Cyclotomic Polynomials

Michel 1 Chapter 0 Welcome to Cyclotomic Polynomials. The following is an expository paper aimed at teaching readers the basic concepts leading to the irreducibility of the cyclotomic polynomials over Z. Carl Fredrick Gauss (1777 – 1855), an epoch-making mathematician, discovered the cyclotomic polynomials at the young age of 19. This discovery was first published as the seventh and final section of his monumental book on number theory

Disquisitiones Arithmeticae. Gauss was able to use these concepts to show exactly which polygons could be constructed by a compass and straight edge, as well as construct a 17sided polygon which, as far as the world knows, had never been constructed before. The scope of this paper, however, will not include polygon construction.

After introducing and defining cyclotomic polynomials, we will look at some examples, and discuss the method of recursive calculation of these polynomials.

Following this are four sections on several properties of the cyclotomic polynomials which contribute to their irreducibility over Z. In the final section of this paper we will discuss and prove their irreducibility.

To understand the proofs the reader is recommended to have had a 300-level abstract algebra course or a solid background in algebra and number theory. Some proofs will only be mentioned in the index since the paper is limited on space. Now begins our journey.

Michel 2 Chapter 1 The purpose of this chapter is to introduce readers to the complex zeros of x n − 1 .

We will focus our attention on two important definitions: the nth Cyclotomic Polynomial and primitive roots. These definitions will be referred to throughout the entire paper. By the end of this chapter readers will be given a theorem which will allow them to recursively calculate the factors of x n − 1 that are irreducible over Q. Recall that an irreducible polynomial over Q is one with coefficients in Q that cannot be reduced further without reaching either real or imaginary coefficients.

Let ω = cos( 2nπ ) + i sin( 2nπ ) and φ( n ) denote the number of positive integers less than or equal to n and relatively prime to n. The following definition is quite possibly the most important as it will be employed again and again. Readers should not proceed without memorizing it. Recall that if G is the cyclic group generated by a where a ∈ G , then

G = a = { a n n ∈ Z} .

(1.0) Definition The complex zeros of x n − 1 are 1, ω,..., ω n −1 . So ω forms a cyclic group of order n under multiplication. The generators of ω of the form ωl , such that 1 ≤ l ≤ n , and

gcd(n,l) = 1, are called primitive roots of x n − 1 .

Now that we know what primitive roots are we can learn another definition.

(1.1) Definition For any positive integer n, let ω1 , ω2 ,..., ωφ( n ) denote the primitive nth roots of unity. The nth Cyclotomic Polynomial over Q is the polynomial

Φn ( x ) = ( x − ω1 )( x − ω2 ) ⋅ ⋅ ⋅ ( x − ωφ( n ) ).

Notice Φn is monic (has leading coefficient 1) and deg Φn = φ( n ) .

Michel 3 Here are a few examples:

(1.2) Example Φ1 ( x ) = x − (cos( 21π ) + i sin( 21π )) = x − 1 ; this makes sense since 1 is the only zero of x – 1.

(1.3) Example Φ2 ( x ) = x − (cos( 22π ) + i sin( 22π )) = x + 1 ; this makes sense since x 2 − 1 has zeros ± 1 , and x – 1 is the only primitive root.

Let’s look at a few more examples as we are becoming familiar with the above definitions.

(1.4) Example Let’s try to compute Φ3 ( x ) . Φ3 ( x ) = ( x − ω )( x − ω 2 ) = ( x − (cos( 23π ) + i sin( 23π )))( x − (cos( 23π ) + i sin( 23π )) 2 ) 3 2

[Notice cos( 23π ) + i sin( 23π ) = − 12 +

i]

So Φ3 ( x ) = ( x − ( − 12 +

3 2

i ))( x − ( − 12 +

3 2

i )2 )

= ( x − ( − 12 +

3 2

i ))( x − ( − 12 −

3 2

i ))

= x 2 − x( − 12 − = x 2 − x(( − 12 −

3 2

i ) − x( − 12 +

3 2

i ) + ( − 12 +

3 2

3 2

i ) + ( − 12 + i )) + 14 + (

3 4

3 2

i )( − 12 −

−

3 4

3 2

i)

)i + 34

= x 2 − x( −1 ) + 1 = x2 + x +1.

Such tedious solving of equations will soon be regarded by readers as a waste of energy.

Michel 4 Notice there is a pattern:

Φ1 ( x ) = x − 1

x − 1 = Φ1

Φ2 ( x ) = x + 1

x 2 − 1 = Φ2 ⋅ Φ1

Φ3 ( x ) = x 2 + x + 1

x 3 − 1 = Φ3 ⋅ Φ1

Φ4 ( x ) = x 2 + 1

x 4 − 1 = Φ4 ⋅ Φ2 ⋅ Φ1

•

•

•

•

•

•

So on...

So on...

Each x n − 1 splits into a product of Φk1 Φk 2 ⋅ ⋅ ⋅ where k i is always a positive divisor of n.

Here is a theorem regarding cyclic groups we will need along the way.

(1.5) Theorem Let a be an element of order n in a group and let k ∈N. Then k a = a gcd( n ,k ) and a k = gcd(nn ,k ) .

The proof of (1.5) is simple and has been left to the reader to prove. Now that (1.5) can help us prove several theorems about cyclotomic polynomials, we will start using it immediately in the proof of the following lemma. After the reader has become familiar with the following lemma as well as the proof of it, it will be used to prove a theorem about cyclotomic polynomials. Since the interdependence of proofs is observed throughout the rest of the paper, readers are suggested not to skip around. It is likely that each statement proved will be used in a proof that follows.

Michel 5 (1.6) Lemma Let t i denote all positive divisors of n. Then n

∪ { generators of ω ti } = { ω k 0 ≤ k ≤ n − 1 } . ti n

Proof: m

Call

{ generators of ∪ i

ω } = Д.

=1

⊇ : NTS If τ ∈ { ω k 0 ≤ k ≤ n − 1 } , then τ ∈Д.

If τ = ω j for some j ∈ N, 0 ≤ j ≤ n − 1 , then gcd(j,n) is a divisor of n.

