aa r X i v : . [ m a t h . N T ] A ug THE INFIMUM IN THE METRIC MAHLER MEASURE
CHARLES L. SAMUELS
Abstract.
Dubickas and Smyth defined the metric Mahler measure on themultiplicative group of non-zero algebraic numbers. The definition involvestaking an infimum over representations of an algebraic number α by otheralgebraic numbers. We verify their conjecture that the infimum in its definitionis always achieved as well as establish its analog for the ultrametric Mahlermeasure. Introduction
Let K be a number field and v a place of K dividing the place p of Q . Let K v and Q p denote the respective completions. We write k · k v for the unique absolutevalue on K v extending the p -adic absolute value on Q p and define | α | v = k α k [ K v : Q p ] / [ K : Q ] v for all α ∈ K . Define the Weil height of α ∈ K by H ( α ) = Y v max { , | α | v } where the product is taken over all places v of K . Given this normalization of ourabsolute values, the above definition does not depend on K , and therefore, H is awell-defined function on Q . Clearly H ( α ) ≥
1, and by Kronecker’s Theorem, wehave equality precisely when α is zero or a root of unity. It is obvious that if ζ is aroot of unity then(1.1) H ( α ) = H ( ζα ) , and further, if n is an integer then it is well-known that(1.2) H ( α n ) = H ( α ) | n | . Also, if α, β ∈ Q × then H ( αβ ) ≤ H ( α ) H ( β ) so that H satisfies the multiplicativetriangle inequality.We further define the Mahler measure of an algebraic number α by M ( α ) = H ( α ) [ Q ( α ): Q ] . Since H is invariant under Galois conjugation over Q , we obtain immediately(1.3) M ( α ) = N Y n =1 H ( α n ) , Mathematics Subject Classification.
Primary 11R04, 11R09.
Key words and phrases.
Weil height, Mahler measure, metric Mahler measure, Lehmer’s problem. where α , . . . , α N are the conjugates of α over Q . Further, it is well-known that(1.4) M ( α ) = | A | · N Y n =1 max { , | α n |} , where | · | denotes the usual absolute value on C . While the right hand side of (1.4)appears initially to depend upon a particular embedding of Q into C , any changeof embedding simply permutes the images of the points { α n } so that (1.4) remainsunchanged.It follows, again from Kronecker’s Theorem, that M ( α ) = 1 if and only if α iszero or a root of unity. As part of an algorithm for computing large primes, D.H.Lehmer [5] asked whether there exists a constant c > M ( α ) ≥ c in allother cases. The smallest known Mahler measure greater than 1 occurs at a rootof ℓ ( x ) = x + x − x − x − x − x − x + x + 1which has Mahler measure 1 . . . . . Although an affirmative answer to Lehmer’sproblem has been given in many special cases, the general case remains open. Thebest known universal lower bound on M ( α ) is due to Dobrowolski [1], who provedthat(1.5) log M ( α ) ≫ (cid:18) log log deg α log deg α (cid:19) whenever α is not a root of unity.Recently, Dubickas and Smyth [2] defined the metric Mahler measure of an al-gebraic number α by(1.6) M ( α ) = inf ( N Y n =1 M ( α n ) : N ∈ N , α n ∈ Q × , α = N Y n =1 α n ) . Here, the infimum is taken over all ways to represent α as a product of elements in Q × . It is easily verified that M ( αβ ) ≤ M ( α ) M ( β )for all α, β ∈ Q × , and further, M is