2014 Algebra SEP
∼
Fields Day 3 Solutions, by E. Dummit
1. (Jan-08.3): Let
that
be a primitive 16th root of unity and
with
E = Q[]
and
f (x) = x8 + 16;
observe
f (α) = 0.
√
(a) Show that
2 ∈ Q[2 ].
f (x)
(b) Show that
splits in
G = Gal(E/Q),
Q[x].
(c) For
E[x].
show that no nonidentity element of
Solution:
G
xes
α.
Conclude that
f (x)
is irreducible in
√
√
2
2
1
+
i, so 2 = 2 + 2 = 2 + 14 .
; then =
2
2
√
√
2k+1 2, for 0 ≤ k ≤ 7. Since 2 ∈ E by part (a) we
√
a)
Without loss of generality let
b)
By the usual, the 8 roots of
roots of
c)
√
α = 2,
f
lie in
iπ/8
=e
f (x)
are
2
see that all
E.
√
√
√
√
√
σ ∈ Gal(E/Q) xes α = 2. Observe that σ( 2) = ± 2: if σ( 2) = 2, then σ(α)√= α implies
√
σ() = , but since generates E , σ xes all of E hence is the identity. If it were true that√σ( 2) = − 2,
2
2
2, contrary to
then we would need σ() = −, but then σ( ) = , so by part (a), σ would have to x
√
√
the assumption that σ( 2) = − 2. Hence σ is the identity, and by the usual argument this implies f is
Suppose
irreducible.
Remark
f is Q[α]: this
0 ≤ k ≤ 7 and from the fact
√
Q[α] = Q[], since 2 ∈ Q[2 ], meaning that
Another approach to this problem is to show directly that the splitting eld of
follows from the observation that the other roots of
f
are
α · 2k
for
1 2
α ∈ Q[α]. Then in fact we can see that
2
√
2 lies in both Q[α] and Q[]. Since the minimal polynomial of has degree ϕ(16) = 8, we see that
|Q[] : Q| = |Q[α] : Q| = 8, meaning that f is irreducible, and that G = Gal(E/Q) is simply the Galois
group of f .
2 =
that
E/F be an extension of char-0 elds with E = F [α]
E = E[] where is a primitive pth root of unity.
2. (Jan-11.3): Let
p.
Let
(a) Show that
(b) If
for some
α
with
αp ∈ F
and some prime
∗
E
E∗
is a Galois extension of
is Galois over
F,
show
E=F
or
F.
E = E∗.
(c) Show by example that it is possible to have
E = E∗
without having
∈ F.
Solution:
a) E ∗
is the splitting eld of
Hence it is Galois, since
b)
xp − αp ∈ F [x], since the
F has characteristic zero.
σ
does not x
σ(α) = α · k for some
∈ E , whence E = E ∗ .
hence
c)
We can take
α · k
for
0 ≤ k ≤ p − 1.
α, then α ∈ F hence E = F . Otherwise, if
xp − αp ∈ F [x] we see that σ(α) ∈ E is also a root of xp − αp ,
σ(α)
1 ≤ k ≤ p − 1. Hence k =
∈ E , so since k is invertible mod p, we get
α
Consider the action of Gal(E/F ) on
some element
roots of this polynomial are
α,
α:
if every element xes
then since
F = Q, α = = ζ3 , p = 3:
then
E ∗ = E = Q(ζ3 ),
but
6∈ F .
L/K be
contains K and
3. (Jan-14.5): Let
properly
(a) Show that
L/K
a eld extension of degree 4.
is properly contained in
We say
K0
is intermediate between
K
if
K0
has at most 3 intermediate elds.
L
and
K.
(c) Give an explicit example to show that there can be 0 intermediate elds between
L
and
K.
In fact, as it was stated on the qualifying exam, part (a) is false unless we also assume
separable. A spectacular counterexample is the following:
an innite eld of characteristic 2 and
x
y
and
K = k(x2 , y 2 )
are indeterminates. Then
and
L/K
L = k(x, y),
(It is reasonably easy to check that both
L/Kα
and
Kα /K
L/K
k
where
is
is
has degree 4, but there
are in fact innitely many intermediate subelds, one family of which is given by
α ∈ k.
