bou] N. Bourbaki, Algèbre, Chapitre 9: Formes sesquilinéaires et formes quadratiques, Actualités Sci. Ind. no. 1272 Hermann, Paris, 1959, 211 pp. MR0107661.

[Ru] W. Rudin, Real and complex analysis, Third edition, McGraw-Hill, 1987. MR0924157.

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