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```									MAT 444 – H. Barcelo
Spring 2004
Homework 13 – Solutions
Chapter 13
Section 13.1
Question:
4.  Let F be a field containing exactly eight elements. Prove or disprove: The characteristic of
F is 2.

Let F be a field with | F | = 8 since F is a finite field, we have that char ( F ) = p for some
prime integer p. This means that p is the least integer such that1F + 1F + + 1F = 0
p times

Let G be the subgroup generated by 1F under addition. Clearly, | G | = p but G ≤ ( F , +)
which implies that p divides 8 so p = 2.

Section 13.2
Question:
3. Determine the irreducible polynomial for α = 3 + 5 over each of the following fields.
(a.)                 (b)    ( 5)           (c)    ( 10 )     (d)     ( 15 )
a ) Let x = 3 + 5 ⇒ x 2 = 3 + 2 3                 5+5
⇒ x2 − 8 = 2 3       5
⇒ ( x 2 − 8) 2 = x 4 − 16 x 2 + 64 = 60
⇒ x 4 − 16 x 2 + 4 = 0
Thus f ( x) = x 4 − 16 x 2 + 4 is a polynomial in [ x] with 3 + 5 has a root. Is is
irreducible? f ( x) is irreducible if either it has a root in     or it factors as two
irreducible quadratics poly. Using the rational root test, one sees that f ( x) has no
roots in . Next, if f ( x) = p ( x) q ( x) with p( x) = x 2 + ax + b and q( x) = x 2 + cx + d ,
then using the quadratic formula we see that this cannot occur.

b)         ( 5) : x =      3+ 5 ⇒x− 3= 5
⇒ x2 − 2 5 x + 5 = 3
⇒ x2 − 2 5 x + 2 = 0
MAT 444 – Barcelo
Homework 13                                                                                                        Page 2

So g ( x) = x 2 − 2 5 x + 2 has                      3 + 5 as one of its roots. g ( x) is reducible over
( 5 ) only if it has a root in ( 5 ) .
But the roots of g ( x) are ± ( 5 + 3 ) , so it suffices to show that
3 + 5 ∉ ( 5) .
First we show that 3 ∉ ( 5 ) . Since 5 is algebraic over            then

( 5 ) = [ 5] . So if 3 ∈ ( 5 ) ⇒ 3 = a + b 5 a, b ∈ . But then
3 = a 2 + 5b 2 + 2ab 5
3 − a 2 − 5b 2
⇒ 5=
2ab
contradicting the fact that 5 is irrational. Hence                      3∉    ( 5 ) and
3 + 5∉               ( 5 ) since         5∉          ( 5).
c) Since 10 cannot be obtained as linear combination over                              , or product of   3 and     5,
then x − 16 x + 4 is an irreducible polynomial over
4             2
and over        10 with   3+ 5
as one of its roots.

(               )                                                                         ( 15 ) with
2
d)          3+ 5                = 8 + 2 15 . Thus x 2 − 8 − 2 15 is a polynomial in
3 + 5 as one of its roots. An argument similar to the one given in (b) reveals that
x 2 − 8 − 2 15 is irreducible over                        ( 15 ) .
Question:
5. Let K = F (α ) , where α is a root of the irreducible polynomial
f ( x) = x n + an + an −1 x n −1 + + a1 x + a0 . Determine the element α −1 explicitly in terms of α
and of the coefficients ai . First note that a0 ≠ 0 , otherwise f ( x) would factor.
Since α n + an −1 α n −1 + + a1 α1 + a0 = 0
⇒ α n + an −1 α n −1 +                 + a1 α = − a0
αn           an −1 n −1              a1
⇒−                 −        α −            −      α =1
a0            a0                     a0
⎛ α n −1                          a1 ⎞
⇒α ⎜        −                  −        ⎟ =1
⎝ − a0                            a0 ⎠
α n −1       α n−2               a1
⇒ α −1 =                   −           −       −
− a0          a0                 a0

Typeset by Georgeann Lorentz
MAT 444 – Barcelo
Homework 13                                                                                                           Page 3

Section 13.3

Question:
4. Let γ n = e 2π i / n . Determine the irreducible polynomial over                        (γ 3 ) of
(a) γ 6 , (b) γ 9 , (c) γ 12 .

