Intro Abstract Algebra - Math User Home Pages

200 Pages·2001·1.19 MB·English

Intro Abstract Algebra
c1997-8, Paul Garrett, [email protected]
(cid:13)
http://www.math.umn.edu/~garrett/
1
Contents
(1)Basic Algebra of Polynomials
(2)Induction and the Well-ordering Principle
(3)Sets
(4)Some counting principles
(5)The Integers
(6)Unique factorization into primes
(7)(*) Prime Numbers
(8)Sun Ze’s Theorem
(9)Good algorithm for exponentiation
(10)Fermat’s Little Theorem
(11)Euler’s Theorem, Primitive Roots, Exponents, Roots
(12)(*) Public-Key Ciphers
(13)(*) Pseudoprimes and Primality Tests
(14)Vectors and matrices
(15)Motions in two and three dimensions
(16)Permutationsand Symmetric Groups
(17)Groups: Lagrange’sTheorem, Euler’s Theorem
(18)Rings and Fields: de(cid:12)nitions and (cid:12)rst examples
(19)Cyclotomic polynomials
(20)Primitive roots
(21)Group Homomorphisms
(22)Cyclic Groups
(23)(*) Carmichael numbers and witnesses
(24)More on groups
(25)Finite (cid:12)elds
(26)Linear Congruences
(27)Systems of Linear Congruences
(28)Abstract Sun Ze Theorem
(29)(*) The Hamiltonian Quaternions
(30)More about rings
(31)Tables
2
1. Basic Algebra of Polynomials
Completing the square tosolvea quadraticequationis perhapsthe (cid:12)rst reallygoodtrickinelemen-
tary algebra. It depends upon appreciating the form of the square of the binomial x+y:
2 2 2 2 2
(x+y) =x +xy+yx+y =x +2xy+y
Thus, running this backwards,
2 2 a 2 a a 2 a 2
x +ax=x +2( )x=x +2( )x+( ) ( )
2 2 2 (cid:0) 2
a 2 a 2
=(x+ ) ( )
2 (cid:0) 2
Then for a=0,
6 2
ax +bx+c=0
can be rewritten as
0 2 b c b 2 c b 2
0= =x +2 x+ =(x+ ) + ( )
a 2a a 2a a (cid:0) 2a
Thus,
b 2 b 2 c
(x+ ) =( )
2a 2a (cid:0) a
b b c
2
x+ = ( )
2a (cid:6) 2a (cid:0) a
r
b b c
2
x= ( )
(cid:0)2a (cid:6) 2a (cid:0) a
r
from which the usual Quadratic Formula is easily obtained.
For positive integers n, we have the factorial function de(cid:12)ned:
n!=1 2 3 ::: (n 2) (n 1) n
(cid:1) (cid:1) (cid:1) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
Also, we take 0!=1. The fundamental property is that
(n+1)!=(n+1) n!
(cid:1)
And there is the separate de(cid:12)nition that 0! = 1. The latter convention has the virtue that it works out in
practice, in the patterns in which factorials are most often used.
The binomial coe(cid:14)cients are numbers with a special notation
n n!
=
k k!(n k)!
(cid:18) (cid:19) (cid:0)
Thenamecomesfromthefactthatthesenumbersappearinthebinomialexpansion(expansionofpowers
of the binomial (x+y)):
n
(x+y) =
n n n(cid:0)1 n n(cid:0)2 2 n 2 n(cid:0)2 n n(cid:0)1 n
x + x y+ x y +:::+ x y + xy +y
1 2 n 2 n 1
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:0) (cid:19) (cid:18) (cid:0) (cid:19)
3
n n(cid:0)i i
= x y
i
0(cid:20)i(cid:20)n (cid:18) (cid:19)
X
Notice that
n n
= =1
n 0
(cid:18) (cid:19) (cid:18) (cid:19)
Therearestandard identitieswhichareusefulinanticipatingfactorizationofspecialpolynomialsand
special forms of numbers:
2 2
x y =(x y)(x+y)
(cid:0) (cid:0)
3 3 2 2 3 3 2 2
x y =(x y)(x +xy+y ) x +y =(x+y)(x xy+y )
(cid:0) (cid:0) (cid:0)
4 4 3 2 2 3
x y =(x y)(x +x y+xy +y )
(cid:0) (cid:0)
5 5 4 3 2 2 3 4
x y =(x y)(x +x y+x y +xy +y )
(cid:0) (cid:0)
5 5 4 3 2 2 3 4
x +y =(x+y)(x x y+x y xy +y )
(cid:0) (cid:0)
and so on. Note that for odd exponents there are two identities while for even exponents there is just one.