Suppose gcd(j,n) = k. Then n ti

n k

is a divisor of n. Thus there exists i such that t i = nk , and

=k. n

n

By theorem (1.5), ω j = ω k = ω ti . Therefore ω j is a generator of ω ti and τ ∈ Д. ⊆ : NTS If τ ∈ Д, then τ ∈ { ω k n

Suppose τ ∈ Д. So τ ∈ ω ti

0 ≤ k ≤ 1} n

for some positive divisor t i . Therefore τ = ( ω t i )l for some

l ∈ Z, and since ω = n , τ = ω k for some k, 0 ≤ k ≤ n − 1 .

Then τ ∈ { ω k Hence Д = { ω k

0 ≤ k ≤ n − 1} .

0 ≤ k ≤ n − 1} by double inclusion. Q.E.D.

Now that we have used and observed some basic definitions about cyclotomic polynomials as well as proved a statement, readers are probably able to endure the last proof of this chapter. Using (1.5) and (1.6) we will show there is a useful and easy way to recursively calculate the rational polynomials, the factors of each x n − 1 .

Michel 6 (1.7) Theorem For every n ∈ N, x n − 1 = ∏ Φk ( x ) , where the product runs over all positive k n

divisors k of n.

Proof: It is clear that x n − 1 and

∏ Φ ( x ) both have leading coefficient 1. k

kn

(A) WTS The roots of Φ s are distinct from those of Φ t if t and s are distinct divisors of n.

Let δ be a root of Φ s . Then δ is a primitive sth root of unity. So δ = s and all roots of

Φ s are nth roots of unity of order s.

Similarly all roots of Φ t are nth roots of unity of order t.

Thus Φ s and Φ t have no roots in common since t ≠ s.

[Notice the roots of Φ k all have multiplicity 1 by (1.1).]

(B) WTS x n − 1 and

∏ Φ ( x ) have the same zeros. k

kn

Let ω = cos( 2nπ ) + i sin( 2nπ ) .

By (1.0) we know ω forms a cyclic group containing the nth roots of unity, and from (1.5) we know ω l divides n for all l, 1 ≤ l ≤ n .

So every root of x n − 1 is a root of

∏ Φ ( x ) where Φ ( x ) = ( x − ω k

k

k1

) ⋅ ⋅ ⋅ ( x − ωk φ ( k ) )

k n

( ωk is a primitive kth root of unity, i.e. ωk = k and { ωk1 ,...,ωk φ( k ) } contains all primitive kth roots of unity).

Conversely, if x − γ is a factor of Φ l for l n , then γ l = 1 by (1.0). Thus γ n = 1 since

l n.

Thus x − γ is a factor of x n − 1 , and every root of

∏ Φ ( x ) is a root of k

kn

x n −1.

Michel 7

(C) WTS deg(

∏ Φ ( x ) ) = deg( x

n

k

− 1 ).

kn

So deg(

∏ Φ ( x ) ) = deg( Φ k

kn

1

⋅ Φ n ⋅Φ n ⋅ ⋅ ⋅Φ n ) t1

t2

tm

= 1 + φ( tn1 ) + φ( tn2 ) + ⋅ ⋅ ⋅ + φ( tnm ) (where φ( i ) denotes the number of positive integers less than or equal to i, and relatively prime to i) = n by (1.6) (where it was shown that m

{ generators of ∪ i

n

ω t i } = { ω k 0 ≤ k ≤ n − 1 } ).

=1

= deg( x n − 1 ) .

Therefore both expressions have equal degree.

Hence, (A), (B), and (C) imply x n − 1 = ∏ Φk ( x ) . Q.E.D. kn

Now we have a technique for recursively calculating Φn , and this brings an end to chapter one.

Michel 8 Chapter 2 Chapter two will help readers understand techniques for inquiring about coefficients of polynomials. We will discuss the consequences of the cyclotomic polynomials being monic and irreducible in Q[x], and we will arrive at our results in such a way that we can be certain all cyclotomic polynomials have integer coefficients without having to calculate each Φ n . Examples will be given after definitions and theorems to aid readers in understanding these concepts. To begin, here is an important definition.

(2.0) Definition Let f(x) be a polynomial. We say f(x) is monic if the leading coefficient of f(x) is 1.

(2.1) Example The following polynomials are monic: • f ( x ) = x 10 − 25 x 4 + 15 x 2 − 12 x + 2 •

f ( x ) = x 2 + .7928x − π

•

The following polynomials are not monic: g( x ) = 12 x10 − x 4 + x 2 − 15

•

g( x ) = πx 2 + ex − 71 Now that we have discussed what it means for a polynomial to be monic, we are

ready for an amazing inductive proof about monic polynomials that live in Q[x]. Readers should carefully read the theorem below and brace themselves for the proof that follows.

(2.2) Theorem Let g(x) and h(x) belong to Z[x] and let h(x) be monic. If h(x) divides g(x) in Q[x], then h(x) divides g(x) in Z[x] .

Proof: Let g(x), h(x) ∈ Z[x], h(x)

g(x) in Q[x] and h(x) be monic.

Then there exists p( x ) ∈ Q[x] such that h(x)p(x) = g(x).

WTS h(x)

g(x) in Z[x] , i.e. p(x) ∈ Z[x].

Call h( x ) = x n + η n −1 x n −1 + ⋅ ⋅ ⋅ + η0 , g( x ) = ρ m x m+ n + γ m+ n −1 x m+ n −1 + ⋅ ⋅ ⋅ + γ 0 and call

p( x ) = ρ m x m + ρ m −1 x m −1 + ⋅ ⋅ ⋅ + ρ 0 . (Notice g and p have the same leading coefficient because h is monic.)

Michel 9 Also, let k m ( x ) = ( p( x ) − ρ m x m )h( x ) = g( x ) − ρ m x m h( x ) . Therefore k ( x ) ∈ Z[x] .

Define S = { j ∈ N ∪ { 0 }

ρ m− j ∈ Z}.

Part 1: ρ m ∈ Z since g( x ) ∈ Z[x] . Therefore 0 ∈ S .

Part 2: Assume j ∈ S for all j, 0 ≤ j ≤ q . WTS q +1∈ S .