Let γ n = e 2π i / n , k = 3 j for j ∈ . Then                     ⊂ (γ 3 ) (γ k ) . So
[ (γ 3 ) (γ k ): ] = [ (γ 3 ) (γ k ): (γ 3 )][              (γ 3 ): ]
and
[ (γ k ) (γ 3 ): ] = [ (γ k ) (γ 3 ): (γ k )][              (γ k ): ]
since (γ 3 ) (γ k ) = (γ 3 ) (γ k ) , we have that               [ (γ k ) (γ 3 ):   (γ k )] [ (γ k ):   ]=
[ (γ 3 ) (γ k ): (γ 3 )][ (γ 3 ): ]
⎛ j⎞
2π i ⎜ ⎟
Next, since k = 3 j , γ k = e
j             ⎝k⎠
= e 2π i / 3 = γ 3 . Thus x − γ k j is the monic irreducible
polynomial of γ 3 in           (γ k ) , and [ (γ k )(γ 3 ) :      (γ k )] = 1 . Moreover, γ 3 3 − 1 = 0 so γ 3 is a
root of β ( x) = x3 − 1 . But x3 − 1 = ( x −1) ( x 2 + x + 1) and x 2 + x + 1 is irreducible in                     .
Since γ 3 − 1 ≠ 0 ⇒ γ 32 − γ + 1 = 0 . So
[ (γ 3 ) : ] = 2 ⇒ [ (γ k ): ] = 2[ (γ 3 ) (γ k ): (γ 3 )] .

a)      γ 6 is a root of x 6 − 1 = ( x − 1) ( x + 1) ( x 4 + x 2 + 1) since γ 3 is a root of
x 6 − 1 ⇒ x 2 + x + 1 (the irreducible polynomial for γ 3 over                 ) must also be a factor.
Thus, x 6 − 1 = ( x + 1) ( x − 1) ( x 2 + x + 1) ( x 2 − x + 1) . Easy computations show that
γ 62 − γ 6 + 1 = 0 and since x 2 − x + 1 is also irreducible (rational root test) over ,
⇒ [ (γ 6 ): = 2 and [ (γ 3 ) (γ k ): (γ 3 )] = 1 ⇒ γ 6 ∈ (γ 3 ) .Indeed
γ 6 = − γ 32 ⇒ x + γ 32 = 0 .Clearly, x + γ 32 is the monic irreducible polynomial over
(γ 3 ) associated to γ 6 .

b)      γ 9 is also a root of x 6 − 1 = ( x + 1) ( x − 1) ( x 2 + x + 1) ( x 2 − x + 1) yielding
[ (γ 3 ) (γ 9 ) : (γ 3 )] = 3 . Thus we are looking for an irreducible (monic) polynomial over
(γ 3 ) of degree 3. But γ 93 = γ 3 , so x 3 − γ 3 is the irreducible (monic) polynomial
associated to γ 9 in (γ 3 ) .

c)      x 4 − x 2 + 1 is the irreducible polynomial for γ 12 over . So [ (γ 12 ) : ] = 4 and
[ (γ 3 ) (γ 12 ) : (γ 3 )] .Note that γ 12 = γ 6 = γ 32 ⇒ x 2 − γ 62 is the monic irreducible
2

polynomial for γ 12 over (γ 3 ) .

Typeset by Georgeann Lorentz
MAT 444 – Barcelo
Homework 13                                                                                                        Page 4

Question:
7. Decide whether or not i is in the field (a)                    (     )
−2 , (b)    (   4
)
−2 , (c)         (α ), where
α3 + α + 1 = 0 .

a)      We have that x 2 + 2 = 0 has                 −2 has a root and degree ⎡
⎣          (     )
−2 :   ⎤ = 2 , with every
⎦
element of          (               )
−2 of the form: a + b −2 with a, b ∈         .

If i ∈      (               )
−2 ⇒ i a + b −2

⇒ i 2 = −1 = a 2 − 2b 2 + 2ab −2
which is clearly impossible.
Thus i ∉        (
−2 .             )
b)      We have that x 4 + 2 = 0 has A −2 for root. Note that f ( x) = x 4 + 2 is irreducible over
since 5 does not divide 1 and residue f ( x) = x 4 + 2 modulo 5 is irreducible.
So if i ∈           (   4
)
−2 ⇒ i = a + bα + cα 2 + dα 3
⇒ i 2 = −1 = (a + bα + cα 2 + dα 3 ) 2 .
Straightforward but tedious calculations show that this is impossible ⇒ i ∉ (α ) .

c)      Note that x3 + x 1 is irreducible over [ x] since ( x 3 + x + 1) is irreducible modulo 2.
Thus, [ (α ) : ] has degree 3. But (i ) has degree 2 and since 2 does not divide 3,
i ∉ (α ) .

Typeset by Georgeann Lorentz

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