Table Of ContentInformal Notes on Algebra
R. Boyer
Contents
1 Rings 2
1.1 Examples and Deflnitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 Integral Domains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2 Ring Homomorphisms 4
2.1 Basic Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2.2 Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
3 Chinese Remainder Theorem 7
3.1 Field of Quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
4 Review of Basic Number Theory 9
5 Euclidean Domains 11
5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
5.2 Prime Factorization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
5.3 Gaussian Integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
5.4 Other Examples of Euclidean Domains . . . . . . . . . . . . . . . . . . . . . . . . . . 17
6 Example of PID that is not a Euclidean Domain 18
6.1 How to show something is not a Euclidean Domain . . . . . . . . . . . . . . . . . . . 18
6.2 How to show a ring is a PID. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
7 Ring of Polynomials 19
7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
7.2 Irreducible Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
7.3 Construction of Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
8 Continuation of Polynomials 26
8.1 Irreducible Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
8.2 Existence of Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
9 Introduction to Finite Fields 27
1
1 RINGS 2
10 Formal Derivatives 30
11 Extension Fields 31
12 Iterated Field Extensions 31
13 Splitting Fields 32
14 Galois Group 34
15 More Field Extension Results 40
16 Discussion Questions about Fields 45
17 Galois Correspondence 47
18 Galois Theory for Cubic Polynomials 50
18.1 Solving the Cubic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
19 Galois Theory of Quartic Polynomials 52
20 Solvability of Polynomials by Radicals 54
20.1 Insolvability of the Quintic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
21 Symmetric Functions 57
22 Other Descriptions of Galois Theory 57
Abstract
These are informal notes taken from a variety of sources on basic ring theory and Galois
theory.
1 Rings
1.1 Examples and Deflnitions
Deflnition 1.1. A non-empty set R with two binary operators, written as addition and multiplica-
tion, is a ring if it satisfles:
(1) R is an abelian group under addition;
(2) (Closure) if a;b R, then ab R;
2 2
(3) (Associativity) if a;b;c R, then (ab)c=a(bc);
2
(4) (Distributivity) if a;b;c R, then a(b+c)=ab+ac and (b+c)a=ba+ca.
2
Example 1.1. The integers, rational numbers, real numbers, and complex numbers are all rings
under the usual operations.
Example 1.2. The integers modulo n is a ring.
1 RINGS 3
Example 1.3. Matrices whose entries come from any of the previous examples are rings.
Deflnition 1.2. If R ;R ;:::;R are all rings, then their direct product R R is a ring
1 2 n 1 n
£¢¢¢£
under componentwise addition and multiplication.
Proposition 1.1. Let a;b R where R is a ring. Then:
2
(1) a0=0a=0;
(2) ( a)b=a( b)= ab;
¡ ¡ ¡
(3) ( a)( b)=ab;
¡ ¡
(4) (n a)(m b)=nm (ab) for all integers m;n.
¢ ¢ ¢
Deflnition 1.3. Let R be a ring. Then
(1) R is a commutative ring if ab=ba for all a;b R;
2
(2) R is a ring with unity if there is an element denoted by 1 R such that 1a = a1 = a for all
2
a R.
2
Deflnition 1.4. A nonempty subset S of a ring R is a subring of R if for all a;b S, we have
2
ab S and a b S.
2 ¡ 2
Observation 1.1. A subring S of a ring R is a ring.
Example 1.4. Consider the ring Z[i] = a+bi : a;b Z , called the ring of Gaussian integers.
f 2 g
Note: Z[i] is a subring of the complex numbers C.
Example 1.5. The set Q(p2)= a+bp2:a;b Q is a ring with the usual operations.
f 2 g
1.2 Integral Domains
Deflnition 1.5. If a;b R 0 where R is a ring, then we call a and b zero divisors if ab=0.
2 nf g
Proposition 1.2. A nonzero element a Zn is a zero divisors if and only if a is not relatively
2
prime to n.
Corollary 1.1. Zp has no zero divisors if and only if p is prime.
