Some Applications of Representation Theory to Classical Number Theory

B.Sury Indian Statistical Institute Bangalore, India E-mail: [email protected]

These are the notes of a lecture on ‘Unity of Mathematics’ Lemma. Let Inv(G) denote the set of involutions of a finite delivered on June 24, 2010 in an advanced instructional school group G. Then, on Representation Theory at the Indian Statistical Institute ν (χ) dim(χ) = 1 +|Inv(G)| Bangalore. 2 χ∈ Irr(G) The applications of representation theory to number theory where ν (χ) = 0, 1,or−1 according as to whether the char- is a subject so vast that it may be said to include the whole 2 acter χ is not real-valued, or, is real-valued and afforded by of the Langlands program. We do not discuss the Langlands a real representation or, is real-valued but not afforded by a program here but only talk briefly about the following three real representation. This ν is the so-called Frobenius-Schur topics: 2 indicator function. (I) The theory of Partitions In fact, for each n, expressing the class function (II) Zeta and L-functions over number fields (III) Kronecker-conjugacy of integer polynomials n θn(g) =|{x : x = g}| as θ = ν (χ)χ obtains 1. The Partition Function n χ irr n 1 ν (χ) =θ ,χ= χ(gn) (Everyone knows that) Ramanujan made outstanding contri- n n O(G) g butions to the theory of partitions. Ramanujan’s first letter for each irreducible χ. In the special case n = 2, we have to Hardy mentions an approximate formula for p(n) and Rademacher published such an exact formula soon (according |Inv(G)|+1 = θ2(1) = ν2(χ)χ(1) to Selberg, if Hardy had been less of an analyst, Ramanujan’s χ irr approximate formula might have been realized to quickly lead The values 0, 1, −1ofν2(χ) are obtained by decomposing the to an exact formula). Hardy and Ramanujan developed the so- representation space of χ into its symmetric and antisymmetric called circle method and published asymptotic formulae like parts. √ 2 nζ(2) e As ν2(χ) =θ2,χ,wehave p(n) ∼ √ . 4 3n √ 1 +|Inv(G)|= ν2(χ)χ(1) =θ2,χreg Here, the crucial thing to note is the exponent 2 ζ(2) which χ irr is approximately 2.56. Therefore, from the Cauchy-Schwarz inequality we have From a point of view of the actual values of p(n), for small n, these values are much smaller than the asymptotic values. |Inv(G)| < r(G) O(G) Representation theory especially of the symmetric group plays where r(G) is the number of irreducible characters afforded a role here to provide easy but close lower bounds for p(n). by real representations. In particular, since r(G) ≤ k(G), the Let us discuss this first. All relevant background information number of conjugacy classes of G,wehave on partition theory as well as much more historical and other O(G) material for the interested reader can be found in [1], and [2]. < k(G) 2 For the basic results on representation theory, one may refer to O(CG(g)) [9] or the classic [15]. Let us start with an elementary lemma. where g is any involution.

Mathematics Newsletter -1- Vol. 25 #3, September & December 2014 √ e2 n The proof for n<190 can be checked by a computer. For Corollary. cn < p(n) for some constant c. slightly bigger values 190 ≤ n<760 also, it can be easily √ = n! 2 n+0.5 ≥ Proof. Apply the lemma to Sn.Nowf (n, t) 2t t!(n−2t)! is the checked that p(n)>e .Forn 760, the above theorem number of involutions of Sn which are products of t disjoint along with induction, gives transpositions. The lemma gives √ √ p(n)>p([[n/2]/2])2 >e4 [[n/2]/2]+1 >e2 n+0.5. 2 |Inv(Sn)| = √ p(n) k(Sn)> 2.5 n n! We end by quoting the sharper lower bound p(n) > e 13n [n/2] 2 [n/2] 2 which can be deduced from the following consequence of the = f (n, t) f (n, t) = t 1 > . theory of blocks for Sn ([11]). n! t=1 n! f (n,t) = 2(t+1) ≤ = n w − We can easily see from f (n,t+1) (n−2t)(n−2t−1) that f (n, t) Theorem. p(n) t=0 4w+t(t+1)=2n l=0 p(l)p(w l). + − 2 ≥ + f (n, t 1) if and only if (n 2t) n 2. Therefore,√ the largest value of f (n, t) for t ≤ n/2 is when t = [(n+2− n + 2)/2]. 2. Zeta Functions on Number Fields Using Stirling’s formula we get the assertion of the corollary. A more careful argument on the above lines shows that one √ The Dedekind zeta function of an algebraic number field is can take c = e3 2π 3. an invariant which plays an important role in density theo- Remarks. rems for ramification of primes like the Frobenius density theorem and the Chebotarev density theorem. Thus, it may (i) The above proof can be combinatorially viewed via the be natural to expect the Dedekind zeta function to deter- Robinson-Schensted algorithm. This algorithm provides a mine the number field and it comes as a surprise that it does bijection between S and pairs of standard tableaux of the n not! In fact, a simple result from the representation theory of same shape. In particular, it gives n! = a2 where λ n λ finite groups provides a method to construct non-isomorphic a is the number of tableaux of shape λ n. Under λ number fields with the same zeta function. These simple this correspondence, elements of order ≤ 2 are in corres- methods also provide a footing to discuss and prove special pondence with pairs of tableaux with identical entries. cases of Dedekind’s conjecture asserting that for number fields = +| | Hence λ n aλ 1 Inv(Sn) . The arithmetic mean – K ⊂ L, the ratio ζL(s)/ζK (s) is an entire function of s. quadratic mean inequality for the a ’s gives the result now. λ We discuss this method here following an approach due to (ii) This estimate is close to the general size of p(n) because Robert Perlis ([13]). most irreducible character degrees of Sn are nearly equal. Dedekind zeta function and Gassmann equivalence

