Abstract Abstract Algebra Algebra Chapter 13 - Field Theory David S. Dummit and Richard M. Foote Solution Manual by positrón0802
Contents 13 Field Theory 13.1 Basic Theory Theory and Field Extensio Extensions ns . . . . . . . . . 13.2 Algeb Algebraic raic Extensions Extensions . . . . . . . . . . . . . . . . . 13.3 Class Classical ical Straightedge Straightedge and Compas Compasss Const Constructi ructions ons 13.4 Split Splitting ting Fields and Algebraic Algebraic Closures . . . . . . . 13.5 Separ Separable able and Insep Inseparable arable Extension Extension . . . . . . . . 13.6 Cyclo Cyclotomic tomic Polynomials Polynomials and Exten Extensions sions . . . . . .
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1 1 4 12 13 15 19
Field ield Theor heory y
§13.1 §13.1
Basic Basic The Theory ory and Field Field Ext Extens ension ionss
Exercise 1. Show that p(x) = x 3 + 9x 9x + 6 is irreducible in Q[x]. Let θ be a root of p(x). Find the inverse of 1 of 1 + θ Q(θ) Solution: p(x) = x3 + 9x + 6 is irreducible in Z[x] by Eisenstein Criterion with p = 3. By Gauss Lemma, then it is irreducible in Q[x]. To find find (1 + θ) θ )−1 , we apply the Euclidean algorithm algorithm (long division) division) to p( p (x) and 1 + x. We find
∈
x3 + 9x 9x + 6 = (1 + x)(x )(x2 Evaluating at θ at θ,, we find ( find (11 + θ)(θ )(θ2
− x + 10) − 4.
− θ + 10) = 4.4. Therefore θ 2 − θ + 10 (1 + θ)−1 = . 4
Exercise 2. Show that x3 2x 2 is irreducible over Q and let θ be a root. root. Compu Compute te 1+θ (1 + θ)(1 + θ + θ2 ) and in Q(θ). 1 + θ + θ2 Solution: Let f ( f (x) = x3 2x 2. f is f is irreducible over Z by Eisenstein Criterion with p = p = 2, hence over Q by Gauss Lemma. Now, if θ if θ is a root of f of f ,, then θ then θ 3 = 2θ + 2. 2 . Hence
− −
− −
(1 + θ)(1 + θ + θ2 ) = 1 + 2θ 2θ + 2θ 2 θ2 + θ3 = 3 + 4θ 4θ + 2θ 2 θ2 . For computing we obtain
1+θ , first we compute (1 + θ + θ2 )−1 . Applying the Euclidean algorithm, 2 1 + θ + θ x3
− 2x − 2 = (x2 + x + 1)(x 1)(x − 1) − 2x − 1,
and x
3
− 2x −
x2 2 = (2x (2x + 1)( 2 1
− x4 − 78 ) − 98 .
13.1 13.1 Basic Basic Theory Theory and and Field Field Extens Extension ionss Evaluating at θ at θ,, from this equalities we obtain 2
(θ + θ + 1)(θ 1)(θ
− 1) = 2θ 2θ + 1
and
8 θ (2θ (2θ + 1) −1 = (
2
9 2
− θ4 − 78 ).
Combining these two equations we obtain 8 2 (θ + θ + 1)(θ 1)(θ 9 So, (θ
2
2
− 1)( θ2 − θ4 − 78 ) = 1.
8 θ + θ + 1) −1 = (θ − 1)(
2
9 2 where we used θ 3 = 2θ + 2 again. Therefore,
θ 4
− −
7 )= 8
1+θ 2θ2 θ 5 = (1 + θ)( + + ) = 1 + θ + θ2 3 3 3
−
−
−
2θ2 θ 5 + + , 3 3 3
θ 2 2θ 2 θ 1 + + . 3 3 3
Exercise 3. Show that x that x 3 + x + 1 is 1 is irreducible over F2 and let θ be a root. Compute the powers of θ θ in F2 (θ). Solution: Since 0 Since 0 3 + 0 + 1 = 1 and 1 1 + 1 + 1 = 1 in F2 , then x then x 3 + x + 1 is irreducible over F2 . Since θ Since θ is root of x x 3 + x + 1, then θ then θ 3 = θ 1 = θ + 1. Hence, the powers of θ of θ in F2 (θ) are
−−
θ, θ 2 , θ 3 = θ + θ + 1, 1 , θ 4 = θ 2 + θ, θ 5 = θ 2 + θ + 1, 1 , θ 6 = θ 2 + 1, 1, and θ 7 = 1.
√ → a − b√ 2 is an isomorphism of Q(√ 2)
Exercise 4. Prove directly that the map a + b 2 with itself. Solution: Denote this map by ϕ. ϕ . Then
√
√
√ √ √ √ − b 2 − d 2 = ϕ( ϕ (a + b 2) + ϕ(c + d 2), 2), and √ √ √ ϕ((a ((a + b 2) · (c + d 2)) = ϕ = ϕ((ac + 2bd 2bd + (ad (ad + bc) bc) 2) √ = ac + 2bd 2bd − (ad + bc) bc) 2 √ √ = (a − b 2)(c 2)(c − d 2) √ √ = ϕ( ϕ(a + b 2)ϕ 2)ϕ(c + d 2), 2), √ √ hence ϕ hence ϕ is an√ homomorphism. Moreover, if ϕ if ϕ((a + b 2) = ϕ( ϕ (c + d 2), 2), then a then a√ − b√ 2 =√ c − d√ 2, ∈ √ Q) a = b and c = d, so ϕ is injective. Q( 2), hence (since 2 injective. Also, given given a + b + b 2 ∈ 2), then √ √ so ϕ is surjective. Therefore, ϕ is an isomorphism of Q( 2) with 2) with itself. ϕ(a − b 2) = a + b 2, so ϕ ϕ(a + b 2 + c + d 2) = a + c
Exercise 5. Suppose α is a rational root of a monic polynomial in Z[x]. Prove Prove that that α is an integer. Solution: Let α Let α = p/q = p/q be be a root of a monic polynomial p( p (x) = x n + with gcd( gcd( p, q ) = 1. Then p p ( )n + an−1 ( )n−1 + q q
· · · + a1x + a0 over Z ,
· · · + a1 pq + a0 = 0.
Multiplying this equation by q n one obtains pn + an−1 pn−1 q + +
· · · + a1 pq n−1 + a0q n = 0 ⇒ q (an−1 pn−1 + · · · + a1 pq n−2 + a0q n−1) = − pn. 2
13.1 Basic Theory and Field Extensions Thus, every prime that divides q divides pn as well, so divides p. Since gcd( p, q ) = 1, there is no prime dividing q , hence q = 1. The result follows.
±
Exercise 6. Show that if α is a root of an xn + an−1 xn−1 + + a1 x + a 0 then an α is a n n 1 n 2 − − root of the monic polynomial x + an−1 x + an an−2 x + + ann−2 a1 x + ann−1a0 . Solution: This is straightforward. If
···
an αn + an−1 αn−1 + then
···
··· + a1α + a0 = 0,
(an α)n + an−1 (an α)n−1 + an an−2 (an α)n−2 +
··· + ann−2a1(anα) + ann−1a0 = a nn αn + ann−1 an−1 αn−1 + ann−1 an−2 αn−2 + ··· + ann−1 a1 α + ann−1 a0 = a nn−1 (an αn + an−1 αn−1 + an−2 αn−2 + ··· + a1 α + a0 ) = 0. Exercise 7. Prove that x 3 − nx + 2 is irreducible for n = −1, 3, 5. Solution: If x 3 − nx + 2 is reducible it must have a linear factor, hence a root. By Rational Root Theorem, if α is a root of x3 − nx + 2, then α must divide its constant term, so the possibilities are α = ±1, ±2. If α = −1 or 2, then n = 3; if α = −1, then n = 1; and if α = 2, then n = 5. Therefore, x 3 − nx + 2 is irreducible for n = −1, 3, 5. Exercise 8. Prove that x 5 − ax − 1 ∈ Z[x] is irreducible unless a = 0, 2 or −1. The first two correspond to linear factors, the third corresponds to the factorization (x2 − x +1)(x3 + x2 − 1).
Solution: We subdivide this exercise in cases and subcases. If x 5 ax 1 is reducible then it has a root (linear factor) or is a product of two irreducible polynomials of degrees 2 and 3. Case 1. If x 5 ax 1 has a root, then, by Rational Root Theorem, it must be α = 1. If α = 1 is a root, then a = 0. If α = 1 is a root, then a = 2. Case 2. Now, suppose that there exists f (x) and g(x) irreducible monic polynomials over Z of degrees 2 and 3 respectively, such that x 5 ax 1 = f (x)g(x). Write f (x) = x 2 + bx + c and g(x) = x 3 + rx2 + sx + t, where b,c,r, s, t Z. Then
− −
− −
±
−
− − ∈
x5
− ax − 1 = (x2 + bx + c)(x3 + rx2 + sx + t)
= x 5 + (b + r)x4 + (br + c + s)x3 + (bs + cr + t)x2 + (bt + cs) + tc.
Equating coefficients leads to b + r = 0 br + c + s = 0 bs + cr + t = 0 bt + cs =
−a ct = −1. From ct = −1 we deduce (c, t) = (−1, 1) of (c, t) = (1, −1), which give us two cases. Case 2.1. First suppose (c, t) = (−1, 1). Then the system of equations reduces to b + r = 0 br
− 1 + s = 0 bs − r + 1 = 0 b − s = −a. Now, put b = r into second and third equations to obtain r2 1 +s = 0 and rs r +1 = 0, that is, r 2 + 1 s = 0 and rs + r 1 = 0. Adding these last two equations we obtain
− −
− −
−
3
− −
13.2 Algebraic Extensions r2 + rs + r s = 0. Thus r 2 + rs + r + s = 2s, so (r +1)(r + s) = 2s. Now, from r 2 + 1 s = 0 we have r 2 = s 1, so then r 2 + rs + r s = 0 becomes rs + r = 1, that is, r(s + 1) = 1. Hence, r = 1 and s = 0, or r = 1 and s = 2. If r = 1 and s = 0, then (r + 1)(r + s) = 2s leads to 2 = 0, a contradiction. If r = 1 and s = 2, it leads to 0 = 4, another contradiction. Therefore, (c, t) = ( 1, 1) is impossible. We now pass to the case (c, t) = (1, 1). Case 2.2. Suppose that (c, t) = (1, 1). The system of equations reduces to
−
−
− −
− − −
−
−
−
−
−
b + r = 0 br + 1 + s = 0 bs + r
−1=0 −b + s = −a.
Adding the second and third equation we obtain b(r + s) + r + s = 0, so that (b +1)(r + s) = 0. Then b = 1 or r = s, so one more time we have two cases. If r = s, then br + 1 + s = 0 becomes br + 1 r = 0. Hence, b = r and br + 1 r = 0 gives r 2 + r 1 = 0. By Rational Root Theorem, this equation has no roots on Z. Since r Z, we have a contradiction. Now suppose b = 1. From b = r we obtain r = 1, so, from br + 1 + s = 0 we obtain s = 0. Finally, from b + s = a we obtain a = 1. Therefore, the solution (b,c,r,s,t) = ( 1, 1, 1, 0, 1) is consistent and we obtain the factorization
−
−
−
x5
§13.2
−
−
−
−
−
−
−
∈
−
−
−
−
− ax − 1 = (x2 + bx + c)(x3 + rx2 + sx + t) = (x2 − x + 1)(x3 + x2 − 1).
