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Discr(a0 + a1 X + a2 X 2 + X 3 ) = ’27a2 + 18a0 a1 a2 + a2 a2 ’ 4a3 a0 ’ 4a3 .
n = 3: 0 12 2 1


Discr(a0 + a1 X + a2 X 2 + a3 X 3 + X 4 ) = 18a3 a3 a2 ’ 6a2 a2 a0 ’ 192a3 a1 a2 ’ 27a4
n = 4: 1 31 0 1
+ 144a2 a2 a2 + 144a0 a2 a2 + 256a3 ’ 4a3 a3 ’ 128a2 a2 + 16a4 a0 ’ 4a3 a2
30 1 0 31 20 2 21
+ 18a3 a1 a2 a0 ’ 80a3 a1 a2 a0 ’ 27a4 a2 + a2 a2 a2 ’ 4a3 a2 a0 .
3 2 30 231 23


Discr(a0 + a1 X + a2 X 2 + a3 X 3 + a4 X 4 + X 5 ) = 2250a4 a2 a3 ’ 36a0 a3 a3 ’ 128a2 a4
n = 5: 30 41 31
+ 2000a2 a3 a2 ’ 900a1 a3 a2 ’ 2500a3 a4 a1 ’ 50a2 a2 a2 ’ 900a4 a3 a2 ’ 27a4 a4 ’ 3750a3 a2 a3
0 1 30 0 041 20 41 0
+ 356a3 2 a2 2 a4 a1 a0 + 560a3 a2 a2 a2 ’ 2050a3 a2 a2 a4 a1 ’ 80a2 a2 a4 a1 3 ’ 630a3 3 a2 a4 a0 2
240 0 3
+ 825a2 a2 a2 + 16a3 a3 a0 + 2000a2 a2 a3 ’ 6a2 a2 a3 ’ 128a2 a4 a2 + 16a4 a3 a0 ’ 4a3 a3 a2
320 32 40 241 240 24 241
+ 108a5 a2 + 108a5 a0 ’ 746a3 a2 a0 a4 2 a1 2 ’ 27a4 a2 + 256a5 a3 ’ 4a3 a2 a2 + 144a3 a2 a3
30 2 21 40 321 21
+ 144a2 a4 a3 + 3125a4 + 256a5 ’ 72a4 a2 a1 a0 + 18a3 a2 a3 a3 + 560a2 a2 a0 a2 + 16a4 a3
41 0 1 3 41 3 1 31
+ 18a3 a2 3 a4 a1 2 ’ 72a3 a2 4 a4 a0 + 144a3 2 a2 a4 3 a0 2 ’ 192a4 4 a1 a3 a0 2 ’ 630a3 a2 3 a1 a0
+ 24a2 3 a4 2 a1 a0 + a3 2 a2 2 a4 2 a1 2 ’ 6a4 3 a1 2 a3 2 a0 ’ 80a3 a2 2 a4 3 a1 a0 ’ 4a3 2 a2 3 a4 2 a0
+ 2250a1 a2 a2 ’ 1600a3 a3 a3 ’ 192a4 a4 a2 ’ 1600a0 a3 a2 ’ 4a3 a3 a2 ’ 27a4 a2 a2
20 40 1 1 314 340
+ 1020a4 2 a3 2 a0 2 a1 + 18a3 3 a2 a4 2 a0 a1 + 160a2 a4 3 a0 2 a1 + 144a2 a4 4 a0 a1 2
+ 24a4 a1 2 a3 3 a0 + 1020a0 a4 a2 2 a1 2 + 160a0a4 a1 3 a3 .
So for example,
Discr(X 5 + a4 X 4 + a0 ) = a3 (3125a0 + 256a5 ), Discr(X 5 + a1 X + a0 ) = 256a5 + 3125a4 .
0 4 1 0

Proposition 6.41. For every σ ∈ Gal(Ef /K), σ(Discr(f (X))) = Discr(f (X)). Hence
Discr(f (X)) ∈ K.
Proof. For σ ∈ Gal(Ef /K) Sn , we have
« 2

