Test 4
April 20, 2006
Math 522
Name
Student Number
Direction: You are required to complete this test by Monday (April 24, 2006). In order to
receive full credit, answer each problem completely and must show all work. This being a
takehome test, I expect that you do the test by yourself and do not discuss the problems or
your solutions with other students until after the due date. The test should be returned to me
by 2:00 pm. Good Luck!
1. (15 points) Prove or disprove Q
√
√ √
3 + i = Q 3, i . Here i = −1 and Q denotes the
field of rationals.
√
Answer: The set Q( 3, i) is given by
√
√
Q( 3, i) = Q( 3)(i)
n √ o
= p( 3) p(x) ∈ Q(i)[x]
( n
)
X
√ k =
ak ( 3) ak ∈ Q(i)
k=0
o
n
√ = A1 + A2 3 A1 , A2 ∈ Q(i)
n
o
√
√ = a + b i + c 3 + d i 3 a, b, c, d ∈ Q
Similarly, we get
n √
o
√
Q( 3 + i) = p( 3 + i) p(x) ∈ Q[x]
( n
)
X
√
k =
ak ( 3 + i) ak ∈ Q
k=0
o
n
√
√ = α + β i + γ 3 + δ i 3 α, β, γ, δ ∈ Q
√
√
Hence Q( 3, i) = Q( 3 + i).
2. (15 points) Find the splitting fields of the following polynomials in Q[x]:
(a) f (x) = x4 − 9,
Answer: Since f (x) = x4 − 9 can be factored as
f (x) = (x −
√
3)(x +
√
√
√
3)(x − i 3)(x + i 3),
√
the splitting field of f (x) over Q is Q( 3, i).
(b) f (x) = x4 − 10x2 + 1,
Answer: Since f (x) = x4 − 10x2 + 1 can be factored as
f (x) = (x −
√
3−
√
2)(x −
√
3+
√
2)(x +
√
3−
√
2)(x +
√
3+
√
2),
√ √
the splitting field of f (x) over Q is Q( 3, 2).
(c) f (x) = x2 + 2x + 2,
Answer: Since f (x) = x2 + 2x + 2 can be factored as
f (x) = (x + i − 1)(x − i + 1),
the splitting field of f (x) over Q is Q(i).
(d) f (x) = x5 − 1
Answer: Since f (x) = x5 − 1 can be factored as
f (x) = (x − w0 )(x − w)(x − w2 )(x − w3 )(x − w4 )
where w = e
2πi
5
(the 5th root of unity), the splitting field of f (x) over Q is Q(w).
3. (15 points) Find the basis for each of the given vector spaces over the given field. What is
the degree of each field extension?
(a) Q
√ 2 over Q
Answer: The basis of the vector space Q
√ √ Q 2 over Q, that is [Q 2 : Q] = 2.
√ √
2 over Q is {1, 2}. The degree of extension of
√ (b) Q i 3 over Q
√ √
Answer: The basis of the vector space Q i 3 over Q is {1, i 3}. The degree of extension
√ √ of Q i 3 over Q, that is [Q i 3 : Q] = 2.
(c) Q
√
3, i over Q
√
√ √
Answer: The basis of the vector space Q 3, i over Q is {1, i, 3, i 3}. The degree of
√
√ extension of Q 3, i over Q, that is [Q 3, i : Q] = 4.
(d) Q
√
√ 3 + i over Q 3
√
√
√
Answer: Recall that Q 3 + i = Q 3, i . The basis of the vector space Q 3 + i over
√
√ √
√ Q 3 is {1, i}. The degree of extension of Q 3 + i over Q 3 , that is [Q 3 + i :
√ Q 3 ] = 2.
(e) Q (π ) over Q(π 2 )
Answer: Since π is transcendental over Q, then π 2 is also transcendental over Q. Therefore
Q (π ) ' Q π 2 ' Q(x). Moreover, since Q π 2 ⊆ Q (π ), therefore Q π 2 = Q (π ). Hence,
the basis of the vector space Q (π ) over Q π 2 is {1}. The degree of extension of Q (π ) over
Q π 2 , that is [Q (π ) : Q π 2 ] = 1.
4. (15 points) Which of the following polynomials are irreducible in Q[x] (You must justify
your answer to receive full credit):
(a) f (x) = x3 + 4x2 − 3x + 5
Answer: We will use the irreducibility test mod 2. Since f (x) = x3 + 4x2 − 3x + 5, we get
f˜(x) = x3 + x + 1 in Z2 [x]. Moreover, since f˜(0) = 1 = f˜(1) and deg(f˜(x)) = 3, the polynomial
f˜(x) is irreducible in Z2 [x] and therefore f (x) is irreducible in Q[x].