L
L.
(b) Give an explicit example to show that there can be 3 intermediate elds between
Remark
and
Kα = K(x + αy)
for
are degree-2 extensions.) The reason
L/K has nitely many
L is obtained by adjoining a
that L ∼
= K[x]/p(x)). The point
this example exists is due to the following theorem: A nite-degree extension
intermediate elds if and only if
single element to
K
L
is a primitive extension of
K
that is, if
(i.e., there exists an irreducible polynomial
p
such
here is that this inseparable extension cannot be generated by a single element.
Solution:
L/K is separable of degree 4, consider its
G = Gal(L̂/K) and also let Gal(L̂/L) = H .
If
Denote
a)
normal closure
L̂/K ,
which is a Galois extension.
L/K , we get an embedding
G as a transitive subgroup of the symmetric group S4 . There are ve possibilities, up to conjugacy
in G: (i) G = h(1 2 3 4)i is cyclic of order 4, (ii) G = h(1 2)(3 4), (1 3)(2 4)i is the Klein 4-group, (iii)
G = h(1 2 3 4), (1 3)i is the dihedral group of order 8, (iv) G is the alternating group A4 , (v) G is S4 . By
0
the Galois correspondence, the intermediate elds K correspond to index-2 subgroups of G containing
a point stabilizer H (in other words: H is the subgroup of G that xes 1). The cases have the following
numbers of such subgroups: (i) one subgroup h(1 3)(2 4)i, (ii) three subgroups each generated by an
element of order 2, (iii) one subgroup h(1 2 3 4)i, (iv) no subgroups, (v) no subgroups. In each case, there
From its action on the roots of the minimal polynomial of a generator of
of
are at most 3 intermediate elds.
a-alt)
By hypothesis,
containing
H
left cosets of
b)
H
is a subgroup of
having index 2 in
H,
G.
G
of index 4, and intermediate subelds correspond to subgroups
Such a subgroup is necessarily the union of
and another of the 3
L/K
√ √
L = Q( 2, 3).
From the analysis in (a), there can be 3 intermediate elds only when the extension
with Galois group the Klein 4-group. One such example over
c)
H
so there are at most 3 intermediate extensions.
K=Q
is
From the analysis in (a), there can be 0 intermediate elds when the extension
with Galois closure the alternating group
degree-4 polynomial in
K[x]
Remark
Q
and
q(x) = x4 − x + 1
p
and
q
p
is
212 34
S4
i.e., if
L = K[α]
where
L/K is a degree-4 extension
α is a root of an irreducible
A4 or S4 . It is not too likely that one would know a
S4 ohand, but p(x) = x4 + 8x + 12 has Galois group A4
group S4 over Q.
A4
has Galois
One can prove these statements about
discriminant of
or
whose Galois group is
quartic whose Galois group is provably
over
A4
itself is Galois
or
p and q
by observing that they are both irreducible, that the
(a square), that the discriminant of
q
is 229 (not a square), and then factoring
modulo primes not dividing the discriminant to see that the irreducible-factor degrees yield all
of the possible cycle types in
An
or
Sn
respectively. For example,
q
is irreducible modulo 2, and modulo
3 it has factors of degrees 1 and 3. By a standard result of number theory, this means the Galois group
of
q
contains a 4-cycle and a 1,3-cycle, but the only such subgroup of
S4
is
S4
itself. Similarly,
p
modulo
5 has irreducible factors of degrees 1 and 3, and since its discriminant is a square, its Galois group must
be
bc-alt)
A4 .
L/K with arbitrary
G: let G be embedded as a subgroup of a symmetric group Sn , and let L = C(x1 , · · · , xn )
and F = C(σ1 , · · · , σn ) where σi is the elementary symmetric function in the xj of degree i. Then
Gal(L/F ) = Sn : it contains Sn (each of the
permutation on the xj ), but L is also
Qσi is xed by
P index n−j
the splitting eld of the polynomial p(x) =
σn−j xj of degree n, so |L : F | ≤ n!,
i (x − xi ) =
j (−1)
hence equality must hold. Now let K be the xed eld of G inside L; then by the Galois correspondence,
L/K is Galois and Gal(L/K) ∼
= G.