Algebraic Extensions
Exercise 1. Let F be a finite field of characteristic p. Prove that F = p n for some positive integer n.
||
Solution: Since the characteristic of F is p, its prime subfield is (isomorphic to) F p = Z/pZ. We can consider F as a vector space over F p . Since F is finite, then [ F : F p ] = n for some n Z+ . Therefore F = F p [F:Fp ] = p n .
∈
|| | | Exercise 2. Let g(x) = x 2 + x − 1 and let h(x) = x 3 − x + 1. Obtain fields of 4, 8, 9 and
27 elements by adjoining a root of f (x) to the field F where f (x) = g(x) of h(x) and F = F2 of F3 . Write down the multiplication tables for the field with 4 and 9 elements and show that the nonzero elements form a cyclic group.
Solution: Note that g and h are irreducible over F2 and F3 . Now, is θ is a root of g , then F2 (θ) = F2 /(g(x)) has 4 elements and F3 (θ) = F3 /(g(x)) has 9 elements. Furthermore, is θ 2 is a root of h, then F2 (θ2 ) = F2 /(h(x)) has 8 elements and F3 (θ2 ) = F3 /(h(x)) has 27 elements. The multiplication table for F2 /(g(x)) is
∼
∼
∼
·
0 1 x x+1
0 0 0 0 0
∼
1 0 1 x x+1
The multiplication table for F3 /(g(x)) is
4
x 0 x x+1 x
x+1 0 x+1 x x
13.2 Algebraic Extensions
·
0 1 2 x x+1 x+2 2x 2x + 1 2x + 2
0 0 0 0 0 0 0 0 0 0
1 0 1 2 x x+1 x+2 2x 2x + 1 2x + 2
2 0 2 1 2x 2x + 2 2x + 1 x x+2 x+1
x 0 x 2x 2x + 1 1 x+1 x+2 2x + 2 2
x+1 0 x+1 2x + 2 1 x+2 2x 2 x 2x + 1
x+2 0 x+2 2x + 1 x+1 2x 2 2x + 2 1 x
2x 0 2x x x+2 2 2x + 2 2x + 1 x+1 1
2x + 1 0 2x + 1 x+2 2x + 2 x 1 x+1 2 2x
2x + 2 0 2x + 2 x+1 2 2x + 1 x 1 2x x+2
In both cases, x is a generator of the cyclic group of nonzero elements. Exercise 3. Determine the minimal polynomial over Q for the element 1 + i.
∈ Q, its minimal polynomial is of degree at least 2. We try conjugation, (x − (1 + i))(x − (1 − i)) = x 2 − 2x + 2, which is irreducible by Eisenstein with p = 2. Therefore, the minimal polynomial is x 2 − 2x + 2. √ √ √ Exercise 4. Determine the degree over Q of 2 + 3 and 1 + 2 + 4. √ 3)2 = 4 + 4√ 3 + 3 = 7 + 4√ 3. Let θ = 2 + √ 3. Then Solution: First, note that (2 + √ √ θ2 − 4θ = 7 + 4 3 − 8 − 4 3 = −1, hence θ is a root of x2 − 4x + 1. Moreover, x2 − 4x + 1 √ 2 Q (because θ is irreducible over ∈ Q), so x − 4x + 1 is the minimal polynomial of 2 + 3. √ Therefore, 2 + 3 has degree 2 over Q. √ Now, let α = 2 and β = 1 + α + α2 . Then β ∈ Q(α), so Q ⊂ Q(β ) ⊂ Q(α). We have Solution: Since 1+i and obtain
3
3
3
[Q(α) : Q] = [Q(α) : Q(β )][Q(β ) : Q]. Note that [ Q(α) : Q] = 3 since α has minimal polynomial x3 2 over Q, so [Q(β ) : Q] = 1 or 3. For a contradiction, suppose [Q(β ) : Q] = 1, that is, Q(β ) = Q so β Q. Then
−
∈
β 2 = (1 + α + α2 )2 = 1 + 2α + 3α2 + 2α3 + α4 = 5 + 4α + 3α2 , where we used α 3 = 2. So β 2
− 3β = 5 + 2α + 3α2 − 3(1 + α + α2) = 2 − α, hence α = −β 2 + 3β + 2 ∈ Q(β ) = Q, a contradiction. Therefore, [ Q(β ) : Q] = 3. Exercise 5. Let F = Q(i). Prove that x 3 − 2 and x 3 − 3 are irreducible over F .
Solution: Since the polynomials have degree 3, if they were reducible they must have a linear factor, hence a root in F . Note that every element of F is of the form a + bi, where a, b Q. The roots of x 3 2 are 2, ζ 2 and ζ 2 2, where ζ is the primitive 3’rd root of unity, √ 3 1 i.e., ζ = exp(2πi/3) = cos(2π/3) + i sin(2π/3) = 2 + 2 . Since 3 Q, none of this elements is in F , hence x 3 2 is irreducible over F . Similarly, the roots of x 3 3 are 3, ζ 3 and ζ 2 3, and by the same argument non of this elements is in F . Hence x 3 3 is irreducible over F .
∈
−
√ √ 3
√ −
3
3
−
√ ∈ − −
√ √ 3
3
√ 3
Exercise 6. Prove directly from the definitions that the field F (α1 , α2 , . . . , αn ) is the composite of the fields F (α1 ), F (α2 ), . . . , F ( αn ). Solution: We have to prove that F (α1 , . . . , αn ) is the smallest field containing F (α1 ), . . . , F ( αn ). Clearly F (αi ) F (α1 , . . . , αn ) for all 1 i n. Now let K be a field such that F (αi ) K for all i. If θ is an element of F (α1 , . . . , αn ), then θ is of the form θ = a1 α1 + + a n αn , where a1 , . . . , an F . Every ai αi is in K , hence θ K . Thus F (α1 , . . . , αn ) K . Therefore,
⊂ ∈
≤ ≤
∈
5
⊂
···
⊂
13.2 Algebraic Extensions F (α1 , . . . , αn ) contains all F (αi ) and is contained in every field containing all F (αi ), hence F (α1 , . . . , αn ) is the composite of the fields F (α1 ), F (α2 ), . . . , F ( αn ).
√ √
√ √ √ √
Exercise 7. Prove that Q( 2 + 3) = Q ( 2, 3) [one inclusion is obvious, for the other consider ( 2+ 3)2 , etc.]. Conclude that [ Q( 2+ 3) : Q] = 4. Find an irreducible polynomial satisfied by 2 + 3.
√ √ √ √
√ √
√ √ √ √ √ 3 √
√ √ ⊂ √ √ √ √ √ √ √ √ 4
Solution: Since 2 + 3 is in Q( 2, 3), clearly Q( 2 + 3) Q( 2, 3). For the other direction we have to prove that 2 and 2 are in Q( 2 + 3). Let θ = 2 + 3. Then θ2 = 5 + 2 6, So
θ = 11 2 + 9 3
and
θ = 37 + 15 6.
√
√ √ − 9θ), 3 = 12 (11θ − θ3) and θ4 = 49 + 20 6. √ Q(θ) and √ 3 ∈ Q(θ), so Q(√ 2, √ 3) ⊂ Q(√ 2 + √ 3). The equality follows. Therefore √ 2 ∈ √ √ √ Hence, [ Q( 2 + 3) : Q] = [Q( 2, 3) : Q] = 4. 1 2 = (θ3 2
We also have
√ √ − 10θ2 = (49 + 20 6) − 10(5 + 2 6) = −1, so θ4 − 10θ2 + 1 = 0. √ √ √ √ Since [ Q( 2+ 3) : Q] = 4, then x 4 − 10x2 +1 is irreducible over Q, and is satisfied by 2+ 3. θ4
Exercise 8. Let F be a field of characteristic = 2. Let D 1 and D 2 be elements of F , neither of which is a square in F . Prove that F ( D1 , D2 ) is of degree 4 over F if D1 D2 is not a square in F and is of degree 2 over F otherwise. When F ( D1 , D2 ) is of degree 4 over F the field is called a biquadratic extension of F .
√ √
√ √
√ √ D ) can be written in the form
Solution: The elements of F ( D1 ,
2
∈ F. [F ( D , D ) : F ] = [F ( D , D ) : F ( D )][F ( D ) : F ]. a + b D1 + c D2 + d D1 D2 , where a,b,c,d
We have
√
1
2
√ √ √ √ √ √ ∈ √
1
2
1
1
√ √ −
Since [F ( D1 ) : F ] = 2, then [F ( D1 , D2 ) : F ] can be 2 or 4. Now, [F ( D1 , D2 ) : F ] = 2 if and only if [F ( D1 , D2 ) : F ( D1 )] = 1, and that occurs exactly when x 2 D2 is reducible in F ( D1 ) (i.e., when D2 F ( D1 )), that is, if there exists a, b F such that
√
(a + b D1 )2 = D 2 ,
∈
so that a2 + 2ab D1 + b2 D12 = D 2 .
√ ∈
Note that ab = 0 as ab = 0 implies D1 F , contrary to the hypothesis. Then a = 0 or b = 0. If b = 0, then D2 is a square in F , contrary to the hypothesis. If a = 0, then b 2 D1 = D 2 , and thus D 1 D2 = ( Db )2 , so D 1 D2 is a square in F . So, x 2 D2 is reducible in F ( D1 ) if and only if D 1 D2 is a square in F . The result follows.
√
−
2
Exercise 9. Let F be a field of characteristic = 2. Let a, b be elements of the field F with b not a square in F . Prove that a necessary and sufficient condition for a + b = m + n for some m and n in F is that a2 b is a square in F . Use this to determine when the field Q( a + b)(a, b Q) is biquadratic over Q.
√ √
√
− ∈ √ √ b = m +n+2√ mn. √ √ Solution: Suppose a + b = m+ n for some m, n ∈ F , then a + √ √ √ √ √ Since b is not a square in F , this means b = 2 mn. We also have a + b − n = m, so
√
√
√ √ √ b = 2 n( a + b − n). 6
13.2 Algebraic Extensions Hence,
√
√ b) − 2n √ √ ( b + 2n)2 = 4n(a + b) √ √ b + 4n b + 4n2 = 4n(a + b) b + 4n2 − 4na = 0 √ 4a ± 16a2 − 16b n = b = 2
⇒ ⇒ ⇒ ⇒ ⇒
n(a +
8
a − b = ± 2n . 2
a √ Therefore, since a and n are in F , a2 − b is in F . √ 2 − b is a square in F , so that Now suppose that a a2 − b ∈ F . We prove that there exists √ √ √ m, n ∈ F such that a + b = m + n. Let √ √ a + a2 − b a − a2 − b m = and n = . 2
2
√ √ √ Note that m and n are in F as char(F ) = 2. We claim a + b = m + n. Indeed, we have √ √ √ √ √ (a + b) + 2 a − b + (a − b) a+ b+ a− b 2
m =
and n =
=(
4
√ √ √ (a + b) − 2 a2 − b + (a − b)
=(
4
Thus
√ m =
a + √ b + a − √ b 2
and
)2 ,
2
a + √ b − a − √ b 2
)2 .
a + √ b − a − √ b
√ n =
2
.
