MATH 819
HOMEWORK 4
SOLUTIONS
1. (Lang Ch. VI, Ex. 1) What is the Galois group of the following polynomials?
(a) X 3 − X − 1 over Q.
Solution. Since ±1 is not a root of X 3 − X − 1, we see that X 3 − X − 1
has no roots over Q (integral root test). Therefore it is irreducible over
Q. The discriminant is ∆ = (−4)(−1) − 27 · (−1)2 = −23, not a square
in Q. So the Galois group is S3 .
(b) X 3 − 10 over Q.
Solution. The polynomial is irreducible by Eisenstein’s criterion at
p = 2. The discriminant is ∆ = −27 · 102 , not a square in Q. So the
Galois group is S3 .
√
(c) X 3 − 10 over Q( 2).
√
Solution. The polynomial is irreducible over Q( 2) since any root has
3
degree√3 over Q (by the previous problem)
√ and so X − 10 has no roots
in Q( 2). ∆ < 0 is not a square in Q( 2) (every square in this field is
positive, since it’s a subfield of the real numbers). So the Galois group
is again S3 .
√
(d) X 3 − 10 over Q( −3).
3
Solution. By the same
√ reasoning as the previous problem,
√X −210 is
2
irreducible over
√ Q( −3). We have ∆ = −27 · 10 = (30 −3) is a
square in Q( −3). So the Galois group is A3 .
√
(e) X 3 − X − 1 over Q( −23).
Solution. From (a), the polynomial
√ is irreducible over Q. Then the
polynomial is irreducible over Q( −23) since√any root will have degree
3 over Q, and so X 3 −X −1√has no root in Q( −23). The discriminant
∆ = −23 is a square in Q( −23) and so the Galois group is A3 . √
√
(f) X 4 − 5 over Q, Q( 5), Q( −5), Q(i).
Solution. The exact same argument as in Lang, Example 3, shows that
the Galois group over Q is the dihedral group D8 , generated by σ and
τ , where σ(α) = iα, σ(i) = i and τ (α) = α, τ (i) = √
−i and √
α is a real
root of X 4 − 5. An easy calculation shows that Q( 5), Q( −5), Q(i)
are the fixed fields of {1, σ 2 , τ, σ 2 τ }, {1, σ 2 , στ, σ 3 τ }, and {1, σ, σ 2 , σ 3 },
respectively. It follows that these groups are also the Galois groups of
X 4 − 5 over the corresponding fields. The last one is isomorphic to
1
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MATH 819 HOMEWORK 4 SOLUTIONS
the cyclic group of order 4 and the other two are clearly isomorphic to
Z/2Z ⊕ Z/2Z (the “Klein 4-group”).
(g) X 4 − a where a is any integer 6= 0, 6= ±1 and is square free. Over Q.
Solution. We will show that it is the dihedral group D8 . If a > 0 this
follows essentially from the same argument as in Example 3 in Lang.
So suppose a < 0. Let α be a root of f (X) = X 4 − a. Then the
four distinct roots of f are ±α, ±iα. So the splitting field is clearly
k = Q(α, i). Let β = (1 + i)α. Then β 4 = −4a > 0 and if we choose α
appropriately then β is real. Note that since a 6= 0, ±1, if a 6= −2 then
x4 + 4a is irreducible by Eisenstein’s criterion
√ applied√at a prime p 6= 2
dividing a. If a = −2 then note that Q( 4 8) = Q( 4 2) has degree 4
over Q. In any case, [Q(β) : Q] = 4, β ∈ k, and β is real. Since α and
i are not real, this immediately implies that we must have [k : Q] = 8.
Applying the same argument as in Example 3 to β and i we see that
the Galois group is again D8 .
(h) X 3 − a where a is any square-free integer ≥ 2. Over Q.
Solution. The polynomial is irreducible by Eisenstein (at some prime
dividing a). The discriminant is −27(−a)2 = −27a2 , which is not a
square in Q. So the Galois group is S3 .
(i) X 4 + 2 over Q, Q(i).
Solution. From (g) the Galois group over Q is D8 . Let α be a root
of X 4 + 2. Then since [Q(α, i) : Q(i)] = 4, X 4 + 2 is irreducible over
Q(i). So there is an automorphism σ sending α to iα and fixing i. This
element has order 4 and so we see that the Galois group over Q(i) is
the cyclic group of order 4 (generated by σ).
(j) (X 2 − 2)(X 2 − 3)(X 2 − 5)(X 2 − 7) over Q.
Solution. See (k).
(k) Let p1 , . . . , pn be distinct prime numbers. What is the Galois group of
(X 2 − p1 ) · · · (X 2 − pn ) over Q?
√
Solution. Let Ki = Q( pi ). We claim that Ki+1 ∩ K1 · · · Ki = Q.