Deflnition 1.6. A ring R is called an integral domain if
(1) R is a commutative ring;
(2) R has a unity;
(3) R has no zero divisors.
Example 1.6. Z;Zp;Q;R are all integral domains; so are the Gaussian integers Z[i], and Q(p2).
Example 1.7. Z Z is not an integral domains.
£
1.3 Fields
Deflnition 1.7. In a ring with unity 1, an element a R is called a unit if a has a multiplicative
2
inverse in R.
2 RING HOMOMORPHISMS 4
Proposition 1.3. Let R be a (commutative) ring with unity 1. Let
U(R)= a R:a is a unit in R (1)
f 2 g
the set of all units in R. Then U(R) is a group under multiplication of R.
Proposition 1.4. In Zn, we flnd U(Zn) = U(n), the group of all positive integers which are both
less than n and relatively prime to n under multiplication.
Deflnition 1.8. A ring R is called a fleld if
(1) R is commutative; (2) R has a unity 1; (3) Every nonzero element in R is a unit.
Example 1.8. Q;R;C are all flelds.
Proposition 1.5. Every fleld is an integral domain.
Proposition 1.6. Every flnite integral domain is a fleld.
Corollary 1.2. Zp is a fleld if and only if p is prime.
Example 1.9. Q(p2) is a fleld.
Example 1.10. Z3[i] and Z7[i] are flnite flelds. In general, we will show below that Z[i]= < p >,
where p Z is prime, is a fleld if and only if p 3mod4. The key condition is whether p can be
2 ·
represented as the sum of two squares.
Deflnition 1.9. A ring R with unity 1 such that every nonzero element a R is a unit is called a
2
division ring.
Deflnition 1.10. In a ring R, the characteristic of R, denoted char R, is the least positive integer
such that n a=0 for all a R. If no such n exists, we say the characteristic of R is 0.
¢ 2
Proposition 1.7. Let R be a ring with unity 1. Then
1. char R=0 if 1 has inflnite order under addition;
2. char R=n if 1 has order n under addition.
Proposition 1.8. Let D be an integral domain. Then either its characteristic is 0 or is a prime p.
2 Ring Homomorphisms
2.1 Basic Properties
Proposition 2.1. Let `:R R be a homomorphism between two rings R and R . Then:
0 0
!
1. `(0)=0;
2. `( a)= `(a);
¡ ¡
3. `(na)=n`(a);
2 RING HOMOMORPHISMS 5
4. ` is injective if and only if Ker(`)= 0 ;
f g
5. `(a)n =`(an), for all n>0;
6. Ker(`) is a subring of R.
Example 2.1. Consider the polynomial equation 2x3 5x2+7x 8=0. Claim: this equation has
¡ ¡
no integer solutions.
We argue by contradiction. Let ` : Z Z3 be the usual map x xmod3. Suppose this equation
! 7!
does have an integral solution, say a. Then: 0=`(a)=2a3 5a2+7a 8. Note: 5 7 8
¡ ¡ ¡ · ·¡ ·
1mod3. In other words,
2`(a)3 `(a)2+7`(a) 8=2`(a)3+`(a)2+`(a)+1: (2)
¡ ¡
So, if the original equation has a solution a then there must be a solution to 2b3+b2+b+1=0 for
some b Z3. By exhaustive checking, we flnd there is no such element b.
2
Observation: Let ` be a homomorphism between two rings, say R and R . Set a = `(1). Then
1 2
it is easy to check that a2 =a. One can use this to show that the only ring homomorphisms of Zn
into itself is either the zero homomorphism or the identity.
2.2 Ideals
Deflnition 2.1. Let R be a ring and I a non-empty subset of R. Then I is an ideal of R if
1. I is a subring of R;
2. For all r R and x I, we have rx I and xr I.
2 2 2 2
Proposition 2.2. Let ` : R R be a homomorphism between two rings. Then its kernel is an
0
!
ideal of R.
Deflnition 2.2. Let R be a commutative ring and let a R. Then the principal ideal generated by
2
a, denoted by <a>, is the set ra:r R .
f 2 g
Proposition 2.3. Every ideal of the ring of integers Z is principal.
