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(vi) If f (x ) is a polynomial over a ﬁeld K whose factorization into
a constant and monic irreducible polynomials in K [x ] is f (x ) =
e
e
ap11 · · · pnn , and if all the coefﬁcients of f (x ) and of the polynoe
e
mials pi (x ) lie in some subﬁeld k ⊆ K , then f (x ) = ap11 · · · pnn
is also the factorization of f (x ) in k [x ] as a product of a constant
and monic irreducible polynomials.
Solution. True.
f f e
e
3.83 In k [x ], where k is a ﬁeld, let g = p11 · · · pmm and h = p1 1 · · · pmm , where
the pi ’s are distinct monic irreducibles and ei , f i ≥ 0 for all i . Prove that
g  h if and only if ei ≤ f i for all i .
Solution. The proof for polynomials is essentially the same as for integers.
3.84
(i) If f (x ) ∈ R [x ], show that f (x ) has no repeated roots in C if and
only if ( f , f ) = 1.
Solution. By part (ii), f (x ) has no repeated roots in C [x ] if and
only if ( f , f ) = 1 in C [x ]. By Corollary 3.75, one can check this
criterion by computing this gcd in R [x ]. (ii) Prove that if p (x ) ∈ Q [x ] is an irreducible polynomial, then p (x )
has no repeated roots.
n
n
Solution. If p (x ) = i =0 ai x i , then p (x ) = i =1 iai x i −1 . It
follows that deg( p ) = n − 1, and so p is not the zero polynomial.
Since p(x ) is irreducible, the gcd ( p , p ) = 1, and so part (ii)
applies to show that p (x ) has no repeated roots.
Note that this argument may fail if Q is replaced by F p , where p
is prime.
3.85 Let ζ = e2π i / n .
(i) Prove that
x n − 1 = (x − 1)(x − ζ )(x − ζ 2 ) · · · (x − ζ n −1 )
and, if n is odd, that
x n + 1 = (x + 1)(x + ζ )(x + ζ 2 ) · · · (x + ζ n −1 ).
Solution. The n numbers 1, ζ, ζ 2 , . . . , ζ n −1 are all distinct. But
they are all roots of x n − 1, and so Theorem 3.50 gives the ﬁrst
equation:
x n − 1 = (x − 1)(x − ζ )(x − ζ 2 ) · · · (x − ζ n −1 ).
If n is odd, then replace x by −x to get
(−x )n − 1 = (−x − 1)(−x − ζ )(−x − ζ 2 ) · · · (−x − ζ n −1 )
= (−1)n (x + 1)(x + ζ )(x + ζ 2 ) · · · (x + ζ n −1 ). ...
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 Fall '11
 KeithCornell

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