Therefore,
√ m + √ n =
a + √ b + a − √ b a + √ b − a − √ b +
2
=
2
a+
√
b,
as claimed. Now, we this to determine when the field Q( a + b), a, b Q, is biquadratic over Q. If 2 a b is a square in Q and b is not, we have Q( a + b) = Q( m + n) = Q ( m, n), so by last exercise Q( a + b) is biquadratic over Q when a 2 b is a square in Q, and neither b, m, n or mn are squares in Q. Since
−
√ √
√
mn =
a+
∈√
√ √
−
√ a2 − b a − √ a2 − b 2
√
2
=
b , 4
√ then mn is never a square when b isn’t. Thus, ( a + b) is biquadratic over exactly when a − b is a square in and neither b, m nor n is a square in . √ Exercise 10. Determine the degree of the extension ( 3 + 2 2) over . √ √ Solution: Note that 3 + 2 2 = 3 + 8. Recalling last exercise with a = 3 and b = 8, we have a − b = 9 − 8 = 1 is a square in and b = 8 is not. Hence, we find (m = 2 and n = 1 √ √ √ 2) = (√ 2) and the degree of from last exercise) 3 + 8 = 2 + 1. Therefore, ( 3 + 2 √ the extension ( 3 + 2 2) over is 2. Q
2
Q
Q
Q
Q
2
Q
Q
Q
Q
Q
7
Q
13.2 Algebraic Extensions
√ √ − √ √ −
Exercise 11. (a) Let 3 + 4i denote the square root of the complex number 3 + 4i that lies in the first quadrant and let 3 4i denote the square root of 3 4i that lies in the fourth quadrant. Prove that [ Q( 3 + 4i + 3 4i) : Q] = 1. (b) Determine the degree of the extension Q( 1 + 3+ 1 3).
− √ − − √ − Solution: (a) First, note that the conjugation map a + bi → a − bi is an isomorphism of C , so it takes squares roots to square roots, and maps numbers of the first quadrant to the fourth √ (and reciprocally). Since 3 + 4i is the square root of 3 + √ 4i in the first quadrant, its conjugate √ √ 3 − 4i is the square of root of 3 − 4i in the fourth quadrant, so is 3 − 4i. Hence 3 + 4i and √ √ are conjugates each other. Now, we use Exercise 9 again. Note that 3 + 4i = 3 + −16. With a = 3 and b = −16, we have a2 −√ b = 25 is a square in Q and b = −16 is not. Hence, √ we m = 1 and n = − 4 and thus √ 3 + 4i = 1 + − 4 = 1 + √ 2i. Furthermore, we find √ 3 find √ √ − 4i = 1 − 2i. Therefore, 3 + 4i√ + 3 − 4i = 4, i.e., 3 + 4i + 3 − 4i ∈ Q. √ (b) Let θ = 1 + −3 + 1 − −3. Then 2
θ =( Since x 2
√ 2 √ √ √ √ 1 + −3 + 1 − −3) = (1 + −3) + (2 1 + 3) + (1 − −3) = 6.
− 6 is irreducible over Q (Eisenstein p = 2), then θ has degree 2 over Q.
Exercise 12. Suppose the degree of the extension K/F is a prime p. Show that any subfield E of K containing F is either K or F . Solution: Let E be a subfield of K containing F . Then [K : F ] = [K : E ][E : F ] = p. Since p is prime, either [K : E ] = 1 or [E : F ] = 1. The result follows. 13. Suppose F = Q(α1 , α2 , . . . , αn ) where α i2 ∈ Q for i = 1, 2, . . . , n. Prove that √ 2 Exercise ∈ F . Solution: Note that, for all 1 ≤ k ≤ n, we √ have [ Q(α1 , . . . , αk ) : Q√ (α1 , . . . , αk−1 )] = 1 or 2. √ m ⊂ Q( 2) ⊂ F , so [ Q( 2) : Q ] Then [F : Q ] = 2 for some m ∈ N. Suppose 2 ∈ F . Then Q √ m divides [F : Q], that is, 3 divides 2 , a contradiction. Hence, 2 ∈ F . 3
3
3
3
3
Exercise 14. Prove that if [F (α) : F ] is odd then F (α) = F (α2 ).
Solution: Since α 2 F (α), clearly F (α2 ) F (α). Thus we have to prove α F (α2 ). For this purpose, consider the polynomial p(x) = x 2 α2 , so that p(α) = 0. Note that α F (α2 ) if and only if p(x) is reducible in F (α2 ). For a contradiction, suppose p(x) is irreducible in F (α2 ), so that [F (α) : F (α2 )] = 2. Thus
∈
⊂ −
∈
∈
[F (α) : F ] = [F (α) : F (α2 )][F (α2 ) : F ] = 2[F (α2 ) : F ], so [F (α) : F ] is even, a contradiction. Therefore, p(x) is reducible in F (α2 ) and α
∈ F (α2).
Exercise 15. A field F is said to be formally real if 1 is not expressible as a sum of squares in F . Let F be a formally real field, let f (x) F [x] be an irreducible polynomial of odd degree and let α be a root of f (x). Prove that F (α) is also formally real. [Pick α a counterexample of minimal degree. Show that 1 + f (x)g(x) = ( p1 (x))2 + + ( pm (x))2 for some pi (x), g(x) F [x] where g(x) has odd degree < deg f . Show that some root β of g has odd degree over F and F (β ) is not formally real, violating the minimality of α.]
∈
−
−
∈
···
Solution: We follow the hint. Suppose there exists a counterexample. Let α be of minimal degree such that F (α) is not formally real and α having minimal polynomial f of odd degree, say deg f = 2k + 1 for some k N. Since F (α) is not formally real, then 1 can be express as
∈
−
8
13.2 Algebraic Extensions
∼
a sum of squares in F (α) = F [x]/((f (x))). Then, the exists polynomials p 1 (x), . . . , pm (x), g(x) such that 1 + f (x)g(x) = ( p1 (x))2 + + ( pm (x))2 .
−
···
As every element in F [x]/((f (x)) can be written as a polynomial in α with degree less than deg f , we have deg pi < 2k + 1 for all i. Thus, the degree in the right hand of the equation is less than 4k + 1, so deg g < 2k + 1 as well. We prove that the degree of g is odd by proving that the degree of ( p1 (x))2 + + ( pm (x))2 is even, because then the equation 1 + f (x)g(x) = ( p1 (x))2 + + ( pm (x))2 implies the result. Let d be the maximal degree over all p i , we prove that x 2d is the leading term of ( p1 (x))2 + + ( pm (x))2 . Note that x 2d is a sum of squares (of the leading coefficients of the p i ’s of maximal degree). Now, since F is formally real, 0 can’t be −1 (ai/al )2 = 1. Therefore expressed as a sum of squares in F . Indeed, if li=1 a2i = 0, then li=1 x2d = 0, so the degree of ( p1 (x))2 + + ( pm (x))2 is 2d, as claimed. Hence, the degree of g must be odd by the assertion above. Then g must contain an irreducible factor of odd degree, say h(x). Since deg g < deg f , we have deg h < deg f as well. Let β be a root of h(x), hence a root of g (x). Then
···
···
−
···
···
−
−1 + h(x) f (x)g(x) = ( p1 (x))2 + ··· + ( pm (x))2 , h(x) so −1 is a square in F [x]/((h(x)) ∼ = F (β ), which means F (β ) is not formally real.
Therefore, β is a root of an odd degree polynomial h such that F (β ) is not formally real. Since deg h < deg f , this contradicts the minimality of α. The result follows. Exercise 16. Let K/F be an algebraic extension and let R be a ring contained in K and containing F . Show that R is a subfield of K containing F . Solution: Let r R be nonzero. Since r is algebraic over F , there exist an irreducible polynomial p(x) = a 0 + a1 x + + xn F [x] such that p(r) = 0. Note that a0 = 0 since p is 1 n−1 irreducible. Then r −1 = a− + + a1 ). Since a i F R and r R, we have r −1 R. 0 (r
∈
−
···
∈ ···
∈ ⊂
∈
∈
Exercise 17. Let f (x) be an irreducible polynomial of degree n over a field F . Let g(x) be any polynomial in F [x]. Prove that every irreducible factor of the composite polynomial f (g(x)) has degree divisible by n. Solution: Let p(x) be an irreducible factor of f (g(x)) of degree m. Let α be a root of p(x). Since p is irreducible, then [F (α) : F ] = deg p(x) = m. Now, since p(x) divides f (g(x)), we have f (g(α)) = 0 and thus g(α) is a root of f (x). Since f is irreducible, this means n = [F (g(α)) : F ]. Note that F (g(α)) F (α). Therefore,
⊂
n = [F (g(α)) : F ] = [F (g(α)) : F (α)][F (α) : F ] = [F (g(α)) : F (α)] m,
·
so m divides n, that is, deg p divides deg f . Exercise 18. Let k be a field and let k(x) be the field of rational functions in x with P (x) coefficients from k. Let t k(x) be the rational function with relatively prime polynomials Q(x) P (x), Q(x) k[x], with Q(x) = 0. Then k(x) is an extension of k(t) and to compute its degree it is necessary to compute the minimal polynomial with coefficients in k (t) satisfied by x. (a) Show that the polynomial P (X ) tQ(X ) in the variable X and coefficients in k(t) is irreducible over k(t) and has x as a root. [By Gauss Lemma this polynomial is irreducible in (k(t))[X ] if and only if it is irreducible in (k[t])[X ]. Then note that (k[t])[X ] = (k[X ])[t].] (b) Show that the degree of P (X ) tQ(X ) as a polynomial in X with coefficient in k(t) is the maximum of the degree of P (x) and Q(x). P (x) (c) Show that [k(x) : k(t)] = [k(x) : k( )] = max(degP (x), degQ(x)). Q(x)
∈
∈
−
−
9
13.2 Algebraic Extensions Solution: (a) We follow the hint. Since k[t] is an UFD and k(t) is its field of fractions, then, by Gauss Lemma, P (X ) tQ(X ) is irreducible in k((t))[X ] is and only if it is irreducible in (k[t])[X ]. Note that (k[t])[X ] = (k[X ])[t]. Since P (X ) tQ(X ) is linear in (k[X ])[t], is clearly irreducible in (k[X ])[t] (i.e., in (k[t])[X ]), hence in (k(t))[X ]. Thus, P (X ) tQ(X ) is irreducible P (x) in k (t). Now, x is clearly a root of P (X ) tQ(X ) since P (x) tQ(x) = P (x) Q(x) = Q(x) P (x) P (x) = 0. (b) Let n = max degP (x), degQ(x) . Write
−
−
−
−
{
−
−
−
}
P (x) = a n xn +
··· + a1x + a0
Q(x) = b n xn +
and
··· + b1x + b0,
where a i , bi k for all i, so at least one of a n or b n is nonzero. The degree of P (X ) tQ(X ) is clearly n, we prove is n. If an or bn is zero then clearly deg (P (X ) tQ(X )) = n. Suppose an , bn = 0. Then an , bn k, but t k (as t k(x)), it cannot be that an = tbn . Thus n (an tbn )X = 0, so the degree of P (X ) tQ(X ) is n. (c) Since P (X ) tQ(X ) is irreducible over k(t) and x is a root by part (a), then [k(x) : k(t)] = degP (X ) tQ(X ), and this degree equals max degP (x), degQ(x) by part (b).
−
≤
∈
∈
−
∈
−
−
∈
−
−
{
}
Exercise 19. Let K be an extension of F of degree n. (a) For any α K prove that α acting by left multiplication on K is an F -linear transformation of K . (b) Prove that K is isomorphic to a subfield of the ring of n n matrices over F , so the ring of n n matrices over F contains an isomorphic copy of every extension of F of degree n.
