Math 121 Homework 1 Solutions
Section 13.1 (Page 519)
Problem 1. Show that p(x) = x3 + 9x + 6 is irreducible in the polynomial
ring Q[x]. Let θ be a root of p(x). Find the inverse of 1 + θ in Q(θ).
First we address the irreducibility of Q. We give two proofs, one from
scratch, one using some important theorems from Chapter 9.
Solution. First we show that if p is not irreducible then it has a root in
Q. (The argument would apply to any polynomial of degree 2 or 3.) If p(x)
factors, p(x) = g(x)h(x), where g, h ∈ Q[x]. We have deg(g) = deg(h) =
deg(p) = 3. So one of g, h has degree 1 and the other has degree 2. Without
loss of generality, g has degree 1, so g(x) = ax − b and so its root b/a is a
root of p. This proves our claim that p has a root in Q.
Now we may obtain a contradiction. Let r = b/a be a rational root of
p. We may assume that a, b ∈ Z and that they are coprime, and that b > 0.
Thus b3 + 9ba2 + 6a3 = 0. If p is any prime divisor of a, then p must divide
b3 = −(9ba + 6a2 )a, so it divides b, which is a contradiction since a and b
are coprime. Thus a is a positive integer with no prime divisors, that is,
a = 1. Thus the root r = b is an integer. However p0 (x) = 9 + 3x2 is always
positive, so p is monotone increasing, and has exactly one real root r. Since
p(−1) = −4 and p(0) = 6, we have −1 < r < 0, and r cannot be an integer.
This is a contradiction.
Second Solution. The polynomial x3 + 9x + 6 is what is called an Eisenstein polynomial . By Proposition 13 on page 309 of Dummit and Foote it is
irreducible in the polynomial ring Z[x]. (Take P to be the prime ideal (3) of
Z.) But it follows from Gauss’ Lemma (Proposition 5 on page 303) that if it
can be factored in Q[x] then it can be factored in Z[x], which it can’t be, so
it is irreducible in Q[x].
1
Since τ = θ + 1 is a root of the polynomial p(x − 1) = x3 − 3x2 + 12x − 4,
we have
1
1
τ (τ 2 − 3τ + 12) = 4,
τ −1 = (τ 2 − 3τ + 12) = (θ2 − θ + 10).
4
4
√
√
Problem
4. Prove directly that a + b 2 7−→ a − b 2 is an isomorphism
√ of Q 2 to itself.
√ First Solution. Note that every element of Q 2 can be written uniquely
√
√ √ as a + b 2 with a, b ∈ Q, so this map γ : Q 2 −→ Q 2 defined by
√
√ γ a + b 2 = a − b 2 is a bijection. We have to show that c is a ring
homomorphism.√ We will just check
√ that γ(αβ) = γ(α)γ(β). Indeed if we
write α = a + b 2 and β = c + d 2 then
√
αβ = (ac + 2bd) + (ad − bc) 2,
so
√ √ √
γ(αβ) = (ac + 2bd) − (ad − bc) 2 = a − b 2 c − d 2 = γ(α)γ(β).
Second Solution. Apply Theorem 8. In the notation of the theorem take
F = F 0 = Q and let p = p0 be the irreducible polynomial x2 − 2. Since
2
α, β are both roots of the irreducible polynomial
√ 2, Theorem 8
√ p(x) = x −
guarantees that there is an isomorphism Q 2 −→ Q − 2 that sends
√
√
√
√
2 to − 2. So this isomorphism sends a + b 2 to a − b 2.
Since the wording of the problem is “prove directly,” I think the first
solution is what they want. However, the second solution is important since
this is how you should really think about this problem.
Problem 5. Suppose that α is a rational root of a monic polynomial in Z[x].
Prove that α is an integer.
Solution. Suppose that α = a/b where a, b are coprime integers with b > 0.
Assume also that α is a root of
f (x) = xn + cn−1 xn−1 + . . . + c0
where ci are in Z. Then
an + cn−1 an−1 b + . . . + c0 bn = 0.
We will show that b = 1. If b > 1 then let p be a prime dividing b. Then p
divides an = −(cn−1 an−1 b + . . . + c0 bn ), so p|a. But this is a contradiction
because a, b are coprime. Thus b = 1 and α = a ∈ Z.
2
Section 13.2 (Page 529)
Problem 1. Let F be a finite field of characteristic p. Prove that |F| = pn
for some positive integer n.