As the irreducible (complex) representations of Sn are The main aim of this section is to discuss a method to produce parametrized by partitions, let χλ denote the character corresponding to a partition λ of n. Using the theory of two non-isomorphic number fields with the same Dedekind blocks – especially using the recent generalization ([10]) of zeta function. Let N/Q denote a finite Galois extension and = Nakayama’s conjecture connecting combinatorial blocks to write G Gal(N/Q).IfK and K are intermediate fields, our

the blocks of modular representation theory – one can prove goal is to express the equality ζK (s) = ζK (s) in terms of the = = the following result on partitions. We do not go into its proof groups G, H : Gal(N/K) and H : Gal(N/K ). here. The Dedekind zeta function of an algebraic number field K is the function of the complex variable s defined in the 2 d s Theorem. For all d ≤ n, we have p(n) ≥ p([n/d ]) . region Re(s) > 1 by the series ζK (s) = I 1/N(I) where √ √ I varies over non-zero integral ideals of K and N(I) denotes If d = [ n/2], this gives p(n) ≥ 2[ n/2] but one can derive the absolute norm (the cardinality of O /I). The Dedekind the stronger consequence: K zeta function has a meromorphic continuation to Re(s) > 1 − √ e2 n = Corollary. p(n) > 14 . 1/[K : Q] and has only a simple pole at the point s 1.

Mathematics Newsletter -2- Vol. 25 #3, September & December 2014 The residue at s = 1 contains information about K like the If A = (f1,f2,...,fg) is the splitting of a prime number class number, regulator etc.: p in K, then we claim that there is a bijection between the set 2r1 (2π)r2 h(K) Reg(K) H\G/C of double cosets of G mod H,C and the set of prime lim (s − 1)ζK (s) = √ . s→1+ |μ(K)| |disc(K)| ideals of K above p. Indeed, this is given by

For Re(s) > 1, there is an Euler product expansion HσC → σP ∩ K. ζ (s) = (1 − N(P)−s)−1. K So A is the coset type of G mod H,C. By this, we mean that 0 =P prime the following holds good. Here, the product is over non-zero prime ideals in the ring of Writing G = h Ht C,wehaveh = g and |Ht C|= −s/2 1−s i=1 i i integers. Let G1(s) = (π) (s/2), G2(s) = (2π) (s) |H|fi. and let r1,r2 denote, respectively, the numbers of real and This is so because HtiC corresponds to tiP ∩K (say Pi) and complex places of K. Then, the completed zeta function −1 = r1 r2 − |H||t Ct | |H||C| ZK (s) G1(s) G2(s) ζK (s) is analytic in the whole plane |Ht C|=|Ht Ct 1|= i i = i i i | ∩ −1| | ∩ −1| except for simple poles at s = 0, 1 and satisfies the functional H tiCti H tiCti equation | |= while C fi fi where fi ,fi are, respectively, inertial −1 1 | ∩ |=| −s degrees of tiP over Pi and Pi over p and H tiCti ZK (s) =|DK | 2 ZK (1 − s) |= decomposition group of tiP over K fi . where DK denotes the discriminant of K over Q. So, we conclude that p has the same splitting type in K as well as in K if and only if the coset type of G mod H,C = Splitting of primes coset type of G mod H ,C. By the Frobenius density theorem, every cyclic subgroup For a prime p, we will define the splitting type of p in a e C of G occurs as a decomposition group for infinitely many O = e1 ··· g number field E as follows. Let p E P1 Pg be the O primes. Hence, we have: decomposition of a prime p into prime ideals in E and . PK (A) = PK (A) for all A ⇔ coset type of G mod H,C = fi = [OE/Pi : Z/pZ] be the inertial degree of Pi over p. coset type of G mod H ,C for all C. We number the fi’s in such a way that fi ≤ fi+1 and call Two subgroups of a finite group are said to be Gassmann (f1,f2,...,fg) the splitting type of p in E. For every such equivalent if the permutation representations of the big group tuple A,wehaveaset on the two coset spaces are equivalent. Note that by looking PE(A) :={p ∈ Z : p has splitting type A in E}. at the corresponding characters, this is equivalent to the state- g = ment that each conjugacy class in the big group intersects both As i=1 eifi [E : Q], PE(A) is empty except for finitely subgroups in the same number of elements. This property was many A. . first studied by F. Gassmann ([8]). We write PE(A) = PE (A) if the two sets differ by at most The relation of this notion to double coset type is given a finite number of elements. In particular, we can exclude by the: ramified primes as there are only finitely many of those in a number field. Lemma. Two subgroups H and H of a finite group G are