∈
×
×
≤
Solution: (a) Fix α in K . Since K is (in particular) a commutative ring, we have α(a+b) = αa + αb and α(λa) = λ(αa) for all a, b, λ K . If, in particular, λ F , we have the result. (b) Fix a basis for K as a vector space over F . By part (a), for every α K we can associate a F -linear transformation T α . Denote by T α the matrix of T α with respect to the basis fixed above. Then define ϕ : K M n (F ) by ϕ(α) = T α . We claim ϕ is an isomorphism. Indeed, if α, β K , then T (α+β) (k) = (α + β )(k) = αk + βk = T α (k) + T β (k) for every k K , hence T (α+β) = T α + T β . We also have T (αβ) (k) = (αβ )(k) = αkβk = T α (k)T β (k) for every k K , so T (αβ) = T α T β . Thus ϕ(α + β ) = ϕ(α) + ϕ(β ) and ϕ(αβ ) = ϕ(α)ϕ(β ) (since the basis is fixed), so ϕ is an homomorphism. Now, if ϕ(α) = ϕ(β ), then αk = βk for every k K , so letting k = 1 we find that ϕ is injective. Therefore, ϕ(K ) is isomorphic to a subfield of M n (F ), so the ring M n (F ) contains an isomorphic copy of every extension of F of degree n.
∈
→
∈
∈
∈
∈
∈
∈
≤
Exercise 20. Show that if the matrix of the linear transformation "multiplication by α" considered in the previous exercise is A then α is a root of the characteristic polynomial for A. This gives an effective procedure for determining an equation of degree n satisfies by a element α in an extension of F of degree n. Use this procedure to obtain the monic polynomial of degree 3 satisfied by 2 and by 1 + 2 + 4.
√
√ √
3
3
3
Solution: The characteristic polynomial of A is p(x) = det(Ix A). For every k K , we have (Iα A)k = αk Ak = αk αk = 0, so det(Iα A) = 0 in K . Therefore, p(α) = 0. Now, consider the field Q( 2) with basis 1, 2, 4 over Q. Denote the elements of this basis by e1 = 1, e 2 = 2 and e3 = 4. Let α = 2 and β = 1 + 2 + 4. Then α(e1 ) = e 2 , α(e2 ) = e3 and α(e3 ) = 2e1 . We also have β (e1 ) = e1 + e 2 + e 3 , β (e2 ) = 2e1 + e 2 + e 3 and β (e3 ) = 2e1 + 2e2 + e3 . Thus, the associated matrices of the their linear transformations are, respectively, 0 0 2 1 2 2 Aα = 1 0 0 and Aβ = 1 1 2 . 0 1 0 1 1 1
−
− √
√ −
− √ √ { √ }
3
3
3
√ 3
10
∈
3
3
−
√ √ 3
3
13.2 Algebraic Extensions The characteristic polynomial of Aα is x3 2, hence is the monic polynomial of degree 3 satisfied by α = 2. Furthermore, the characteristic polynomial of A β is x 3 3x2 3x 1, hence is the monic polynomial of degree 3 satisfied by β = 1 + 2 + 4.
−
√ 3
−
√ √ 3
√ √
3
− −
√
Exercise 21. Let K = Q( D) for some squarefree integer D. Let α = a + b D be an element of K . Use the basis 1, D for K as a vector space over Q and show that the matrix of the linear transformation "multiplication by α" on K considered in the previous exercises has a bD a bD the matrix . Prove directly that the map a + b D is an isomorphism of b a b a the field K with a subfield of the ring of 2 2 matrices with coefficients in Q. Solution: The matrix of the linear transformation "multiplication by α" on K is found by acting of α in the basis 1, D. We have α(1) = α = a + b D and α( D) = a D + bD. Hence a bD a bD the matrix is . Now let ϕ : K M 2 (Q) be defined by ϕ(a + b D) = . b a b a We have
√ →
×
√
√
√
√
ϕ(a+b D+c+d D) =
√
√
√
→
a + c
(b + d)D b+d a+c
a bD c dD =
b
a
+
d
c
√
√
= ϕ(a+b D)+ϕ(c+d D),
and
√
√
ϕ((a + b D) (c + d D)) =
·
ac + bdD ad + bc
(ad + bc)D ac + bdD
a bD c dD =
b
a
√
·
d
c
√
= ϕ(a + b D)ϕ(c + d D), so ϕ is an homomorphism. Since K is a field, its ideals are 0 and K , so ker(ϕ) is trivial or K . Since ϕ(K ) is clearly non-zero, then ker(ϕ) = K and thus ker(ϕ) = 0 . Hence, ϕ is injective. Therefore, ϕ is an isomorphism of K with a subfield of M 2 (Q).
{}
{}
Exercise 22. Let K 1 and K 2 be two finite extensions of a field F contained in the field K . Prove that the F -algebra K 1 F K 2 is a field if and only if [K 1 K 2 : F ] = [K 1 : F ][K 2 : F ]. Solution: Define ϕ : K 1 K 2 K 1 K 2 by ϕ(a, b) = ab. We prove that ϕ is F -bilinear. Let a, a1 , a2 K and b, b1 , b2 K 2 . Then
∈
∈
⊗ × →
ϕ((a1 , b) + (a2 , b)) = ϕ(a1 + a2 , b) = (a1 + a2 )b = a 1 b + a2 b = ϕ(a1 , b) + ϕ(a2 , b), and ϕ((a,1 b) + (a, b2 )) = ϕ(a, b1 + b2 ) = a(b1 + b2 ) = ab 1 + ab1 = ϕ(a, b1 ) + ϕ(a, b2 ). We also have, for r F , ϕ(ar,b) = (ar)b = a(rb) = ϕ(rb). Therefore, ϕ is a F -bilinear map. Hence, ϕ induces a F -algebra homomorphism Φ : K 1 F K 2 K 1 K 2 . We use Φ to prove both directions. Note that K 1 F K 2 have dimension [K 1 : F ][K 2 : F ] as a vector space over F . First, we suppose [K 1 K 2 : F ] = [K 1 : F ][K 2 : F ] and prove K 1 F K 2 is a field. In this case K 1 F K 2 and K 1 K 2 have the same dimension over F . Let L = Φ(K 1 F K 2 ). We claim L = K 1 K 2 , i.e. Φ is surjective. Note that L contains K 1 and K 2 . Since L is a subring of K 1 K 2 containing K 1 (or K 2 ), then L is a field (Exercise 16). Hence, L is a field containing both K 1 and K 2 . Since K 1 K 2 is the smallest such field (by definition), we have L = K 1 K 2 . Therefore Φ is surjective, as claimed. So, Φ is an F -algebra surjective homomorphism between F -algebras of the same dimension, hence is an isomorphism. Thus, K 1 F K 2 is a field. Now suppose that K 1 F K 2 is a field. In this case Φ is a field homomorphism. Therefore, Φ is either injective or trivial. It is clearly nontrivial since Φ(1 1) = 1, so it is injective. Hence, [K 1 : F ][K 2 : F ] [K 1 K 2 : F ]. As we already have [K 1 K 2 : F ] [K 1 : F ][K 2 : F ] (Proposition 21 of the book), the equality follows.
∈
⊗
⊗
→
⊗
⊗
⊗ ⊗
⊗
≤
11
≤
⊗
13.3 Classical Straightedge and Compass Constructions
§13.3
Classical Straightedge and Compass Constructions
Exercise 1. Prove that it is impossible to construct the regular 9-gon. Solution: Suppose the 9-gon is constructible. It has angles of 40◦ . Since we can bisect an angle by straightedge and compass, the angle of 20◦ would be constructible. But then cos 20◦ and sin 20◦ would be constructible too, a contradiction (see proof of Theorem 24). Exercise 2. Prove that Archimedes’ construction actually trisect the angle θ. [Note the isosceles triangles in figure to prove that β = γ = 2α.]
Solution: Let O,P,Q and R be the points marked in the figure below.
Then α = ∠QP O, β = ∠RQO, γ = ∠QRO, and θ is an exterior angle of P RO. Since P QO is isosceles, then α = ∠QP O = ∠QOP . Since β is an exterior angle of P QO, it equals the sum of the two remote interior angles, i.e., equals ∠QP O + ∠QOP . This two angles equals α, hence β = 2α. Now, since QRO is isosceles, then β = γ . Finally, since θ is an exterior angle of P RO, equals the sum of the two remote interior angles, which are α and γ . Therefore, θ = α + γ = α + β = 3α.
Exercise 3. Prove that Conway’s construction indicated in the text actually constructs 2k 1/3 and 2k 2/3 . [One method: let (x, y) be the coordinates of the point C , a the distance from y 1 k2 B and C and b the distance from A to D; use similar triangles to prove (a) = , (b) 1 1+a x b+k y 1 k2 = , (c) = , and also show that (d) (1 k2 ) + (b + k)2 = (1 + a)2 ; solve a 1+a x k 3k these equations for a and b.] Solution: We follow the hint. The distances a, b, x, y and x k are marked in the figure below.
−
√ −
√ −
−
−
12
13.4 Splitting Fields and Algebraic Closures From the figure, using similar triangles for (a), (b) and (c), and Pythagoras Theorem for (d), the 4 relations are clear. Hence, we have y =
√ 1 − k2 1+a
,
b+k x = a , 1+a
y x
−k
=
√ 1 − k2 3k
and
(1
− k2) + (b + k)2 = (1 + a)2.
√ 1 − k2 = y(1 + a) = 3ky implies 3k = (x − k)(1 + a). From the equation for x above, we x−k a(b+a) find 3k = ( 1+a − k)(1 + a) = a(b + k) − k(1 + a), so b + k = 4k+ka a . Using this in the last So,
equation and reducing, we get
(1
⇒ ⇒ ⇒ ⇒
− k2) + (b + k)2 = (1 + a)2
− k2) + ( 4k +a ka )2 = (1 + a)2 a2 (1 − k 2 ) + (4k + ka)2 = a 2 (1 + a)2 (1
a2
− (ka)2 + (4k)2 + 8k2a + (ka)2 = a2 + 2a3 + a4 a4 + 2a3 − 8k 2 a − 16k 2 = 0.
We let a = 2y to obtain h4 + h3
− k2h − k2 = 0. We find h = k 2/3 , hence a = 2k 2/3 . From b = 4k+ka − k, we find b = 2k1/3. Therefore, we can a construct 2k 1/3 and 2k2/3 using Conway’s construction.
Exercise 4. The construction of the regular 7-gon amounts to the constructibility of cos(2π/7). We have shall see later (Section 14.5 and Exercise 2 of Section 14.7) that α = 2 cos(2π/7) satisfies the equation x 3 + x2 2x 1 = 0. Use this to prove that the regular 7-gon is not constructible by straightedge and compass. Solution: Let p(x) = x 3 + x2 2x 1 and α = 2 cos(2π/7). By Rational Root Theorem, if p has a root in Q, it must be 1 since it must divide its constant term. But p(1) = 1 and p( 1) = 1, so p is irreducible over Q. Therefore, α is of degree 3 over Q, hence [Q(α) : Q] cannot be a power of 2. Since we can’t construct α, the regular 7-gon is not constructible by straightedge and compass.