√
Since Ki+1 = Q( pi+1 ) and [Ki+1 : Q] = 2, it suffices to show that
√
√
√
pi+1 6∈ K1 · · · Ki = Q( p1 , . . . , pi ). This is an elementary calpQ
culation I’ll leave to you (use that 1 and the elements
i∈I pi ,
√
√
I ⊂ {1, . . . , i}, form a basis for Q( p1 , . . . , pi ) over Q and take
squares of both sides of an appropriate equation). Then by Corollary 1.15 the Galois group is (Z/2Z)n .
√
(l) (X 3 − 2)(X 3 − 3)(X 2 − 2) over Q( −3).
Solution. The splitting field is clearly
√
√
√
√
√
√
√
3
3
3
3
K = Q( 2, ζ3 , 3, 2) = Q( 2, −3, 3, 2).
√
√
√
√
Let k = Q( −3). Let K1 = k( 3 2), K2 = k( 3 3), K3 = k( 2). Note
that ∆(x3 − 2) = −27 · 42 is a square in k. So G(K1 /k) is cyclic of
MATH 819
HOMEWORK 4
SOLUTIONS
3
order 3 (note that X 3 −2 is irreducible over k). √Similarly for G(K2 /k).
Of course, G(K3 /k) is cyclic of order 2 (since 2 is real k 6= K3 ).
We claim
that K1 ∩ K2 = k. If not, then
√
β = 3 2. Then
(1)
√
3
3 ∈ K2 . Let α =
√
3
3 and
α = a + bβ + cβ 2
for some a, b, c ∈ k. Let σ ∈ G(K2 /k), σ(β) = ζ3 β. Since α ∈
6 k and
is a root of X 3 − 3 = 0, we must have σ(α) = ζ3i α for either i = 1 or
i = 2. Applying σ to (1), we find
ζ3i α = a + bζ3 β + cζ32 β 2 = ζ3i (α) = ζ3i (a + bβ + cβ 2 ).
This implies that either a = c = 0 or a = b = 0. But this is impossible.
For instance, α = bβ implies 3 = 2b3 which is plainly impossible.
So K1 ∩ K2 = k. For simple degree reasons, K3 ∩ K1 K2 = k. But then
by √
Corollary 1.15, the Galois group of (X 3 − 2)(X 3 − 3)(X 2 − 2) over
Q( −3) is Z/3Z ⊕ Z/3Z ⊕ Z/2Z.
(m) X n − t, where t is transcendental over the complex numbers C and n
is a positive integer. Over C(t).
Solution. f (X) = X n −t is irreducible by Eisenstein’s criterion (at the
prime t in C[t]). Let α be a root of f . The n distinct roots of f are
ζni α, i = 0, . . . , n − 1. There is an automorphism σ ∈ G(C(α)/C(t))
satisfying σ(α) = ζn α. Since ζn ∈ C, σ(ζn ) = ζn . But then we see
that σ has order n (and [C(α) : C(t)] = n) and so the Galois group
must be cyclic of order n, generated by σ.
(n) X 4 − t, where t is as before. Over R(t).
Solution. Let α be a root of X 4 − t. The four distinct roots are
±α, ±iα. So the splitting field is C(α) and [C(α) : R] = [C(α) :
C][C : R] = 4 · 2 = 8 from the previous problem. Nearly the same
proof as in Example 3 shows that the Galois group is D8 .
2. (Lang Ch. VI, Ex. 2) Find the Galois groups over Q of the following polynomials.
(a) X 3 + X + 1.
(b) X 3 − X + 1.
(c) X 3 + 2X + 1.
(d) X 3 − 2X + 1.
(e) X 3 − X − 1.
(f) X 3 − 12X + 8.
(g) X 3 + X 2 − 2X − 1.
Solution. By the integral root test, the polynomials in (a), (b), (c),
(e), (f), and (g) are irreducible. For (d) we have the factorization
X 3 − 2X + 1 = (X − 1)(X 2 + X − 1)
and so the Galois group is Z/2Z. The discriminants of (a), (b), (c),
(e), (f), and (g) are −31, −23, −59, −23, 5184, 49, respectively (after a
7
translation the last polynomial is X 3 − 37 X − 27
). Only the last two
4
MATH 819 HOMEWORK 4 SOLUTIONS
are squares in Q. So the last two Galois groups are cyclic of order 3
and the other Galois groups are S3 .
3. (Lang Ch. VI, Ex. 5) Let k be a field of characteristic 6= 2, 3. Let f (X), g(X) =
X 2 − c be irreducible polynomials over k, of degree 3 and 2 respectively.
Let D be the discriminant of f . Assume that
[k(D1/2 ) : k] = 2 and k(D1/2 ) 6= k(c1/2 ).
Let α be a root of f and β a root of g in an algebraic closure. Prove:
(a) The splitting field of f g over k has degree 12.