(uσ(j) ’ uσ(i) )2 =  (uσ(j) ’ uσ(i) ) .
σ(Discr(f (X))) =
16i<j6n 16i<j6n

Now for each pair i, j with i < j,
σ(uj ’ ui ) = uσ(j) ’ uσ(i) ,
and by Equation (6.3)
(6.6) (uσ(j) ’ uσ(i) ) = sgn σ (uj ’ ui ) = (±1) (uj ’ ui ).
16i<j6n 16i<j6n 16i<j6n
Gal(Ef /K)
Hence σ(Discr(f (X))) = Discr(f (X)). Since Ef = K, we have Discr(f (X)) ∈ K.
Now let
δ(f (X)) = (uj ’ ui ) ∈ Ef .
16i<j6n
Then δ(f (X))2 = Discr(f ), so the square roots of Discr(f (X)) are ±δ(f (X)). Now consider the
e¬ect of σ ∈ Gal(Ef /K) on δ(f (X)) ∈ Ef . By Equation (6.6),
σ(δ(f )) = sgn σ δ(f ) = ±δ(f ).
If δ(f (X)) ∈ K, this means that sgn σ = 1. On the other hand, if δ(f ) ∈ K then
/
Gal(Ef /K)©An
K(δ(f (X))) = Ef .
6.7. SYMMETRIC FUNCTIONS 79

Of course | Gal(Ef /K)/ Gal(Ef /K) © An | = 2.
Proposition 6.42. The Galois group Gal(Ef /K) Sn is contained in An if and only if
Discr(f (X)) is a square in K.
Example 6.43. For the polynomials of Examples 6.36 and 6.38 we obtain
Discr(X 5 ’ 35X 4 + 7) = ’4611833296875 = ’33 · 56 · 74 · 29 · 157,
√ √
δ(X 5 ’ 35X 4 + 7) = ±53 · 3 · 72 · 3 · 29 · 157 i = ±18375 13659 i ∈ Q;
/
Discr(X 5 + 20X + 16) = 1024000000 = 216 · 56 ,
δ(X 5 + 20X + 16) = ±28 53 ∈ Q.

6.7. Symmetric functions
Let k be a ¬eld. Consider the polynomial ring on n indeterminates k[X1 , . . . , Xn ] and its
¬eld of fractions K = k(X1 , . . . , Xn ). Each permutation σ ∈ Sn acts on k[X1 , . . . , Xn ] by
σ · f (X1 , . . . , Xn ) = f σ (X1 , . . . , Xn ) = f (Xσ(1) , . . . , Xσ(n) ).
Viewed as a function σ· : k[X1 , . . . , Xn ] ’’ k[X1 , . . . , Xn ] is a ring isomorphism; this extends to
a ring isomorphism σ· : k(X1 , . . . , Xn ) ’’ k(X1 , . . . , Xn ). Varying σ we obtain actions of the
group Sn on k[X1 , . . . , Xn ] and k(X1 , . . . , Xn ) by ring isomorphisms ¬xing k and in the latter
case it is by ¬eld automorphisms ¬xing k.
Definition 6.44. The ¬eld 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) ⇐’ ∀σ ∈ Sn f (X1 , . . . , Xn ) = f (Xσ(1) , . . . , Xσ(n) ).
Theorem 6.45. The extension k(X1 , . . . , Xn )/ Symn (k) is a ¬nite 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 = Xi1 Xi2 · · · Xik ,
i1 <i2 <···<ik
where 1 k n. It is easy to see that for every σ ∈ Sn , eσ = ek , so ek ∈ Symn (k). Working in
k
the ring k(X1 , . . . , Xn )[Y ] we have
fn (Y ) = Y n ’ e1 Y n’1 + · · · + (’1)n’1 en’1 Y + (’1)n en = 0,
hence the roots of this polynomial are the Xi . So k(X1 , . . . , Xn ) is the splitting ¬eld 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 .
Remark 6.46. 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 ].
Corollary 6.47. If n 5, the extension k(X1 , . . . , Xn )/ Symn (k) is not solvable.
80 6. A GALOIS MISCELLANY

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 ¬eld 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.
6.3. Find all values of n 1 for which •(n) | 4. Using this, determine which roots of unity lie
in the following ¬elds:
√ √ √
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 e¬ect of each element on Q(ζ24 ) and Q(cos(π/12)) under the
action described in Theorem 6.2.
(b) Determine the group (Z/20)— and describe the e¬ect of each of its elements on Q(ζ20 ) and
Q(cos(π/10)) under the action described in Theorem 6.2.
6.5. Let n 1.
(a) What can you say about sin(2π/n) and Gal(Q(sin(2π/n))/Q))?
(b) Determine sin(π/12) and Gal(Q(sin(π/12))/Q)).