well-defined on the quotient group G = Q × / Tor( Q × ). This implies that the map ( α, β ) log M ( αβ − ) defines a metricon G which induces the discrete topology if and only if there is an affirmative answerto Lehmer’s problem.Also in [2], Dubickas and Smyth conjecture that the infimum in the definition of M is always achieved. We verify this conjecture as well as explicitly determine aset in which the infimum must occur.If K is any number field letRad( K ) = n α ∈ Q × : α r ∈ K for some r ∈ N o , the set of all roots of points in K . Also, for the remainder of this paper, we write K α for the Galois closure of Q ( α ) over Q . Theorem 1.1. If α is a non-zero algebraic number then there exist α , . . . , α N ∈ Rad( K α ) such that α = α · · · α N and M ( α ) = M ( α ) · · · M ( α N ) . HE INFIMUM IN THE METRIC MAHLER MEASURE 3
Motivated by the work of Dubickas and Smyth, Fili and the author [4] defined anon-Archimedean version of M by replacing the product in (1.6) by a maximum.That is, define the ultrametric Mahler measure by M ∞ ( α ) = inf ( max ≤ n ≤ N M ( α n ) : N ∈ N , α n ∈ Q × , α = N Y n =1 α n ) . It easily verified that M ∞ satisfies the strong triangle inequality M ∞ ( αβ ) ≤ max { M ∞ ( α ) , M ∞ ( β ) } for all non-zero algebraic numbers α and β . It is further shown in [4] that M ∞ iswell-defined on the quotient group G . We can now establish the obvious analog ofTheorem 1.1 for M ∞ . Theorem 1.2. If α is a non-zero algebraic number then there exist α , . . . , α N ∈ Rad( K α ) such that α = α · · · α N and M ∞ ( α ) = max { M ( α ) , . . . , M ( α N ) } . The remainder of this paper is organized in the following way. Section 2 containsthe core of our argument in which we show that computing M ( α ) and M ∞ ( α ) re-quires only the use of elements in Rad( K α ). In section 3, we finish the proofs ofTheorems 1.1 and 1.2 by showing, essentially, that there are only finitely manyvalues for the Mahler measure in Rad( K α ). Finally, section 4 contains some ap-plications of these results, giving the location of the algebraic numbers M ( α ) and M ∞ ( α ). 2. Reducing to simpler representations
The main idea in both proofs involves a method for replacing an arbitrary rep-resentation of α by a potentially smaller representation containing only points inRad( K α ). This technique is summarized by the following result. Theorem 2.1. If α, α , . . . , α N are non-zero algebraic numbers with α = α · · · α N then there exists a root of unity ζ and algebraic numbers β , . . . , β N satifying(i) α = ζβ · · · β N ,(ii) β n ∈ Rad( K α ) for all n ,(iii) M ( β n ) ≤ M ( α n ) for all n . It is worth noting that we are unaware of an example in which computing M ( α )or M ∞ ( α ) requires the use of elements outside K α . Hence, it seems reasonable tobelieve that we can, in fact, choose the points β n to belong to K α . Unfortunately,our proof suggests no way to verify this.The proof of Theorem 2.1 is based on the following lemma. Lemma 2.2.
Suppose that K is Galois over Q . If γ is an algebraic number then (2.1) [ K ( γ ) : K ] = [ Q ( γ ) : K ∩ Q ( γ )] . Moreover, we have that (2.2) N Y n =1 γ n ∈ K ∩ Q ( γ ) , where γ , . . . , γ N are the conjugates of γ over K . C.L. SAMUELS
Proof.