Let us look at a Galois extension N and let K,K be inter- Gassmann equivalent if, and only if, the coset type of G mod mediate fields. Consider p ∈ Z which is unramified in N. Let H,C = coset type of G mod H ,Cfor all cyclic subgroups C C be a decomposition group in G = Gal(N/Q) over p in G of G. i.e., C = GP for some prime P of N lying above p. Note C is | |=| | cyclic (generated by a so-called Frobenius automorphism) as Proof. Note that each of the conditions implies H H . = ; ∈ p is unramified in N. Let C c c G. Then |H||C| Look at one of K,K (say K) and let us see how a splitting | |=| −1|= HgC HgCg − . type in K reflects group-theoretically in terms of G and its |H ∩ gCg 1|

subgroups H = Gal(N/K) and C = GP . Look at the cardinalities li of the sets {g ∈ G : |HgC|=|H |i}.

Mathematics Newsletter -3- Vol. 25 #3, September & December 2014 ∞ Then, we have s ζK (s) = A (n)/n , for Re(s) > 1. n=1 l =|{g ∈ G : |HgC| divides |H |i}| d d|i Letting s →∞,wegetA(1) = A (1). Cancel this term =|{g ∈ G : H ∩gcg−1⊇gcig−1}| from both zeta functions, multiply by 2s and let s →∞;

− we get A(2) = A (2). Repeating this argument, we have =|{g ∈ G : gcig 1 ∈ H }|. A(n) = A (n) for all n by induction.

Call the last quantity ki. Now, clearly the splitting type of p in K is determined by = the number B(pf ) of prime ideals of K of norm pf .Onthe Then, by the Mobius¨ inversion formula li d|i kd μ(i/d). So, the ki’s and li’s determine one another. As the same holds other hand, for k and l when H is replaced by H , it follows that the i i B(pf ) = A(pf ) − A(pa1 )A(pa2 ) ···A(pat ) double cosets have the same decomposition types for all cyclic = ⇔ = ≥ +···+ = subgroups of G if and only if li li ki ki.Loand where the sum is over t 2,a1 at f. Thus the split- behold, this happens for every cyclic subgroup C if and only ting types coincide for p in K and in K . Hence, (a) implies (b). if the subgroups H and H of G are Gassmann-equivalent. (b) implies (c) is a tautology. T. Sunada constructed isospectral manifolds which are not (d) implies (a): isometric, using this property of Gassmann equivalence. This is sometimes expressed in the colourful language “one cannot Let C be decomposition group of the real place of Q. Since hear the shape of a drum.” The main theorem in our number- it has order 1 or 2, it is cyclic. So, the numbers r1,r2 of real theoretic context here is: and complex places of K are equal to the number of double cosets HtiC of cardinality |H| and of 2|H| respectively. Thus,

Theorem. Let K,K be number fields contained in a Galois (d) implies that the numbers r1,r2 are the same for K and K . = = r1 extension N over Q. When G Gal(N/Q), H Gal(N/K) From the completed zeta function ZK (s) = G1(s)

= r2 and H Gal(N/K ), the following are equivalent: G2(s) ζK (s) and its functional equation