So ρ m , ρ m -1 ,..., ρ m− q ∈ Z. Then k m− q ( x ) = ( p( x ) − ρm x n − ρm −1 x m−1 − ⋅ ⋅ ⋅ − ρm −q x m− q )h( x ) = ( ρ m − q −1 x m − q −1 + ⋅ ⋅ ⋅ + ρ 0 )h( x )

= ( ρm −q −1 x m−q−1 + ⋅ ⋅ ⋅ + ρ0 )x n + ( ρm− q−1 x m −q −1 + ⋅ ⋅ ⋅ + ρ0 )( h( x ) − x n )

= ( ρ m−( q+1 ) x m−( q+1 ) + ⋅ ⋅ ⋅ + ρ0 )x n + ( ρ m −( q +1 ) x m −( q +1 ) + ⋅ ⋅ ⋅ + ρ 0 )( h( x ) − x n ) = ρ m −( q +1 ) x m −( q +1 )+ n + ( ρ m −( q + 2 ) x m −( q + 2 )+ n + ⋅ ⋅ ⋅ + ρ 0 x n )

+ ( ρm−( q +1 ) x m−( q+1 ) + ⋅ ⋅ ⋅ + ρ0 )( h( x ) − x n ) ∈ Z[x] by our inductive hypothesis.

Since deg (h(x) - x n ) < n, deg(( ρ m −( q +1 ) x m −( q +1 ) + ⋅ ⋅ ⋅ + ρ0 )( h( x ) − x n )) < m − ( q + 1 ) + n by the degree rule.

Therefore ρ m −( q +1 ) x m −( q +1 )+ n is the leading term of k m − q ( x ) ∈ Z[x] and ρ m −( q +1 ) ∈ Z and

q + 1 ∈S. Therefore p( x ) ∈ Z[x] by induction and h( x ) | g(x) in Z[x]. Q.E.D. If the reader is reading this sentence then they are truly a survivor, and will be glad to know that it doesn’t stop here. We still must show that all cyclotomic polynomials are in Z[x] . Below is the highlight of chapter two.

Michel 10 (2.3) Theorem For every positive integer n, Φn ( x ) ∈ Z[x] .

Proof: Define S = { n ∈ N

Φn ∈ Z[x]}.

(A) Suppose n = 1. Obviously Φ1 ( x ) = x − 1 ∈Z[x].

Therefore 1 ∈ S .

(B) Assume k ∈ S for all k ≤ n . So Φk ( x ) ∈ Z[x] for all k ≤ n .

Then h( x ) =

∏Φ

k

( x ) ∈ Z[x]. Thus x n+1 − 1 = Φn+1 ( x )h( x ) by (1.7).

k ≤ n ,k n +1

Since h(x) is monic, and h( x )

( x n − 1 ) in Q[x] by (1.0), h( x )

( x n − 1 ) in Z[x] by

(2.2).

Hence n + 1 ∈ S , and by induction, Φn ∈ Z[x] for every n ∈ N. Q.E.D. (2.4) Example

n 1 2 3

Φ 1( x ) = x − 1 Φ 2( x ) = x +1

4

Φ 4( x ) = x2 +1

5

Φ 5 ( x ) = x 4 + x3 + x 2 + x + 1

6

Φ6( x ) = x2 − x +1

7

Φ 7 ( x ) = x6 + x5 + x 4 + x3 + x 2 + x + 1

8

Φ 8( x ) = x 4 + 1

9

Φ9( x ) = x6 + x3 +1

Φn

Φ 3( x ) = x 2 + x + 1

10 Φ 10 ( x ) = x 4 + x 3 + x 2 − x + 1 Image taken from: http://images.search.yahoo.com/ search/images?ei=UTF-8&fr=sfp&p=Gauss

Michel 11 Chapter 3 Section A

From this section forward, we will focus our attention on the irreducibility of the cyclotomic polynomials ÎŚn over Z. The following two sections will discuss definitions and theorems which contribute to the irreducibility of the cyclotomic polynomials, and in the final section we will prove their irreducibility over Z. In order to do all of this we must broaden our understanding of integral domains. Shall we begin?

Recall that an integral domain is a commutative ring with unity and no zero divisors. However, even with a restriction excluding zero divisors there still may be some non-unit elements of integral domains. Here is a general categorization of integral domain elements.

(3.0) Definition Elements x and y of an integral domain D are called associates if x = vy, where v is a unit of D. A nonzero element x of an integral domain D is called irreducible if x is not a unit and, whenever y and z are in D with x = yz, then y or z is a unit. A nonzero element x of an integral domain D is called prime if x is not a unit and x yz implies

x y or x z (Gallian 312).

Also, in an integral domain every prime is irreducible (Gallian 314). Gaussâ&#x20AC;&#x2122; last pupil, Dedekind (1831 â&#x20AC;&#x201C; 1916), was a pioneer in the theory of rings and fields, and a first discoverer of unique factorization. The idea of unique factorization will be employed in our final section, and since it follows nicely from our categorization of the elements of integral domains, here is the definition (James 341).

(3.1) Definition An integral domain D is a unique factorization domain if, a) All nonzero elements of D that are not units can be written as products of irreducibles in D, and b) Factorization into irreducibles is unique up to associates and the order in which the factors appear (Gallian 318).

(3.2) Example Notice that in the integral domain Z, 45 can be factored more than one way: 45 = ( 3)(3)( 5) and 45 = ( -3)(5)(-3 ).

Michel 12

(3.3) Example Since every nonzero integer other than 1 and -1 can be written as a product of irreducible integers, Z satisfies (3.1.a). Let u1n1 u 2n2 ⋅ ⋅ ⋅ u rnr and v1m1 v 2m2 ⋅ ⋅ ⋅ v sms be two different factorizations of some integer as a product of irreducible integers. No two u i s are associates as well as two vi s and r = s. Thus each u ini is an associate of a single

vini and Z satisfies (3.1.b).

(3.4) Example Consider 2 + i ∈ Z 3 [ i ] . Notice i( 1 + i) = 2 + i and i(2i)(i( 1 + i) = 2 + i. Thus Z 3 [ i ] is not a unique factorization domain.