Solution. Since [k(D1/2 ) : k] = 2, the splitting field L of f has degree 6
over k. Note that k(D1/2 ) is the unique subfield of L that is quadratic
over k (the Galois group of f (over k) ∼
= S3 has a unique index two
subgroup). Since k(D1/2 ) 6= k(c1/2 ), L ∩ k(c1/2 ) = k. It follows that
the splitting field of f g is the compositum of L and k(c1/2 ), which has
degree 12 over k ([L(c1/2 ) : L] = 2 and [L : k] = 6).
(b) Let γ = α + β. Then [k(γ) : k] = 6.
Solution. Note that γ ∈ k(α, β) and [k(α, β) : k] = 6 (since [k(α) : k] =
3 and [k(β) : k] = 2 and these are coprime integers). So [k(γ) : k] ≤ 6.
Let α1 , α2 , α3 be the three roots of f in k a . The six embeddings of
k(α, β) in k a (over k) are given by sending α to one of α1 , α2 , α3 and
β to ±β. These six embeddings restrict to embeddings of k(γ). To
show that [k(γ) : k] = 6 it suffices to show that αi ± β are distinct for
i = 1, 2, 3 and the two choices of the signs (for then k(γ) will have at
least 6 distinct embeddings over k). That αi +β, i = 1, 2, 3 are distinct
is immediate. Suppose that αi + β = αj − β. Then αj − αi = 2β. But
this implies that k(β) ⊂ L, the splitting field of f , which contradicts
our earlier remarks. So [k(γ) : k] = 6.
4. (Lang Ch. VI, Ex. 15) Let K/k be a Galois extension, and let F be an
intermediate field between k and K. Let H be the subgroup of Gal(K/k)
mapping F into itself. Show that H is the normalizer of Gal(K/F ) in
Gal(K/k).
Solution. Let σ ∈ Gal(K/k) and suppose that σ(F ) ⊂ F . Since F is an
algebraic extension of k, we have in fact that σ(F ) = F (and σ −1 (F ) = F ).
Let τ ∈ Gal(K/F ). Let x ∈ F . Since τ fixes F and σ −1 (x) ∈ F , we
have that στ σ −1 (x) = σ(σ −1 (x)) = x. So στ σ −1 fixes F and so must be an
element of Gal(K/F ). This implies that σ is in the normalizer of Gal(K/F )
in Gal(K/k).
Let N be the normalizer of H = Gal(K/F ) in Gal(K/k). Let σ ∈ N .
Then σHσ −1 = H. Let τ ∈ H. Then τ = σψσ −1 for some ψ ∈ H. Let
x ∈ F . Then τ (σ(x)) = σψ(x) = σ(x). So τ fixes σ(x). Since τ ∈ H was
arbitrary, σ(x) is in the fixed field of H and so σ(x) ∈ F . So σ(F ) ⊂ F , as
desired.
5. (Lang Ch. VI, Ex. 28) Let E be an algebraic extension of k such that every
non-constant polynomial f (X) in k[X] has at least one root in E. Prove
that E is algebraically closed.
MATH 819
HOMEWORK 4
SOLUTIONS
5
Solution. We need to show that every nonconstant polynomial g(X) ∈
E[X] has at least one root in E. Let E a be an algebraic closure of E
and let α ∈ E a be a root of g(X). We need to show that α ∈ E. Since α
is algebraic over E and E is algebraic over k, α is algebraic over k. Let K
be a normal closure (in E a ) of k(α) over k (note that K/k is again finite).
Then it suffices to show that K ⊂ E. Let G be the group of automorphisms
of K over k. Let K0 be the maximal separable subextension of K over k.
Then by Proposition 6.11, K = K G K0 and K G is purely inseparable over
k, while K0 is separable and normal over k (Corollary 6.8). It suffices to
show that K G ⊂ E and K0 ⊂ E.
First, consider K0 . Since K0 /k is separable, by the Primitive Element
Theorem, K0 = k(β) for some β ∈ K0 . Let f (X) be the minimal polynomial
of β over k. By assumption, f (X) has a root γ with γ ∈ E. Since K0 /k is
normal, k(γ) ⊂ K0 and, in fact, k(γ) = K0 (look at the degrees of k(γ)/k
and k(β)/k). Since γ ∈ E, K0 ⊂ E as desired.
Now consider K G . Let β ∈ K G . Then β is purely inseparable over k.
Let f (X) be the minimal polynomial of β over k. Then (Lang, p. 249: P.
n
n
Ins. 3), f (X) = X p − a = (X − β)p for some a ∈ k and some integer n.
By assumption, f (X) has a root in E. But f (X) has only one root, β, and
so we must have β ∈ E. So K G ⊂ E.
It follows that E is algebraically closed.