1 (a) ζK = ζK . −s ZK (s) =|DK | 2 ZK (1 − s) (b) PK (A) = PK (A)∀ tuples A. . (c) PK (A) = PK (A)∀ tuples A. we see that = = − (d) H Gal(N/K) and H Gal(N/K ) are Gassmann- ζK (s) 1 −s ζK (1 s) =|DK /DK | 2 . equivalent. ζK (s) ζK (1 − s) Moreover, if the above conditions hold, then the degree But, the equivalence of (c) and (d) and the definition of the and the discriminants over Q and the numbers of real and Dedekind zeta function as an Euler product when Re(s) > 1

complex places of K and K coincide. Also, the two fields implies that the left hand side above is a finite product. That determine the same normal closures, the same normal is, using (c) we have cores (largest normal sub-extensions) and, the unit groups m −s ζ (s) = (1 − d ) of K,K are isomorphic as well. K = j 1 j n − . − s ζK (s) j=1(1 cj ) Note that the equivalence of (c) and (d) is what we esta- By analytic continuation above is valid for all complex s.Sothe blished above. conclusion would follows from the following easily proved Proof of (a) implies (b). Let A(n), A (n) denote the numbers fact asserting that finite products can not satisfy a general kind of functional equation: of integral ideals of norm n in the rings of integers of K,K τ1(s) m −s Let τ(s) = where τ1(s) = = (1 − c ) and respectively. Then τ2(s) j 1 j τ (s) = n (1 − d−s) with c ,d real and > 1. Note that ∞ 2 j=1 j j j s τ (s), τ (s) have no poles. Let f(s)be a meromorphic func- ζK (s) = A(n)/n , for Re(s) > 1 1 2 n=1 tion whose zeroes and poles do not lie among the zeroes of

Mathematics Newsletter -4- Vol. 25 #3, September & December 2014 either τ1(s) or τ2(s).Ifτ(s) = f(s)τ(1 − s) for all s, then A class of examples

τ1(s) = τ2(s) and f(s)= 1 for all s. Here is an infinite family of examples of fields which are arith- Hence we have shown that the four statements are equi- metically equivalent but are not isomorphic. valent. Let H and H be two non-isomorphic abstract groups having Let us now assume they are true and deduce the rest of the same number of elements of each order – let us say then that the assertions. If any of the above conditions holds then

the pair H,H satisfies the condition (∗). There are infinitely |H |=|H | which implies [K : Q] = [K : Q]. Also, while many such pairs. For example, if H is an abelian group of type proving the equivalence of (d) and (a), we have shown already

(p,p,p)and H is the semi direct product of an abelian group that the numbers of real and complex places match for K,K . a,b of type (p, p) and a cyclic group c of order p with Also, near the end of that proof, we observed that cac−1 = a and cbc−1 = ab for some odd prime p, then H,H

ζK (s) 1 − ζK (1 − s) satisfies (∗). =| | 2 s DK /DK . ζK (s) ζK (1 − s) When H,H is a pair satisfying (∗), then both H and H can be embedded in S via their left regular representations, Applying the above assertion on non-existence of a general n where n is their common order. Note that H is not conjugate to functional equation for finite products, in the case of the 1 −s H because they are not isomorphic. However, let us note that function f(s) =|DK /DK | 2 .Wehavef(s) to be identi- they are Gassmann-equivalent using the following lemma: cally equal to 1 and so |DK |=|DK |. But the sign of the discriminant is given by the number of complex Lemma. Elements h, h ∈ H ∪ H of the same order are places and, therefore, the discriminants themselves are conjugate in Sn. equal. Proof. As an element of S , each element of H acts by multi- Now, the normal closure of K over Q is the fixed field n −1 ∈ −1 |{ −1 plying the elements of H on the left and is then the product of σ∈G σHσ . So, h σ ∈G σHσ implies ghg : − − of n/i disjoint cycles of length i where O(h) = i. The same g ∈ G}|=|{ghg 1 : g ∈ G}∩H |=|{ghg 1 : g ∈ G}∩H | − − holds for h . So, h and h have same cycle structure and are so that h ∈ σH σ 1. In other words, σHσ 1 ⊆ σ ∈G σ ∈G − thus necessarily conjugate in S . σH σ 1. By symmetry, the two intersections are equal n σ ∈G Now, to show that H,H are Gassmann-equivalent, we need and thus the normal closures of K,K over Q are the to show that |{gcg−1 : g ∈ G}∩H|=|{gcg−1 : g ∈ G}∩H |. same. If both intersections are empty then equality trivially holds. We will show now that the normal cores are the same as So, let h ∈{gcg−1 : g ∈ G}∩H. By condition (∗), there is well. Note that the normal core of K over Q is the fixed field some h ∈ H of the same order as that of h;soh and h are of the subgroup generated by all the conjugates of H in G. −1 conjugate in S and so h ∈{gcg : g ∈ G}∩H . Thus, But, if h ∈ H , then the number of its conjugates in H equals n − the number in H (which is thus non-zero). In other words, {gcg 1 : g ∈ G}∩H ={h ∈ H : O(h) = O(c)} some conjugate of h is in H . This gives that the subgroups and generated by the conjugates of H and of H are the same. Thus, {gcg−1 : g ∈ G}∩H ={h ∈ H : O(h ) = O(c)} the normal cores are equal. O∗ ∗ Finally, the unit group K is the direct product of a free which gives by condition ( ) that H,H are Gassmann- group of finite rank r1 + r2 − 1 and the finite cyclic group equivalent. generated by the largest root of unity in K. The free parts are Finally, (by Hilbert’s irreducibility theorem for instance),

isomorphic as r1,r2 are equal. Now, we observe that K and there exists a Galois extension N of Q such that Gal(N/Q) =

