4 Divisiblity of the Class number by p

Combining the results of sections 1 and 2 we have shown that any counter-example to Fermat's Last theorem for a prime p $\geq$ 5 leads to a non-trivial representation

$$\rho$$ : Gal($$\overline{K}$$/K)$$\to$$$$\Bbb$$F_p

which is unramified everywhere; here K denotes the subfield of complex numbers generated by the p-th roots of unity. Kummer called primes which admit such representations irregular. He showed that there are indeed such primes (p = 37 is one such) and hence this particular attempt to prove Fermat's Last theorem fails. We now wish to show how one goes about checking whether a prime is irregular. We apply the results of Section 3 in the special case where K is the prime cyclotomic field of section 1 and also to the totally real subfield L. First of all we use the divisibility of the class number h of R by the class number h₊ of S to write h = h₊ ^. h_- for some integer h_-. Let W denote the (finite cyclic) group of roots of unity in K. Then we have U = W ^. U₊, where U₊ denotes the group of units in S and so #(U/U₊) = #(W/{±1}) = p. We have the natural inclusion L $\otimes_{{\Bbb Q}}^{}$ $\Bbb$R $\hookrightarrow$ K $\otimes_{{\Bbb Q}}^{}$ $\Bbb$R from which we obtain the isomorphism

(K $$\otimes_{{\Bbb Q}}^{}$$ $$\Bbb$$R)^*₁/(L $$\otimes_{{\Bbb Q}}^{}$$ $$\Bbb$$R)^*₁ = ($$\Bbb$$C^*₁/$$\Bbb$$R^*₁)^{(p - 1)/2}

since (p - 1)/2 is the degree of L over $\Bbb$Q. From this we deduce that

$$\nu$$((K $$\otimes_{{\Bbb Q}}^{}$$ $$\Bbb$$R)^*₁/U) = $${\frac{1}{p}}$$ ^. $$\nu$$($$\Bbb$$C^*₁/$$\Bbb$$R^*₁)^{(p - 1)/2 .} $$\nu$$((L $$\otimes_{{\Bbb Q}}^{}$$ $$\Bbb$$R)^*₁/U₊)

The formula for computing discriminants yields

$$\mu$$(K $$\otimes_{{\Bbb Q}}^{}$$ $$\Bbb$$R/R) = $$\mu$$(L $$\otimes_{{\Bbb Q}}^{}$$ $$\Bbb$$R/S)^{2 .} p^1/2

since p is the norm of the relative discrimant. Thus the class number formulas for K and L then give a formula for h_-

$${\frac{h_{-} \cdot \nu({\Bbb C}^{*}_1 / {\Bbb R}^{*}_1)^{(p-1)/2} }{p^{3/2} \cdot \mu(L\otimes _{{\Bbb Q}}{\Bbb R}/S) }}$$ = $$\prod_{\chi(-1)=-1}^{}$$L(1,$$\chi$$)

Hence h_- can be computed explicitly and in closed form. In particular, the divisibility of h_- by p is an easily computable criterion. The divisibility of h₊ by p is more complicated. As remarked earlier, the term $\nu$((L $\otimes_{{\Bbb Q}}^{}$ $\Bbb$R)^*₁/U₊) is difficult to compute. However, we have the subgroup U_{+ , cycl} = U₊ $\cap$ U_cycl and one can compute $\nu$((L $\otimes_{{\Bbb Q}}^{}$ $\Bbb$R)^*₁/U_{+ , cycl}). In fact one shows that

$$\nu$$((L $$\otimes_{{\Bbb Q}}^{}$$ $$\Bbb$$R)^*₁/U_{+ , cycl}) = $$\mu$$(L $$\otimes_{{\Bbb Q}}^{}$$ $$\Bbb$$R/S) ^. $$\prod_{\chi\text{ even }}^{}$$L(1,$$\chi$$)

where the product runs over all non-trivial characters $\chi$ such that $\chi$(- 1) = 1. The class number formula for h₊ becomes

h₊ = [U₊ : U_{+ , cycl}] = [U : U_cycl].

This is the first coincidence that makes Kummer's calculations possible. From the above identity we see that if p divides h₊ then we have a real unit u such that its p-th power is a cyclotomic unit but u is not itself cyclotomic. Hence v = u^p is a cyclotomic unit which is congruent to an integer modulo pS. If we find a w $\in$ U_cycl such that v = w^p then one shows easily that u is itself a cyclotomic unit. Let Q denote the quotient group (S/pS)^*/($\Bbb$Z/p$\Bbb$Z)^*. We obtain a natural homomorphism

m : U_cycl $$\otimes$$ ($$\Bbb$$Z/p$$\Bbb$$Z)$$\to$$Q

which is represented by a square matrix with entries from $\Bbb$F_p. The preceding remarks imply that p| h₊ only if det(m) = 0. The second coincidence that makes Kummer's calculation work is that det(m) $\cong$ h_-(mod p). Thus we see that p| h if and only if p| h_-. Hence we can easily check which primes are regular. to3em