kummerflt.pdf

8/11/2019 kummerflt.pdf

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2 K. H. PARANJAPE

representation. As he showed, there are indeed such primes and thushis proof works only for regular primes.

In section 1 we recall some computations in the cyclotomic field ofp-th roots of unity. In section 2 we show how a counter-example toFermats Last Theorem (if it exists) can be used to construct a cyclicextension of order p of the cyclotomic field which is unramified every-where. We review the Class number formula in section 3. Finally, inSection 4 we use this formula to check when such unramified extensionsdo indeed exist.

Most of the material in this note can be found in more detail (thoughin a more classical presentation) in the book of H. M. Edwards [1].This re-examination of Kummers proof was inspired by some remarksmade by V. Kumar Murty during his lecture on the work of Wiles atthe TIFR. I would like to thank A. Raghuram for his careful readingof the manuscript and numerous suggestions.

We fix a prime p 5 throughout the discussion.

1. Arithmetic of prime cyclotomic fields

Let R denote the subring of complex numbers generated by =exp(2/p); let Kdenote the quotient field ofR, which is called thecyclotomic field of p-th roots of unity. We review some well-knownfacts about the ring Rand the field Kmostly without proof.

The ring R is isomorphic to Z[X]/(p(X)), where

p(X) =Xp1 + + X+ 1 = (Xp 1)/(X 1)

is an irreducible polynomial (a simple application of the Eisenstein cri-terion). The field K is a Galois extension ofQ with Galois group Fp ;this is a cyclic group of order (p1). We usefor a fixed choice of gen-erator. We use to denote (p1)/2() since (p1)/2 is the restrictionof complex conjugation to R.

The ringR is a Dedekind domain, i. e. unique factorization holds forideals. The prime ideals in this ring are described as follows:

(1) Ifq Zis a prime number different from p. Then let fbe theorder ofq inFp and let g = (p 1)/f. Then there are g prime

ideals Q1, . . . , Qg in Rsuch that their norms are qf.

(2) The element = 1 is prime in R and p1 = (unit) p.

A closed form expression for the generators of the group U of unitsofR is not known. However, the numbers

uj =j()/= 1 + + + j1


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KUMMERS PROOF 3

are in R and are units there. The subgroupUcycl of the group U ofunits ofR generated by the uj forj = 2, . . . , (p 1) is called the group

of cyclotomic units. Ifu is a unit in R, thenu/uis a root of unity inR.The roots of unity inRare all of the form j for somej = 0, . . . , p1.An element ofR is a p-th power only if it is congruent to an integermodulopR. It follows that u/u = j for somej (i. e. there is no minussign).

LetL denote the subfield ofKfixed by complex conjugation; letS=L R. ThenL is a Galois extension ofQ with Galois group Fp/{1}.No complex embeddings ofK have image within real numbers whileall complex embeddings ofL have image within real numbers; in otherwords, K is purely imaginary and L is totally real. Again, S is aDedekind domain and its ideals are described as follows:

(1) Ifq Zis a prime number different fromp. Then let f be theorder ofqinFp/{1}and let g

= (p 1)/2f. Then there are

g prime ideals Q1, . . . , Qg inR such that their norms are qf.

(2) The element = 1 (+ 1) is prime in R and (p1)/2 =(unit) p.

Ifu is a unit inR, then we have seen thatu/u = r for some integerr.But then r 2s (mod p) for some integer s; hence u1 =

suis in S.Hence, any unit in R is the product of a root of unity and a unit in S.

If I is any ideal in S then IR is principal in R if and only if I isprincipal inS. Hence the homomorphism from the class group ofStothat ofR is injective. In particular the order hof the class group ofRis divisible by the order h+ of the class group ofS.

If we have a unit u in R such that it is congruent to an integermodulopRand ifu is itselfnota p-th power, then the field extensionofKobtained by adjoining a p-th root ofu is a cyclic extension ofKof order pwhich is unramified everywhere.

Finally we have a fact from Class Field theory. If there is an idealI in R such that Ip is principal and Iis not principal, then there is acyclic entension ofKof order pwhich is unramified everywhere. Thisfollows from the identification of the class group ofR with the Galois

group of the maximal unramfied abelian extension ofK. Now we usethe fact that if an abelian group has an element of order p, then it hasa non-trivial character of order p.

2. Construction of cyclic cover

The aim is to show that if we have a counter-example to FermatsLast Theorem, then there is a cyclic extension of order p ofKwhichis unramified everywhere. As is usual we can assume that the given


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4 K. H. PARANJAPE

counter-example (X, Y , Z) has the property that these are mutuallyco-prime integers.

Case 1: p|XY Z. First of all we see easily that (X, Y , Z) are not allcongruent modulo p. If not, we have

3X X+ Y + Z Xp + Yp + Zp 0 (mod p)

Now, we are assuming that p 5 and so we obtain X 0 (mod p);this contradicts our hypothesis for Case 1.