Another interesting form integral domains can take is that of a principle ideal domain. This will come up again in our final chapter.

(3.5) Definition A principle ideal domain is an integral domain D in which every ideal has the form x = { rx r ∈ R } for some x in D , and x is called a principal ideal (Gallian 289).

(3.6) Example Recall that Q[x] is an integral domain because Q is a field. Choose any nonzero ideal Aideal ≤ Q[x] . If you chose an ideal containing the element x 3 + x 2 , then you must have read my mind, somehow. Notice x 3 + x 2 = x( x 2 + x ) ∈ x 2 + x ⊂ x ⊂ 1 = Q[x] , where x and x 2 + x are principle ideals.

(3.7) Example Consider Z[x] and let A denote the subset of Z[x] containing all polynomials with a constant term that is a multiple of 3. Then A = x, 3 and is not a principle ideal.

While we’re filling our arsenal with ammunition for proving the irreducibility of the cyclotomic polynomials, here is a very useful theorem about principle ideal domains.

(3.8) Theorem Let F be a field, then F[x] is a principal ideal domain.

The proof of (3.13), in the next section of this chapter, can readily be adapted to (3.8).

Michel 13

(3.9) Example For any Z p where p is prime, it follows that Z p [ x ] is a principle ideal domain.

(3.10) Example Example 3.6 is consistent with theorem 3.6; since Q is a field, Q[x] is a principle ideal domain.

Let’s recall example 3.6 again. Remember that x 2 + x ⊂ x ⊂ 1 = Q[x]. This is a strictly increasing chain of ideals, and whenever this occurs, whether or not the chain is finite depends on where the ideals come from.

(3.11) Theorem In any principal ideal domain, a strictly increasing chain of ideals must be finite in length, i.e. A1 ⊂ A2 ⊂ ⋯ ⊂ An (Gallian 319).

Proof: Let A1 ⊂ A2 ⊂ ⋅ ⋅ ⋅ be a chain of strictly increasing ideals in an integral domain D. Let A = A1 ∪ A2 ∪ ⋅ ⋅ ⋅ . WTS Aideal ≤ D .

(1) We know A ≠ ∅ since A is a union of strictly increasing chain members, and A1 ⊆ A ⊆ D .

(2) Let a ,b ∈ A . WTS a - b ∈ A .

Suppose a ∈ Am and b ∈ An ( m , n ∈ N). WLG let m > n .

Then An ⊂ Am . Therefore a ,b ∈ Am , and a − b ∈ Am by closure under addition.

Hence, since Am ⊆ A , a − b ∈ A .

(3) WTS A absorbs, i.e. ra and ar are in A whenever a ∈ A and r ∈ D .

Let a ∈ A and r ∈ D . Then a ∈ An for some n ∈ N. Since Anideal ≤ D , ar and ra are in An , and An ⊂ A .

Therefore ar and ra are in A whenever a ∈ A and r ∈ D .

Since D is a principle ideal domain there is an element a in D such that A = a .

Michel 14

Then, since a ∈ A , a ∈ Am for some m ∈ N. So for any i, Ai ⊆ A = a ⊆ Am and Am must be the last member of the chain. Q.E.D. This concludes (3.11), which brings us to the highlight of this section. Proving that the cyclotomic polynomials are irreducible over Z can be abridged by knowing that Z p [ x ] is a unique factorization domain, thus we will need the following theorem.

(3.12) Theorem Principle ideal domains are unique factorization domains (Tignol 259).

Proof: Let a 0 ∈ D and a 0 ≠ 0 (and not a unit) and D is a principle ideal domain.

WTS a 0 has at least one irreducible factor.

If a 0 is irreducible, we are done, so assume a 0 = α1 β1 where α1 ≠ 0 and neither α1 nor

β1 are units.

If a1 is irreducible we are done, so assume a1 is reducible, then α1 = α 2 β 2 where neither α 2 nor β 2 are units and α 2 ≠ 0 .

Repeating this process produces a sequence of βi , we get β1 , β 2 ,... are not units in D. Also, we obtain α0 ,α1 ,... in D, all nonzero and not units.

(1) Notice α n = α n +1 β n +1 for all n ∈ N. Thus α 0 ⊂ α1 ⊂ ⋅ ⋅ ⋅ is a strictly increasing chain of ideals by definition, and by (3.11) it is finite.

At some point there must exist an a n such that a n is irreducible and a 0 has an irreducible factor.

WTS a0 factors into irreducibles.

Say α0 = p1 β1 , where p1 is irreducible and β1 is not a unit.

If β1 is irreducible we are done, so assume β1 is reducible, then β1 = p 2 β 2 where p 2 is irreducible and β 2 is not a unit.

Michel 15 Similar to (1) we obtain α 0 ⊂ p1 ⊂ ⋅ ⋅ ⋅ which is a finite chain by (3.9).

Suppose that for all j ∈ N ∪ { 0 } , p j ⊂ p k (so the chain ends at p k ).

Then β k is irreducible and α 0 = p1 p 2 ⋅ ⋅ ⋅ p k β k where each pi is also irreducible.

Hence, every nonzero, nonunit element of a principle ideal domain is a product of irreducibles.

NTS factorization is unique up to associates and the order in which the factors appear.

Suppose a = p1 p 2 ⋅ ⋅ ⋅ p r = q1 q 2 ⋅ ⋅ ⋅ q s where p i and q j are irreducible (repetition allowed).

Define S = { n ∈ N if r = n, then s = n and, WLG p i is an associate of q i for all i, 1 ≤ i ≤ n } (may require reordering).

(2) WTS 1 ∈ S . Suppose r = 1. Call a = p1 = q1 q 2 ⋅ ⋅ ⋅ q s . So p1 ( q1 q 2 ⋅ ⋅ ⋅ q s ) .

But p1 is irreducible, and thus prime. Then there must exist some q t such that p1

qt .

WLG let t = 1. Therefore p1 = ( up1 )q 2 ⋅ ⋅ ⋅ q s . So 1 = uq 2 ⋅ ⋅ ⋅ q s .

This gives us that q 2 is a unit →← Hence s = 1 and p1 is an associate of q1 .

Therefore 1 ∈ S .