Proposition 2.4. Let R be a commutative ring with unity. Then R is a fleld if and only if 0 and
f g
R are the only ideals in R.
Proposition 2.5. Let I be an ideal of R. Then the quotient R=I is a ring with multiplication
(a+I)(b+I)=ab+I: (3)
Theorem 2.1. First Isomorphism Theorem Let ` : R R be a surjective homomorphism
0
!
between two rings. Then R=I =R, where I is the kernel of `.
» 0
Deflnition 2.3. A nontrivial proper ideal I of R in a commutative ring R is called a prime ideal if
ab I implies either a I or b I for all a;b R.
2 2 2 2
2 RING HOMOMORPHISMS 6
Deflnition 2.4. A nontrivial proper ideal I of R in a ring R is called a maximal ideal if the only
ideals J in R such that I J R are either I or R.
(cid:181) (cid:181)
Example 2.2. Let R be the ring of integers. Let U be an ideal of R.
CLAIM: U is maximal if and only if U =<p>, where p is prime.
Example 2.3. Let R be the ring of all continuous functions on the unit interval [0;1]. Let M be
the ideal of all continuous functions that vanish at the flxed point, say x [0;1].
0
2
CLAIM: M is a maximal ideal of R.
Proposition 2.6. Let R be a commutative ring with unity, and let I be an ideal in R. Then
1. I is a prime ideal if and only if R=I is an integral domain
2. I is a maximal ideal if and only if R=I is a fleld.
Proof. (1) Suppose R=I is an integral domain and ab I. Then (a+I)(b+I)=ab+I =I, which
2
is the zero in the quotient ring. Hence, either a+I or b+I must equal I. In other words, either
a I or b I, which is the condition for a prime ideal.
2 2
Next assume I is a prime ideal. Consider (a+I)(b+I) = I in the quotient R=I. Then ab I so
2
either a I or b I. That is, either a+I or b+I must be I.
2 2
(2) Suppose R=I is a fleld and J is an ideal of R that properly contains I. Choose b J I. Then
2 n
b+I must be a non-zero element of R=I. Hence, there is an element c R so (b+I)(c+I)=1+I.
2
Note that 1 bc I. On the other hand, bc J since b J. We flnd 1 J which implies J =R.
¡ 2 2 2 2
Suppose I is a maximal ideal and b R I. Then we need to show that b+I has a multiplicative
2 n
inverse. Consider J = br+a:r R;a I , which is an ideal that contains both a and I. Then J
f 2 2 g
is an ideal of R that properly contains I. Since I is maximal, we flnd J = R. In particular, 1 J
2
so we may flnd r R and a A so 1=br +a. Hence (b+I)(r +I)=1+I.
0 0 0 0 0
2 2
Corollary 2.1. In a commutative ring R with unity, every maximal ideal is prime.
Proof. Let I be a maximal ideal. Then R=I is a fleld; in particular, it is an integral domain.
Example 2.4. Later, we will use maximal ideals of polynomial rings to construct flelds. Informally,
consider the quotient of A = R[x]= < x2 +1 > which will be isomorphic to the fleld of complex
numbers.
Let g(x) R[x]. Then the coset g(x)+ < x2+1 > can be represented as a1x+a0+ < x2+1 > by
2
division of polynomials. Further x2 = 1 in A (verify!).
¡
We can flnd the multiplicative inverse, say b x+b +<x2+1>, of a non-zero element a x+a +<
1 0 1 0
x2+1>. Set
(a x+a +<x2+1>)(b x+b +<x2+1>)=1+<x2+1>: (4)
1 0 1 0
Then
a b a b =1; a b +a b : (5)
0 0 1 1 1 0 0 1
¡
To solve for a ;a , we consider two cases: either b =0 or b =0. In both cases, we will flnd that
0 1 0 1
6 6
b b
0 1
a = ; a = ¡ : (6)
0 b2+b2 1 b2+b2
0 1 0 1
Further, we observe that the quotient ring is isomorphic to the fleld of complex numbers.