6.6. In this question, work in the cyclotomic ¬eld Q(ζ5 ) where ζ5 = e2πi/5 .
(a) Describe the Galois group Gal(Q(ζ5 )/Q) and its action on Q(ζ5 ).
(b) Determine the minimal polynomial of cos(2π/5) over Q. Hence show that

’1 + 5
cos(2π/5) = .
4
For which other angles θ is cos θ a root of this minimal polynomial? What is the value
of sin(2π/5) ?
(c) Find the tower of sub¬elds of Q(ζ5 ) and express them as ¬xed ¬elds of subgroups of
Gal(Q(ζ5 )/Q).

6.7. In this question, let p be an odd prime and let ζp = e2πi/p ∈ Q(ζp ) C.
(a) Consider the product
(p’1)/2
r ’r
ξ= (ζp ’ ζp ) ∈ Q(ζp ).
r=1

Show that
p’1
2 (p’1)/2 r
ξ = (’1) (1 ’ ζp ).
r=1

(b) Deduce that
p if p ≡ 1 (mod 4),
ξ2 =
’p if p ≡ 3 (mod 4).
(c) Conclude that

±p if p ≡ 1 (mod 4),
ξ= √
± piif p ≡ 3 (mod 4).
√ √
and also p ∈ Q(ζp ) if p ≡ 1 (mod 4) and p i ∈ Q(ζp ) if p ≡ 3 (mod 4).
81

6.8. Prove Lemma 6.37. [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 6.16.
Let E/K be a Galois extension with cyclic Galois group Gal(E/K) = σ of order n.
(a) Show that the function
T (u) = u + σ(u) + σ 2 (u) + · · · + σ n’1 (u),
T : E ’’ E;
takes values in K and use this to de¬ne a K-linear mapping TrE/K : E ’’ K.
(b) If v ∈ E has TrE/K (v) = 0, show that there is a w ∈ E such that v = w ’ σ(w).
[Hint: Show that there is an element t ∈ E for which TrE/K t = 0, then consider
1
vσ(t) + (v + σ(v))σ 2 (t) + · · · + (v + σ(v)σ 2 (t) + · · · + σ n’2 (v))σ n’1 (t)
w=
(TrE/K t)
and adapt the proof of Hilbert™s Theorem 90 in Theorem 6.16, using TrE/K in place of NE/K .]
6.10. Show that f (X) = X 3 ’ 3X + 1 ∈ Q[X] is irreducible over Q, and show that its
discriminant is a square in Q. Prove that the Galois group of f (X) over Q is cyclic.
n, the k-th power sum sk ∈ k[X1 , . . . , Xn ]Sn is de¬ned to be
6.11. For n 1 and 1 k
Xik .
sk =
16i6n
Prove the formula
sk = e1 sk’1 ’ e2 sk’2 + · · · + (’1)k’1 ek’1 s1 + (’1)k kek .

n, the total symmetric function hk ∈ k[X1 , . . . , Xn ]Sn is de¬ned
6.12. For n 1 and 1 k
to be
hk = Xj1 Xj2 · · · Xjk ,
6j2 6···6jk
j1
i.e., the sum of all the monomials in the Xi of degree k.
(a) For large values of n, express h1 , h2 , h3 in terms of the elementary symmetric functions
e1 , e2 , e3 .
(b) Show that the power sum functions sk of the previous question satisfy
sk = ’(h1 sk’1 + h2 sk’2 + · · · + hk’1 s1 ) + khk .
Bibliography

[1] E. Artin, Galois Theory, Dover Publications (1998); ISBN 0 486 62342 4.
[2] J-P. Esco¬er, Galois theory, Springer-Verlag, New York (2001); ISBN 0-387-98765-7. Highly recommended,
especially for its historical notes]
[3] J. B. Fraleigh, A First Course in Abstract Algebra, Addison Wesley (1999); ISBN 0 201 33596 4. [Highly
recommended ]
[4] S. Lang, Algebra, Addison Wesley (1993); ISBN 0 201 55540 9.
[5] R. Lidl & H. Niederreiter, Finite ¬elds, Cambridge University Press (1997); ISBN 0 521 39231 4.
[6] J. Rotman, Galois Theory, Springer-Verlag (1998); ISBN 0 387 98541 7.
[7] I. Stewart, Galois Theory, Chapman and Hall (1989); ISBN 0 412 34550-1.




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