(3) Assume i ∈ S for all i, 1 ≤ i < r and a = p1 p 2 ⋅ ⋅ ⋅ p r = q1 q 2 ⋅ ⋅ ⋅ q s . s

Then we get p1

∏q

k

, which means p1 ql for some l, 1 ≤ l ≤ s , since all q k are

k =1

irreducible.

WLG assume p1 q1 . Then q1 = vp1 ( v is a unit since q1 may be an associate of p i ).

So vp1 p 2 ⋅ ⋅ ⋅ p r = vq1 q 2 ⋅ ⋅ ⋅ q s (original hypothesis). = q1 ( vq2 ) ⋅ ⋅ ⋅ q s (multiplication is commutative in D).

So vp1 = q1 , and p 2 ⋅ ⋅ ⋅ p r = ( vq2 ) ⋅ ⋅ ⋅ q s by cancellation.

Michel 16 But r – 1 < r so r = s and p i is an associate of qi for all i, 2 ≤ i ≤ s by inductive hypothesis.

Thus r = s and p i is an associate of qi for all i, 1 ≤ i ≤ s .

Thus factorization is unique up to associates and the order in which factors appear. Q.E.D.

This concludes the proof of (3.12) as well as this section.

Michel 17 Chapter 3 Section B

This section contains the last set of definitions and theorems to learn in order to show the cyclotomic polynomials are irreducible over Z. We have learned important statements about integral domains, but since cyclotomic polynomials live not only in integral domains, but in polynomial fields as well, we must learn some important statements about fields. Recall that a field is a commutative ring with unity in which every nonzero element is a unit. Becoming familiar with the content below will make us ready to conquer the proof that lies in the following and final chapter. We will start with a definition.

(3.13) Definition Let H be an extension field of F and let a ∈ H . If a is a zero of some nonzero polynomial in F[x], we say a is algebraic over F.

(3.14) Example • The irrational number π is in C, and it is also in R. Since C is an extension field of R and π is a root of the nonzero polynomial x − π in R[x], we say π is algebraic over R (any a ∈ F is algebraic over F) •

The irrational number 2 is in R and, of course, not in Q. Since R is an extension field of Q and 2 is the root of a nonzero polynomial ( x − 2 )( x + 2 ) = x 2 − 2 in Q[x] , we say 2 is algebraic over Q. Knowing what it means for an element to be algebraic will come up in the

following two theorems concerning fields.

(3.15) Theorem Let F be a field, I a nonzero ideal in F[x] , and g(x) an element of F[x] . Then A = g( x ) if and only if g(x) is a nonzero polynomial of minimal degree in A.

Proof: Let F be a field and let A be a nonzero ideal of F[x].

⇒ : Suppose A = g( x ) . NTS g( x ) ≠ 0 and g(x) has minimal degree in A. Suppose g(x) = 0 →← (A is a nonzero ideal).

Suppose α( x ) ≠ 0 and α( x ) ∈ g ( x ) .

Michel 18 Then α( x ) = f ( x ) g ( x ) where deg f ≥ 0. So deg a ≥ deg g. Hence g(x) has minimal degree.

⇐ : Suppose g(x) ≠ 0 and g(x) ∈ A has minimum degree. Since g( x ) ∈ A , g( x ) ⊆ A , so suppose κ ( x ) = A (since A is a principle ideal domain).

Then κ ( x ) = g( x )h( x ) + r ( x ) by the Division Algorithm where r(x) = 0 or deg r < deg g (choice of g(x)).

But r(x) = κ( x ) − g ( x )h( x ) ∈ A . Therefore r(x) = 0 since r(x) cannot have degree less than deg g.

Hence κ ( x ) ∈ g ( x ) and A = κ ( x ) ⊆ g ( x ) . So A = g( x ) by double inclusion.

Q.E.D. This concludes (3.15). The proof of the following theorem can be adapted from (3.15), and is left to the reader to prove.

(3.16) Theorem If a is algebraic over a field F, then there is a unique, monic, irreducible polynomial p(x) in F[x] such that p(a) = 0. We say the polynomial p(x) is minimal for a over F .

(3.17) Example • The monic, irreducible polynomial p( x ) = x − π in R[x] (where π is algebraic over R) is minimal for π over R. •

The monic, irreducible polynomial p( x ) = x 2 − 2 in Q[x] (where algebraic over Q) is minimal for

2 is

2 over Q.

Keeping in mind the idea of minimal, here is another theorem regarding polynomial division.

(3.18) Theorem Let a be algebraic over a field K, and let p(x) be minimal for a over K. If f ( x ) ∈ K [ x ] and f(a) = 0, then p( x ) f ( x ) in K[x] .

Proof: Suppose a is algebraic over F.

Michel 19 Let φ : F [ x ] → F [ a ] , defined by φ( f ( x )) = f ( a ) . Then φ is a homomorphism (Gallian 363).

Then ker φ ≠ { 0 } since ker φ = { f ( x ) ∈ F [ x ] φ( f ( x )) = 0 } and p(x) ∈ ker φ . By (3.17), since p(x) has minimum degree (because it is monic and irreducible), p( x ) ∈ F [ x ] such that ker φ = p( x ) .

So p(x) = g(x)h(x) for some h(x) in F[x]. But p is irreducible, so h(x) has a unit and, WLG g(x) = p(x) (so g( x ) = p( x ) ).

Suppose f ( x ) ∈ F [ x ] such that f(a) = 0 and f ( x ) ≠ 0 .

Then f ( x ) ∈ ker φ . So f(x) = p(x)k(x) for some k(x) in F[x].

Hence p( x )

f ( x ) in F[x]. Q.E.D.

This brings us to the next set of definitions and theorems about fields.

(3.19) Definition Let f(x) and g(x) belong to F[x]. If there does not exist a polynomial of positive degree in F[x] that divides both f(x) and g(x) , we say f(x) and g(x) are relatively prime.

(3.20) Example The following polynomials are relatively prime: • x 2 + 2 x + 1 and x 2 + 4 x + 4 (no factors in common of positive degree) • 3x and 3x + 3 (common factor of 3 does not have positive degree)

•

The following polynomials are not relatively prime: x 2 + 8 x + 15 and x 2 + 12x + 35 (common factor of x – 5 has positive degree)

Now that we know the meaning of relatively prime, we are ready to introduce the following theorem.