We see clearly that K ( γ ) is the compositum of K and Q ( γ ). Since K isGalois over Q , it follows (see [3], p. 505, Prop. 19) that [ K ( γ ) : K ] = [ Q ( γ ) : K ∩ Q ( γ )], verifying (2.1). We also observe that( K ∩ Q ( γ ))( γ ) ⊆ ( Q ( γ ))( γ ) = Q ( γ ) ⊆ ( K ∩ Q ( γ ))( γ )so we conclude from (2.1) that(2.3) [ K ( γ ) : K ] = [( K ∩ Q ( γ ))( γ ) : K ∩ Q ( γ )] . Let f be the monic minimal polynomial of γ over K ∩ Q ( γ ) so that f has degree D equal to both sides of (2.3). Now write f ( x ) = x D + · · · + a x + a and note that f is, of course, a polynomial over K . In fact, f is the monic minimalpolynomial of γ over K because it vanishes at γ and has degree [ K ( γ ) : K ]. Since γ , . . . , γ N are the conjugates of γ over K we conclude that N Y n =1 γ n = ± a which belongs to K ∩ Q ( γ ). (cid:3) It is worth observing that if Q ( γ ) is Galois over Q , then Lemma 2.2 becomestrivial. Indeed, γ · · · γ N certainly belongs to K by definition. But also, if Q ( γ ) isGalois, then Q ( γ ) contains all conjugates of γ over Q . In particular, it contains γ n for all n , so it contains their product as well. Of course, the proof of Theorem 2.1does not permit such a hypothesis, so we require the above lemma.Additionally, we cannot omit the hypothesis that K be Galois over Q . Forexample, let γ , γ and γ be the roots of a third degree, irreducible polynomialover Q having Galois group S . This means that Q ( γ ) ∩ Q ( γ ) = Q . Further, weobserve that γ must have degree 2 over Q ( γ ) implying that its conjugates overthis field are γ and γ . But if γ · γ ∈ Q ( γ ) then γ ∈ Q ( γ ), a contradiction. Proof of Theorem 2.1 . Suppose that α = α · · · α N and let E be a Galois extensionof K α containing α n for all n . Let G = Gal( E/K α ), G n = Gal( E/K α ( α n )) and S n a set of left coset representatives of G n in G . We have that α [ E : K α ] = Norm E/K α ( α )= N Y n =1 Norm
E/K α ( α n )= N Y n =1 Y σ ∈ G σ ( α n )= N Y n =1 Y σ ∈ S n Y τ ∈ G n σ ( τ ( α n ))= N Y n =1 Y σ ∈ S n σ ( α n ) | G n | HE INFIMUM IN THE METRIC MAHLER MEASURE 5 so we conclude that(2.4) α [ E : K α ] = N Y n =1 Y σ ∈ S n σ ( α n ) ! [ E : K α ( α n )] . For each n , we select an element β n ∈ Q such that(2.5) β [ K α ( α n ): K α ] n = Y σ ∈ S n σ ( α n )so that, in view of (2.4), we obtain(2.6) α [ E : K α ] = N Y n =1 β [ E : K α ] n . This implies the existence of a root of unity ζ such that α = ζβ · · · β N . Furthermore, the set { σ ( α n ) : σ ∈ S n } is precisely the set of conjugates of α n over K α so that Y σ ∈ S n σ ( α n ) ∈ K α . It then follows from (2.5) that β n ∈ Rad( K α ) for each n as well.It remains to show that M ( β n ) ≤ M ( α n ) for all n . To see this, we note that(2.5) yields immediately(2.7) deg( β n ) ≤ [ K α ( α n ) : K α ] · deg Y σ ∈ S n σ ( α n ) ! . Once again, the elements σ ( α n ) for σ ∈ S n are precisely the conjugates of α n over K α . Hence, we may apply Lemma 2.2 (2.1) to find that Y σ ∈ S n σ ( α n ) ∈ K α ∩ Q ( α n ) . Combining this with (2.7), we obtain(2.8) deg( β n ) ≤ [ K α ( α n ) : K α ] · [ K α ∩ Q ( α n ) : Q ] . Then we find that M ( β n ) ≤ H ( β n ) [ K α ( α n ): K α ] · [ K α ∩ Q ( α n ): Q ] = H Y σ ∈ S n σ ( α n ) ! [ K α ∩ Q ( α n ): Q ] ≤ H ( α n ) [ K α ( α n ): K α ] · [ K α ∩ Q ( α n ): Q ] where the last inequality follows since the Weil height is invariant under Galoisconjugation and satisfies the triangle inequality. But also K α ( α n ) is the compositumof K α and Q ( α n ) so that [ K α ( α n ) : K α ] = [ Q ( α n ) : K α ∩ Q ( α n )] by (2.1) in Lemma2.2. This yields M ( β n ) ≤ H ( α n ) [ Q ( α n ): K α ∩ Q ( α n )] · [ K α ∩ Q ( α n ): Q ] = M ( α n )which completes the proof. (cid:3) C.L. SAMUELS Proofs of Theorems 1.1 and 1.2
In view of Theorem 2.1, it is enough, in the definitions of M and M ∞ , toconsider only representations α = α · · · α N having α n ∈ Rad( K α ) for all n . Anyrepresentation that fails to have this property may simply be replaced a smallerrepresenation that does. The remainder of our proofs of both Theorem 1.1 and1.2 require us to show that such representations yield only finitely many differentvalues for max ≤ n ≤ N M ( α n ) and N Y n =1 M ( α n ) . The following lemma provides the starting point for this argument.
