Math 503, Spring 2015: Homework Assignment #1
1. Show that p(x) = x3 + 9x + 6 is irreducible over Q. Let α be a root of p(x). Find
the (multiplicative) inverse of 1 + α in Q(α). (Hint: there is a version of the Euclidean
Algorithm for Q[x]. How can you make that work for you?)
Proof. First, p(x) is irreducible over Q using Eisenstein’s Criterion with p = 3. If α is a
root of p(x), then Q(α) ∼
= Q[x]/hx3 + 9x + 6i. Then via this isomorphism, (1 + α)g(α) = 1
in Q(α) if and only if (x + 1)g(x) = 1 + hx3 + 9x + 6i in Q[x]/hx3 + 9x + 6i for some
g(x) ∈ Q[x] with degree less than 3. Equivalently, we need
(x + 1)g(x) + (x3 + 9x + 6)f (x) = 1
for some f (x) ∈ Q[x]. Dividing x3 + 9x + 6 by x + 1 in Q[x] yields
x3 + 9x + 6 = (x + 1)(x2 − x + 10) − 4.
Solving for 4 and then dividing by 4 yields
1 2
1
(x − x + 10)(x + 1) − (x3 + 9x + 6) = 1.
4
4
1
2
Thus the inverse of 1 + α in Q(α) is 4 (α − α + 10).
p
√
2. Show that α := 3 + 2 2 is algebraic over Q. What is the minimal polynomial for α
over Q? What is [Q(α) : Q]?
p
√
√
√
Proof. Let α := 3 + 2 2. Then α2 = 3+2 2, so α2 −3 = 2 2, and squaring both sides
yields α4 −6α2 +9 = 8. Therefore, α is a root of the polyonomial f (x) = x4 −6x2 +1 ∈ Q[x].
However, we can factor f (x) = (x2 − 2x − 1)(x2 + 2x − 1) ∈ Q[x], and we calculate
that α
p
√
2
is a √
root of x − 2x − 1. (Alternatively, you can note at the start that α = 3 + 2 2 =
1 + 2 and then use similar steps to the ones above to arrive at this quadratic.) Then
p(x) = x2 − 2x − 1 is irreducible over Q by the Mod 3 test, since p(x) = x2 + x + 2 is of
degree 2 with no roots in Z3 . Thus [Q(α) : Q] = 2.
3. Let E ⊂ C and
√ suppose that [E : Q] = 2. Prove that there exists a squarefree d ∈ Q such
that E = Q( d).
Proof. Let E ⊂ C and suppose that [E : Q] = 2. Then E is a vector space over Q with
basis {1, α} for some α ∈ E with α ∈
/ Q. Then α2 = c1 α + c0 for some c0 , c1 ∈ Q, hence
2
α is a root of the
√ quadratic x − c1 x + c0 ∈ Q[x]. Since E ⊂ C, the quadratic formula
c1 ±
c2 +4c0
1
gives us α =
, where c21 + 4c0 is not the square of a rational number, otherwise
2
p
α ∈ Q. Now
E
=
Q(α)
and
we
claim
that
Q(α)
=
Q(
c21 + 4c0p
). The containment
p
2
2
Q(α) ⊆ Q( c1 + 4c
p0 ) is clear by the formula for α. Then we have p c1 + 4c0 = ±(2α
√−
2
2
c1 ) ∈ Q(α), so Q( c1 + 4c0 ) ⊆ Q(α). Finally, we show that Q( c1 + 4c0 ) = Q( d)
for some squarefree d ∈ Q. Let c21 + 4c0 = ab with a, b ∈ Z and b 6= 0. Then by the
Fundamental Theorem of Arithmetic, we canqwrite a = s2 t and b = u2 v with s, t, u, v ∈ Z
p
and t and v squarefree. Thus Q( ab ) = Q( us vt ) where vt is squarefree. Then since us ∈ Q
q
q
p
t
s
2
we have Q(α) = Q( c1 + 4c0 ) = Q( u v ) = Q( vt ).
4. Prove that if [F (α) : F ] is odd then F (α) = F (α2 ).
Proof. We have α2 ∈ F (α), so F ⊆ F (α2 ) ⊆ F (α). Then
[F (α) : F ] = [F (α) : F (α2 )][F (α2 ) : F ].
Now if F (α) 6= F (α2 ), then x2 − α2 is irreducible over F (α2 ), so it is the minimal polynomial for α over F (α2 ), and therefore [F (α) : F (α2 )] = 2. But then [F (α) : F ] is even.
Therefore if [F (α) : F ] is odd then F (α) = F (α2 ).
√
5. Let ζn := e2πi/n , a primitive√nth root of unity. What is the degree of Q(ζ6 , ζ3 3 2) over Q?
Give a Q-basis for Q(ζ6 , ζ3 3 2).
√
√
show that
Q(ζ6 , ζ3 3 2)√= Q(ζ6 , 3 2). We
Proof.
First√we note that ζ3√= ζ62 , so we will
have
√
√
√
√
3
3
3
3
3
3
−2
2 3
= ζ6 2, so Q(ζ6 , ζ3 2) ⊆ Q(ζ6 , 2), and 2 = (ζ6 )(ζ3 2), so Q(ζ6 , ζ3 2) ⊇
ζ3 2 √
Q(ζ6 , 3 2). Now we have
√
√
√
√
3
3
3
3
[Q(ζ6 , 2) : Q] = [Q(ζ6 , 2) : Q( 2)][Q( 2) : Q].
√
3
2 over Q is x3 − 2, which is irreducible by Eisenstein’s
The minimal polynomial for
√
3
criterion with p = 2, so [Q( 2) : Q] = 3. Then ζ6 is a root of the cyclotomic
polynomial
√
3
2
Φ6 (x) = x − x + 1. The two roots of this
quadratic are complex
and Q( 2) is totally
√
√
3
3
2
real, so
x − x + 1 is irreducible √
over √
Q( 2). Therefore
[Q(ζ6 , 2) : Q] = 6. A basis for
√
√
√
3
3
3
3
3
2
Q(ζ6 , 2) over Q is given by {1, 2, ( 2) , ζ6 , ζ6 2, ζ6 ( 2)2 }.
6. Let f (x) and g(x) be irreducible polynomials over a field F and let α and β belong to
some extension E/F . If α is a root of f (x) and β is a root of g(x), then show that f (x)
is irreducible over F (β) if and only if g(x) is irreducible over F (α).
Proof. Let f (x) have degree m and let g(x) have degree n. Since f and g are each
irreducible, they must be associates of the minimal polynomials for α and β over Q,
respectively. Then we have
[F (α, β) : F (α)]m = [F (α, β) : F (α)][F (α) : F ] = [F (α, β) : F ]
= [F (α, β) : F (β)][F (β) : F ] = [F (α, β) : F (β)]n.
Let pα (x) ∈ F (β)[x] be the minimal polynomial for α over F (β) and let pβ (x) ∈ F (α)[x]
be the minimal polynomial for β over F (α). Since f (x) ∈ F (β)[x] with f (α) = 0 and
g(x) ∈ F (α)[x] with g(β) = 0, then pα | f and pβ | g. Now suppose that f (x) is irreducible
over F (β). Then deg f (x) = deg pα (x) so [F (α, β) : F (β)] = m. Thus [F (α, β) : F ] = mn,
so we have [F (α, β) : F (α)] = n. Now we see that deg pβ (x) = n. Since we also have
deg g(x) = n and pβ (x) | g(x), then g(x) is an associate of an irreducible polynomial over
F (α), and is therefore also irreducible over F (α). The converse argument is essentially
identical.