(3.21) Theorem Let f ( x ), g ( x ) ∈ F [ x ] , and let f(x) and g(x) be relatively prime. Then there exist polynomials h(x) and k(x) in F[x] such that h(x)f(x) + k(x)g(x) = 1.

Consider { a( x ) f ( x ) + b( x )g( x ) a( x ),b( x ) ∈ F [ x ]} ⊆ F [ x ] .

Call A = { a( x ) f ( x ) + b( x )g( x ) a( x ),b( x ) ∈ F [ x ]} . WTS Aideal ≤ F [ x ] .

Michel 20 We know A ≠ ∅ since 0 ∈ F [ x ] and 0 = 0 ⋅ f ( x ) + 0 ⋅ g( x ) ∈ A . Let c( x ),d ( x ) ∈ A .

Then c( x ) = α( x ) f ( x ) + β ( x ) g( x ) for some α( x ), β ( x ) ∈ F [ x ] , and d ( x ) = δ( x ) f ( x ) + ε( x )g ( x ) for some δ( x ), ε( x ) ∈ F [ x ] .

So c(x) – d(x) = ( α( x ) f ( x ) + β( x )g ( x )) − ( δ( x ) f ( x ) + ε( x )g ( x ))

= α( x ) f ( x ) + β ( x )g( x ) − δ( x ) f ( x ) − ε( x )g( x ) = ( α( x ) − δ( x )) f ( x ) + ( β( x ) − ε( x ))g( x ) ∈ A .

Also, if c( x ) ∈ A and ν( x ) ∈ F [ x ] , c( x )v( x ) = v( x )c( x ) = v( x )( α( x ) f ( x ) + β( x ) g( x )) = ( v( x )α ( x )) f ( x ) + ( v( x )β ( x )) g( x ) ∈ A .

Thus Aideal ≤ F [ x ] . But since F is a field, F[x] is a principle ideal domain by (1.21).

So A = r( x ) for some r ( x ) ∈ F [ x ] , and r ( x )

f(x) (= 1 ⋅ f ( x ) + 0 ⋅ g ( x ) ∈ A ) and

r(x) g(x) (= 0 ⋅ f ( x ) + 1 ⋅ g( x ) ∈ A ). Then r(x) = l, some constant, because f(x) and g(x) are relatively prime.

Consider l = m( x ) f ( x ) + n( x )g ( x ) ∈ A . Dividing both sides of the equation by l gives us 1 = h( x ) f ( x ) + k( x )g ( x ) where l ⋅ h( x ) = m( x ) and l ⋅ k( x ) = n( x ) . Q.E.D. That brings an end to (3.21), and moves us to another theorem regarding common factors in polynomials of positive degree.

(3.22) Theorem A polynomial f(x) over a field F has a multiple zero in some extension H if and only if f(x) and f'(x) have a common factor of positive degree in F[x].

Proof: ⇒: Let f ( x ) ∈ F [ x ] , and let H be an extension of F. Suppose γ is a multiple zero of h(x) for h( x ) ∈ H[x]. So there exists some k(x) in H[x] such that h( x ) = ( x − γ ) 2 k ( x ) .

Then h'(x) = 2( x − γ )k ( x ) + ( x − γ ) 2 k' (x) and h' ( γ ) = 0 .

Clearly x − γ is a factor of both h(x) and h' (x) in H[x] and deg ( x − γ ) = 1 > 0.

Michel 21 ⇐: Assume h(x) and h' (x) have a common factor of positive degree. Let γ be the zero of the common factor.

Thus γ is a zero of h( x ) and h' (x). So there exists a polynomial β( x ) such that h( x ) = ( x − γ ) β( x ) .

Then h'(x) = ( x − γ ) β ' ( x ) + β( x ) and 0 = h' ( γ ) = β ( γ ) .

Hence x − γ is a factor of β( x ) and thus γ is a multiple zero of h(x) . Q.E.D. This brings us to an end of the proof of (3.22) and to this chapter.

Michel 22 Chapter 4 We have finally arrived at the acme of this book. We have learned the necessary statements to perform a truly elegant proof. The entire set of statements contained in chapter three will now be strung together in an irrefutable argument about the cyclotomic polynomials Φn .

(4.0) Theorem The cyclotomic polynomials Φn are irreducible in Z[x].

Proof: Let α( x ) ∈ Z[x] be monic and irreducible over Z such that α( x ) is a factor of Φn ( x ) .

WTS if η is a zero of α( x ) , then it is a zero of Φn ( x ) .

We know α( x )

Φn ( x ) from our assumption so Φn ( x ) = α( x )δ( X ) for some

δ( x ) ∈ Z[x] .

Therefore Φn ( η ) = α( η )δ( η ) = 0 ⋅ δ( η ) = 0. Thus η is a zero of Φn ( x ) .

Therefore η is a primitive nth root of unity that is a zero of α( x ) .

Also, Φn ( x )

( x n − 1 ) by (2.3), thus α( x )

( x n − 1) .

Then there exists β( x ) ∈ Z[x] such that x n −1 = α( x ) β( x ) .

STS α( x ) is minimal for η over Q. We know η is algebraic over Q by (1.1), and we also know α( x ) is monic and irreducible by assumption.

Since η is a zero of α( x ) by assumption, α( x ) is minimal over Q by (3.16).

Let p be prime such that p does not divide n. Then from (1.5) we know that 0 = ( η p ) n − 1 = α ( η p )β ( η p ) (substitution).

Suppose α( η p ) ≠ 0 . Then β( η p ) = 0 .

By (3.18) we get that α( x )

α( x )

β( x p ) in Q[x] . We also know from (2.2) that

β( x p ) in Z[x] .

Michel 23

Suppose β( x p ) = α( x )υ( x ) where υ( x ) ∈ Z[x] . Call the polynomials in Z p [ x ] −

-

-

obtained from α( x ), β( x ) and υ( x ) ; α( x ), β( x ) and υ( x ) respectively. −

−

−

−

By (6.1) and (6.2) we get ( β( x )) p = β( x p ) = α( x ) γ( x ) (since we reduced the coefficients modulo p)1. −

−

−

We now know that β( x ) p = α( x ) υ( x ) in Z p [ x ] .