±
−
− − − −
−
Exercise 5. Use the fact that α = 2cos(2π/5) satisfies the equation x2 + x 1 = 0 to conclude that the regular 5-gon is constructible by straightedge and compass. Solution: Let p(x) = x2 + x 1 = 0 and α = 2cos(2π/5). By Rational Root Theorem, if p has a root in Q, it must be 1. Since p(1) = 1 and p( 1) = 1, p is irreducible over Q. Hence, α is of degree 2 over Q, so it is constructible. We can bisect an angle by straightedge and compass, so β = cos(2π/5) is also constructible. Finally, as sin(2π/5) = 1 cos2 (2π/5), sin(2π/5) is also constructible. Therefore, the regular 5-gon is constructible by straightedge and compass.
−
− ±
−
−
−
§13.4
Splitting Fields and Algebraic Closures
Exercise 1. Determine the splitting field and its degree over Q for x 4
− 2.√ √ √ √ −i 2. Hence,√ the Solution: Let f (x) = x4 − 2. The roots of f are 2, − 2, i 2 and √ √ splitting field of f is Q(i, 2). So, the splitting field of f has degree [ Q(i, 2) : Q] = [Q(i, 2) : √ √ √ Q( 2)][Q Q, ( 2) : Q] over Q. Since 2 is a root of √ the irreducible polynomial x4 − 2 over√ √ 2 ∈ Q( 2), then x√ + 1 is irreducible over Q( 2) then [Q( 2) : Q] = 4. Furthermore, since i √ √ having i as a root, so [ Q(i, 2) : Q( 2)] = 2. Therefore, [ Q(i, 2) : Q] = 8. 4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
Exercise 2. Determine the splitting field and its degree over Q for x 4 + 2. 13
13.4 Splitting Fields and Algebraic Closures Solution: Let f (x) = x 4 + 2. Let K be the splitting field of f and let L be the splitting field of x4 2, that is,√ L = Q (i, 2) (last exercise). We claim K = L, so that [K : Q ] = 8 by √ last exercise. Let ζ = 22 + i 22 . First we prove ζ L and ζ K , then we prove K = L. We prove ζ L. This is easy. Let θ = 2. Since θ L, then θ 2 = 2 L. We also have i L, so 2, i L implies ζ L. We prove ζ K . We have to prove i K and 2 K . Let α be a root of x 4 + 2, so that α4 = 2. Let β be a root of x4 1, so that β 4 = 1. Then (αβ )4 = α4 β 4 = 2, hence αβ is also a root of x 4 + 2. Since the roots of x 4 1 are 1, i, the roots of x 4 + 2 are α and iα. Since K is generated over Q by there roots, then iα/α = i K . Now let γ = α 2 K . Since γ 2 = α 4 = 2, then γ is a root of x 2 + 2. Since the roots of x2 + 2 are i 2 and i 2, then γ is one of this roots. In either case γ /i K , which implies 2 K . Therefore, ζ K . Now we prove L = K proving both inclusions. Let α be a root of x 4 + 2 and θ be a root of x 4 2. Then α 4 = 2 and θ 4 = 2. Note that ζ 2 = i, so ζ 4 = 1. Hence, (ζθ)4 = ζ 4 θ4 = 2, so ζθ is a root of x4 + 2. Then, as we proved earlier, the roots of x4 + 2 are ζθ and iζθ. We also have (ζα)4 = ζ 4 α4 = 2, so ζ α is a root of x4 2. Then, by last exercise, the roots of x4 2 are ζα and iζα. Now, since ζ and α are in K , we have ζα K . We also have i K , so all roots of x4 2 are in K . Since L is generated by these roots, then L K . Similarly, ζ and θ are in L, so ζ θ L; since i L, then all roots of x4 + 2 are in L. Since K is generated by these roots, we have K L. Therefore, K = L, so [K : Q] = 8.
√ 4
−
∈
−
√ ∈ ∈ ∈ √ ∈ ∈
∈
−
−
−
± ±
∈
−
−
√ ∈
4
√ ∈∈ ∈
±
∈
±
⊂
−
∈ √ ∈
−
−
∈
∈
−
−
∈
√
± ± ∈√ − ∈
±
± −
⊂
Exercise 3. Determine the splitting field and its degree over Q for x 4 + x2 + 1. Solution: Let f (x) = x 4 +√ x2 + 1. Note that f (x) = (x2 + x +1)(x2 x + 1), so the roots of √ f are 12 i 23 . Let w = 21 i 23 , so this roots are w, w,w, w, where w denotes the complex
± ±
−
−
√ 3
−
−
conjugate of w (i.e., w = 21 + i 2 ). Hence, the splitting field of f is Q(w, w). Since w + w = 1, then Q(w, w) = Q (w). Furthermore, w is a root of x2 x + 1, that is irreducible over Q since w Q. Therefore, the degree of the splitting field of f is [ Q(w) : Q] = 2.
−
∈
Exercise 4. Determine the splitting field and its degree over Q for x 6
− 4.
−
√
− √ √ − √ √ − √ √ √ − √
√
Solution: Let f (x) = x 6 4. Note that f (x) = (x3 2)(x3 + 2). The roots of x 3 2 are 2, ζ 2 and ζ 2 2, where ζ is the primitive 3’rd root of unity, i.e., ζ = exp(2πi/3) = cos(2π/3) + √ 3 1 i sin(2π/3) = 2 + 2 . Furthermore, the roots of x3 +2 are 2, ζ 2 and ζ 2 2. Therefore, the splitting field of f is Q(ζ, 2). Then [ Q(ζ, 2) : Q] = [Q(ζ, 2) : Q( 2)][Q( 2) : Q]. We have that 2 is a root of the irreducible polynomial x 3 2 over Q , so 2 has degree 3 over Q . Furthermore, ζ is a root of x2 + x + 1, irreducible over Q( 2), so ζ has degree 2 over Q( 2). Hence, the degree of the splitting field of f is [ Q(ζ, 2) : Q] = 6.
√ 3
−
√ − √ 3
−
3
√ 3
√ 3
3
3
3
3
3
3
3
3
3
√
3
3
Exercise 5. Let K be a finite extension of F . Prove that K is a splitting field over F if and only if every irreducible polynomial in F [x] that has a root in K splits completely in K [x]. [Use Theorems 8 and 27.] Solution: We follow the hint. First suppose that K is a splitting field over F . Hence, there exists f (x) F [x] such that K is the splitting field of f . Let g(x) be an irreducible polynomial in F [x] with a root α K . Let β be any root of g. We prove β K , so that g splits completely ∼ F (β ) such that ϕ(α) = β . in K [x]. By Theorem 8, there is an isomorphism ϕ : F (α) Furthermore, K (α) is the splitting field for f over F (a), and K (β ) is the splitting field for f ∼ K (β ). Since over F (β ). Therefore, by Theorem 28, ϕ extends to an isomorphism σ : K (α) K = K (α), then [K : F ] = [K (α) : F ] = [K (β ) : F ], so K = K (β ). Thus, β K . Now suppose that every irreducible polynomial in F [x] that has a root in K splits completely in K [x]. Since [K : F ] is finite, then K = F (α1 , . . . , αn ) for some α1 , . . . , αn . For every 1 i n, let pi be the minimal polynomial of αi over F , and let f = p 1 p2 pn . Since every
∈
∈
∈ −→
∈
≤ ≤
···
14
−→
13.5 Separable and Inseparable Extension αi is in K , every p i has a root in K , hence splits completely in K . Therefore, f splits completely in K and K is generated over F by its roots, so K is the splitting field of f (x) F [x].
∈
Exercise 6. Let K 1 and K 2 be finite extensions of F contained in the field K , and assume both are splitting fields over F . (a) Prove that their composite K 1 K 2 is a splitting field over F . (b) Prove that K 1 K 2 is a splitting field over F . [Use the preceding exercise.]
∩
Solution: (a) Let K 1 be the splitting field of f 1 (x) F [x] over F and K 2 the splitting field of f 2 (x) F [x] over F . Thus, K 1 is generated over F by the roots of f 1 , and K 2 is generated over F by the roots of f 2 . Therefore, f 1 f 2 splits completely in K 1 K 2 and K 1 K 2 is generated over F by its roots, hence is the splitting field of f 1 f 2 (x) F [x]. (b) We follow the hint. By last exercise, we have to prove that every irreducible polynomial in F [x] that has a root in K 1 K 2 splits completely in (K 1 K 2 )[x]. So, let f (x) be an irreducible polynomial in F [x] that has a root, say α, in K 1 K 2 . By last exercise, f splits completely in K 1 and splits completely in K 2 . Since K 1 and K 2 are contained in K , by the uniqueness of the factorization of f in K , the roots of f in K 1 must coincide with its roots in K 2 . Hence, f splits completely in (K 1 K 2 )[x].
∈
∈
∈
∩
∩ ∩
∩
§13.5
Separable and Inseparable Extension
Exercise 1. Prove that the derivative Dx of a polynomial satisfies Dx (f (x) + g(x)) = Dx (f (x))+Dx (g(x)) and Dx (f (x)g(x)) = D x (f (x))g(x)+Dx (g(x))f (x) for any two polynomials f (x) and g(x). Solution: Let f (x) = an xn + + a 1 x + a 0 and g(x) = bm xm + + b 1 x + b 0 be two polynomials. Suppose, without any loss of generality, that n m. Thus, we can write g(x) = bn xn + + b1 x + b0 , where some of the last coefficients b i could be zero. We have f (x) + g(x) = (an + bn )xn + + (a1 + b1 )x + (a0 + b0 ), so
···
···
···
≥
···
Dx (f (x) + g(x)) = n(an + bn )xn−1 +
··· + 2(a2 + b2)x + (a1 + b1) = Dx(f (x)) + Dx(g(x)). n k=0 k n k
a b − , so that a x )( b x ) = ( a b − )x = c x . f (x)g(x) = (
Now we prove the formula for the product. Let c n = n
n
k
k=0
2n
k
k
l
2n
k
l
k l k
k=0
l=0 k=0
l
l
l=0
Hence, 2n
2n 1
− D (f (x)g(x)) = D ( cx)= (l + 1)c x
x
l
l
l=0
l l+1 x ,
l=0
so the coefficient of x l in D x (f (x)g(x)) is (l + 1)cl+1. Now, we have Dx (f (x)) = na n xn−1 + +2a2 x+a1 , and Dx (g(x)) = nb n xn−1 + So (recall product of polynomials, page 295 of the book)
···
n
n
ka x − )( b x ) D (f (x))g(x) = ( − − = ( (k + 1)a x )( b x )= ( (k + 1)a x
k
k=1 n 1
k 1
k
k
k=0
n
k+1
k=0
···+2b2x+b1.
k
2n 1
k
k=0
15
l
k
l=0 k=0
l k+1 bl k )x ,
−
13.5 Separable and Inseparable Extension and n
n
f (x)D (g(x)) = ( a x )( kb x − ) − =( a x )( (k + 1)b x
k
k
k=0 n
k
k 1
k=1 n 1
k
k
k=0
2n 1
k+1
k=0
l
− x )= ( a (l − k + 1)b − k
l k+1 )x
k
l
.
l=0 k=0
Therefore, the coefficient of x l in D x (f (x))g(x) + Dx (g(x))f (x) is l
l
( (k + 1)a
b − ) + ( (l − k + 1)a b − ) − = (l + 1)a b + ( (k + 1)a b − ) + ( (l − k + 1)a b − ) + (l + 1)a b = (l + 1)a b + ( ka b − ) + ( (l − k + 1)a b − ) + (l + 1)a b = (l + 1)a b + ( (l + 1)a b − ) + (l + 1)a b a b − ) = (l + 1)c . = (l + 1)( k+1 l k
k l k+1
k=0
k=0
l 1
l+1 0
l
k+1 l k
k=0 l
l+1 0
k=1
l
k l k+1
0 l+1
k l k+1
k=1 l
l+1 0
0 l+1
k l k+1
k=1
0 l+1
k l k+1
k=1
l+1
k l k+1
l+1
k=0
Since all their coefficients are equal, we conclude Dx (f (x)g(x)) = D x (f (x))g(x)+ Dx (g(x))f (x). Exercise 2. Find all irreducible polynomials of degree 1, 2 and 4 over F2 and prove that their product is x 16 x.
