rings lect 1

8/6/2019 Rings Lect 1

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1 Commutative Rings with Identity

You have already encountered rings in the rst-year course Groups, Rings and Fields. Inthis course we will concentrate on an important subclass of rings: commutative rings with

identity. In this setting we will be able to generalize in a more abstract framework resultsand notions from the familiar case of the integers, Z . For example we will generalize Euclidsalgorithm, the notion of prime numbers and we will prove a more general version of thefundamental theorem of arithmetic.

Whenever you are in doubt about what a theorem means, then the example to keepreturning to is the ring of integers, Z .

1.1 The Denition

Denition 1. A commutative ring with identity is a non-empty set R , equipped with two operations + : R R R, : R R R satisfying the following axioms:

R1 (R, +) is a commutative group (and we denote by 0 its zero element).R2 a b = b a for all a, bR . [ is commutative] R3 a (b c) = ( a b) c for all a,b,c R . [ is associative] R4 There exists an element 1 = 0 such that a 1 = a for all aR . [identity for ] R5 a (b + c) = a b + a c for all a,b,c R . [ distributes over + ]

1.1.1 Notation

We denote by a the inverse element of a , so a + ( a ) = 0.We write

ab for a ba b for a + ( b).

1.1.2 CommentsNote that the axioms are satised in Z and capture (we hope) the algebraic essence of theintegers. (What we have not attempted to build in is the order enjoyed by the integers.)

In the rst year rings were dened as sets with + , satisfying (R1), (R3) and (R5) wherewe postulate also that ( b + c) a = b a + c a .Recall the axioms dening abelian groups: ( A, +) is an abelian group if the followinghold:

(A1) a + b = b + a for all a, bA(A2) a + ( b + c) = ( a + b) + c for all a,b,c A(A3) There is an element 0 A such that a + 0 = a for all aA

(A4) For anya

Athere is an element

asuch that

a+ (

a) = 0.We dont propose to repeat work done in the rst year Groups course; for example we

will use without fuss such facts as the zero element is unique.In a similar vein we will not repeat work done in the rst year Analysis I course; much

of what we did when we investigated the real numbers from an axiomatic point of viewcan be used here too. For example, we dont mean to fuss at all when we use facts like(b + c) a = b a + c a .

We will also follow the practice we have learned in Groups and Vector Spaces: all zeroelements will be denoted by 0, and all identity elements by 1.

Some authors do not require that 1 = 0. Note that there is only one ring that does notsatisfy this: the trivial ring {0}.

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We state in the following lemma some familiar computational rules that hold also forrings.

Lemma 1.1.1. Let R be a commutative ring with identity. Then for any a, b R the following hold

1. a0 = 0a = 0 .

2. (a ) = a.3. a (b) = ( a )b = (ab).4. (a )(b) = ab .5.(1)a = a

Proof. The proofs are quite straightforward. We give some hints below. To show 1 note that

a0 = a (0 + 0) = a0 + a0a0 = 0

Assertion 2 was proven last year in the groups course. For 3 we have

a (b) + ab = a (b + ( b)) = a0 = 0hence a (b) is the additive inverse of ab , ie a (b) = (ab). We leave 4,5 as exercises.

1.1.3 An example

We will deal with examples later, but here is an example rather different in avour from theintegers Z . For R take the set of diagonal n n matrices with real entries; for the operationstake the usual matrix operations. Then we have a commutative ring with identity.

1.2 Two important classes

We begin with two denitions.

Denition 2. A non-zero element z of a commutative ring with identity R is called a zero-divisor if there exists a non-zero element wR such that zw = 0 .

For example, in the commutative ring with identity consisting of the 2 2 diagonalmatrices with real entries the element 1 00 0 and its friend

0 00 1 are both zero-divisors.

More generally, the zero-divisors are precisely thea 00 d with

a = 0 or d = 0.

Denition 3. An element u of a commutative ring with identity R is called a unit if thereexists an element vR such that uv = 1 . In this case we say that v is the inverse of u .We denote the set of units by R , which we call the (multiplicative) group of units of R .

We see easily that ( R , ) is indeed an abelian group. We remark that if u is a unit thenu is not a zero divisor. Indeed if u is a unit then there is some v such that uv = 1. If au = 0then ( au )v = 0 va (uv ) = 0 a = 0. So u is not a zero divisor.For a trivial example, in any R the identity is a unit. For a more elaborate example takeagain for R the 2 2 diagonal matrices with real entries. Then the units are precisely thea 0

0 d withad = 0.

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1.2.1 Integral Domains

We can now dene this important class of rings.