3 CHINESE REMAINDER THEOREM 7
Example 2.5. Consider A = Q[x]= < x2 2 >. Then in the quotient x2 = 2. The cosets can
¡
be represented as b x + b + < x2 2 >. Again we may check that every non-zero coset has a
1 0
¡
multiplicative inverse. In this calculation, we need to use that p2 is irrational. It is interesting to
write out the isomorphism between A and the fleld Q(p2).
Example 2.6. Let A = Z[i]= < 2 i >. Note: in A we flnd 2 = i; more precisely 2+ < 2 i >=
¡ ¡
i+<2 i>. Hence, every coset representative can be written as a+<2 i> where a Z. In fact,
¡ ¡ 2
there are further restrictions since 22+ < 2 i >= i2+ < 2 i >= 1+ < 2 i > in A. Hence,
¡ ¡ ¡ ¡
there are only flve distinct cosets <2 i>;1+<2 i>;2+<2 i>;3+<2 i>;4+<2 i>
¡ ¡ ¡ ¡ ¡
(verify!). In fact, one can show that A is isomorphic to Z5.
Deflnition 2.5. An integral domain R is called a principal ideal domain or a PID if every ideal of
R has the form <a>.
We saw that the ring Z is a PID.
Proposition 2.7. In a PID R every prime ideal is maximal.
Proof. Let <p> be a non-zero prime ideal in R. Let I =<m> be any ideal that contains <p>.
We must show either I =<p> or I =R. Now p <m> so p=rm for some element r R. Since
2 2
<p> is prime and rm <p> either r <p> or m <p>. When m <p>, the ideal I agrees
2 2 2 2
with<p>. Whenr <p>,writer =sp,wheres R,sop=spm. SinceRisanintegraldomain,
2 2
we may cancel out the common factor of p to obtain 1=sm; that is, m is invertible so I =R.
3 Chinese Remainder Theorem
Thereisageneralizationtoarbitrarycommutativeringswithunityoftheconceptofrelativelyprime
integers m and n. In Z this is equivalent to being able to solve the equation mx+ny =1. This in
turn is equivalent to nZ+mZ=Z as ideals.
We shall call two ideals I and J of a ring R comaximal if A+B =R.
Recall that the product AB of two ideals is the ideal that consists of all flnite sums of the form
a b where a A and b B. Moreover, when A and B are principal ideals, say A=<a> and
j j j 2 2
B =<b> we flnd AB =<ab>.
P
Proposition 3.1. (Chinese Remainder Theorem) Let A ;A ;:::;A be ideals in R. Consider the
1 2 k
mapping
R R=A R=A R=A by r (r+A ;r+A ;:::;r+A ) (7)
1 2 k 1 2 k
! £ £¢£ 7!
is a ring homomorphism with kernel A A ::: A . If for each i;j 1;2;:::;k with i=j the
1 2 k
\ \ \ 2f g 6
ideals A and A are comaximal, then the map is surjective and A A ::: A =A A :::A .
i j 1 2 k 1 2 k
\ \ \
Hence, we have the natural isomorphism
R=(A A :::A )=R=(A A ::: A )=R=A R=A R=A : (8)
1 2 k 1\ 2\ \ k » 1£ 2£¢¢¢ k
Proof. We flrst show the case when k =2. Consider the map `:R R=A R=A deflned by
1 2
! £
`(r)=(r modA ;r modA ): (9)
1 2
3 CHINESE REMAINDER THEOREM 8
Thismapisaringhomomorphismsincer r modA isjustanalternativenotationforthenatural
1
7!
projectionofaringontoitsquotient. Furthermore,thekernelof`mustconsistofallelementsr R
2
such that r A and r A ; that is, r A A . Note: all this holds without any restrictions on
1 2 1 2
2 2 2 \
the ideals A and A .
1 2
To complete the proof, we now impose the condition A and A are also comaximal. We must
1 2
establish that (1) ` is surjective, and (2) A A =A A .
1 2 1 2
\
TheconditionA andA arecomaximalforcesA +A =R. Inparticular,theremustexistelements
1 2 1 2
x A and y A such that x+y =1. So `(x)=(0;1) and `(y)=(1;0) (verify).