K have the same roots of unity. This is because we can adjoin Sn. Hence, the fixed fields K and K of H and H are non- largest root of unity in K to Q to produce a normal exten- isomorphic but have the same zeta function. sion of Q in K as the normal cores are the same. Thus, the An interesting group-theoretic result of Bart De Smit & unit groups are isomorphic as well. Hence the theorem is H. W. Lenstra Jr. from 2000 ([3]) implies the following proved. beautiful number-theoretic theorem:

Mathematics Newsletter -5- Vol. 25 #3, September & December 2014 Call a natural number n special if pqr|n for (not necessarily p has splitting field with Galois group cyclic of degree q. distinct) primes p, q, r such that p|q(q−1). Let K be a number In other words, f mod p is irreducible for infinitely many field of degree n which is solvable by radicals. Suppose n is primes. not special. Then, K is determined up to isomorphism by its Interestingly, it turns out that for every composite degree n Dedekind zeta function. one may find monic irreducible integer polynomials of degree Conversely, for each special n, there are two solvable, non- n which are reducible modulo any natural number. isomorphic number fields of degree n which have the same However, we shall return now to the other aspect of integer Dedekind zeta function. polynomials which we started the section with. This is also A key result used in the above theorem is: analyzed using similar ideas. Towards that, we state the Let (n, φ(n)) = 1 and let X is a set of size n on which a following lemma which can be proved using the Frobenius finite, solvable group G acts transitively. Suppose G also acts density theorem: on another finite set Y such that |Xg|=|Y g| for all g ∈ G Lemma. (Frobenius). Let h ∈ Z X be monic, and assume whose order is divisible only by primes dividing n. Then, X [ ] that h(X) ≡ p, has a solution in Z for almost all non- and Y are isomorphic as G-sets. 0 mod zero primes p. Then every element of the Galois group of h(X) over Q fixes at least one root of h. 3. Value Sets of Integer Polynomials Conversely, let h be monic and assume that every element of the Galois group of h(X) over Q fixes at least one root of h. We briefly discuss the relation of permutation representa- Then h(X) ≡ 0 (mod p) has a solution in Z for every non-zero tions with value sets of integer polynomials. It turns out that prime p. the concrete number-theoretic problem of deciding when two integer polynomials take the same values modulo almost all To prove this lemma, let us recall the following weaker primes, is equivalent to a group-theoretic problem. Let us start version of the Frobenius density theorem: with a related problem. The set of primes p modulo which a monic integral, Can we have irreducible integer polynomials which are irreducible polynomial f has a given decomposition type

reducible modulo every positive integer? n1,n2,...,nr , has density equal to N/O(Gal(f )) where

More precisely, what is the relation between Galois groups N =|{σ ∈ Gal(f ) : σ has a cycle pattern n1,n2,...,nr }|. of integer polynomials and the reducibility modulo primes of Look at the first part. Assume that h is irreducible of degree the polynomials? > 1, if possible. The Frobenius Density Theorem shows that

If K is the splitting field of a monic irreducible polynomial every σ has a cycle pattern of the form 1,n2,... This means f of degree n over Z, then look at any prime p which does that every element of Gal(h) fixes a root, say β. Since h is not divide the disrciminant of f .Iff is irreducible modulo irreducible, the group Gal(h) acts transitively on the roots of p, then the corresponding Gal(K/Q) contains an element of h. Thus, this group would be the union of the conjugates of its order n (a decomposition group is of order n). In other words, subgroup H consisting of those elements which fix the root β. if Gal(K/Q) does not contain an element of order n, then f But a finite group cannot be the union of conjugates of a proper must be reducible modulo p. subgroup; this implies H is the whole group. Hence Gal(h) In prime degrees, one cannot have monic irreducible integer fixes each root of h and is therefore trivial. So we get h to be polynomials which are reducible modulo all but finitely many a linear polynomial, a contradiction. primes. This is seen by an application of the Chebotarev The converse is proved as follows.

