** Next:** Bibliography
** Up:** Kummer's proof of Fermat's
** Previous:** 3 Transcendental computation of

Combining the results of sections 1 and 2 we have shown that
any counter-example to Fermat's Last theorem for a prime *p* 5
leads to a non-trivial representation
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* *R* *K* *R* from which we obtain
the isomorphism
(

*K* *R*)

^{*}_{1}/(

*L* *R*)

^{*}_{1} = (

*C*^{*}_{1}/

*R*^{*}_{1})

^{(p - 1)/2}
since (*p* - 1)/2 is the degree of *L* over *Q*. From this we deduce
that
((

*K* *R*)

^{*}_{1}/

*U*) =

^{ . }(

*C*^{*}_{1}/

*R*^{*}_{1})

^{(p - 1)/2 . }((

*L* *R*)

^{*}_{1}/

*U*_{+})

The formula for computing discriminants yields
(

*K* *R*/

*R*) =

(

*L* *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*_{-}
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
((*L* *R*)^{*}_{1}/*U*_{+}) is
difficult to compute. However, we have the subgroup
*U*_{+ , cycl} = *U*_{+} *U*_{cycl} and one can compute
((*L* *R*)^{*}_{1}/*U*_{+ , cycl}). In fact one
shows that
((

*L* *R*)

^{*}_{1}/

*U*_{+ , cycl}) =

(

*L* *R*/

*S*)

^{ . }*L*(1,

)

where the product runs over all non-trivial characters such
that
(- 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* *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*)^{*}/(*Z*/*p**Z*)^{*}. We obtain a natural homomorphism
which is represented by a square matrix with entries from
*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*) *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

** Next:** Bibliography
** Up:** Kummer's proof of Fermat's
** Previous:** 3 Transcendental computation of
Kapil Hari Paranjape
2002-11-22