1 2
2 2
Now let r = (r modA ;r modA ) be an arbitrary element of the product R=A R=A . We
1 1 2 2 1 2
£
claim that the element r x+r y is mapped to r. Consider:
2 1
`(r x+r y) = `(r )`(x)+`(r )`(y) (10)
2 1 2 1
= (r modA ;r modA )(0;1)+(r modA ;r modA )(1;0) (11)
2 1 2 2 1 1 1 2
= (0;r modA )+(r modA ;0) (12)
2 2 1 1
= (r modA ;r modA ): (13)
1 1 2 2
Hence, the ring homomorphism ` is surjective.
ItremainstoshowthatA A =A A . Now,theidealA A isalwayscontainedintheintersection
1 2 1 2 1 2
\
A A . If A and A are comaximal and x A and y A are chosen as above, then for any
1 2 1 2 1 2
\ 2 2
c A A , we have
1 2
2 \
c=c1=cx+cy A A : (14)
1 2
2
Hence A A A A .
1 2 1 2
\ ‰
The general case follows by induction from the case of two ideals by using A = A and B =
1
A A A once we know that A and A A A are comaximal.
2 3 k 1 2 3 k
¢¢¢ ¢¢¢
Corollary 3.1. Suppose that a and b are relatively prime integers. Let fi;fl Z. Then there exists
2
an integer x such that x fi(moda); x fl(modb).
· ·
Proof. Let a and b be relatively prime integers so Z =< a > + < b > so Z=(< a > < b >) is
\
ring isomorphic to Z= < a > Z= < b >. Hence, given any elements of the rings Z= < a >, say
'
fi+ < a >, and Z= < b >, say fl+ < b >, there must exist an element x Z that is mapped to
2
(fi+<a>;fl+<b>, by the homomorphism of the proposition.
Observation We can also rephrase these results as a structure theorem about the ring Zm. As
preparation, suppose m=pq where p and q are distinct primes. Then Zm has ideals I =<p> and
J =<q > with zero intersection. Hence Zm is isomorphic to the direct product of Zp and Zq.
3.1 Field of Quotients
Let D be an integral domain. Then there exists a fleld F consisting of elements written as a=b,
wherea;b D withb=0. Moreover,wecanidentifyeveryelementa D withtheelementa=1 F
2 6 2 2
so D becomes a subring of F. Every element has the form a=b=ab 1 where a;b D with b=0.
¡
2 6
Any fleld with these properties is called the fleld of quotients of D. Further, any two such flelds are
isomorphic.
The explicit construction was outlined in class.
4 REVIEW OF BASIC NUMBER THEORY 9
4 Review of Basic Number Theory
Deflnition 4.1. We say that c Z+ is the greatest common divisor of integers a and b if:
2
1. ca and cb,
j j
2. any common divisor of a and b is a divisor of c.
Observation 4.1. the greatest common divisor is unique, if it exists.
Proposition 4.1. 1. If a;b Z are not both zero, then their greatest common divisor exists.
2
2. The greatest common divisor may be written in the form: m a+n b.
0 0
Proof: Let S be the set: S = ax+by :x;y Z : Then S must contain a positive integer (verify!)
f 2 g
Claim: GCD(a;b)=c=min ax+by >0:x;y Z :
f 2 g
Now, any common divisor – of a and b must divide z =ax+by. In particular, – c.
j
Next, we must show that ca and cb. This will follow by showing that cz, or c(ax+by).
j j j j
Now, z =qc+r, where 0 r <c. That is, r =z qc=ax+by qc.
• ¡ ¡
Hence, r S and 0 r <c. We obtain a contradiction to the minimality of c unless r =0.
2 •
We conclude that cz. In particular, ca and cb for proper choices of x and y.
j j j
Deflnition 4.2. We call a and b relatively prime if GCD(a;b)=1.
Corollary 4.1. GCD(a;b)=1 if and only if 1=ax+by for some choices of x and y.
Deflnition 4.3. Call p>1 prime if its only positive divisors are 1 and p.