6 6
#1.1 Factor x y in two di(cid:11)erent ways.
(cid:0)
2 2 3 3 3 3
#1.2While we mostly know that x y has a factorization, that x y has a factorization, that x +y
(cid:0) (cid:0) 4 4
has,and soon,thereisafactorizationthatseldomappearsin‘highschool’: x +4y hasafactorizationinto
two quadratic pieces, each with 3 terms! Find this factorization. Hint:
4 4 4 2 2 4 2 2 2 2 2 2
x +4y =(x +4x y +4y ) 4x y =(x +2y ) (2xy)
(cid:0) (cid:0)
4
2. Induction and the Well-ordering Principle
The meaning of the word ‘induction’ within mathematics is very di(cid:11)erent from the colloquial sense!
First,letP(n)be astatementinvolvingtheintegern,whichmaybetrueorfalse. That is,atthispoint
we have a grammatically correct sentence, but are making no general claims about whether the sentence is
true, true for one particular value of n, true for all values of n, or anything. It’s just a sentence.
Now we introduce some notation that is entirely compatible with our notion of function, even if the
present usage is a little surprising. If the sentence P(n) is true of a particular integer n, write
P(n)= true
and if the sentence asserts a false thing for a particular n, write
P(n)= false
That is, we can view P as a function, but instead of producing numbers as output it produces either ‘true’
or ‘false’ as values. Such functions are called boolean.
This style of writing, even if it is not what you alreadyknew or learned, is entirely parallel to ordinary
English, is parallel to programming language usage, and has many other virtues.
Caution: Thereis ananother, oldertradition ofnotationin mathematics whichis somewhat di(cid:11)erent,
whichisandwhichishardertoreadandwriteunlessyouknowthetrick, sinceitisnotlikeordinaryEnglish
at all. In that other tradition, to write ‘P(n)’ is to assert that the sentence ‘P(n)’ is true. In the other
tradition, to say that the sentence is false you write ‘ P(n)’ or ‘ P(n)’.
: (cid:24)
So, yes, these two ways of writing are not compatible with each other. Too bad. We need to make a
choice, though, and while I once would havechosen what I call the ‘older’ tradition, now I like the (cid:12)rst way
better, for several reasons. In any case, you should be alert to the possibility that other people may choose
one or the other of these writing styles, and you have to (cid:12)gure it out from context!
Principle of Induction
If P(1)=true, and
(cid:15)
if P(n)=true implies P(n+1)=true for every positive integer n,
(cid:15)
then P(n)=true for every positive integer n.
(cid:15)
Caution: The second condition does not directly assert that P(n) = true, nor does it directly assert
that P(n+1)=true. Rather, it only asserts a relative thing. That is, more generally, with some sentences
A and B (involving n or not), an assertion of the sort
(A implies B)=true
does not assert that A=true nor that B =true, but rather can be re-written as conditional assertion
if (A=true) then B =true
In other words we prove that an implication is true.
5
That is, pushing this notation style a little further, we usually prove
(A=true) implies B =true)=true
In the more traditional notation, the assertion of Mathematical Induction is
If P(1), and
(cid:15)
if P(n) implies P(n+1) for every positive integer n,
(cid:15)
then P(n) forevery positive integern. Even though I am accustomed to this style of writing, in the end
(cid:15)
I think it is less clear!
Another Caution: Whateverthe notationwe use, the statements abovedo not indicate the waythat
we usually go about proving something by induction. Rather, what we use is Practical Paraphrase of
‘Principle of Induction’:
First, prove P(1)=true.
(cid:15)
Second, assume P(n)=true and using this prove P(n+1)=true (for every positive integer n)
(cid:15)
Then conclude P(n)=true for every positive integer n.
(cid:15)
Theseconditeminthisprocedureiswhatisusuallycalledthe induction step. Ourparaphrasemakes
it look a little di(cid:11)erent than the more o(cid:14)cial version: in the o(cid:14)cial version, it looks like we have to prove
thatanimplicationiscorrect,whereasbycontrastinourmodi(cid:12)edversionweinsteadassumesomethingtrue
and see if we can then prove something else.
The most popular traditional example is to prove by induction that
1
1+2+3+4+:::+(n 2)+(n 1)+n= n(n+1)
(cid:0) (cid:0) 2
Let P(n) = true be the assertion that this formula holds for a particular integer n. So the assertion
P(1)=true is just the assertion that
1
1= 1(1+1)
2
which is indeed true. To do the induction step, we assume that
1
1+2+3+4+:::+(n 2)+(n 1)+n= n(n+1)
(cid:0) (cid:0) 2
is true and try to prove from it that
1
1+2+3+4+:::+(n 2)+(n 1)+n+(n+1)= (n+1)((n+1)+1)
(cid:0) (cid:0) 2
is true.