Lemma 3.1.
Let K be a Galois extension of Q . If γ ∈ Rad( K ) then there existsa root of unity ζ and L, S ∈ N such that ζγ L ∈ K and M ( γ ) = M ( ζγ L ) S . In particular, the set { M ( γ ) : γ ∈ Rad( K ) , M ( γ ) ≤ B } is finite for every B ≥ .Proof. Suppose that γ r ∈ K , so that each conjugate of γ over K must be a root of x r − γ r ∈ K [ x ]. Therefore, we may assume that γ has conjugates ζ γ, . . . , ζ L γ over K for some roots of unity ζ , . . . , ζ L . By Lemma 2.2 we conclude that(3.1) ζ · · · ζ L γ L = ζ γ · · · ζ L γ ∈ K ∩ Q ( γ ) . Since K is Galois, Lemma 2.2 also implies that L = [ K ( γ ) : K ] = [ Q ( γ ) : K ∩ Q ( γ )].Hence, we find that M ( γ ) = H ( γ ) [ Q ( γ ): Q ] = H ( γ ) [ Q ( γ ): K ∩ Q ( γ )] · [ K ∩ Q ( γ ): Q ] = H ( γ ) L · [ K ∩ Q ( γ ): Q ] . Since L is a positive integer and ζ · · · ζ L is a root of unity, we conclude from (1.1)and (1.2) that(3.2) M ( γ ) = H ( ζ · · · ζ L γ L ) [ K ∩ Q ( γ ): Q ] . By (3.1) we know that there exists a positive integer S such that[ K ∩ Q ( γ ) : Q ] = S · [ Q ( ζ · · · ζ L γ L ) : Q ]and so (3.2) yields M ( γ ) = H ( ζ · · · ζ L γ L ) S · [ Q ( ζ ··· ζ L γ L ): Q ] = M ( ζ · · · ζ L γ L ) S . Taking ζ = ζ · · · ζ L we have that ζγ L ∈ K by (3.1) and M ( γ ) = M ( ζγ L ) S whichestablishes the first statement of the lemma.Further, we note that (3.2) implies that M ( γ ) = H (( ζγ L ) [ K ∩ Q ( γ ): Q ] ) . HE INFIMUM IN THE METRIC MAHLER MEASURE 7
But ( ζγ L ) [ K ∩ Q ( γ ): Q ] ∈ K implying that(3.3) { M ( γ ) : γ ∈ Rad( K ) , M ( γ ) ≤ B } ⊆ { H ( α ) : α ∈ K × , H ( α ) ≤ B } . It follows from Northcott’s Theorem [6] that the right hand side of (3.3) is finite,completing the proof. (cid:3)
The proof of Theorem 1.2 is somewhat simpler than that of Theorem 1.1 so weinclude it here first.
Proof of Theorem 1.2.