Denition 4. We say that a commutative ring with identity R is an integral domain if

there are no zero-divisors.For example, Z is an integral domain; other examples appear below.

1.2.2 Fields

Even more specialized are the elds.

Denition 5. We say that the commutative ring with an identity is a eld if every non-zeroelement is a unit.

Since units are not zero divisors elds are integral domains.

For example,R

is a eld.Note that this denition of eld (setting elds in a more general picture) is completelyconsistent with the denition used in the Linear Algebra course.

We note that

{elds} {integral domains } {commutative rings with identity }If R is an integral domain one can dene the eld of fractions K of R . This is a

generalization of the construction of Q from Z . We outline this in the appendix.

1.3 Examples, Non-examples and Nearly Examples

1.3.1 The integersWe repeat: the integers, with the usual operations form a commutative ring with identity.

1.3.2 The integers mod n

The set of integers modn , Z n is a commutative ring with identity.

1.3.3 Polynomials over a eld

Let K be any eld. Then the set of polynomials K [X ], with the usual polynomial denitionsof addition and multiplication forms a commutative ring with identity.

Next toZ

these are our most important examples of commutative rings with identity.1

1.3.4 Some elds

Here are examples of elds that we have met already: the rational eld, Q , the real eld R ,the complex eld C .

There are other, more exotic elds: many we will construct later as examples of theoremswe prove. For the moment you may like to check that Q [2] := {a + b2 | a, b Q}is aeld; and that C (X ) := {

f (X )g(X ) | f, g polynomials with complex coefficients, g = 0}is a eld.

1 Z is what number theorists study, the polynomial rings are what the geometers study; the similaritybetween the structures goes very deep.

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1.3.5 Not quite examples

The set of n n matrices with entries from a eld K is not a commutative ring with identity;it fails the commutative requirement. But suitably adapted much of what we say and provecan be adapted to this situation. Various subsets, however, yield useful examples.

The set of even integers is not a commutative ring with identity; it fails the identityrequirement. Again some of what we say can be suitably adapted to this sort of situation.

1.4 Subrings

Denition 6. Let R be a commutative ring with identity. A subset A R is said to be a subring [more properly, a sub-(commutative ring with identity)] if it contains the identity and is a commutative ring with identity under the same operations.

For example, Z is a subring of Q .Just as for subgroups we have a

Proposition 1.4.1 (Test for Subringhood) . Let R be a commutative ring with identity. Then A

R is a sub-(commutative ring with identity) if and only if

(i) 1A;

(ii) if a, bA then (a b)A;(iii) if a, bA then abA .

Proof. The proof is just as for groups or vector spaces; these criteria guarantee that theoperations restrict to operations on A and then the fact that the axioms which hold for allelements of R certainly hold in A .

As an application: the only sub-(commutative ring with identity) of Z is Z .

Note 1. Note that if by ring we mean, as most authors do, a system satisfying our axioms(R1), (R3) and distributivity for , + , then there are many subrings of Z : for each dZ theset dZ := {dr | r Z}is a sub-ring, but has no identity. It is therefore sometimes important to adopt the tedious sub-(commutative ring with identity) language.1.5 Direct Products

This is an easy recipe to make new rings from old.

Proposition 1.5.1. Let R 1 and R 2 be commutative rings with identity. Then the set R 1 R 2 := {(x1, x 2) | x i R i , i = 1 , 2} is a commutative ring with identity under thecoordinatewise operations:

(i) the zero element is (0, 0);

(ii) (a 1, a 2) := ( a 1, a 2);(iii) (a 1, a 2) + ( b1, b2) := ( a 1 + b1, a 2 + b2);

(iv) the identity element is (1, 1);

(v) (a 1, a 2)

(b1, b2) := ( a 1

b1, a 2,

b2).

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Proof. Trivial.

We usually denote this ring by R 1R 2 (or sometimes just R 1 R 2).For an example, take RR . This is a commutative ring with identity. Considered justas an additive group it is isomorphic to C ; but as rings they are very different, C has nozero-divisors, but every R 1R 2 has.

