Question on cyclotomic polynomial We know that the cyclotomic polynomial phi p (

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Question on cyclotomic polynomial

We know that the cyclotomic polynomial phi p (x) = x (p ? 1) / (x-1) = X^(p-l) + x (p-2) + ... +x+1 is irreducible over Q for every prime p. Let omega be a zero of phi p (x), and consider the field Q(omega). Show that omega, omega 2 ,..., omega p?1 are distinct zeros of omega p (x), and conclude that they are all the zeros of omega p (x). Show that G(Q(omega)/Q) is abelian of order p?1. Show that the fixed field of G(Q(omega)/Q) is Q. HINTS... Clearly omega, omega 2 ,...,omega (p?1) are distinct since omega /= 1 or 0. To show that omega ^i is a zero of omega p , calculate omega p (omega i ). The conjugates of omega are omega, omega 2 ,..., omega (p?l) . Define a map phi i : Q(omega) rightarrow Q(omega i ) by phi i (aO + al omega + + ap?2 omega^(p-2) ) = aO + a1 omega^ i + + cp?2 (omega i) (p-2) , where ai Q. Prove that omega i is an isomorphism of fields. Show that omega 2 generates G(Q(omega)/Q). Show that {omega, omega 2 ,..., omega (p?l) } is a basis for Q(omega) over Q, and consider which lineal combinations of omega, omega^2 ,..., omega (p?l) are left fixed by all elements of G(Q(omega)/Q)

Explanation / Answer

b 6= 0 in a eld k, the exponent of b is the smallest positive integer n (if it exists) such that b

n = 1.

That is, b is a root of x

n 1 but not of x

d 1 for any smaller d. We construct polynomials n(x) 2 â©[x]

such that

n(b) = 0 if and only if b is of exponent n

These polynomials n are cyclotomic polynomials.

[2.0.1] Corollary: The polynomial x

n 1 has no repeated factors in k[x] if the eld k has characteristic

not dividing n.

Proof: It suces to check that x

n 1 and its derivative nxn1 have no common factor. Since the

characteristic of the eld does not to divide n, n 1k 6= 0 in k, so has a multiplicative inverse t in k,

and

(x

n 1) (tx) (nxn1

) = 1

and gcd(x

n 1; nxn1

) = 1. ===

Dene the n

th cyclotomic polynomial n(x) by

1(x) = x 1

and for n > 1, inductively,

n(x) = x

n 1

lcm of all x

d 1 with 0 < d < n, d dividing n

with the least common multiple monic.

[2.0.2] Theorem:

n is a monic polynomial with integer coecients. â¬âœ»âª

For in the eld k, n() = 0 if and only if

n = 1 and

t 6= 1 for all 0 < t < n.

gcd(m(x);n(x)) = 1 for m < n with neither m nor n divisible by the characteristic of the eld k.

The degree of n(x) is '(n) (Euler's phi-function)

Another description of n(x):

n(x) = x

n 1 Q

1d<n;djn d(x)

â¬âœ»âª More properly, if the ambient eld k is of characteristic 0, then the coecients lie in the copy of â© inside the

prime eld â—— inside k. If the ambient eld is of positive characteristic, then the coecients lie inside the prime eld

(which is the natural image of â© in k). It would have been more elegant to consider the cyclotomic polynomials as

polynomials in â©[x], but this would have required that we wait longer.

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