(or even the) Frobenius density theorem in the following Let L be a splitting field of h(X) over Q. Let OL be the sense: ring of integers in L, and let P be a prime ideal of OL lying

A cyclic subgroup of Gal(K/Q) can be realized as a decom- over p. The roots of h lie in OL by the assumption about position group over infinitely many primes. h. Let D and I be the decomposition and inertia group of Therefore, if f is monic irreducible of prime degree q over P respectively. Then D/I is cyclic, and maps isomorphi- integers, there are infinitely many primes p so that f mod cally to the Galois group of the extension of residue fields.

Mathematics Newsletter -6- Vol. 25 #3, September & December 2014 Let d ∈ D such that the coset dI generates D/I.Bythe Galois group Gal(f (X) − t) over K(y) fixes at least one assumption, d fixes a root z of h. Thus dI fixes the image of z root. This Galois group is just the induced action of V on

in OL/P .AsdI generates the full Galois group of the residue the roots of f(X)− t (but V need not act faithfully). Hence field extensions, there is an integer b which is congruent to z every element in V fixes a root. But these roots are the modulo P . This gives h(b) ∈ P ∩ Z = pZ, and the assertion conjugates of x whose stabilizer is U. So every element of follows. V lies in some conjugate of U. By symmetry, the result follows. Kronecker-conjugacy Conjecture. Over a field of characteristic 0, two polynomials We describe in outline some deep work of M. Fried f and g are Kronecker-conjugate if and only if, U and V are ([4], [5], [6]) and later work by Peter Muller¨ ([12]) relating to Gassmann equivalent; that is, the permutation representations G G value sets of integer polynomials. IndU 1 and IndV 1 are equivalent. ∈ For f Z[X] and, p prime, consider the value set It should be noted that such a result is not purely group- theoretic because there are examples of abstract finite groups G Val p(f ) ={f(a) mod p; a ∈ Z}. −1 = −1 and subgroups U,V such that g∈G gUg g∈G gVg

Call f, g ∈ Z[X]tobeKronecker-conjugate if Valp(f ) = but U,V are not Gassmann equivalent.

Val p(g) for all but finitely many primes p. In order to state a group-theoretic criterion for Kronecker-conjugacy, we need to 4. Dedekind’s entirety conjecture fix some notations. ∈ Given f, g Z[X] which are non-constant, consider a Dedekind conjectured that for number fields K ⊂ L, the ratio Galois extension K of the field Q(t) of rational functions in ζL(s)/ζK (s) is an entire function of s. The Dedekind conjec- = a variable t such that K contains a root x of f(x) t and a ture remains open in general. Actually, there is a more general = root y of g(y) t. Let G denote Gal(K/Q(t)) and let U,V conjecture due to Artin which we describe first. A proof of denote the stabilizers of x,y respectively. the Brauer-Aramata theorem appears in the classical text by Fried’s Theorem. J.-P. Serre [15] but we mention here a proof of Foote and Kumar Murty. The interested reader can also look ∈ Given f, g Z[X] which are non-constant, consider at [14]. t,K,G,x,y,U,V as above. Then, f, g are Kronecker- Let L/K be a Galois extension of number fields and let G conjugate if and only if denote the Galois group. For any prime ideal P of OK , look e eg at the factorization P O = P 1 ···P . Now, the decompo- gUg−1 = gVg−1. L 1 g ={ ∈ = } g∈G g∈G sition groups DPi σ G : σ(Pi) Pi are mutually ={ ∈ ≡ conjugate and the inertia subgroups IPi σ G : σ(x) ∀ ∈ O } Idea of Proof. x mod Pi x L are the kernels of the natural surjections O O from DPi to Gal (( L/Pi)/ K /P )). Suppose that f, g are Kronecker-conjugate. Write f = One also denotes by FrP , the Frobenius at P – this is a n + ··· uX be of degree n. As Kronecker-conjugacy is conjugacy class in G. n−1 preserved when we replace f(X) and g(X) by u f (X/u) Artin associated to any finite-dimensional representation n−1 and u g(X) respectively, we may assume that f is ρ : G → GL(V ),anL-function defined as monic. Now, for any integer a, the hypothesis gives that −s IP −1 f(X)≡ g(a)(mod p)has a root for almost all non-zero primes L(s, ρ; L/K) = det(1 − ρ(FrP )NK/Q(P ) |V i ) . p. Hilbert’s irreducibility theorem tells us that the Galois P

− − IPi groups Gal(f (X) g(y) over K(y) and Gal(f (X) g(a)) Here V is the subspace fixed by IPi for any i and the defini- over K are isomorphic as permutation groups for infinitely tion makes sense as FrP is a conjugacy class. many a. Recall that g(y) = t. Thus every element of the This Artin L-function has the following properties.