There exists ε > α = α · · · α N with α n ∈ Rad( K α ) and M ∞ ( α ) ≤ max { M ( α ) , . . . , M ( α N ) } ≤ M ∞ ( α ) + ε then M ∞ ( α ) = max { M ( α ) , . . . , M ( α N ) } . Otherwise, we get a sequence { x m } ⊆ Rad( K α ) such that { M ( x m ) } is strictly decreasing, contradicting Lemma 3.1.By definition, there exists a representation α = γ · · · γ N with M ∞ ( α ) ≤ max { M ( γ ) , . . . , M ( γ N ) } ≤ M ∞ ( α ) + ε. By Theorem 2.1, there exists a representation α = ζα · · · α N such that ζ is a rootof unity, α n ∈ Rad( K α ) and and M ( α n ) ≤ M ( γ n ) for all n . This yields M ∞ ( α ) ≤ max { M ( α ) , . . . , M ( α N ) } ≤ M ∞ ( α ) + ε so that M ∞ ( α ) = max { M ( α ) , . . . , M ( α N ) } by our earlier remarks. (cid:3) We note that the above proof is not sufficient to establish Theorem 1.1. Indeed,Lemma 3.1 does not prevent the product M ( α ) · · · M ( α N ) from having infinitelymany values between M ( α ) and M ( α ) + ε unless we can bound N uniformly fromabove by a function of α .In order to do this, we introduce an additional definition. We say that a repre-sentation α = α · · · α N is B -restricted if the following three conditions hold.(i) M ( α ) · · · M ( α N ) ≤ B (ii) α n ∈ Rad( K α ) for all n (iii) At most one element α n is a root of unity.We write R B ( α ) to denote the set of all N -tuples, for all N ∈ N , of non-zeroalgebraic numbers that form B -restricted representations of α . Further, set q ( α ) = inf n H ( x ) : x ∈ K × α \ Tor( Q × ) o and note that, by Northcott’s Theorem [6], this quantity is always strictly greaterthan 1. Using these definitions, we obtain the result we need to finish the proof ofTheorem 1.1. Lemma 3.2.
Let α be a non-zero algebraic number and B ≥ . If α = α · · · α N is an B -restricted representation of α then N ≤ B log q ( α ) . Moreover, the set ( N Y n =1 M ( α n ) : N ∈ N , ( α , . . . , α N ) ∈ R B ( α ) ) is finite. C.L. SAMUELS
Proof.
Suppose that α = α · · · α N is an B -restricted representation. By assump-tion, at least N − α n in our representation are not roots of unity.Assume α n is one such element. Lemma 3.1 implies that there exists a point γ n ∈ K α such that(3.4) M ( α n ) = H ( γ n ) . Since α n is not a root of unity, neither side of (3.4) equals 1, so that γ n is not aroot of unity either. Therefore, we find that M ( α n ) ≥ q ( α ) for N − { α , . . . , α N } . This yields B ≥ M ( α ) · · · M ( α N ) ≥ q ( α ) N − . We know that q ( α ) > q ( α ) to obtain(3.5) N ≤ B log q ( α )verifying the first statement of the lemma. We now find that ( N Y n =1 M ( α n ) : N ∈ N , ( α , . . . , α N ) ∈ R B ( α ) ) = ( N Y n =1 M ( α n ) : ( α , . . . , α N ) ∈ R B ( α ) , N ≤ B log q ( α ) ) ⊆ ( N Y n =1 M ( α n ) : N ≤ B log q ( α ) , M ( α n ) ≤ B, α n ∈ Rad( K α ) ) which is finite by Lemma 3.1. (cid:3) Proof of Theorem 1.1.
By Lemma 3.2, we may select
B > M ( α ) such that( M ( α ) , B ) \ ( N Y n =1 M ( α n ) : N ∈ N , ( α , . . . , α N ) ∈ R M ( α )+1 ( α ) ) = ∅ . Of course, we may choose B ≤ M ( α ) + 1 which gives ( N Y n =1 M ( α n ) : N ∈ N , ( α , . . . , α N ) ∈ R B ( α ) ) ⊆ ( N Y n =1 M ( α n ) : N ∈ N , ( α , . . . , α N ) ∈ R M ( α )+1 ( α ) ) , and therefore,(3.6) ( M ( α ) , B ) \ ( N Y n =1 M ( α n ) : N ∈ N , ( α , . . . , α N ) ∈ R B ( α ) ) = ∅ . By definition of M , there exists a representation α = γ · · · γ L such that M ( α ) ≤ M ( γ ) · · · M ( γ L ) < B. Theorem 2.1 implies that there exists a representation α = ζβ · · · β L with ζ a rootof unity, each element β ℓ belonging to Rad( K α ) and M ( β ℓ ) ≤ M ( γ ℓ ) for all ℓ . Thisyields M ( α ) ≤ M ( ζ ) M ( β ) · · · M ( β L ) < B. HE INFIMUM IN THE METRIC MAHLER MEASURE 9
