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Á " Sn is a product of 2-cycles, so it suffices to show that every 2-cycle (a b) " Sn is a product
of 2-cycles of the form (r + 1 r + 2). Assuming that a
(a b) = (b - 1 b) · · · (a + 2 a + 3)(a + 1 a + 2)(a a + 1)(a + 1 a + 2)(a + 2 a + 3) · · · (b - 1 b),
and this is in H. Hence H = Sn.
6.9. (a) For each u " E,
Ã(T (u)) = Ã(u + Ã(u) + Ã2(u) + · · · + Ãn-1(u))
= Ã(u) + Ã2(u) + · · · + Ãn(u)
= Ã(u) + Ã2(u) + · · · + Ãn-1(u) + u = T (u),
so T (u) is fixed by à and all its powers, hence by Gal(E/K). Therefore T (u) is in EGal(E/K) = K.
It is straightforward to verify that the resulting function TrE/K : E -’! K is K-linear.
(b) Let v " E and suppose that TrE/K(v) = 0. By Artin s Theorem 6.15, the linear combination
of characters id +Ã + · · · + Ãn-1 must be linearly independent, so there is an element t " E for
which
TrE/K t = t + Ã(t) + · · · + Ãn-1(t) = 0.
Then
u = vÃ(t) + (v + Ã(v))Ã2(t) + · · · + (v + Ã(v)Ã2(t) + · · · + Ãn-2(v))Ãn-1(t)
satisfies
u - Ã(u) = v Ã(t) + Ã2(t) + · · · + Ãn-1(t) - Ã(v) + · · · + Ãn-1(v) t
= v t + Ã(t) + Ã2(t) + · · · + Ãn-1(t) - v + Ã(v) + · · · + Ãn-1(v) t
= (TrE/K t)v - (TrE/K v)t = (TrE/K t)v.
So we obtain
1 1
v = u - Ã u .
TrE/K t TrE/K t
6.10. (a) This can be proved by induction on n. Write
r
e[m] = Xi1 · · · Xir, s[m] = Xi .
r r
i1
106 SOLUTIONS
Then we easily find that
r
e[m] = e[m-1] + e[m-1]Xm, s[m] = s[m-1] + Xm.
r r r-1 r r
Notice also that e[m] = 0 whenever r > m. The desired result is that for all n 1 and k 1,
r
s[n] = e[n]s[n] - e[n]s[n] + · · · + (-1)k-1e[n] s[n] + (-1)kke[n].
2
k 1 k-1 k-2 k-1 1 k
r
When n = 1 we have s[1] = X1 and e[1] = X1 from which the result follows. Now suppose that
r
1
k
the result is true for some n 1. Then s[n+1] = s[n] + Xn+1, while
k k
e[n+1]s[n+1] - e[n+1]s[n+1] + · · · + (-1)k-1e[n+1]s[n+1] + (-1)kke[n+1] =
1 k-1 2 k-2 k-1 1 k
k-1 k-2
(e[n] + Xn+1)(s[n] + Xn+1) - (e[n] + e[n]Xn+1)(s[n] + Xn+1) + · · ·
1 k-1 2 1 k-2
+ (-1)k-1(e[n] + e[n] Xn+1)(s[n] + Xn+1) + (-1)kk(e[n] + e[n] Xn+1)
k-1 k-2 1 k k-1
k-1 k-2
= s[n] + (e[n]Xn+1 - e[n]Xn+1 + · · · + (-1)k-1e[n] Xn+1)
k 1 2 k-1
+ (s[n] - e[n]s[n] + · · · + (-1)k-1e[n] s[n] + (-1)kke[n] )Xn+1
1
k-1 k-2 k-2 1 k-1
k k-1 2
+ (Xn+1 - e[n]Xn+1 + · · · + (-1)k-1e[n] Xn+1)
1 k-2
k
= s[n] + Xn+1 = s[n+1],
k k
which demonstrates the inductive step.
(b)(i) We have h1 = e1, h2 = e2 - e2 and h3 = e3 - 2e1e2 + e3.
1 1
(ii) This can be done by induction on n in a similar way to part (a). [ Pobierz całość w formacie PDF ]