1.6 Polynomial Rings

This is another recipe to make new rings from old. Let R be a commutative ring withidentity. A polynomial over R in the indeterminate X is a formal expression of the form

a 0 + a 1X + ... + a n X n

where a 0, a 1, . . . ,a n R . The elements a 0, a 1, . . . ,a n are called coefficients of the polynomial.If p(X ) = a 0 + a 1X + ... + a n X n and a n = 0 we say that the degree of p(X ) is n (if p(X )is the zero polynomial then the degree is not dened). We add and multiply polynomials in

the familiar way; if p(X ) =

n

i=0

a i X i , q(X ) =k

i=0

bi X i

then we dene their sum by

p(X ) + q(X ) =

i=0

(a i + bi )X i

where by convention a i = 0 if i > n and bi = 0 if i > k . We dene the product p(X )q(X ) tobe the polynomial

r (X ) =n + k

t =0

ct X t

where

ct =t

i=0

a i bt i

It is easy to see that with this operations the set of polynomial with coefficients in Rbecomes a commutative ring with identity denoted by R [X ]. We give in an appendix to thissection a more formal denition of R [X ], which has the advantage that it gets rid of themysterious indeterminate X .

We can repeat the process, and manufacture for example R [X ][Y ], which we usuallyabbreviate to R [X, Y ]. The study of real plane curves is essentially the study of this ring.

1.6.1 Power Series Rings

If R is a commutative ring with identity we can consider also innite formal expressions of the form

i=0

a i X i , a iR

Such an expression is called a power series over R . Note that here convergence is irrelevant,these are just formal expressions. Addition and multiplication are dened again in theobvious way. The set of all power series over R is a commutative ring with identity, denotedby R [[X ]].

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1.7 Important: Notation

All the rings in this course are commutative rings with identity. We will fromnow on usually just say ring. We will say subring and mean sub-(commutativering with identity) and later we will speak of ring homomorphism and meanhomomorphism of commutative rings with identity And so on.

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1.8 Appendix

Integral domains and elds

Let R be an integral domain. We consider the set of pairs:

S = {(a, b ) : a, bR, b = 0}We want to see these pairs as fractions in R . However we know from the example of Q thatdifferent fractions may represent the same number. So we dene an equivalence relation:

(a, b )(c, d ) if ad = bc

It is easy to see that is indeed reexive, symmetric and transitive. We denote the equiva-lence class of (a, b ) by ab and we consider the set

K = {ab

: (a, b )S }We dene now addition and multiplication on K .

ab

+cd

=ad + bc

bd

and ab

cd

=acbd

It is easy to check that these operations are well dened on equivalence classes. Oneveries easily that K is a commutative ring with identity, for example we dene

0 :=01 , 1 :=

11

and we check that axioms R1-R5 hold. One sees further that K is a eld as

ab

ba

= 1

One can see R as a subring of K via the identication

a =a1

Polynomials over rings

We give here a different denition of the ring of polynomials over a ring in order todemystify the unknown.

Proposition 1.8.1. Let R be a commutative ring with identity. Then the set of sequences

(a k )

k =0 : a k R, only a nite number of the entries a k non-zero

is a commutative ring with identity under the operations:

(i) the zero element is (0, 0, . . . );

(ii) (a k )k =0 := ( a k )k =0 ;7


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(iii) (a k )k =0 + ( bk )

k =0 := ( a k + bk )

k =0 ;

(iv) the identity element is (1, 0, 0, . . . );

(v) (a k )k =0

(bk )k =0 := ( r + s = k a r bs )

k =0 .

What has this got to do with polynomials in X ? Well, write X := (0 , 1, 0, . . . 0),and note that X 2 = (0 , 0, 1, 0, . . . ) and so on. With that in place we then recover all polynomials: for example, (a 0, a 1, a 2, a 3, 0, 0, . . . ) = a 0 + a 1X + a 2X 2 + a 3X 3.

With this choice of name X for (0, 1, 0, 0, . . . ) we call this new ring R [X ]. If we called (0, 1, 0, . . . ) by the name Y wed call the new ring R [Y ].

We can perform the same construction on the set of all sequences. In that case we get the power series ring denoted by R [[X ]]. Note that this is algebra, theres no question of convergence.

Proof. It is easy to check that the set is closed under the operations. The addition axioms

are trivial. For this convolution multiplication the axioms are slightly tedious to check, butnot difficult.

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Which of the following are true?

1. If a,b,c are non zero elements of a ring R then ab = ac implies that b = c.

2. If a,b,c are non zero elements of an integral domain R then ab = ac implies that b = c.3. If R is a nite integral domain and aR, a = 0 then for some n 1, a n = 1.4. If a, b are non zero elements of an integral domain R then the equation ax = b has

exactly one solution.

5. Any ring has nitely many units.

6. Every subring of a eld is a eld.

7. If R is an integral domain then R [X ] is also an integral domain.

8. Z [X ] has innitely many subrings.9. If p(X ), q(X )

Z 6[X ] and deg p(X ) = 2 , deg q(X ) = 3 then deg( p(X )q(X )) = 5.

10. If A, B are commutative rings with identity and AB then A is a subring of B .

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rings lect 1

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