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TitleFermat's Last Theorem
LanguageEnglish
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Total Pages167
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Page 83

(d) RQn can be topologically generated by r elements as a O-algebra.

Proof: Take r = dimkH
1
∅ (Q, ad

0ρ̄(1)). It suffices to find a set Qn with the
following properties.

(a) If q ∈ Qn then q ≡ 1 mod `n.

(b) If q ∈ Qn then ρ̄ is unramified at q and ρ̄(Frob q) has distinct eigenvalues.

(c) H1∅ (Q, ad
0ρ̄(1))

∼→


q∈Qn
H1(Fq, ad

0ρ̄(1)).

As each H1(Fq, ad
0ρ̄(1)) is one dimensional we may replace the last condition

with
H1∅ (Q, ad

0ρ̄(1)) ↪→


q∈Qn

H1(Fq, ad
0ρ̄(1)).

Thus what we need show is that for each non-zero class [ψ] ∈ H1∅ (Q, ad
0ρ̄(1))

there is a prime q (depending on [ψ]) such that

(a) q ≡ 1 mod `n,

(b) ρ̄ is unramified at q and ρ̄(Frob q) has distinct eigenvalues,

(c) resq[ψ] ∈ H1(Fq, ad0ρ̄(1)) is nontrivial.

Using the Chebotarev density theorem we see that it will do to find σ ∈ GQ
such that

(a) σ|Q(ζ`n ) = 1,

(b) ad0ρ̄(σ) has an eigenvalue other than 1,

(c) ψ(σ) 6∈ (σ − 1)ad0ρ̄(1).

For m ≥ 0, let Fm denote the extension of Q(ζ`m) cut out by ad0ρ̄; i.e.,
the field fixed by the kernel of the representation ad0ρ̄ restricted to GQ(ζ`m ).
We will show that ψ(GFn) is non-trivial. For this it suffices to prove that
H1(Gal (Fn/Q), ad

0ρ̄(1)) = (0). Consider the inflation-restriction exact se-
quence

(0)→ H1(Gal (F0/Q), (ad0ρ̄(1))GF0 )→ H1(Gal (Fn/Q), ad0ρ̄(1))
→ H1(Gal (Fn/F0), ad0ρ̄(1))GQ .

83

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