−
Solution: The polynomials x and x + 1 are the only (non-constant, i.e. = 0, 1) polynomials F2 [x] of degree 2 is of degree 1 over F2 ; they are clearly irreducible. A polynomial f (x) irreducible over F2 if and only if it does not have a root in F2 , that is, exactly when f (0) = f (1) = 1. Hence, the only irreducible polynomial of degree 2 over F2 is x2 + x + 1. Now, for a polynomial f (x) F2 [x] of degree 4 to be irreducible, it must have no linear or quadratic factors. We can also apply the condition f (1) = f (0) = 1 to discard the ones with linear factors. Furthermore, f must have an odd number of terms (or it will be 0), and must have constant term 1 (or x will be a factor). We are left with
∈
∈
x4 + x3 + x2 + x + 1
x4 + x3 + 1
x4 + x2 + 1
x4 + x + 1.
For any of this polynomials to be irreducible, it can’t be factorized as two quadratic irreducible factors. Since x2 +x+1 is the only irreducible polynomial of degree 2 over F2 , only (x2 +x+1)2 = x4 + x2 + 1 of this four is not irreducible. Hence, the irreducible polynomials of degree 4 over F2 are x 4 + x3 + x2 + x + 1, x 4 + x3 + 1 and x 4 + x + 1. Now, since x + 1 = x 1 in F2 , we have (x + 1)(x4 + x3 + x2 + x + 1) = x 5 1. We also calculate (x2 + x + 1)(x4 + x + 1)(x4 + x 3 + 1) = x10 + x 5 + 1. So, the product of all this irreducible polynomials is
−
−
x(x + 1)(x2 + x + 1)(x4 + x + 1)(x4 + x3 + 1)(x4 + x3 + x2 + x + 1) = x(x5
− 1)(x10 + x5 + 1) = x16 − x. 16
13.5 Separable and Inseparable Extension Exercise 3. Prove that d divides n if and only if xd n = qd + r then x n 1 = (xqd+r xr ) + (xr 1).]
−
−
−
− 1 divides xn − 1.
[Note that if
Solution: We follow the hint. Suppose d divides n, so that n = qd for some q Z . Then 1 = x qd 1 = (xd 1)(xqd−d + xqd−2d + . . . + xd + 1). So x d 1 divides x n 1. Conversely, suppose d does not divide n. Then n = qd + r for some q, r Z with 0 < r < d. Thus x n 1 = (xqd+r xr ) + (xr 1) = x r (xqd 1) + (xr 1) = x r (xd 1)(xqd−d + xqd−2d + . . . + xd + 1 ) + (xr 1). Since x d 1 divides the first term, but doesn’t divide x r 1 (as r < d), then x d 1 does not divide x n 1. xn
−
−
− −
−
− −
−
− − −
−
−
∈ −
∈
−
−
Exercise 4. Let a > 1 be an integer. Prove for any positive integers n, d that d divides n if and only if ad 1 divides an 1 (cf. the previous exercise). Conclude in particular that F pd F pn if and only if d divides n.
−
⊆
−
Solution: The first assertion is analogous to the last exercise. d Now, F pd is defined as the field whose p d elements are the roots of x p x over F p . Similarly is defined F pn . Take a = p. So, d divides n if and only if p d 1 divides p n 1, and that occurs d n exactly when x p −1 1 divides x p −1 1 (by last exercise). Thus, if d divides n, any root of d d n n x p x = x(x p −1 1) must be a root of x p x = x(x p −1 1), hence F pd F pn . Conversely, d n if F pd F pn , then x p −1 1 divides x p −1 1, so d divides n.
− −
− ⊆
−
−
−
− −
−
−
−
⊆
Exercise 5. For any prime p and any nonzero a F p prove that x p x + a is irreducible and separable over F p . [For the irreducibility: One approach prove first that if α is a root then α + 1 is also a root. Another approach suppose it’s reducible and compute derivatives.]
∈
−
−
−
Solution: Let f (x) = x p x + a. Let α be a root of f (x). First we prove f is separable. Since (α + 1) p (α + 1) + a = α p + 1 α 1 + a = 0, then α + 1 is also a root of f (x). This gives p distinct roots of f (x) given by α + k with k F p , so f is separable. Now we prove f is irreducible. Let f = f 1 f 2 f n where f i (x) F p [x] is irreducible for all 1 i n. Let 1 i < j n and let α i be a root of f i and α j be a root of f j , so that f i is the minimal polynomial of αi and f j the minimal polynomial of α j . We prove deg f i = deg f j . Since αi is a root of f i , it is a root of f , hence there exists k 1 F p such that α i = α + k1 . Similarly, there exists k2 F p such that α j = α+k2 . Thus, αi = α j +k1 k2 , so f i (x+k1 k2 ) is irreducible having α j as a root, so it must be its minimal polynomial. Hence, f i (x + k1 k2 ) = f j (x), so deg f i = deg f j , as claimed. Since i and j were arbitrary, then all f i are of the same degree, say q . Then p = deg f = nq , so n = 1 or n = p (as p is prime). If n = p, then all roots of f are in F p , so α F p and thus 0 = α p α + a = a, contrary to the hypothesis. Therefore, n = 1, so f is irreducible.
−
−
≤ ≤
≤
− −
···
≤
∈
∈
∈
∈
∈
−
− −
−
n
Exercise 6. Prove that x p −1 1 = α∈F n (x α). Conclude that p the product of the nonzero elements of a finite field is +1 if p = 2 and odd and n = 1 derive Wilson’s Theorem: ( p 1)! 1 (mod p).
−
−
×
− ≡−
−
×
−
n
Solution: By definition, F pn is the field whose pn elements are the roots of x p n n F p . Since x p 1 = x(x p −1 1), clearly
−
−
n
x p
−1 − 1 =
(x − α).
α F× pn
∈
17
n
p so α Fpn α = ( 1) 1 is p is odd. For p
∈
− x over
13.5 Separable and Inseparable Extension Set x = 0. Then
−1 =
pn 1
(−α) = (−1)
α F× pn
α F× pn
∈
n
( 1) p
⇒
−
α
−
∈
−1 (−1) = (−1) pn−1 (−1) pn −1
−
n
∈
α F× pn
∈
Hence, the product of the nonzero elements is +1 if p = 2 and n = 1, we have 1= α,
−1 is p is odd. For p odd and
−
so taking module p we find [1][2]
α F× pn
α.
( 1) p =
⇒
α
α F× p
∈
··· [ p − 1] = [−1], i.e., ( p − 1)! ≡ −1 (mod p).
Exercise 7. Suppose K is a field of characteristic p which is not a perfect field: K = Prove there exist irreducible inseparable polynomials over K . Conclude that there exist inseparable finite extensions of K . Solution: Let a K such that a = b p for every b K . Let f (x) = x p a. We prove that f is irreducible and inseparable. If α is a root of x p a, then x p a = (x α) p , so α is a multiple root of f (with multiplicity p), hence f is inseparable. Now, let g(x) be an irreducible factor of f (x). Note that α K , otherwise a = α p , contrary to the hypothesis. Then g(x) = (x α)k for some k p. Using the binomial theorem, we have
K p .
∈
∈
−
−
−
−
∈
≤
−
g(x) = (x
− α)k = xk − kαxk−1 + ··· + (−α)k .
Therefore, kα K . Since α K , then k = p, so g = f . Hence, f is irreducible. We conclude that K (α) is an inseparable finite extension of K .
∈
∈
Exercise 8. Prove that f (x) p = f (x p ) for any polynomial f (x) F p [x]. Solution: Let f (x) = an xn + . . . + a1 x + a0 F p [x]. Since F p has characteristic p, then (a + b) p = a p + b p for any a, b F p . Easily we can generalize this to a finite number of terms, so that (x1 + + xn ) p = x p1 + + x pn for any x 1 , , xn F p . Furthermore, by Fermat’s Little Theorem, a p = a for every a F p . So, over F p , we have
∈ ··· ∈
∈ ··· ∈
···
∈
f (x) p = (an xn + . . . + a1 x + a0 ) p = a pn xnp + . . . + a p1 x p + a p0 = a n xnp + . . . + a1 x p + a0 = f (x p ). pi Exercise 9. Show that the binomial coefficient pn pi is the coefficient of x in the expansion of (1 + x) pn . Working over F p show that this is the coefficient of (x p )i in (1 + x p )n and hence n prove that pn pi i (mod p). Solution: By the binomial theorem, we have
≡
pn
pn (1 + x) = x, i . in the expansion of (1 + x) is pn
i
i=0
pn pn so the coefficient of x pi pi Since F p has characteristic p, we have (1 + x) pn = 1 + x pn = (1 + x p )n , so over F p this is the coefficient of (x p )i in (1 + x p )n . Furthermore, (1 + x) pn = (1 + x p )n implies pn
n
p n
(1 + x )
n pn = (x ) = x = (1 + x) p i
i=0
i
i
i=0
18
k
pn
13.6 Cyclotomic Polynomials and Extensions over F p , hence
pn pi
n i
≡ (mod p).
Z[x1 , x2 , . . . , xn ] be a polynomial in the variables Exercise 10. Let f (x1 , x2 , . . . , xn ) x1 , x2 , . . . , xn with integer coefficients. For any prime p prove that the polynomial
f (x1 , x2 , . . . , xn ) p
∈
− f (x p1, x p2, . . . , x pn) ∈ Z[x1, x2, . . . , xn]
has all its coefficients divisible by p. Solution: This is equivalent to prove that for any prime number p, we have f (x1 , x2 , . . . , xn ) p = p p f (x1 , x2 , . . . , x pn ) in F p [x1 , x2 , . . . , xn ]. Let
f (x1 , x2 , . . . , xn ) =
γ n aγ ,...,γ n xγ 1 . . . xn 1
1
γ 1 ,...,γ n =0
be an element of F p [x1 , x2 , . . . , xn ]. Since F p has characteristic p, then (x1 + + xn ) p = x p1 + Furthermore, by Fermat’s Little Theorem, a p = a for every a
···
a = a
f (x1 , x2 , . . . , xn ) p = (
γ 1 γ n p γ 1 ,...,γ n x1 . . . xn )
(a ) = a =
γ 1 p γ n p γ 1 ,...,γ n (x1 . . . xn
··· + x pn for any x1, ··· , xn ∈ F p. ∈ F p. Hence, over F p we have
γ 1 γ n p γ 1 ,...,γ n x1 . . . xn )
pγ 1 γ 1 ,...,γ n (x1
p p p n . . . x pγ n ) = f (x1 , x2 , . . . , xn ).
Exercise 11. Suppose K [x] is a polynomial ring over the field K and F is a subfield of K . If F is a perfect field and f (X ) F [x] has no repeated irreducible factors in F [x], prove that f (x) has no repeated irreducible factors in K [x]. Solution: Let f (x) F [x] with no repeated irreducible factors in F [x]. We can suppose f is monic. Then f = f 1 f 2 f n for some monic irreducible polynomials f i (x) F [x]. Since F is perfect, f is separable, hence all f i has distinct roots. Thus, f splits in linear factors in the closure of F , hence splits in linear factors in the closure of K . Therefore, f (x) has no repeated irreducible factors in K [x].