Mathematics Newsletter -7- Vol. 25 #3, September & December 2014 Properties: Indeed, one need only observe that

r ; (I) L(s, ρ L/K) depends only on the character χρ and one L(s, χ; L/K) = L(s, IndG (ψ ); L/K)ni Hi i often writes L(s, χ; L/K) for characters χ. i=1

(II) If χ1 ⊕ χ2 = χ, then L(s, χ; L/K) = L(s, χ1; L/K) r Hi ni = L(s, ψi; L/L ) . L(s, χ2; L/K). i=1 G ; = (III) If H is a subgroup of G, then L(s, IndH (χ) L/K) L(s, χ; L/LH ). Heilbronn character

(IV) L(s, 1; L/K) = ζK (s). The behaviour of an Artin L-function at any point s0 can (V) L(s, χreg; L/K) = ζL(s). be studied through the so-called Heilbronn character,a (VI) (Artin-Takagi factorization) = = χ(1) certain virtual character of G Gal(L/K).Ifn(G, χ) : ζ (s) = ˆ L(s, χ; L/K) . L χ∈G ; Ords=s0 L(s, χ L/K), is the order of (zero/pole of)

L(s, χ; L/K) at s0 the Heilbronn character is: Artin’s Conjecture: G(g) = n(G, χ)χ(g). L(s, χ; L/K) extends to an entire function for any irreducible χ nontrivial character χ of G. The sum is over all irreducible characters of G. Notice that if Artin’s conjecture is true, then this is an actual character when What Artin’s reciprocity law means: s0 = 1. The following result was proved by Heilbronn: Artin’s reciprocity law implies that Artin’s conjecture holds Lemma. For a subgroup H, the restriction of  to H is the for one-dimensional characters. More precisely: G Heilbronn character  of Gal(L/LH ). If L/K is a Galois extension of number fields, and if χ is H a monomial character of Gal(L/K) (that is, is induced from Proof. By the orthogonality of characters, a one-dimensional character of some subgroup) and does not ⎛ ⎞ contain the trivial character, then L(s, χ; L/K) extends to an  |H = n(G, χ) ⎝ χ|H,ψ ψ⎠ entire function of s. G H χ∈Gˆ ψ∈Hˆ Artin’s conjecture implies Dedekind’s entirety conjecture ⎛ ⎞

for ζL(s)/ζK (s) in the case of Galois extensions L/K.How- = ⎝ n(G, χ)χ,IndG ψ ⎠ ψ ever, without using Artin’s conjecture, one can prove the H G ψ∈Hˆ χ∈Gˆ above case of Dedekind conjecture – this is due to Brauer & Aramata. where the second equality follows from the Frobenius reciprocity theorem. Using the basic properties of the Artin Theorem. (Brauer). Let G be any finite group and χ an irre- L-function, we have ducible character of it. Then, there exist nilpotent subgroups = ; H H ,...,H and one-dimensional characters ψ on H and n(H, ψ) Ords=s0 L(s, ψ L/L ) 1 r i i integers n such that χ = r n IndG (ψ ). i i=1 i Hi i = G ; Ords=s0 L(s, IndH ψ L/K) In fact, combined with Artin’s reciprocity law, this theorem  G  = ; χ,IndH ψ Ords=s0 L(s, χ L/K) immediately implies the following one: χ =  G  Theorem. (Brauer). Let L/K be a Galois extension of n(G, χ) χ,IndH ψ G χ number fields. Let G denote the Galois group and let χ be an irreducible character of G. Then L(s, χ; L/K) which proves the lemma. admits a meromorphic continuation to the whole The following beautiful inequality was derived by R. Foote plane. and V. Kumar Murty ([7]), and this has several consequences.

Mathematics Newsletter -8- Vol. 25 #3, September & December 2014 Lemma. (Foote-Kumar Murty). Using this lemma, A. Raghuram and M. Ram Murty prove: Let G = Gal(L/K) with L/K, a solvable extension of 2 ≤ 2 n(G, χ) (Ords=s0 ζL(s)) . number fields. Write Lab denote the fixed field under [G, G] χ∈Gˆ and C denote the set of different 1-dimensional characters of G. Then Proof. Now 1 | (g)|2 = ,  = O(G) g∈G G G G G 2 2 ˆ n(G, χ) . ζ (s) χ∈G 2 ≤ L n(G, χ) Ords=s0 We apply Heilbronn’s lemma to the cyclic subgroups of G. 1 =χ∈C ζLab (s) We get We end with a few smatterings of statements which occur in G(g) = g(g) = n(g, ψ)ψ(g). the Langlands program. First, we make a small observation to ψ∈g the effect that Artin’s entirety conjecture needs to be proved As the Artin reciprocity theorem implies that the Artin only when the base is Q. More precisely: L-function is entire for one-dimensional characters, each Artin cojecture enough to prove over Q:   ≥ | |≤   n( g ,ψ) 0. So G(g) ψ∈g n( g ,ψ). Finally, the Artin-Takagi factorization shows that Proposition. If all nontrivial irreducible characters χ of G := Gal(E/Q) are so that L(s,χ,E/Q) extends to an entire Ord = ζ (s) = n(g,ψ) s s0 L function, then Artin’s conjecture holds good. ψ∈g for each element g ∈ G. Hence, Proof. Let L/K be a Galois extension of number fields. Let E be the Galois closure over Q and let G = Gal(E/Q). Then 2 2 | (g)| ≤ (Ord = ζ (s)) ∀ g ∈ G. G s s0 L H = Gal(L/K) is a subquotient of G. Let τ be any irreducible character of H and lift it to a character of Gal(E/K). Denote by Corollary. (Brauer-Aramata Theorem). For any Galois θ the character of G induced from it. By Frobenius reciprocity

