MTH745U/P (2013–14)
Further Topics in Algebra
(Fields and Galois Theory)
2nd December 2013
Solutions 9
√
Question 1. From Question 1 of √
Coursework 7, we know that K = F ( d) or
(for√characteristic
2 only) K = F ( ∗ d) for some d ∈ F . The minimal polynomials
√
of d and ∗ d are X 2 − d and X 2 + X + d respectively. (We are presuming that
|K : F | = 2, and so that these elements are
F .) If σ ∈ Gal(K : F ) then
√ not in√
σ must fix F elementwise, while sending d (or ∗ d) to an element with the
same minimal polynomial. This √
determines the
action of σ on K, as√the typical
√
∗
element
of K has the form a + b d or a + b d where a, b ∈ F
. So d maps
to
√
√
√
√
∗
∗
d
or
−
d
(and
these
are
the
same
in
characteristic
2),
and
d
maps
to
d
or
√
∗
d + 1. So the possible Galois groups are as follows.
√
(a) K = F ( d), 2 6= 0. The Galois group is cyclic
2, consisting of the
√ of order √
identity map and the map that sends a + b d to a − b d for a, b ∈ F .
√
(b) K = F ( d), 2 = 0. The Galois group is trivial (cyclic of order 1), conisting
only of the identity map.
√
(c) K = F ( ∗ d), 2 = 0. The Galois group is cyclic
of order 2, consisting
of the
√
√
∗
∗
identity map and the map that sends a + b d to a + b + b d for a, b ∈ F .
Note that −1 = 1 in Cases (b) and (c). The listed field extensions (being in
characteristic 0) have Galois groups of order 2, and they belong to Case (a).
Question 2. Let H 6 G. Now g m ∈ H if and only if (g m )−1 = g −m ∈ H. So
either H = {g 0 } = {1} = h1i is cyclic, or there exists a unique integer m > 0
such that g m ∈ H but g j ∈
/ H for 0 < j < m. Now suppose that g s ∈ H. In
the latter case, write s = qm + r where q, r ∈ Z and 0 6 r < m. Then we
get g r = g s (g m )−q ∈ H. So r = 0 by the minimal condition on m. Therefore
H = hg m i = { g im : i ∈ Z } is cyclic.
If g has infinite order, then the subgroups h1i and hg m i for m > 0 are pairwise
distinct (having indices ∞ [actually ℵ0 ] and m in G). If g has finite order n > 0,
then h1i = hg n i, so any subgroup H of G has the form H = hg m i for some m > 0.
n
Since 1 = g n ∈ H, we get m | n. The group G has order n, and H has order m
.
n/d ∼
∼
So G = Cn has precisely one subgroup for each divisor d | n, namely hg i = Cd ,
and no other subgroups.
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Question 3. First, we need to find the degrees and roots of the minimal polynomials of ζn over Q, for our n of interest. For n = p prime, the minimal polynomial
is X p−1 + · · · + X + 1 (irreducible by Coursework 4, Question 6) with roots ζpi
for 1 6 i 6 p − 1. Clearly ζ4 = i has minimal polynomial X 2 + 1 (over Q),
whose roots are i and −i. We know that ζ8 has minimal polynomial X 4 + 1 from
Coursework 6, Question 6. The roots of X 4 + 1 are ζ8i for i ∈ {1, 3, 5, 7}. Finally,
ζ9 is clearly a root of X 6 + X 3 + 1, which is its minimal polynomial. To see
this, we note that Q(ζ9 ) has subfields Q(ζ3 ) and Q(γ7 ) [notation of Coursework
7, Question 2] of degrees 2 and 3 over Q. The roots of X 6 + X 3 + 1 are ζ9i for
i ∈ {1, 2, 4, 5, 7, 8}.
Any element φ ∈ Gn = Gal(Q(ζn ) : Q) is determined by the image (ζn φ) of ζn ,
and this must have the same minimal polynomial over Q as ζn does. Thus ζn φ is a
root of the minimal polynomial of ζn over Q. Let φj be the [Q-]automorphism of
Q(ζn ) that takes ζn to ζnj . (This is an automorphism if and only if gcd(n, j) = 1,
and we can restrict j to the range 0 6 j 6 n − 1.) We have φj φk = φjk , where
of course we may reduce jk modulo n.