(b) f (x) = 4x4 − 6x2 + 6x − 12
Answer: Take p = 3 and apply Eisenstein’s test to conclude f (x) is irreducible in Q[x].
(c) f (x) = x5 + x3 + x2 + 1
Answer: Since f (x) = x5 +x3 +x2 +1 = (x2 +1)(x3 +1), the polynomial f (x) is not irreducible
in Q[x].
(d) f (x) = x4 + 2x3 − 6x2 + 7x − 13
Answer: We will use the irreducibility test mod 2. Since f (x) = x4 + 2x3 − 6x2 + 7x − 13, we
get f˜(x) = x4 + x + 1 in Z2 [x]. Moreover, since f˜(0) = 1 = f˜(1) so f˜(x) has no linear factor.
Since f˜(x) has no linear factor, therefore the possible quadratic factors are: x2 + x + 1 and
x2 + 1. But x4 + x + 1 6= (x2 + x + 1)(x2 + x + 1) or x4 + x + 1 6= (x2 + x + 1)(x2 + 1) or
x4 + x + 1 6= (x2 + 1)(x2 + 1). Hence the polynomial f˜(x) is irreducible in Z2 [x] and therefore
f (x) is irreducible in Q[x].
5. (15 points) Draw the subfield lattice of GF 318 and of GF 330 . Make sure that your
lattice diagram is eye-pleasing.
Answer: The subfield lattice of GF 318 and of GF 330 are shown below.
6. (15 points) Suppose F is an arbitrary field and let F3 = {(a, b, c) | a, b, c ∈ F}. Prove that
{(1, 1, 0), (1, 0, 1), (0, 1, 1)} is a basis of the vector space F3 over F provided the characteristic
of F is not 2. When the characteristic of F is 2, show that the set is not a basis.
Answer: Suppose char(F) 6= 2. Let B = {(1, 1, 0), (1, 0, 1), (0, 1, 1)}. We want to show that B
is a basis of the vector space F3 over F. Suppose there exist a, b, c ∈ F such that
a (1, 1, 0) + b (1, 0, 1) + c (0, 1, 1) = (0, 0, 0).
From the last equation, we get (a + b, a + c, b + c) = (0, 0, 0). Solving this, we have a = b = c
and 2a = 0. Since characteristic of F is not 2, therefore 2a = 0 yields a = 0 and hence we have
a = b = c = 0. Thus we see that B consists of a set of linearly independent vectors.
Let (α, β, γ) ∈ F3 . Then we can find constants a, b, c ∈ F such that
a (1, 1, 0) + b (1, 0, 1) + c (0, 1, 1) = (α, β, γ).
Writing the above in matrix notation, we have

1
1
0
1
0
1
   
0
a
α
1b  = β .
1
c
γ
The determinant of the above 3 × 3 matrix is nonzero, therefore it is invertible and we have
 

 
a
1
1 −1
α
 b  = 2−1  1 −1
1 β .
−1
1
1
γ
c
Hence any arbitrary vector (α, β, γ) ∈ F3 can be uniquely represented as a linear combination
of vectors in B.
Hence B is a basis of the vector space F3 over F.
If char(F) = 2, then (1, 1, 0) + (1, 0, 1) = (0, 1, 1). Therefore the vectors in B are not
linearly independent.
7. (15 points) Given that a belongs to some field extension of Q and that [Q(a) : Q] = 5,
prove that Q(a3 ) = Q(a).
Answer: It is easy to see that Q ⊆ Q(a3 ) ⊆ Q(a). Hence we have
[Q(a) : Q] = [Q(a) : Q(a3 )] [Q(a3 ) : Q] = 5.
The last equality implies that either [Q(a) : Q(a3 )] = 1 or [Q(a3 ) : Q] = 1. If the first case is
true, we are done, so let us assume that [Q(a3 ) : Q] = 1 that is, Q = Q(a3 ). Then, a3 ∈ Q and
p(x) = x3 − a3 ∈ Q[x] is such that p(a) = 0. But this implies that [Q(a) : Q] ≤ 3 which is a
contradiction to the fact that [Q(a) : Q] = 5. Therefore, Q(a3 ) = Q(a).
