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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. Answer: 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 ) Answer: 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. Answer: 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 . Answer: 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 . Answer: 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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