The case n = 3. We have Gn = G3 = hφ2 i ∼
= C2 (the cyclic group of order 2),
with subgroups 1 (trivial subgroup) with fixed subfield Q(ζ3 ) [note that ζ3 = ω],
and G3 with fixed subfield Q.
The case n = 4. We have Gn = G4 = hφ3 i ∼
= C2 (the cyclic group of order 2),
with subgroups 1 (trivial subgroup) with fixed subfield Q(ζ4 ) [note that ζ4 = i],
and G4 with fixed subfield Q.
The case n = 5. We have Gn = G5 = hφ2 i = hφ3 i ∼
= C4 (the cyclic group of order
∼
4), with subgroups
with fixed
√ 1 (trivial subgroup)
√subfield 2Q(ζ53), hφ14 i = C2√with
1
4
fixed subfield Q( 5) (we have ζ5 + ζ5 = 2 (−1 + 5) and ζ5 + ζ5 = 2 (−1 − 5)),
and G5 with fixed subfield Q.
The case n = 7. We have Gn = G7 = hφ3 i = hφ5 i ∼
= C6 (the cyclic group of
order 6). Its subgroups are given below:
1,
hφ6 i ∼
= C2 ,
hφ2 i = hφ4 i ∼
= C3 ,
∼
hφ3 i = hφ5 i = C6 ,
with
with
with
with
fixed
fixed
fixed
fixed
field
field
field
field
Q(ζ7 );
Q(ζ√7 + ζ76 );
Q( −7);
Q.
√
We have ζ7 + ζ72 + ζ74 = 12 (−1 + −7), and for 1 6 j 6 6, we also have Q(ζ7j ) =
Q(ζ7 ) and Q(ζ7j + ζ77−j ) = Q(ζ7j + ζ7−j ) = Q(ζ7 + ζ76 ).
The case n = 8. We have Gn = G8 = {φ1 , φ3 , φ5 , φ7 } ∼
= C2 ×C2 ∼
= 22 , all of whose
elements square to the identity (φ1 ). This is not the cyclic group of order 4. It
is possible to generate G8 by any two of its non-trivial elements. Its subgroups
2
are given below:
√
field Q(ζ√8 ) = Q(i, 2);
field Q(√−2);
field Q(√−1);
field Q( 2);
field Q.
√
√
√
We have ζ82 = ζ4 = i = −1, ζ8 + ζ87 = 2 and ζ8 + ζ83 = −2, and also
ζ8 + ζ85 = 0.
The case n = 9. We have Gn = G9 = hφ2 i = hφ5 i ∼
= C6 (the cyclic group of
order 6). Its subgroups are given below:
1,
∼
hφ3 i = C2 ,
hφ5 i ∼
= C2 ,
∼
hφ7 i = C2 ,
hφ3 , φ5 i ∼
= 22 ,
with
with
with
with
with
fixed
fixed
fixed
fixed
fixed
1, with fixed field Q(ζ9 );
∼
hφ8 i = C2 , with fixed field Q(ζ√9 + ζ98 );
hφ4 i = hφ7 i ∼
= C3 , with fixed field Q( −3);
∼
hφ2 i = hφ5 i = C6 , with fixed field Q.
√
We have ζ93 = ζ3 = ω = 12 (−1 + −3), and for j ∈ {1, 2, 4, 5, 7, 8}, we also have
Q(ζ9j ) = Q(ζ9 ) and Q(ζ9j + ζ99−j ) = Q(ζ9j + ζ9−j ) = Q(ζ9 + ζ98 ).
The case n = 11. We have Gn = G11 = hφ2 i = hφ6 i = hφ7 i = hφ8 i ∼
= C10 (the
cyclic group of order 10). Its subgroups are given below:
1,
∼
hφ10 i = C2 ,
hφ3 i = hφ4 i = hφ5 i = hφ9 i ∼
= C5 ,
hφ2 i = hφ6 i = hφ7 i = hφ8 i ∼
= C10 ,
with fixed field Q(ζ11 );
10
with fixed field Q(ζ√11 + ζ11
);
with fixed field Q( −11);
with fixed field Q.