But Z p [ x ] is a unique factorization domain by (3.12). −

−

Then α( x ) and β( x ) share an irreducible factor in Z p [ x ] , call this factor w(x) . −

−

Thus α( x ) = k1 ( x )w( x ) and β( x ) = k 2 ( x )w( x ) for some k1 and k 2 in Z p [ x ] . −

−

Let us take x n − 1 in Z p [ x ] so we have α( x ) β ( x ) = k1 ( x )k 2 ( x )( w( x ))2 .

Then x n − 1 has a multiple zero in some extension of Z p [ x ] . But p does not divide n and so the derivative nx n −1 of x n − 1 is not 0.

Thus nx n −1 and x n − 1 do not have a common factor of positive degree in Z p [ x ] →← (this contradicts (3.19)) Therefore α ( η p ) = 0 .

Then if η is some positive nth root of unity that is a zero of α( x ) and p is any prime that does not divide n, then η p is a zero of α( x ) .

Let c be any integer such that 0 ≤ c ≤ n and gcd(c, n) = 1. Then c = p1 ⋅ ⋅ ⋅ p t where each p i is a prime that does not divide n (repetition allowed).

Then each η ,η p1 ,( η p1 ) p2 ,...,( η p1 ⋅⋅⋅ p 2 ) pt = η c is a zero of α( x ) .

Since η is a primitive nth root of unity, every other primitive nth root of unity has the form η c such that 0 ≤ c ≤ n and gcd(c, n) = 1.

1

Theorems 6.0, 6.1, and 6.2 are found in the appendix following this chapter.

Michel 24 Therefore every zero of Φn ( x ) is a zero of α( x ) . Q.E.D. This concludes our conquest; with one swift stroke we have shown Φn to be irreducible over Z for all n ∈ N .

Michel 25 Chapter 5 Since this paper is a dedication to the brilliant Carl Friedrich Gauss, it would be appropriate to mention that what we have presented is no match even for one out of seven sections of Gauss’s Disquisitiones. Since we barely scratch the surface of a topic Gauss understood so well, we should at least try to imagine the enormous breadth of the man’s mathematical conquest.

The photos below were taken at Brunswick University in Brunswick Germany, where Gauss once carried out geodedic (pertaining to the study of the size of the earth) experiments as a student 213 years ago. The photo on the right is of the geodedic point used by Gauss himself. The photo on the left is of the plaque hanging directly above Gauss’s trigonometric point in the main entrance of Brunswick University.

Photos taken by Jerod Michel

Michel 26 The plaque reads as follows: Carl Friedrich Gauβ,

Carl Friedrich Gauss,

geboren am 30.4.1777 in Braunschweig, gestorben am 23.2.1855 in Göttingen,

born April 30, 1777 in Brunswick, died February 23, 1855 in Göttingen,

studierte von 1792 bis 1795 am Collegium Carolinum in Braunschweig.

Er führte als Privatgelehrter von 1802 bis 1807 an dieser Stelle mit seiner Methode der Dreiecksmessung (Triangulation) geodäthische Untersuchungen mit einem Sextanten durch.

studied from 1792 to 1795 at the Collegium Carolinum in Brunswick.

He executed at this juncture from 1802 to 1807 geodetic investigations as a private scholar with his Method of Triangulation, using a sextant.

Aus Anlaβ seines 200. Geburtstages am 30.4.1977 wurde dieser Trigonomitrische Punkt vom Institut für Vermessungskunde der Technischen Universität Braunschweig rekonstruiert und markiert.

Thank you for reading.

On the occasion of his 200th birthday, on April 30, 1977, this Trigonometric Point was reconstructed and marked by the Institute for Geodetics of the Brunswick Technical University (translated by Maggie Gemmell).

Michel 27 Chapter 6 Appendix

We will show that the process of reducing the coefficients modulo p is a homomorphism from Z[x] to Z p [ x ] by proving the following theorem.

(6.0) Theorem If φ : E → F is a homomorphism, define φ0 : E [ x ] → F [ x ] by

φ0 ( a n x n + a n−1 x n−1 + ⋅ ⋅ ⋅ + a 0 ) = φ( a n )x n + ⋅ ⋅ ⋅ + φ( a 0 ) . Then φ0 is a homomorphism.

Proof: Let τ be a homomorphism.

WTS φ0 preserves multiplication.

Let ε( x ),ω( x ) ∈ E [ x ] where ε( x ) = en x n + en −1 x n −1 + ⋅ ⋅ ⋅ + e0 and

ω( x ) = wm x m + wm−1 x m −1 + ⋅ ⋅ ⋅ + w0 .

Consider φ0 ( ε( x )ω( x )) = φ0 ( rk x k + rk −1 x k −1 + ⋅ ⋅ ⋅ + r0 ) where k = m + n and ri = ei w0 + ⋅ ⋅ ⋅ + e0 wi .

So φ0 ( ε( x )ω( x )) = φ( rk ) x k + φ( rk −1 ) x k −1 + ⋅ ⋅ ⋅ + φ( r0 ) (definition of φ0 ) = φ( e k w0 + ⋅ ⋅ ⋅ + e0 wk ) x k + φ( ek −1 w0 + ⋅ ⋅ ⋅ + e0 wk −1 ) x k −1 + ⋅ ⋅ ⋅ + φ( e0 w0 )

= ( φ( e k w0 ) + ⋅ ⋅ ⋅ + φ( e0 wk )) x k + ( φ( ek −1 w0 ) + ⋅ ⋅ ⋅ + φ( e0 wk −1 )) x k −1 + ⋅ ⋅ ⋅ + ( φ( e0 w0 )) (since φ is a homomorphism)

= ( φ( e k )φ( w0 ) + ⋅ ⋅ ⋅ + φ( e 0 )φ( wk )) x k + ( φ( ek −1 )φ( w0 ) + ⋅ ⋅ ⋅ + φ( e0 )φ( wk −1 )) x k −1 + ⋅ ⋅ ⋅ + φ( e0 )φ( wk ) (since φ is a homomorphism) = ( φ( e n ) x n + ⋅ ⋅ ⋅ + φ( e0 ))( φ( wm ) x m + ⋅ ⋅ ⋅ + φ( w0 )) (since k = m + n) = φ0 ( ε( x ))φ0 ( ω( x )) Therefore φ0 preserves multiplication.