Secondly, we see that (X+jY) are mutually co-prime inRasj runsover 0, . . . , p 1. If not, then we have a prime idealP inR containing(X+jY, X+kY). Then this idealP contains (1 jk)Y. Nowfrom the factorisation

(Z)p =Xp + Yp = (X+ Y)(X+ Y) (X+ p1Y)

we see thatP containsZ. Hence, by the assumption that (X, Y , Z) aremutually co-prime we see thatPcontains (1l) for some 0 l p1.By the description of prime ideals in R as in section 1 we see thatP =R. But thenZis a multiple ofpwhich contradicts our hypothesisin Case 1.

By the above paragraph and unique factorization of ideals we seethat we have ideals Ij ofRsuch that I

pj = (X+

jY)R. Assume I1 isprincipal; then we have an equation

(X+ Y) =u p

for some R and u a unit in R. Applying complex conjugation weobtain

(X+ 1Y) =u p

By the results mentioned in section 1 we have ru = u for some r.Moreover,p is congruent to an integer modulo pR and hence is con-gruent to its own complex conjugate. Thus we obtain an equation

X+ Y rX r1Y 0 (mod p)

Now it follows from the description ofRgiven in Section 1 that it is a

free abelian group with basis consisting of any (p 1) elements of theset {1, , . . . , p1}. From this and the fact that X and Y are primeto pit follows that r= 1 and X Y (mod p).

By similar reasoning interchanging the roles of Y and Z we canconclude that there is an idealJ1 such thatJ

p1 = (X+ Z). Assuming

J1 is principal we see by an argument like the one above that XZ(mod p). But as seen above the two congruences

X Y (mod p) and X Z (mod p)


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KUMMERS PROOF 5

contradict the hypothesis of Case 1. Hence, either I1 or J1 must benon-principal. But then by the principal result of Class Field theory

as mentioned in section 1 we have required cyclic extension ofK.

Case 2: p|XY Z. We may assume that Z=pkZ0 and (p, X, Y, Z0) aremutually co-prime. By writing p = (unit) (p1) in the ring R, weobtain an equation of the form

Up + Vp + (unit)mpWp = 0 with m >0

where (U, V , W) are in R so that (U,V,W,) are mutually co-prime.Let (U, V , W) be a collection of elements of R that satisfy such anequation with m the least possible. Then divides one of the factors(U+ jV). But then we have

(U+ jV) (U+ kV) =j(1 kj)V= (unit) Vand thus, divides all the factors (U+ jV). Moreover, sinceV is co-prime to p and thus as well, we see that (U+jV)/ have distinctresidue classes moduloR. But then, by the pigeon-hole principle thereis at least one 0 j (p 1) such that (U+jV) is divisible by 2

inR. ReplacingV byjVwe may assume that (U+ V) is divisible byl for some l >1. Hence we may write

U+ V = la0

U+ kV = ak; for k >0

where all theak are elements ofR that are co-prime to and with eachother (as in the previous case). This gives us the identityl + (p 1) =mpor equivalentlyl= (m 1)p+ 1. Since l 2 we have m 2.

Now by unique factorisation of ideals inRwe see that there are idealsIj inR such that I

pj =ajR. Assume thatI0,I1 and Ip1 are principal,

then we have the equations

U+ V = l u bp0U+ V = v bp1

U+ 1V = w bp1

for some unitsu,v and w inR and some elements b0,b1 and b1 inR.

Eliminating U and V from these equations we obtain

l u bp0 v bp1=( w b

p1

l u bp0)

which becomesbp1+ v1 b

p1+

l1 v2bp0= 0

where v1 and v2 are units (we use here the fact that 1 + is a unitin R). Modulo pR the last term on the left-hand side vanishes sincel p > (p 1). Thus we see that v1 is congruent to a p-th power


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6 K. H. PARANJAPE

and thus an integer modulo pR. By section 1 we have a representationof Galois as required, unless v1 is a p-th power. If v1 = v

p3, then

(U, V , W) = (b1, v3b1, b0) satisfy

Up + Vp + (unit)(m1)pWp = 0

which contradicts the minimality ofm since we have seen that m 2.Thus, either we have constructed a cyclic extension of the required typeor one ofI0,I1, Ip1 is non-principal. But then again by the principalresult of Class Field theory we have a cyclic extension as required.

3. Transcendental computation of the Class number

We first need to introduce the Dedekind zeta function for a number

field K, and its Euler product expansion

K(s) =I

1

N(I)s =Q

1

(1 1N(Q)s

)

where the sum runs over all ideals I ofR and the product runs overall prime ideals Q of R. The two expressions give us two ways ofcomputing lims1(s 1)K(s). The left-hand side is expressed in termsof arithmetic invariants and the right-hand side in terms of invariantsfor the Galois group. The resulting identity gives a way for computingthe Class number h ofK.

The left-hand limit can be computed to be

lims1

(s 1)I

1

N(I)s = lim

r

#{I | N(I) r}

r

The set {I | N(I) r} can be split according to ideal classes. We tryto compute for each ideal class C,

z(C) = limr

#{I C | N(I) r}

r .