For (b) and (c) one can also use this general construction for an extension
Galois group
4. (Aug-05.3): Let
(a) Let
F
(b) With
p, q
be distinct primes and let
be a subeld of
F
R
not containing
as in (a), show that
(c) Show that
µ = q 1/p
µ.
If
and
µn ∈ F
ν = p1/q .
for some
n > 0,
show that
p|n.
|F [µ] : F | = p.
|Q[µ + ν] : Q| = pq .
Solution:
a)
If
b)
Consider the minimal polynomial
p|n we are done so assume n is invertible mod p, say with mn = 1 + kp for some integers m and k .
q −k µnm = µ ∈ F , contradicting the assumption that µ 6∈ F .
Then
q(x) of µ over F : it divides xp − q , so its other roots must be of the
k
form µ · ζp for some integers 1 ≤ k ≤ p − 1. Its constant term q(0) is the product of some of these,
d
f
f
hence is of the form ±µ · ζp . Since F is a subeld of R we see that ζp must be in R, hence must be ±1
d
thus µ ∈ F , so by part (a) we see p|d. Since d ≤ p we see d = p, and so q(x) has degree p, and so
|F [µ] : F | = p.
c)
xp −q and xq −p are irreducible, so |Q(µ) : Q| = p, |Q(ν) : Q| = q , and then |Q(µ, ν) : Q| = pq
since the degrees are coprime, so it is enough to show that Q(µ + ν) = Q(µ, ν). If one of µ, ν is in
F = Q(µ + ν) then the other is also, so suppose neither is: then by part (b) applied to F [µ] = Q(µ, ν)
and then F [ν] = Q(µ, ν), we would have |Q(µ, ν) : F | = p and also |Q(µ, ν) : F | = q , which is impossible.
By Eisenstein,
Q ⊆ K ⊆ E ⊆ C be elds
E . Assume E is Galois over Q.
5. (Aug-06.3): Let
unity in
with
E = Q[α]
with
αn ∈ Q,
and
K
is generated by all roots of
(a) Show that Gal(E/K) is cyclic.
(b) If the restriction
τ
of complex conjugation to
E
is in the center of Gal(E/Q), prove that
2
|α| ∈ Q.
Solution:
a)
Let
σ∈
Gal(E/K) and observe that
b)
Let
σ∈
Gal(E/Q) and as above observe that
αn ∈ Q, so σ(α) = α · ζσ for some nth
root of unity ζσ . Since E is Galois, all Galois conjugates of α lie in E , so ζ ∈ E hence by denition of
K , ζ ∈ K . But now we see that the map π : Gal(E/K) 7→ µn (the group of nth roots of unity) dened
by sending σ 7→ ζσ is a well-dened homomorphism from the Galois group to the group of nth roots of
unity, which is cyclic. The map is also injective since if π(σ) = 1, then σ(α) = α, so that σ xes E , hence
is the identity in Gal(E/K). Since subgroups of cyclic groups are cyclic, we see that Gal(E/K) is cyclic.
σ
xes
since
σ(α) = α · ζn for some nth root of unity. But then since τ
2
2
2
2
σ(|α| ) = σ(α · τ (α)) = σ(α) · σ(τ (α)) = σ(α) · τ (σ(α)) = |σ(α)| = |αζn | = |α| .
2
2
|α| for every σ , meaning |α| ∈ Q.
is central, we have
Hence
σ(α)n = σ(αn ) = αn ,
6. (Jan-05.3): Let
α
of
f (x)
F
be a eld and
such that
Solution:
f (α2 ) = 0.
f (x) ∈ F [x] be irreducible.
f splits over E .