By combining all roots of unity in the representation into a single element, weobtain a new representation α = α · · · α N having α n ∈ Rad( K α ), at most one rootof unity, and M ( α ) · · · M ( α N ) = M ( β ) · · · M ( β L ) . Therefore, we see that(3.7) M ( α ) ≤ M ( α ) · · · M ( α N ) < B which implies, in particular, that ( α , . . . , α N ) ∈ R B ( α ). Then by (3.6) we get that(3.8) M ( α ) · · · M ( α N ) ( M ( α ) , B ) . Finally, combining (3.7) and (3.8) we obtain M ( α ) = M ( α ) · · · M ( α N ). (cid:3) The location of M ( α ) and M ∞ ( α )We now apply Theorems 1.1 and 1.2 in order to show that M ( α ) and M ∞ ( α )belong to K α . We begin with M ∞ in which case we are able to prove a slightlystronger result. Theorem 4.1. If α is an algebraic number then there exists β ∈ K α such that M ∞ ( α ) = M ( β ) . In particular, M ∞ ( α ) ∈ K α .Proof. By Theorem 1.2 there exist α , . . . , α N ∈ Rad( K α ) such that α = α · · · α N and M ∞ ( α ) = max { M ( α ) , . . . , M ( α N ) } . For each n , Lemma 3.1 implies that there exists a root of unity ζ n and L n , S n ∈ N such that M ( α n ) = M ( ζ n α L n n ) S n and ζ n α L n n ∈ K α . For simplicity, we write L = N Y n =1 L n and J n = Y k = n L k so that L = L n J n for all n . Then we obtain immediately α L = N Y n =1 α L n J n n so there exists a root of unity ζ such that ζα L = N Y n =1 ( ζ n α L n n ) J n . By Theorem 1.3 of [4] we obtain that M ∞ ( α ) = M ∞ ( ζα L ) ≤ max ≤ n ≤ N { M ( ζ n α L n n ) }≤ max ≤ n ≤ N { M ( ζ n α L n n ) S n } = max ≤ n ≤ N { M ( α n ) } = M ∞ ( α ) . Therefore, we have that M ∞ ( α ) = max ≤ n ≤ N { M ( ζ n α L n n ) } . As we have noted, eachelement ζ n α L n n belongs to K α completing the proof of the first statement.Now we have that M ∞ ( α ) = M ( β ) for some β ∈ K α . Since K α is Galois, itmust contain all conjugates of β over Q , and therefore, it contains the product ofall roots outside the unit circle. This product is a real number so K α must containits absolute value. Hence we get that M ∞ ( α ) ∈ K α . (cid:3) In the case of M , we cannot establish a result as strong as Theorem 4.1, but wecan prove an analog of its second statement. Theorem 4.2. If α is an algebraic number then M ( α ) ∈ K α .Proof. By Theorem 1.1, we know that there exist α , . . . , α N ∈ Rad( K α ) such that α = α · · · α N and M ( α ) = M ( α ) · · · M ( α N ) . According to Lemma 3.1, for each n there exists an algebraic number γ n ∈ K α anda positive integer S n such that M ( α n ) = M ( γ n ) S n . Each conjugate of γ over Q must belong to the Galois extension K α , which implies that M ( γ n ) ∈ K α for all n .It follows that M ( α ) ∈ K α . (cid:3) References [1] E. Dobrowolski,
On a question of Lehmer and the number of irreducible factors of a polyno-mial , Acta Arith. (1979), no. 4, 391–401.[2] A. Dubickas and C.J. Smyth, On the metric Mahler measure , J. Number Theory (2001),368–387.[3] D.S. Dummit and R.M. Foote, Abstract Algebra , Prentice Hall, Englewood Cliffs, NJ (1991).[4] P. Fili and C.L. Samuels,
On the non-Archimedean metric Mahler measure , preprint (2008).[5] D.H. Lehmer,
Factorization of certain cyclotomic functions , Ann. of Math. (1933), 461–479.[6] D.G. Northcott, An inequality on the theory of arithmetic on algebraic varieties , Proc. Cam-bridge Philos. Soc., (1949), 502–509. Max-Planck-Institut f¨ur Mathematik, Vivatsgasse 7, 53111 Bonn, Germany
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