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It is worth remarking that the most extreme version of this occurs when we ask for a Galois
group which is simple. There has been a great deal of research activity on this question in the
past few decades, but apparently not all simple groups are known to occur as Galois groups of
extensions of Q or other finite subextensions of C/Q. Here is an example whose Galois group
is A5; this is verified using Proposition 4.26.
6.42. Example. The Galois group of f(X) = X5+20X+16 over Q is Gal(Q(f(X))/Q) A5,
=
hence it is not solvable.
82 6. A GALOIS MISCELLANY
6.6. Symmetric functions
Let k be a field. Consider the polynomial ring on n indeterminates k[X1, . . . , Xn] and its
field of fractions K = k(X1, . . . , Xn). Each permutation à " Sn acts on k[X1, . . . , Xn] by
à · f(X1, . . . , Xn) = fÃ(X1, . . . , Xn) = f(XÃ(1), . . . , XÃ(n)).
Viewed as a function ÷: k[X1, . . . , Xn] -’! k[X1, . . . , Xn] is a ring isomorphism; this extends to
a ring isomorphism ÷: k(X1, . . . , Xn) -’! k(X1, . . . , Xn). Varying à we obtain actions of the
group Sn on k[X1, . . . , Xn] and k(X1, . . . , Xn) by ring isomorphisms fixing k and in the latter
case it is by field automorphisms fixing k.
6.43. Definition. The field of symmetric functions on n indeterminates is
Symn(k) = k(X1, . . . , Xn)Sn k(X1, . . . , Xn).
So if f(X1, . . . , Xn) " k(X1, . . . , Xn), then
f(X1, . . . , Xn) " Symn(k) Ð!Ò! "Ã " Sn f(X1, . . . , Xn) = f(XÃ(1), . . . , XÃ(n)).
6.44. Theorem. The extension k(X1, . . . , Xn)/ Symn(k) is a finite Galois extension for
which Gal(k(X1, . . . , Xn)/ Symn(k)) Sn.
=
Proof. There are elements of k[X1, . . . , Xn] †" k(X1, . . . , Xn) called elementary symmetric
functions,
ek = Xi1Xi2 · · · Xik,
i1
where 1 k n. It is easy to see that for every à " Sn, eà = ek, so ek " Symn(k). Working in
k
the ring k(X1, . . . , Xn)[Y ] we have
n n-1
fn(Y ) = Y - e1Y + · · · + (-1)n-1en-1Y + (-1)nen = 0,
hence the roots of this polynomial are the Xi. So k(X1, . . . , Xn) is the splitting field of fn(Y )
over Symn(k). Now Sn Gal(k(X1, . . . , Xn)/ Symn(k)), hence
[k(X1, . . . , Xn) : Symn(k)] = | Gal(k(X1, . . . , Xn)/ Symn(k))| |Sn| = n!.
But as every element of Gal(k(X1, . . . , Xn)/ Symn(k)) permutes the roots of fn(Y ) and is de-
termined by this permutation, we also have
n! | Gal(k(X1, . . . , Xn)/ Symn(k))|.
Combining these inequalities we obtain | Gal(k(X1, . . . , Xn)/ Symn(k))| = n! and therefore
Gal(k(X1, . . . , Xn)/ Symn(k)) = Sn.
6.45. Remark. In fact, this proof shows that the extension k(X1, . . . , Xn)/k(e1, . . . , en) is
Galois of degree n!. Since k(e1, . . . , en) Symn(k) we can also deduce that k(e1, . . . , en) =
Symn(k). Hence every element of Symn(k) is a rational function in the ei. Analogous results
are true for polynomials, i.e.,
k[X1, . . . , Xn]Sn = k[e1, . . . , en].
6.46. Corollary. If n 5, the extension k(X1, . . . , Xn)/ Symn(k) is not solvable.
Exercises on Chapter 6
6.1. Let p > 0 be a prime and G a group of order |G| = pn for some n 1. Show by induction
on n that there is a normal subgroup N G with |N| = pn-1. [Hint: what do you know about
the centre of G? Use this information to produce a quotient group of smaller order than G.]
6.2. Let K be a field for which char K = 2 and n 1 be odd. If K contains a primitive n-th
root of unity, show that then K contains a primitive 2n-th root of unity.
83
6.3. Find all values of n 1 for which Õ(n) | 4. Using this, determine which roots of unity lie
in the following fields:
" " "
Q(i), Q( 2 i), Q( 3 i), Q( 5 i).
6.4. (a) Describe the elements of (Z/24)× explicitly and verify that this group is isomorphic
to Z/2 × Z/2 × Z/2. Describe the effect of each element on Q(¶24) and Q(cos(À/12)) under the
action described in Theorem 6.2.
(b) Determine the group (Z/20)× and describe the effect of each of its elements on Q(¶20) and
Q(cos(À/10)) under the action described in Theorem 6.2.
6.5. Let n 1.
(a) What can you say about sin(2À/n) and Gal(Q(sin(2À/n))/Q))?
(b) Determine sin(À/12) and Gal(Q(sin(À/12))/Q)).
6.6. In this question, work in the cyclotomic field Q(¶5) where ¶5 = e2Ài/5.
(a) Describe the Galois group Gal(Q(¶5)/Q) and its action on Q(¶5).
(b) Determine the minimal polynomial of cos(2À/5) over Q. Hence show that
"
-1 + 5
cos(2À/5) = .
4
For which other angles ¸ is cos ¸ a root of this minimal polynomial? What is the value
of sin(2À/5) ?
(c) Find the tower of subfields of Q(¶5) and express them as fixed fields of subgroups of
Gal(Q(¶5)/Q).
6.7. In this question, let p be an odd prime and let ¶p = e2Ài/p " Q(¶p) C.
(a) Consider the product
(p-1)/2
r -r
¾ = (¶p - ¶p ) " Q(¶p).
r=1
Show that
p-1
r
¾2 = (-1)(p-1)/2 (1 - ¶p).
r=1
(b) Deduce that
p if p a" 1 (mod 4),
¾2 =
-p if p a" 3 (mod 4).
(c) Conclude that
"
± p if p a" 1 (mod 4),
¾ =
"
± p i if p a" 3 (mod 4).
" "
and also p " Q(¶p) if p a" 1 (mod 4) and p i " Q(¶p) if p a" 3 (mod 4).
6.8. Prove Lemma 6.41. [Hint: show that every 2-cycle of the form (i i + 1) is in H by
considering elements of the form (1 2 · · · n)r(1 2)(1 2 · · · n)n-r.]
6.9. This question is about an additive version of Hilbert s Theorem 90, see Theorem 6.18.
Let E/K be a Galois extension with cyclic Galois group Gal(E/K) = Ã of order n.
(a) Show that the function
T : E -’! E; T (u) = u + Ã(u) + Ã2(u) + · · · + Ãn-1(u),
takes values in K and use this to define a K-linear mapping TrE/K : E -’! K.
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