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Solution. Since a = 0 and a = ±1, there is a prime p dividing
a ; since a is squarefree, p2 a . As all coefﬁcients of x i for 0 <
i < n are 0, Eisenstein’s criterion applies to show that x n − a is
irreducible in Q [x ].
√
(ii) If a = ±1 is a squarefree integer, prove that n a is irrational.
Solution. Since x n − a is irreducible in Q [x ], it has no rational
roots.
3.91 Let k be a ﬁeld, and let f (x ) = a0 + a1 x + · · · + an x n ∈ k [x ] have degree
n . If f (x ) is irreducible, then so is an + an −1 x + · · · + a0 x n .
Solution. If f ∗ (x ) denotes the polynomial f (x ) with coefﬁcients reversed,
then a factorization f ∗ (x ) = g (x )h (x ) gives a factorization f (x ). One
sees this just by using the deﬁnition of multiplication of polynomials. Let
p
q
g (x ) = i =0 bi x i and h (x ) = j =0 c j x j , where p + q = n . Thus,
an −m = bi c j .
i+ j = m It follows that
b p−i cq − j = an −(n −m ) = ak .
i + j = n −m Therefore, if we deﬁne g ∗ (x ) = i =0 b p−i x i and h ∗ (x ) = j =0 cq − j x j ,
then f (x ) = g ∗ (x )h ∗ (x ), contradicting the irreducibility of f (x ).
Note that f (x ) → f ∗ (x ), which reverses coefﬁcients, is not a welldeﬁned function k [x ] → k [x ], because it is not clear how to deﬁne f ∗ (x )
if the constant term of f (x ) is zero. And even if one makes a bona ﬁde
deﬁnition, the function is not a homomorphism. For example, let f (x ) =
x 5 + 3x 4 ; that is, in sequence notation,
p q f (x ) = (0, 0, 0, 0, 3, 1, 0, . . .).
Let g (x ) = x 3 + x ; in sequence notation,
g (x ) = (0, 1, 0, 1, 0, . . .).
Now f (x )g (x ) = [x 8 + 3x 7 + x 6 + 4x 5 + 3x 4 ; in sequence notation,
f (x )g (x ) = (0, 0, 0, 0, 3, 4, 1, 3, 1, 0, . . .).
Therefore, [ f (x )g (x )]∗ = 3x 4 + 4x 3 + x 2 + 3x + 1, which is a quartic. But f ∗ (x ) = 3x + 1 and g ∗ (x ) = x 2 + 1, so that
f ∗ (x )g ∗ (x ) is a cubic. Therefore, [ f g ]∗ = f ∗ g ∗ . ...
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 Fall '11
 KeithCornell

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