Proposition 4.2. If GCD(a;b)=1 and abc, then ac.
j j
Proof: Write 1=ax+by so c=acx+bcy. Now, abcy and aacx, hence ac.
j j j
Corollary 4.2. If p is a prime and divides a product of integers, then it must divide at least one of
them.
Theorem 4.1. Any positive integer a>1 is a unique product
a=pfi1pfi2 pfik; (15)
1 2 ¢¢¢ k
where p >p >::: are prime and each fi >0.
1 2 i
Proof. (Existence) We use induction. The result holds for a = 2. We now assume the result holds
for all integers less than a.
Now, either a is either prime so the result holds or a=bc, where 1<b;c<a.
By induction b and c are products of primes. Hence, so is a itself.
(Uniqueness) Consider a=pfi1pfi2 pfik =qfl1qfl2 qfl‘ where p >p >::: and q >q >::: are
1 2 ¢¢¢ k 1 2 ¢¢¢ k 1 2 1 2
prime and their exponents are only positive.
Claim: k =‘, p =q and fi =fl for all i.
i i i i
We use induction. The result holds for a=2. We assume the result for all integers less than a.
Since fi >0, we flnd p a so
1 1
j
p qfl1qfl2 qfl‘: (16)
1j 1 2 ¢¢¢ k
4 REVIEW OF BASIC NUMBER THEORY 10
In particular, p q for some i, since p is prime. But q >q =p .
1 i 1 1 i 1
j
On the other hand, q a implies q p . As before, p p q .
1 1 j 1 j 1
j j ‚ ‚
Hence, q p and p q implies p =q .
1 1 1 1 1 1
‚ ‚
Without loss of generality, assume fi b . We cancell out one factor of p . Then
1 1 1
‚
a
p =p1fi1¡1p2fi2¢¢¢pfikk =pfl11¡1q2fl2¢¢¢qkfl‘: (17)
1
But induction, p =q , k =‘ and fi =b for all i.
i i i i
Proposition 4.3. Let p be a prime. Then for any integer a, we have ap a(modp). Moreover, if
·
p does not divide a, we have ap 1 1(modp).
¡
·
Proof. (1) Without loss of generality, we can assume that a is positive. Now, we establish the result
using induction on a. The equivalence clearly holds if a = 1. Assume the result for a. We need to
establishequivalencefora+1. Consider: (a+1)p =1+pa+p(p 1)=2a2+ +ap,bythebinomial
¡ ¢¢¢
theorem. All the intermediate terms are divisible by p so are 0 under congruence by p. That is,
(a+1)p (1+ap)(modp). By induction, ap a(modp). We flnd (a+1)p (a+1)(modp).
· · ·
(2) If p does not divide a, then a is relatively prime to p. In other words, a has a multiplicative
inverse modulo p. Multiply both sides of the identity in part (1) by this inverse to obtain (2).
Observation: The result that ap 1 1(modp) where p is a prime and p does not divide a is called
¡
·
Fermat’s Little Theorem.
Proposition 4.4. ChineseRemainderTheorem Supposethataandbarerelativelyprimeintegers.
Let fi;fl Z. Then there exists an integer x such that
2
x fi(moda); x fl(modb): (18)
· ·
Proof. Weflrstindicatesomereductions. Nowitisenoughtoshowthatthereareintegersm;nsuch
that
fi+ma=fl+nb (19)
since x fi(moda); x fl(modb) is equivalent to fi(moda) fl(modb) which itself is equivalent
· · ·
to the existence of integers m;n such that fi+ma=fl+nb.
To flnd these integers m;n, it is enough to flnd other integers s;t such that as+bt = fi fl since
¡
fi+ma=fl+nb can be written as
fi fl =nb ma: (20)
¡ ¡
Finally, wecanalsoflndsolutionstotheidentityfi fl =nb masinceaandbarerelativelyprime.
¡ ¡
That is, we can flnd integers n and m so that
0 0
1=n b m a: (21)
0 0
¡
We can simply multiply this last equation by fi fl to obtain the desired solution.
¡