extension L/K of number fields, the function ζL(s)/ζK (s) is law, θ does not contain the trivial character. In this case, by the entire. property (II) we recalled earlier, L(s,θ,E/Q) is entire. But, the property (III) shows that L(s,θ,E/Q) = L(s, τ, L/K). Proof. Apply the Foote-Murty lemma and note that We end with the statement of one of the Langlands conjec- = ; ζK (s) L(s, 1 L/K) where 1 is the trivial character tures known as: of G. Other variations have been obtained by M. Ram Murty and The Langlands reciprocity conjecture. collaborators. Let (V, ρ) be an n-dimensional irreducible representation of One such is: a Galois group Gal(L/K) of number fields. Then, there is a Let G = Gal(L/K) with L/K, a solvable extension of cuspidal automorphic representation π of GLn(AK ) such that number fields. Then L(s, ρ, L/K) is the L-function attached to π by Langlands. ζ (s) 2 2 ≤ L We have not defined cuspidal automorphic representations n(G, χ) Ords=s0 ζK (s) 1 =χ∈Gˆ or the corresponding L-function of Langlands but just remark A group-theoretic lemma which they prove in this direction is: that the latter L-functions are known to be entire! Let G be a nontrivial finite, solvable group and H , a sub- group. Consider the derived series {G(i)} of G. Then, for all i, References G G Ind 1 = Ind (i) 1 (i) + Ind θ [1] G. Andrews, The theory of partitions, Encyclopaedia H H HG HG Hj j j of mathematics and its applications, Vol. 2, Addison-

where θj ’s are 1-dimensional characters of some subgroups Wesley, Reading 176 (Reissued: Cambridge University

Hj which depend on H and i. Press 1985 and 1998).

Mathematics Newsletter -9- Vol. 25 #3, September & December 2014 [2] B. Berndt, Ramanujan’s notebooks, Parts I–V,Springer, [9] I. M. Isaacs, Character theory of finite groups, New York (1985, 1989, 1991, 1994, 1998). Academic Press, New York (1976). [3] B.de Smit and H. W. Lenstra, Linearly equivalent [10] B. Kulshammer,¨ J. B. Olsson and G. R. Robinson, actions of solvable groups, J. of Algebra, 228 (2000) Generalized blocks for symmetric groups, Inventiones 278–285. Mathematicae, 151 (2003) 513–552. [4] M. Fried, On a conjecture of Schur, Michigan Math. J., [11] A. Maroti,´ On elementary lower bounds for the partition 17 (1970) 41–55. function, Integers, Electronic Journal of Combinatorial [5] M. Fried, The field of definition of function fields Number Theory, 3 (2003) #A10. and a problem on the reducibility of polynomials [12] P. Muller,¨ Kronecker conjugacy of polyno- in two variables, Michigan Math. J., 17 (1973) mials, Trans. Amer. Math. Soc., 350 (1998) 128–146. 1823–1850. [6] M. Fried, On Hilbert’s irreducibility theorem, J. of [13] R. Perlis, On the class numbers of arithmetically equi- Numb. Theory, 6 (1974) 211–231. valent fields, J. of Numb. Theory, 10 (1978) 489–509. [7] R. Foote and V.Kumar Murty, Zeroes and poles of Artin [14] M. Ram Murty and A. Raghuram, Some variations on L-series, Math. Proc. Camb. Phil. Soc., 105 (1989) the Dedekind conjecture, J. Ramanujan Math. Soc., 15 5–11. no. 4 (2000) 225—245. [8] F. Gassmann, Bemerkungen zu der vorstehenden Arbeit [15] J.-P. Serre, Linear representations of finite groups, von Hurwitz, Math. Z., 25 (1926), 124–143. Springer-Verlag, GTM, 42 (1977).

Mathematics Newsletter -10- Vol. 25 #3, September & December 2014