Well, if we add n+1 to both sides of the assumed equality
1
1+2+3+4+:::+(n 2)+(n 1)+n= n(n+1)
(cid:0) (cid:0) 2
then we have
1
1+2+3+4+:::+(n 2)+(n 1)+n+(n+1)= n(n+1)+(n+1)
(cid:0) (cid:0) 2
6
The left-hand side is just what we want,but the righthand side is not. But we hope that it secretlyis what
we want; that is, we hope that
1 1
n(n+1)+(n+1)= (n+1)((n+1)+1)
2 2
We have to check that this is true.
This raises an auxiliary question, which is easy enough to answer once we make it explicit: how would
a person go about proving that two polynomials are equal? The answer is that both of them should be
simpli(cid:12)ed and rearranged in descending (or ascending) powers of the variable, and then check that corre-
sponding coe(cid:14)cients are equal. (And this description de(cid:12)nitely presumes that we have polynomials in just
one variable.)
In the present example it’s not very hard to do this rearranging: (cid:12)rst, one side of the desired equality
simpli(cid:12)es and rearrangesto
1 1 2 1 1 2 3
n(n+1)+(n+1)= n + n+n+1= n + n+1
2 2 2 2 2
On the other hand, the other side of the desired equality simpli(cid:12)es and rearrangesto
Or we can try to be a little lucky and just directly rearrange one side of the desired equality of poly-
nomials into the other: in simple situations this works, and if you have some luck, but is not the general
approach. Still, we can manage it in this example:
1 1 1
n(n+1)+(n+1)=( n+1)(n+1)= (n+2)(n+1)=
2 2 2
1 1
= (n+1)(n+2)= (n+1)((n+1)+1)
2 2
Thus, we can conclude that if
1
1+2+3+4+:::+(n 2)+(n 1)+n= n(n+1)
(cid:0) (cid:0) 2
then
1
1+2+3+4+:::+(n 2)+(n 1)+n+(n+1)= (n+1)((n+1)+1)
(cid:0) (cid:0) 2
which isthe implication wemust provetocomplete the induction step. Thus, weconcludethat this formula
really does hold for all positive integers n.
In this example, we used the fact that we knew what we were supposed to be getting to help us do the
elementary algebra to complete the induction step. We certainly needed to know what the right formula was
before attempting to prove it! This is typical of this sort of argument!
In some circumstances, a seemingly di(cid:11)erent proof concept works better:
Well-Ordering Principle Every non-empty subset of the positive integers has a least element.
This Well-OrderingPrinciplesoundscompletely innocuous, but it is provablylogically equivalent to the
Principle of Induction. Another logically equivalent variant is:
Let P be a property that an integer may or may not have. If P(1) = true, and if P(m) = true for all
m<n implies that P(n)=true, then P holds for all integers.
In the other notation, this would be written
7
Let P be a property that an integer may or may not have. If P(1), and if P(m) for all m<n implies
that P(n), then P holds for all integers.
#2.3 Prove by induction that
1
1+2+3+:::+n= n(n+1)
2
#2.4 Prove by induction on n that
n n(cid:0)1 n(cid:0)2 n(cid:0)3 2
x 1=(x 1)(x +x +x +:::+x +x+1)
(cid:0) (cid:0)
Hint: To do the induction step, notice that
n+1 n+1 n
x 1=x x+x 1=x(x 1)+(x 1)
(cid:0) (cid:0) (cid:0) (cid:0) (cid:0)
#2.5 Prove by induction that
2 2 2 2 1 3 1 2 1
1 +2 +3 +:::+n = n + n + n
3 2 6
#2.6 Prove by induction the following relation among binomial coe(cid:14)cients:
n n n+1
+ =
k k 1 k
(cid:18) (cid:19) (cid:18) (cid:0) (cid:19) (cid:18) (cid:19)
for integers 0<k n.
(cid:20)
#2.7 (*) Proveby induction that
2 3 3 3 3
(1+2+3+:::+n) =1 +2 +3 +:::+n
k k k k k
#2.8 (**) How would one systematically obtain the \formula" for 1 +2 +3 +:::+(n 1) +n for a
(cid:0)
(cid:12)xed positive integer exponent k?