Fixing an idealI0 C, this latter set is bijective to the set {aR I10 |

N(a) r N(I0)1}. (Here N(a) denotes the modulus of the norm of

a.)We have a natural embeddingK KQ R. The image ofJ=I

10

is a lattice in KQ R. Let denote the image of J {0} in thequotientS= (KQ R)

/U whereUis the image of the group of unitsin R under the above embedding. There is a natural homomorphismN :S R which restricts to the modulus of the norm on the imageofK. We obtain a natural bijection between{aRI10 | N(a) r}and {l | N(l) r}. Let r denote the image of (1/r)J {0} in


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KUMMERS PROOF 7

S, then we have a natural bijection between {l | N(l) rd} and{l r | N(l) 1}, where d denotes the degree ofKoover Q.

Let S1 denote locus ofl Ssuch that N(l) 1. Let denote theHaar measure on KQ R. This is invariant under the action ofU andthus gives a measure also denoted by on S. Since J is a lattice inKQ R we have

limr

#{l r | N(l) 1}

rd =

(S1)

(KQ R/J)

Moreover, the denominator can be re-written

(KQ R/J) =N(J)(KQ R/R).

In particular, we see that the limit z(C) is independent of the class C.Let (KQR)1 denote the kernel of the norm map. This is a groupand thus we have a Haar measure on it. One shows that

(S1) =((KQ R)1/U)

Combining the above calculations one obtains

lims1

(s 1) K(s) =h ((KQ R)

1/U)

(KQ R/R)

This often called the Class number formula for K. Note that thedenominator can be computed in closed form in terms of the discrim-

inant D of the field Kand the number of pairs of conjugate complexembeddings r2 ofK.

(KQ R/R) = 1

2r2

|D|

However, the numerator is in general more complicated since it involvescomputing the group of units ofK.

To expand the right-hand term we restrict our attention to abelianextensions KofQ. The product term on the left can be first groupedaccording to rational primes

Q

1

(1 1N(Q)s )=

qQ|q

1

(1 1N(Q)s )

Now for each rational prime qwhich is unramified in Kwe haveQ|q

1

(1 1N(Q)s

)=

1

(1 (q)qs

)

where runs over all characters of the Galois group and (q) =(Frobq) is the value of on a Frobenius element associated with q.


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8 K. H. PARANJAPE

We define the Dirichlet L-series and their Euler product formulas asfollows

L(s, ) =n

(n)ns

=p

1(1 (p)

ps )

where we set (p) = 0 when is ramified at p. We also define theadditional factor

F(s) =

p ramified

1

(1 1pfp

)gp

where the product runs over all ramified primes and fp denotes theresidue field extension over p and gp the number of distinct primes inK lying over p. The product expansion ofK(s) becomes

K(s) =F(s)

L(s, ).

Thus the computation of the limit can be reduced to the correspond-ing computation for the Dirichlet L-series. For the case of the unitcharacter we get by comparison with the zeta function

lims1

(s 1)F(s)L(s, 1) = 1.

So the right-hand limit gives

lims1(s 1)K(s) ==1

L(1, ).

There is a positive integer m such that is determined on classesmodulo m and is zero on all primes p dividing it; m is called theconductorof. We rewrite theL-function associated with as follows

L(s, ) =

x(Z/mZ)

(x)

nx (mod m)

1

ns

The latter sum can be rewritten using the identity

m1i=0

xi =

0, ifx 0 (mod m)

m, ifx 0 (mod m)

where is a primitive m-th root of unity. The second sum then becomes

nx (mod m)

1

ns =

1

m

n=1

1

ns

m1i=0

(xn)i


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KUMMERS PROOF 9

Thus we obtain

L(s, ) = 1m

m1i=0

x(Z/mZ)

(x)ix

n=1

in

ns

The expression

i() =

x(Z/mZ)

(x)ix

is called the Gaussian sum associated with the integer i and the char-acter . If is not the unit character then 0() = 0. Moreover, ifi = 0 then we have the identity

n=1

in

n = log(1

i

)

Hence, we obtain the formula when is not the unit character

L(1, ) = 1

m

m1i=1

i() log(1 i)

4. Divisiblity of the Class number by p

Combining the results of sections 1 and 2 we have shown that anycounter-example to Fermats Last theorem for a prime p5 leads toa non-trivial representation

: Gal(K/K) Fp

which is unramified everywhere; hereKdenotes the subfield of complexnumbers generated by the p-th roots of unity. Kummer called primeswhich admit such representations irregular. He showed that there areindeed such primes (p = 37 is one such) and hence this particularattempt to prove Fermats Last theorem fails. We now wish to showhow one goes about checking whether a prime is irregular.

We apply the results of Section 3 in the special case where Kis theprime 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 ofR by theclass number h+ ofS to write h = h+ h for some integer h. LetW denote the (finite cyclic) group of roots of unity in K. Then wehave 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 QR KQ R from which we obtain the isomorphism

(KQ R)1/(L Q R)

1= (C

1/R

1)

(p1)/2


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KUMMERS PROOF 11

References

[1] H. M. Edwards,Fermats last theorem, Graduate Texts in Mathematics, vol. 50,Springer-Verlag, New York Berlin Heidelberg, 1977.

School of Mathematics, TIFR, Homi Bhabha Road, Bombay 400 005,

India

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