∈
∈
§13.6
···
∈
Cyclotomic Polynomials and Extensions
Exercise 1. Suppose m and n are relatively prime positive integers. Let ζ m be a primitive m root of unity and let ζ n be a primitive nth root of unity. Prove that ζ m ζ n is a primitive mnth root of unity. Solution: Since (ζ m ζ n )mn = 1, then ζ m ζ n is an mn th root of unity. Now, let 1 k < mn. k ζ k . For a contradiction, suppose ζ k ζ k = 1. Then, ζ k = 1 and ζ k = 1, Then (ζ m ζ n )k = ζ m n m n m n hence m divides k and n divides k, so mn divides k (as m = n). Since 1 k < mn, this is impossible. So, (ζ m ζ n )k = 1 for all 1 k < mn and thus the order of ζ m ζ n is mn. Therefore, ζ m ζ n generates the cyclic group of all mnth roots of unity, that is, ζ m ζ n is a primitive mnth root of unity. th
≤
≤
≤
Exercise 2. Let ζ n be a primitive n th root of unity and let d be a divisor of n. Prove that ζ nd is a primitive (n/d)th root of unity. Solution: Since (ζ nd )(n/d) = ζ nn = 1, then ζ nd is an (n/d)th root of unity. Now let 1 k < (n/d). Then (ζ nd )k = ζ nkd . Since 1 kd < n, then ζ nkd = 1, so (ζ nd )k = 1. Hence, the order of ζ nd is (n/d), so it generates the cyclic group of all (n/d)th roots of unity, that is, ζ nd is a primitive (n/d)th root of unity.
≤
≤
Exercise 3. Prove that if a field contains the nth roots of unity for n odd then it also contains the 2nth roots of unity. 19
13.6 Cyclotomic Polynomials and Extensions Solution: Let F be a field that contains the nth roots of unity for n odd and let ζ be a 2nth root of unity. If ζ n = 1, then ζ F , so suppose ζ n = 1. Since ζ 2n = 1, then ζ n is a root of x 2 1. Since the roots of this polynomial are 1 and 1, and ζ n = 1, then ζ n = 1. Hence, ( ζ )n = ( 1)n (ζ )n = ( 1)n+1 = 1 (since n is odd), so ζ F . Since F is a field, then ζ F .
−
−
∈
−
− − ∈
−
−
∈
Exercise 4. Prove that is n = p k m where p is a prime and m is relatively prime to p then there are precisely m distinct n th roots of unity over a field of characteristic p. Solution: Let F be a field with char F = p. The roots of unity over F are the roots of k k 1 = x p m 1 = (xm 1) p , so are the roots of x m 1. Now, since m is relatively prime to p, so is xm 1 and its derivative mxm−1 , so xm 1 has no multiple roots. Hence, the m different roots of x m 1 are precisely the m distinct n th roots of unity over F . xn
−
− −
−
−
−
−
Exercise 5. Prove there are only a finite number of roots of unity in any finite extension K of Q.
≥ √ ∈
Solution: We use the inequality ϕ(n) n/2 for all n 1. Let K be an extension of Q with infinitely many roots of unity. Let N N. Then, there exits n N such that n > 4N 2 and there exits some nth root of unity ζ K . Then [K : Q] [Q(ζ ) : Q] = ϕ(n) n/2 > N . Since N was arbitrary, we conclude that [K : Q] > N for all N N, so [K : Q] is infinite. Therefore, in any finite extension of Q there are only a finite number of roots of unity.
∈
≥ ≥
∈
≥ √
∈
Exercise 6. Prove that for n odd, n > 1, Φ 2n (x) = Φn ( x).
−
Solution: Since Φ2n (x) and Φn ( x) are irreducible, they are the minimal polynomial of any of its roots. Thus, it is sufficient to find a common root for both. Let ζ 2 be the primitive 2 th root of unity and let ζ n be a primitive n th root of unity. Note that ζ 2 = 1, so that ζ 2 ζ n = ζ n . Since n is odd, 2 and n are relatively prime. So, by Exercise 1, ζ 2 ζ n is a primitive 2nth root of unity, i.e, a root of Φ2n (x). Furthermore, ζ n is clearly a root of Φn ( x). Thus, ζ n is a common root for both Φ 2n (x) and Φ n ( x), hence Φ 2n (x) = Φn ( x).
−
−
−
−
−
−
−
−
Exercise 7. Use the Möbius Inversion Formula indicated in Section 14.3 to prove
Φ (x) = (x − 1) d
n
µ(n/d)
.
dn
|
Solution: The Möbius Inversion Formula sates that if f (n) is defined for all nonnegative integers and F (n) = d|n f (d), then f (n) = d|n µ(d)F ( nd ). So lets start with the formula
xn
− 1 = Φ (x). d
dn
|
We take natural logarithm in both sides and obtain ln(xn
− 1) = ln(
Φ (x)) = ln Φ (x). d
d
dn
dn
|
|
So, we use the Möbius Inversion Formula for f (n) = ln Φn (x) and F (n) = ln(xn ln Φn (x) =
µ(d)ln(x
n/d
dn
|
− 1) =
ln(x
(x
− 1)µ(d) =
dn
|
n/d
− 1) to obtain
− 1)µ(d).
Hence, taking exponentials we obtain Φn (x) = exp(
ln(x
n/d
dn
|
− 1)µ(d)) =
n/d
dn
|
20
(x − 1) d
dn
|
µ(n/d)
.
13.6 Cyclotomic Polynomials and Extensions −1
Exercise 8. Let be a prime and let Φ (x) = xx−1 = x −1 + x−2 + + x + 1 Z[x] be the th cyclotomic polynomial, which is irreducible over Z by Theorem 41. This exercise determines the factorization of Φ (x) modulo p for any prime p. Let ζ denote any fixed primitive th root of unity. (a) Show that if p = then Φ (x) = (x 1)−1 F [x] (b) Suppose p = and let f denote the order of p mod , i.e., f is the smallest power of p with p f 1 mod . Use the fact that F p×n is a cyclic group to show that n = f is the smallest power p n of p with ζ F pn . Conclude that the minimal polynomial of ζ over F p has degree f . (c) Show that F p (ζ ) = F p (ζ a ) for any integer a not divisible by . [One inclusion is obvious. For the other, note that ζ = (ζ a )b where b is the multiplicative inverse of a mod .] Conclude using (b) that, in F p [x], Φ (x) is the product of −f 1 distinct irreducible polynomials of degree f . (d) In particular, prove that, viewed in F p [x], Φ 7 (x) = x 6 + x5 + + x + 1 is (x 1)6 for p = 7, a product of distinct linear factor for p 1 mod 7, a product of 3 irreducible quadratics for p 6 mod 7, a product of 2 irreducible cubics for p 2, 4 mod 7, and is irreducible for p 3, 5 mod 7.
≡
≡
−
···
∈
∈
∈
···
≡
≡
−
≡
Solution: l−1 . (a) Since p is prime, in F p [x] we have (x 1) p = x p 1, so Φ (x) = xx−1 = (xx−−1) 1 = (x 1) (b) Note that ζ has order as being a primitive th root of unity. Since pf 1 mod , f f p 1 q − then p 1 = q for some integer q , hence ζ = ζ = 1, so ζ F pf . Now we prove that f is the smallest integer with that property. Suppose ζ F pn for some n. Then ζ is a root of n x p −1 1, hence divides p n 1 (see Exercise 13.5.3). Since f is the smallest power of p such that p f 1 mod , is the smallest integer such that divides p f 1, so n l, as desired. This in fact proves that F p (ζ ) = F pf , so the minimal polynomial of ζ over F p has degree f . (c) Since ζ a F p (ζ ), clearly F p (ζ a ) F p (ζ ). For the other direction we follow the hint. Let b the the multiplicative inverse of a mod , i.e ab 1 mod . Then (ζ a )b = ζ , so ζ F p (ζ a ) and thus F p (ζ ) F p (ζ a ). The equality follows. Now, consider Φ (x) as a polynomial over F p [x]. Let ζ i for 1 i be distinct primitive th roots of unity. The minimal polynomial of each ζ i has degree f by part (b). Hence, the irreducible factors of Φ (x) have degree f . Since Φ have degree 1, then there must be −f 1 factors, and all of them are different since Φ (x) is separable. (d) If p = 7, then Φ7 (x) = (x 1)6 by part (a). If p 1 mod 7, then f = 1 in (b) and all roots have degree 1, so Φ 7 (x) splits in distinct linear factors. If p 6 mod 7, then f = 2 is the smallest integer such that p f = p 2 36 1 mod 7, so we have 3 irreducible quadratics. If p 2, 4 mod 7, then f = 3 is the smallest integer such that p3 23 , 43 8, 64 1 mod 7, so we have 2 irreducible cubics. Finally, if p 3, 5 mod 7, then f = 6 is the smallest integer such that p 6 36 , 56 729, 15626 1 mod 7, hence we have an irreducible factor of degree 6.
−
−
− ≡
−
−
⊂
≥
≡
∈
≤ ≤ −
−
≡
≡
−
≡
∈
∈
∈ ⊂
≡
−1
−
≡
≡
≡ ≡ ≡
≡
≡
≡
≡
Exercise 9. Suppose A is an n by n matrix over C for which Ak = I for some integer 1 α k 1. Show that A can be diagonalized. Show that the matrix A = where α is an 0 1 element of a field of characteristic p satisfies A p = I and cannot be diagonalized if α = 0.
≥
Solution: Let A be an n by n matrix over C for which Ak = I for some integer k 1. k Then the minimal polynomial of A divides x 1. Since we are working over C, there are k distinct roots of this polynomial, so the minimal polynomial of A can be split in linear factors. Hence, A is diagonalizable. 1 α Now consider A = where α is an element of a field of characteristic p. 0 1 1 nα Computing powers of A, we can prove (by induction) that An = for every positive 0 1
−
21
≥
13.6 Cyclotomic Polynomials and Extensions integer n. Since pα = 0, then A p = I . Now, if A is diagonalizable, there exists some non-singular matrix P such that A = P DP −1 , where D is a diagonal matrix whose diagonal entries are the eigenvalues of A. Since A has only one eigenvalue 1, then D = I , and thus A = P IP −1 = I . So, if A is diagonalizable, it must be α = 0. Exercise 10. Let ϕ denote the Frobenius map x x p on the finite field F pn . Prove that ϕ gives an isomorphism of F pn to itself (such an isomorphism is called an automorphism ). Prove that ϕ n is the identity map an that no lower power of ϕ is the identity.
→
F pn . Then ϕ(a + b) = (a + b) p = a p + b p = ϕ(a) + ϕ(b), and Solution: Let a, b ϕ(ab) = (ab) p = a p b p = ϕ(a)ϕ(b), so ϕ is and homomorphism. Moreover, if ϕ(a) = 0, then a p = 0 implies a = 0. Hence, ϕ is injective. Since F pn is finite, then ϕ is also surjective n so it is an isomorphism. Furthermore, since every element of F pn is a root of x p x, then n n p n ϕ (a) = a = a for all a F pn , so ϕ is the identity map. Now, let m be an integer such that m ϕm is the identity map. Then a p = a for all a F pn , so every element of F pn must be a root m of x p x. Hence, F pn F pm and thus n divides m (Exercise 13.5.4), so n m.