8. (15 points) Find a polynomial p(x) in Q[x] so that Q
p
√ 1 + 2 2 i is isomorphic to
Q[x]/ hp(x)i.
p
√
√
Answer: Let x = 1 + 2 2 i. Then squaring both sides, we get x2 = 1 + 2 2 i. Hence
√
x2 − 1 = 2 2 i. Again squaring both sides of the last equation, we have x4 − 2x2 + 9 = 0.
p
√ Therefore p(x) = x4 − 2x2 + 9 ∈ Q[x] satisfies p
1 + 2 2 i = 0. It is easy to check that
p(x) = x4 − 2x2 + 9 is irreducible in Q[x].
9. (15 points) Suppose that p(x) is irreducible over F and a and a2 are elements of some
extension field of F and both are zeros of p(x). Prove that F(a) = F(a2 ).
Answer: Since a2 ∈ F(a) we have that F(a2 ) ⊆ F(a). By the Corollary to Theorem 20.3, we
obtain F(a) ' F(a2 ) as fields and therefore as vector spaces over F. Thus [F(a) : F] = [F(a2 ) : F].
From
[F(a) : F] = [F(a) : F(a2 )] [F(a2 ) : F]
we see that [F(a) : F(a2 )] = 1. Hence we have F(a) = F(a2 ).
10. (15 points) Construct a field of order 9 and carry out the analysis done in Example 1 (on
page 383), including the conversion table.
Answer: The polynomial p(x) = x9 − x has 9 distinct roots since p(x) = x9 − x and p0 (x) = −1
have no common factors of positive degree. The finite field GF (32 ) consists of these nine roots
of the polynomial p(x).
The polynomial p(x) = x9 − x is not irreducible in Z3 [x]. In fact using maple command
“Factor(x9 − x) mod 3;” we see that
p(x) = x (x + 1) (x + 2) (x2 + 1) (x2 + x + 2) (x2 + 2x + 2).
We can check using maple command “Irreduc(x2 +2x+2) mod 3;” that the factor x2 +2x+2
is irreducible in Z3 [x]. Similarly the factors x2 +1 and x2 +x+2 can be verified to be irreducible
in Z3 [x]. Each of the irreducible factor gives rise to a finite field with 32 elements. Hence we
have the following additional finite fields Z3 [x]/ < x2 + 1 >, Z3 [x]/ < x2 + x + 2 > and
Z3 [x]/ < x2 + 2x + 2 >. However each one is isomorphic to GF (32 ).
Not that GF (32 ) ' Z3 [x]/ < x2 + 1 > but x is not a generator of GF (32 )# . So we use
GF (32 ) ' Z3 [x]/ < x2 + x + 2 >. This time x is a generator of GF (32 )# . The conversion table
is the given below:
Multiplicative
1
x
x2
x3
x4
x5
x6
x7
Conversion Table
Additive
Additive
1
x
2x + 1
2x + 2
2
2x
x+2
x+1
1
2
x
x+1
x+2
2x
2x + 1
2x + 2
Multiplicative
1
x4
x
x7
x6
x5
x2
x3
Bonus Problem. (15 points) (a) Let β be a zero of x3 + x + 1 in some extension field of Z2 .
Show that β + 1 is a zero of x3 + x2 + 1.
Answer: Since β is a zero of x3 + x + 1, therefore β 3 + β + 1 = 0. We show that β + 1 is zero
of p(x) = x3 + x2 + 1 in Z2 [x]. Since
p(β + 1) = (β + 1)3 + (β + 1)2 + 1
= β3 + β2 + β + 1 + β2 + 1 + 1
= β3 + β + 1
= 0,
therefore β + 1 is a zero of x3 + x2 + 1.
(b) Let K be a splitting field of x2 + 2x + 2 over Z3 . Determine order of K (that is, the number
of elements in K).
Answer: Since f (x) = x2 + 2x + 2 = (x + 1 − i) (x + 1 + i). Hence K = Z3 (i). Since
Z3 (i) = {a + b i | a, b ∈ Z3 }, therefore |K| = 9.
(c) Let p be a any prime number. Write xp − x as a product of linear factors over Zp .
Answer: Since by Fermat’s Little theorem, we have ap−1 − 1 = 0 mod p for gcd(a, p) = 1,
therefore ap − a = 0 mod p. Hence every element a ∈ Zp is a root of the polynomial xp − x.
Therefore xp − x = x(x − 1)(x − 2)(x − 3) · · · (x − (p − 1)).