NTS φ preserves addition.

Michel 28 Consider φ0 ( ε( x ) + ω( x )) = φ0 (( ek + wk ) x k + ( e k −1 + wk −1 ) x k −1 + ⋅ ⋅ ⋅ + ( e0 + w0 )) (where k = max(m, n), ei = 0 for i > n, wi = 0 for i > m) = φ( e k + wk ) x k + φ( e k −1 + wk −1 ) x k −1 + ⋅ ⋅ ⋅ + φ( e0 + w0 ) = ( φ( e k ) + φ( wk ))x k + ( φ( e k −1 ) + φ( wk −1 )) x k −1 + ⋅ ⋅ ⋅ + φ( e0 + w0 ) (since φ is a homomorphism) = ( φ( e n ) x n + ⋅ ⋅ ⋅ + φ( e0 )) ( φ( wm ) x m + ⋅ ⋅ ⋅ + φ( w0 )) = φ0 ( ε( x )) + φ0 ( ω( x )) .

Therefore φ0 preserves addition and φ0 is a homomorphism. Q.E.D. Theorem (6.1) Let m be a fixed integer. For any integer a, let a 0 denote a (mod m ) . The mapping φ : Z[x] → Z m [ x ] defined by −

−

−

φ( a n x n + a n−1 x n−1 + ⋅ ⋅ ⋅ + a 0 ) = a n x n + a n −1 x n −1 + ⋅ ⋅ ⋅ + a 0 is a homomorphism.

Proof: Let m ∈ Z and define φ : Z[x] → Z k [ x ] by φ( a n x n + a n −1 x n −1 + ⋅ ⋅ ⋅ + a0 ) −

−

−

−

= a n x n + a n −1 x n −1 + ⋅ ⋅ ⋅ + a0 where a ≡ a(mod k ) .

Consider, for α( x ), β( x ) ∈ Z[x] where α( x ) = a n x n + a n −1 x n −1 + ⋅ ⋅ ⋅ + a 0 and

β( x ) = bm x m + bm −1 x m −1 + ⋅ ⋅ ⋅ + b0 , φ( α( x ) β( x )) = φ( γ m + n x m + n + γ m + n −1 x m + n −1 + ⋅ ⋅ ⋅ + γ 0 ) (where γ k = a k b0 + ⋅ ⋅ ⋅ + a 0 bk ) −

−

−

= γ m + n x m + n + γ m + n −1 + ⋅ ⋅ ⋅ + γ 0 −

−

−

−

−

−

−

−

−

−

= ( a m+ n b 0 + ⋅ ⋅ ⋅ + a 0 b m+n )x m +n + ( a m+n−1 b 0 + ⋅ ⋅ ⋅ + a 0 b m+n−1 )x m+ n−1 + ⋅ ⋅ ⋅ + ( a 0 b 0 ) −

( a ≡ a(mod k ) ) −

−

−

−

−

−

= ( a n x n + a n−1 x n−1 + ⋅ ⋅ ⋅ + a 0 )( b m x m + b m−1 x m−1 + ⋅ ⋅ ⋅ + b 0 ) = φ( a n x n + a n −1 x n −1 + ⋅ ⋅ ⋅ + a 0 )φ( bm x m + bm−1 x m−1 + ⋅ ⋅ ⋅ + b0 )

Michel 29 = φ( α( x ))φ( β( x )) . Therefore φ preserves multiplication.

NTS φ preserves addition.

Consider

φ( α( x ) + β( x )) = φ( a n x n + a n−1 x n−1 + ⋅ ⋅ ⋅ + a0 ) + φ( bm x m + bm−1 x m −1 + ⋅ ⋅ ⋅ + b0 ) = φ(( a k + bk ) x k + ( a k −1 + bk −1 ) x k −1 + ⋅ ⋅ ⋅ + ( a 0 + b0 )) where k = max( m , n ) and a i = 0 for i > n, and bi = 0 for i > n. −

−

−

−

−

−

= ( a k + b k )x k + ( a k −1 + b k −1 )x k −1 + ⋅ ⋅ ⋅ + ( a 0 + b 0 ) −

−

−

−

−

−

= ( a n x n + a n−1 x n−1 + ⋅ ⋅ ⋅ + a 0 ) + ( b m x m + b m−1 x m−1 + ⋅ ⋅ ⋅ + b 0 ) = φ( α( x ) + β ( x )) Therefore φ preserves addition. Q.E.D. (6.2) Corollary of Lagrange’s Theorem For every integer a and every prime p, a p (mod p ) ≡ a(mod p ) .

The proof of (6.2) is simple and left to the reader.

Michel 30 Bibliography Dunnington, Waldo. Carl Friedrick Gauss Titan of Science. Exposition Press Inc. New York New York, 1955.

Gallian, Joseph. Contemporary Abstract Algebra. Houghton Mifflin Company Boston New York, 2002.

Gauss, Carl. Disquisitiones Arithmeticae English Edition. Springer Verlag New York New York, 1986.

Hollingdale, Stewart. Makers of Mathematics. Penguin Books New York New York, 1989.

Hofmann, Joseph. The History of Mathematics to 1800. Littlefield, Adams & Co. Totowa New Jersey, 1967.

James, Ioan. Remarkable Mathematicians from Euler to Neumann. The Mathematical Association of America, Cambridge University Press. New York New York, 2002.

Tignol, Jean-Pierre. Galoisâ&#x20AC;&#x2122; Theory of Algebraic Equations. World Scientific Publishing Co. River Edge New Jersey, 2001.

Shapiro, Harold. Introduction to the Theory of Numbers. John Wiley and Sons New York New York, 1983.

Bourbaki, Nocolas. Elements of Mathematics. Algebra I . Hermann Publishers in Arts and Science, 1943.