α 7→ α2 is
root of f . To
The idea is:
should also be a
Suppose
E/F
is an extension containing a root
Show that
morally an element of the Galois group, so the square of any root of
f
show this without passing to a Galois closure (in case we are in positive
g(x) = f (x2 ) and observe that α is a root of both f
2
and g . Since f is the minimal polynomial, we see that f divides g , and so we can write f (x ) = f (x)·q(x),
2
2
where q has the same degree as f . Then if β is any root of f , we see f (β ) = f (β) · q(β) = 0, so β is
2a
2b
also a root of f . Since f has only nitely many roots, by iterating this we see that β
= β for some
a and b, so either β = 0 (in which case f (x) = x), or β is a primitive nth root of unity for some n. If
k
any roots of f are primitive k th roots of unity for k < n, then gcd(f, x − 1) has positive degree and
divides f ; since f is irreducible, this means its roots are all primitive nth roots of unity. Then f splits
over F [α] ⊆ E since the other roots are powers of α.
characteristic, which could cause problems), we can let
Remark
The ocial solution assumes without explanation the fact that if
other roots are also
nth
K ⊆ F ⊆ E be elds with E = F [α], αn ∈ F for
Let L be a eld with K ⊆ L ⊆ E with L ∩ F = K .
7. (Aug-11.3): Let
root of unity.
(a) If
L
is Galois over
K,
show that
(b) Show by example that
α
is an
nth
root of unity, then all
roots of unity. While this is true, it requires justication!
L
L = K[β]
for some
need not be Galois over
β
with
some
n,
and
K
containing a primitive
nth
βn ∈ K .
K.
Solution:
a)
xn − αn ∈ F [x], as
all its roots
(since K contains a primitive nth root of unity). The Galois group Gal(E/F ) is also
n
n
cyclic: every element of the Galois group xes x − α ∈ F [x], hence σ(α) = α · ζ ∈ E where ζ is some
nth root of unity (necessarily xed by σ since K contains a primitive nth root of unity); conversely since
α generates the extension E/F , any element σ ∈ Gal(E/F ) is determined uniquely by its action on α.
σ(α)
Then the map ϕ : Gal(E/F ) → µn sending σ 7→
is an injective homomorphism, so we conclude
α
that Gal(E/F ) is isomorphic to a subgroup of a cyclic group, hence is cyclic. We conclude that it is
generated by an element τ : α 7→ ζα where ζ is some (possibly non-primitive) nth root of unity.
Then by the sliding-up property of Galois extensions, Gal(L/K) = Gal(L/L ∩ F ) ∼
= Gal(E/F ) because
L∩F = K . We conclude that Gal(L/K) is also a cyclic group generated by a map of the form ϕ : β 7→ βζ
for some β , by standard Kummer theory (or Hilbert's theorem 90, or another method). This element
β is necessarily a generator of L/K (since ϕ generates the Galois group), and ϕ(β n ) = β n hence β n is
xed by all of Gal(L/K) hence is in K .
E
lie in E
First observe that
Remark
is Galois over
F,
E
is the splitting eld of the polynomial
It is a standard fact of Kummer theory that if
K[β 1/n ]
for some
β ∈ K.
is a eld containing the
then any cyclic Galois extension of
K
nth
roots of unity and
of degree
To prove this, if
that the Lagrange resolvent
σ(α, ζn ) = ζn−1 (α, ζn ).
n,
K
n has the form
σ is a generator of the Galois group and α ∈ K , one observes
(α, ζn ) = α + ζn σ(α) + ζn2 σ 2 (α) + · · · + ζnn−1 σ n−1 (α) has the property that
whose characteristic does not divide
b)
since
√
K = Q, L = Q(21/3 ), F = Q(21/3 ζ3 ), E = Q(21/3 , ζ3 ) = F [ −3], for ζ3 a primitive cube
root of unity. Then L is not Galois (its Galois closure is E ), and since |L : Q| = |F : Q| = 3 we see that
[L ∩ F : Q] divides 3: if it were 3, it would necessarily be equal to L and to F , but L and F are not equal
(F contains a nonreal element while E does not) hence we see L ∩ F = Q.
We can take