√
4
5
9
3
We have ζ11 + ζ11
+ ζ11
+ ζ11
+ ζ11
= 21 (−1 + −11), and for 1 6 j 6 10, we also
j
j
11−j
j
−j
10
have Q(ζ11
) = Q(ζ11 ) and Q(ζ11
+ ζ11
) = Q(ζ11
+ ζ11
) = Q(ζ11 + ζ11
).
∼
The case n = 13. We have Gn = G13 = hφ2 i = hφ6 i = hφ7 i = hφ11 i = C12 (the
cyclic group of order 12). Its subgroups are given below:
1,
hφ12 i ∼
= C2 ,
hφ3 i = hφ9 i ∼
= C3 ,
∼
hφ5 i = hφ8 i = C4 ,
hφ4 i = hφ10 i ∼
= C6 ,
∼
hφ2 i = hφ6 i = hφ7 i = hφ11 i = C12 ,
fixed field Q(ζ13 );
12
fixed field Q(ζ13 + ζ13
);
fixed field Q(d13 );
fixed field Q(c√13 );
fixed field Q( 13);
fixed field Q.
√
4
3
12
9
10
We have ζ13 + ζ13
+ ζ13
+ ζ13
+ ζ13
+ ζ13
= 21 (−1 + 13), and for 1 6 j 6 12, we
j
j
13−j
j
−j
12
also have Q(ζ13
) = Q(ζ13 ) and Q(ζ13
+ ζ13
) = Q(ζ13
+ ζ13
) = Q(ζ13 + ζ13
). We
5
8
12
3
9
have c13 = ζ13 + ζ13 + ζ13 + ζ13 [which we have met before] and d13 = ζ13 + ζ13 + ζ13
[which we have not].
3
with
with
with
with
with
with
In general, for p a prime, Gp is cyclic of order p − 1. Its subgroup structure (and
thus the fixed subfields) depends on the factorisation of p − 1.
In all cases, it is clear that the fixed fields of our subgroups are at least as big as
claimed. Note that we have written all our (sub-)fields in the form Q(α) for some
α ∈ Q(ζn ). Also, for all n, Q(ζn ) is a normal extension of Q [why?], so Galois
Correspondence applies (as we are in characteristic 0). Thus given a subgroup
H of Gn , its fixed subfield should have degree |Gn |/|H|. So if the fixed subfield
is claimed to be Q(α) and is visibly fixed by H, one can prove that its degree
is the expected |Gn |/|H| by exhibiting |Gn |/|H| distinct elements αφj (with
gcd(j, n) = 1) such that Q(αφj ) = Q(α). [The effectiveness of this method relies
on all subfields of Q(ζn ) being normal extensions of Q, which here follows from
the fact that Gal(Q(ζn ) : Q) is abelian, and so all its subgroups are normal and
so all the corresponding subfields are normal extensions of Q.] I leave the details
involved in these verifications to the reader, but I shall consider one case, namely
2
6
5
Q(d13 ). We have d13 φ3 = d13 φ9 = d13 . However, d13 , d13 φ2 = ζ13
+ ζ13
+ ζ13
, d13 φ4
and d13 φ8 are distinct, which is quite easily seen as the minimal polynomial of ζ13
has degree 12. The minimal polynomial (over Q) of d13 is X 4 +X 3 +2X 2 −4X +3,
and we have d13 φ2 = 13 (2d313 + 3d213 + 4d13 − 6), d13 φ4 = 13 (−d313 − 2d13 + 3) and
d13 φ8 = 31 (−d313 − 3d213 − 5d13 ).
Question 4. Let α0 , α1 and α2 be the complex roots of an irreducible cubic (f )
over Q, with splitting field F . There are two possibilities.
Firstly, F = K can have degree 3 over Q. So F = K = Q(α0 ) = Q(α1 ) = Q(α2 ),
and so any element of Gal(F : Q) is determined by its action on α0 . The Galois
group must be cyclic of order 3, generated by the 3-cycle (α0 , α1 , α2 ). Note that
all the αi must be real in this case. There are only two subgroups, h(α0 , α1 , α2 )i,
with fixed field Q, and the trivial subgroup, with fixed field F . This is what
happens in Cases (d) and (e), which are concerned with the polynomials f9 and
f7 (respectively) on Coursework 7 [where we determined the degrees of their
splitting fields].