Solution. The field F contains a copy of Fp = Z/pZ. (See the discussion of
the prime field on page 511.) Now we can consider F to be a vector space over
its subfield F
Pp . If β1 , · · · , βn this means every element of F can be written
uniquely as
ci βi with ci ∈ Fp . Therefore the cardinality of F is |Fp |n = pn .
Problem 3. Determine the minimal polynomial over Q for the element 1+i.
Solution. We have
(x − (1 + i))(x − (1 − i)) = x2 − 2x + 2.
This polynomial is irreducible over Q. (Otherwise, it has a linear factor and
hence a root in Q.) So this is the minimal polynomial.
√ √ √ √
Problem 7. Prove that Q 2 + 3 = Q 2, 3 .
√
√ √ √ Solution. It is clear that Q 2 + 3 ⊆ Q 2, 3 . To prove the other
√
√
√
√
inclusion, we must express 2 and 3 in terms of θ = 2 + 3. We comute
√
√
θ3 = 9 3 + 11 2.
So
√
√
1
1
2 = (θ3 − 9θ),
3 = (−θ3 + 11θ).
2
2
√
√ √
√ This shows that both 2, 3 are in Q 2 + 3 , giving us the other inclu√ √ √
√ sion Q 2, 3 ⊆ Q 2 + 3 .
Problem 8. Let F be a field of characteristic 6= 2. Let D1 and√D2 be
√ ele-
D1 , D2
ments of F , neither one of which is a square in F . Prove that F
is of degree√4 over
√ F if D1 D2 is not a square in F and is of degree 2 otherwise.
When F
D1 , D2 is of degree 4 over F , the field is called a biquadratic
extension of F .
Solution. We need:
Lemma 1. Assume that neither
√ D1 nor√D2 is a square in F . Then D1 D2 is
a square in F if and only if D2 ∈ F
D1 .
3
2
Proof.
First √
assume that √
D1 D2 is a√square
in F , say D1 D2 = m . Then
√
D2 = ±m/ D1 and so D2 ∈ F √
D1 .
√ On the other hand, assume
that
D2 ∈ F
D1 . Since D1 is not a
√ D1 has degree√2 over F , and every element of
square
F , the field F
√ in
F
D
may
be
written
uniquely
as a + b D1 with a, b ∈ F . In particular,
√1
√
write D2 = a + b D1 . Squaring this equation,
p
D2 = (a2 + D1 b2 ) + 2ab D1 .
√
Equating the coefficients of D1 on the left and right-hand sides of this
equation, 2ab = 0. Since F does not have characteristic 2, either a = 0
or b = 0. If b = 0, then D2 = a2 which is a contradiction since we are
assuming that D2 is not a square in F . Therefore a = 0 and D2 = D1 b2 . So
D1 D2 = (bD1 )2 is a square in F .
√
The polynomial f (x) = x2 − D1 that
is
satisfied
by
1 is irreducible,
√ √ D√
since if is reducible it factors as x − D1 x + D1 , so D√
F . This
1 ∈ contradicts the assumption that D1 is not a square in F . So F
D1 : F =
2.
We have
i h p
p i h p i
h p
p p F
D1 , D2 : F = F
D1 , D2 : F
D1
D1 , F .
F
Therefore
h p
i
h p
p i
p p F
D1 , D2 : F = 2 F
D1 , D2 : F
D1 .
(1)
√
√
2
Since D2 satisfies a quadratic polynomial
0 which
D1 [x],
is in F
√ x −D1 = √
D
over
F
D
has
degree
1
or 2.
the irreducible
polynomial
satisfied
by
√
√ √2
√ √ 1
Thus F
D1 , D2 : F
D
1 √= 1 if√ D2 ∈ F√ D1 , and 2 otherwise.
By the Lemma, this means F
D1 , D2 : F
D1 = 1 if D1 D2 is a
square in F , 2 otherwise. Combining this with (1) completes the solution.
Problem 14. Prove that if [F (α) : F ] is odd then F (α) = F (α2 ).
Solution. Obviously F (α2 ) ⊆ F (α). We have
[F (α) : F ] = [F (α) : F (α2 )][F (α2 ) : F ].
(2)
Now α satisfies a quadratic polynomial x2 − α2 = 0 over F (α2 ), so [F (α) :
F (α2 )] = 1 or 2, depending on whether this polynomial is irreducible or
not. But if [F (α) : F (α2 )] = 2, then (2) implies that [F (α) : F ] is even,
which contradicts the hypothesis. Therefore [F (α) : F (α2 )] = 1 and so
F (α) = F (α2 ).
4