8
3. Sets
Sets and functions
(cid:15)
Equivalence relations
(cid:15)
3.1 Sets
Herewereviewsomerelativelyelementarybutveryimportantterminologyandconceptsaboutsetsand
functions, in aslightlyabstractsetting. We usethe wordmapasasynonym for\function",as isveryoften
done.
Naively, a set is supposed to be a collection of ‘things’ (?) described by ‘listing’ them or prescribing
them by a ‘rule’. Please note that this is not a terribly precise description, but will be adequate for most of
our purposes. We can also say that a set is an unordered list of di(cid:11)erent things.
There are standard symbols for some often-used sets:
(cid:30)= = set with no elements
fg
Z= the integers
Q= the rational numbers
R= the real numbers
C= the complex numbers
A set described by a list is something like
S = 1;2;3;4;5;6;7;8
f g
which is the set of integers bigger than 0 and less than 9. This set can also be described by a rule by
S = 1;2;3;4;5;6;7;8 = x:x is an integer and 1 x 8
f g f (cid:20) (cid:20) g
This follows the general format and notation
x: x has some property
f g
If x is in a set S, then write x S or S x, and say that x is an element of S. Thus, a set is the
2 3
collection of all its elements (although this remark only explains the language). It is worth noting that the
ordering of a listing has no e(cid:11)ect on a set, and if in the listing of elements of a set an element is repeated,
this has no e(cid:11)ect. For example,
1;2;3 = 1;1;2;3 = 3;2;1 = 1;3;2;1
f g f g f g f g
9
A subsetT ofa set S is aset allof whoseelements areelements of S. Thisis written T S orS T.
(cid:26) (cid:27)
So always S S and (cid:30) S. If T S and T = (cid:30) and T = S, then T is a proper subset of S. Note that
(cid:26) (cid:26) (cid:26) 6 6
the empty set is a subset of every set. For a subset T of a set S, the complementof T (inside S) is
c
T =S T = s S :s T
(cid:0) f 2 62 g
Sets can also be elements of other sets. For example, Q;Z;R;C is the set with 4 elements, each of
f g
which is a familiar set of numbers. Or, one can check that
1;2 ; 1;3 ; 2;3
ff g f g f gg
is the set of two-element subsets of 1;2;3 .
f g
The intersectionoftwosetsA;B is the collectionofall elements which liein both sets, and isdenoted
A B. Two sets are disjoint if their intersection is (cid:30). If the intersection is not empty, then we may say
\
that the two sets meet. The union of two sets A;B is the collection of all elements which lie in one or the
other of the two sets, and is denoted A B.
[
Note that, for example, 1 = 1 , and 1 = 1 . That is, the set a with sole element a is not the
6 f g ff gg6 f g f g
same thing as the item a itself.
An ordered pair (x;y) is just that, a list of two things in which there is a (cid:12)rst thing, here x, and a
0 0 0 0
second thing, here y. Two ordered pairs (x;y) and (x;y ) are equal if and only if x=x and y =y.
The (cartesian) product of two sets A;B is the set of ordered pairs (a;b) where a A and b B.
2 2
It is denoted A B. Thus, while a;b = b;a might be thought of as an unordered pair, for ordered pairs
(cid:2) f g f g
(a;b)=(b;a) unless by chance a=b.
6
2
In case A = B, the cartesian power A B is often denoted A . More generally, for a (cid:12)xed positive
th n (cid:2)
integer n, the n cartesian power A of a set is the set of ordered n-tuples (a1;a2;:::;an) of elements ai
of A.
Some very important examples of cartesian powers are those of R or Q or C, which arise in other
2
contextsaswell: forexample, R isthecollectionoforderedpairsofrealnumbers,whichweusetodescribe
3
points in the plane. And R is the collection of ordered triples of real numbers, which we use to describe
points in three-space.
The power set of a set S is the set of subsets of S. This is sometimes denoted by S. Thus,
P
(cid:30)=
P f;g
1;2 = (cid:30); 1 ; 2 ; 1;2
Pf g f f g f g f gg
Intuitively, a function f from one set A to another set B is supposed to be a ‘rule’ which assigns to
each element a A an element b=f(a) B. This is written as
2 2
f :A B
!
although the latter notation gives no information about the nature of f in any detail.
More rigorously, but less intuitively, we can de(cid:12)ne a ‘function’ by really telling its graph: the formal
de(cid:12)nition is that a function f : A B is a subset of the product A B with the property that for every
! (cid:2)
a A there is a unique b B so that (a;b) f. Then we would write f(a)=b.
2 2 2
This formal de(cid:12)nition is worth noting at least because it should make clear that there is absolutely no
requirement that a function be described by any recognizable or simple ‘formula’.
10