Secondly, F can have degree 6 over Q. So for all i 6= j (0 6 i, j 6 2) we have
F = Q(αi , αj ), while K ∼
= Ki := Q(αi ) has degree 3 over Q (and so F has degree
2 over Ki for all i). This means that K contains exactly one root of f , and it is
generated over Q by this root, which forces Gal(K : Q) to be trivial. The full
Galois group Gal(F : Q) is S3 acting as such on {α0 , α1 , α2 }. The six subgroups
of S3 , and their corresponding fixed fields are as follows.
Subgroup
Fixed field
1 (trivial)
F = Q(α0 , α1 , α2 )
h(α0 )(α1 , α2 )i ∼
K0 = Q(α0 )
= C2
h(α1 )(α0 , α2 )i ∼
C
K1 = Q(α1 )
= 2
∼
h(α2 )(α0 , α1 )i = C2
K2 =√Q(α2 )
∼
h(α0 , α1 , α2 )i = C3
Q( D)
∼
h(α0 , α1 , α2 ), (α0 )(α1 , α2 )i = S3
Q
4
√
Here D ∈ Q, but D ∈
/ Q. Note that either just one of α0 , α1 , α2 is real (in
which case we can choose α0 to be the real root), or they all are. In the case
when only α0 is real, the automorphism (α0 )(α1 , α2 ) corresponds to complex
conjugation. Concrete versions of the above for Cases (a), (b) and (c) are given
in the following paragraphs. (The degrees of their splitting fields were already
determined on Coursework 7.)
√
√
√
In Case (a), with polynomial X 3 −2, we take α0 = 3 2, α1 = ω 3 2 and α2 = ω√3 2,
where√ω and ω are the primitive cube roots of 1, as usual. The subfield Q( D)
is Q( −3) = Q(ω). Only the root α0 is real.
In Case (b), we can take D = −31, and we let α√0 := α be the unique real root
−31
of f . If you must know, we have α1 , α2 = − 12 α ± 62
(6α2 − 9α + 4), as we have
already noted on Coursework 2 (solutions to Question√7).
√
In Case (c), we can take D = 148 or 37 (we have Q( 148) = Q( 37)). In this
case, all 3 roots are real,
and we let α0 = α be any of them. It can be shown
√
37
1
that α1 , α2 = − 2 α ± 74 (12α2 + 9α − 32), but we really do not need to know
this when determining the Galois group. (The factorisation of X 3 − 4X + 2 in
Q(α)[X] is (X − α)(X 2 + αX + α2 − 4).)
i
Question 5. We know that the m maps φi : x 7→ xp (with i taken modulo m)
j
are automorphisms of Fq , and the m/a maps φaj : x 7→ xr fix the subfield Fr
elementwise, since pa = r and Fr 6 Fq is precisely the set of roots of X r − X = 0.
In lectures we showed that the multiplicative group of a finite field is cyclic.
So let λ be an element of order q − 1 in Fq . The elements of Fr are roots of
X r − X = 0, and so the nonzero elements of Fr inside Fq are precisely the powers
q−1
of λ r−1 . So the degree m/a extension Fq : Fr is simple and we have Fq = Fr (λ).
Therefore, the minimal polynomial f (X) ∈ Fr [X] of λ over Fr has degree m/a.
Now if φ is a field automorphism of Fq , then λφ = λj for some integer j (and
without loss of generality 0 6 j < q − 1), since λφ 6= 0. So λk φ = λkj , and
this entirely determines φ. From a lemma proved in lectures, we know that any
Fr -automorphism must permute the m/a roots of f . So there are at most m/a
possibilities for λφ, and thus for φ itself. But we have already seen this many
Fr -automorphisms above, so this must be all of them.
}, which is a cyclic group of order m/a.
Answer: Gal(Fq : Fr ) = { φaj : 0 6 j < m
a
(Compostion of these maps is obvious.) Note that we also proved in lectures
that any automorphism of Fq must fix the prime subfield Fp elementwise. So
Aut(Fp ) = Gal(Fq : Fp ) ∼
= Cm .
Dr John N. Bray, 2nd December 2013
5