∈
−
∈
−
∈
⊂
≤
Exercise 11. Let ϕ denote the Frobenius map x x p on the finite field F pn as in the previous exercise. Determine the rational canonical form over F p for ϕ considered as an F p -linear transformation of the n-dimensional F p -vector space F pn .
→
Solution: Note that the minimal polynomial of ϕ is xn 1, for if ϕ satisfies some polynomial n xn−1 + + a1 x + a0 of degree n 1 (or less) with coefficients in F p , then x p + + a1 x p + a0 for all x F pn , which is impossible. Since F pn has degree n as a vector space over F p , then xn 1 is also the characteristic polynomial of ϕ, hence is the only invariant factor. Therefore, the rational canonical form of ϕ over F p is the companion matrix of x n 1, which is
−
··· ∈
−
−
−1
···
−
0 1 0.. .
0 0 1 0 0 0 1 0 0 .. . . .. .. . . . . 0 0 1 0
··· ··· ··· ···
.
Exercise 12. Let ϕ denote the Frobenius map x x p on the field F pn as in the previous exercise. Determine the Jordan canonical form (over a field containing all the eigenvalues) for ϕ considered as an F p -linear transformation of the n-dimensional F p -vector space F pn .
→
Solution: We’ll work over the algebraic closure of F pn , to ensure the field to contain all eigenvalues. In last exercise we proved that the minimal and characteristic polynomial of ϕ is xn 1. Moreover, the eigenvalues of ϕ are the n th roots of unity. We use Exercise 4 and write k n = pk m for some prime p and some m relatively prime to p, so that xn 1 = (xm 1) p and we get exactly m distinct nth roots of unity, each one with multiplicity pk . Since all the eigenvalues are zeros of both the minimal and characteristic polynomial of multiplicity pk , we get m Jordan blocks of size p k . Now, fix a primitive m th root of unity, say ζ . Then, each Jordan block each of the form ζ i 1 0 0 0 i 0 ζ 1 0 0 i 0 0 ζ 0 0 J i = . . . . .. .. .. . . ... ...
−
−
0 0
··· ··· ···
0 0
0 0
22
··· ···
1
ζ i 0 ζ i
−
13.6 Cyclotomic Polynomials and Extensions for some 0
≤ i ≤ m − 1. We already know the Jordan canonical form is given by J 0 0 ··· 0 0 J 1 ··· 0 0 0 ··· 0 .
...
0
.. . 0
..
.. .
.
J m−1
···
Exercise 13. (Wedderburn’s Theorem on Finite Division Rings ) This exercise outlines a proof (following Witt) of Wedderburn’s Theorem that a finite division ring D is a field (i.e., is commutative). (a) Let Z denote the center of D (i.e., the elements of D which commute with every element of D). Prove that Z is a field containing F p for some prime p. If Z = Fq prove that D has order q n for some integer n [D is a vector space over Z ]. (b) The nonzero elements D × of D form a multiplicative group. For any x D× show that the elements of D which commute with x form a division ring which contains Z . Show that this division ring if of order q m for some integer m and that m < n if x is not an element of Z . (c) Show that the class equation (Theorem 4.7) for the group D × is
∈
r
q − 1 = (q − 1) + n
i=1
q n 1 C D (xi )
−
|
×
|
where x1 , x2 , . . . , xr are representatives of the distinct conjugacy classes in D× not contained in the center of D × . Conclude from (b) that for each i, C D (xi ) = q mi 1 for some m 1 < n. q n 1 (d) Prove that since m is an integer (namely, the index D× : C D (xi ) ) then m i and q i 1 divides n (cf. Exercise 4 of Section 5). Conclude that Φn (x) divides (xn 1)/(xmi 1) and hence that the integer Φ n (q ) divides (q n 1)/(q mi 1) for i = 1, 2, . . . , r. (e) Prove that (c) and (d) imply that Φ n (q ) = ζ primitive (q ζ ) divides q 1. Prove that q ζ > q 1 (complex absolute value) for any root of unity ζ = 1 [note that 1 is the closest point on the unit circle in C to the point q on the real line]. Conclude that n = 1, i.e., that D = Z is a field.
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− −
−
| − |
| |
×
−
−
−
×
−
−
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−
−
Solution: (a) Z is a division subring of D, and it is commutative by definition of the center, so Z is a field. Since it is finite, its prime subfield is F p for some prime p, so it is isomorphic to F pm for some integer m. Let q = p n , so that Z is isomorphic to Fq . Since D is a vector space over Z , then D = q n for some integer n. (b) Let x D × and let C D (x) be the set of the elements in D that commutes with x. Clearly Z C D (x). We prove that every element a C D (x) has an inverse in C D (x). Since a C D (x), then ax = xa. Since D is a division ring, then a−1 D. Moreover, we have 1 1 1 1 1 1 − − − − − − a ax = a xa and thus x = a xa, so xa = a x and a C D (x). Hence, C D (x) is a division ring. Now, since Z C D (x), then C D (x) is a Z -vector space, so C D (x) = q m for some integer m. If x Z , then C D (x) is a proper subset of D and hence m < n. (c) The class equation for the group D × is
⊂
∈
| | ∈
∈
∈
⊂
∈
∈
|
|
r
|D×| = |Z (D×)| +
|D× : C
D× (xi )
|,
i=1
where the xi are the representatives of the distinct conjugacy class. By (a) we have D× = q n 1, q n 1 q n 1 Z (D× ) = q 1 and C D (xi ) = q mi 1. Then D× : C D (xi ) = = m . C D (xi ) q i 1
|
|
−
|
×
|
−
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23
×
|
| | − | | ×
− − −
13.6 Cyclotomic Polynomials and Extensions Replacing in the class equation we obtain r
q − 1 = (q − 1) + n
i=1
|
q n 1 = (q C D (xi )
−
×
|
r
n
− 1) + q − 1 . i=1
q mi
−1
q n 1 (xi ) = m is an integer. Hence, q i 1 q mi 1 divides q n 1, so (Exercise 13.5.4) mi divides n. Since mi < n (no xi is in Z ) , no th mth root of unity. Therefore, as Φn (x) divides xn 1, it must divide i root of unity is a n (xn 1)/(xmi 1) for i = 1, 2, . . . , r.. Letting x = q we have Φ n (q ) divides (q n 1)/(q mi 1) for i = 1, 2, . . . , r. (e) From (d), Φn (q ) divides (q n 1)/(q mi 1) for i = 1, 2, . . . , r, so the class equation in (c) implies Φ n (q ) divides q 1. Now, let ζ = 1 be a n th root of unity. In the complex plane q is closer to 1 that ζ is, so q ζ > q 1 = q 1. Since Φ n (q ) = ζ primitive (q ζ ) divides q 1, this is impossible unless n = 1. Hence, D = Z and D is a field. (d) Since |D× : C D
− −
−
×
(xi )| is an integer, then |D× : C D
− −
|
×
−
−
− − − | − | | − | −
−
−
−
−
Exercise 14. Given any monic polynomial P (x) Z[x] of degree at leat one show that there are infinitely many distinct prime divisors of the integers
∈
P (1), P (2), P (3), . . . , P ( n), . . . [Suppose p1 , p2 , . . . , pk are the only primes dividing the values P (n), n = 1, 2, . . .. Let N be an integer with P (N ) = a = 0. Show that Q(x) = a−1 P (N + ap 1 p2 . . . pk x) is an element of Z[x] and that Q(n) 1 (mod p1 p2 . . . pk ) for n = 1, 2, . . .. Conclude that there is some integer M such that Q(M ) has a prime factor different from p1 , p2 , . . . , pk and hence that P (N + ap1 p2 . . . pk M ) has a prime factor different from p 1 , p2 , . . . pk .] Solution: We follow the hint. Let P (x) = x n + + a1 x + a0 be a monic polynomial of degree 1 over Z. For a contradiction, suppose there are only finitely many primes dividing the values P (n), n = 1, 2, . . ., say p1 , p2 , . . . , pk . Let N be an integer such that P (N ) = a = 0. Let Q(x) = a −1 P (N + ap1 p2 . . . pk x). Then, using the binomial theorem, we have
≡
···
≥
Q(x) = a −1 P (N + ap1 p2 . . . pk x) = a −1 ((N + ap1 p2 . . . pk x)n + = a −1 (N n + an−1 N n−1 + = a −1 (P (N ) + R(x))
··· + a1(N + ap1 p2 . . . pk x) + a0)
··· + a1N + a0 + R(x))
= 1 + a−1 R(x) for some polynomial R(x) Z[x] divisible by ap1 p2 . . . pk . Hence, Q(x) Z[x]. Moreover, for all n Z+ , P (N +ap1 p2 . . . pk n) a (mod p1 , p2 , . . . , pk ), so Q(n) = a −1 P (N +ap1 p2 . . . pk n) a−1 a = 1 (mod p1 , p2 , . . . , pk ). Now let m be a positive integer such that Q(m) > 1, so that Q(m) 1 (mod pi ) for all i. Therefore, none of the pi ’s divide Q(m). Since Q(m) > 1, there exists a prime q = pi for all i such that q divides Q(m). Then q divides aQ(m) = P (N + ap 1 p2 . . . pk m), contradicting the fact that only the primes p1 , p2 , . . . , pk divides the numbers P (1), P (2), . . ..
∈ ≡
∈
∈ |
≡
| |
≡
|
Exercise 15. Let p be an odd prime not dividing m and let Φ m (x) be the m th cyclotomic polynomial. Suppose a Z satisfies Φ m (a) 0 (mod p). Prove that a is relatively prime to p and that the order of a in ( Z/pZ)× is precisely m. [Since
∈
≡
xm
−1=
Φ (x) = Φ (x) Φ (x) m
d
dn
d
dn d
|
|
24
13.6 Cyclotomic Polynomials and Extensions we see first that a m 1 0 (mod p) i.e., a 1 (mod p). If the order of a mod p were less that m, then ad 1 (mod p) for some d dividing m, so then Φd (a) 0 (mod p) for some d < m. but then x m 1 would have a as a multiple root mod p, a contradiction.]
− ≡
≡ −
≡
≡
Solution: We follow the hint. Since Φm (a) 0 (mod p), then am 1 (mod p). Hence, m 1 − there exist b such that ba 1 mod p (indeed, b = a ), so a is relatively prime to p. We prove that the order of a is m. For a contradiction, suppose ad 1 (mod p) for some d dividing m, so that Φd (a) 0 (mod p) for some d < m. Thus, a is a multiple root of xm 1, so is also a root of its derivative ma m−1 . Hence, ma m−1 0 mod p, impossible since p does not divide m nor a. Therefore, the order of a in ( Z/pZ)× is precisely m
≡
≡
≡
≡
≡
−
≡
Exercise 16. Let a Z. Show that if p is an odd prime dividing Φm (a) then either p divides m of p 1 (mod m).
≡
∈
Solution: Let p be an odd prime dividing Φ m (a). If p does not divide m, then, by (c), a is relatively prime to p and the order of a in F p× is m. Since F p× = p 1, this implies m divides p 1, that is, p 1 (mod m).
−
| |
≡
Exercise 17. Prove there are infinitely many primes p with p
−
≡ 1 (mod m).
Solution: By Exercise 14, there are infinitely many primes dividing Φm (1), Φm (2), Φm (3), . . .. Since only finitely of them can divide m, then, by Exercise 16, there must exists infinitely many primes p with p 1 (mod m).
≡
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