Root lattices in number fields

We explore whether a root lattice may be similar to the lattice $\mathscr O$ of integers of a number field $K$ endowed with the inner product $(x, y):={\rm Trace}_{K/\mathbb Q}(x\cdot\theta(y))$, where $\theta$ is an involution of $K$. We classify all pairs $K$, $\theta$ such that $\mathscr O$ is similar to either an even root lattice or the root lattice $\mathbb Z^{[K:\mathbb Q]}$. We also classify all pairs $K$, $\theta$ such that $\mathscr O$ is a root lattice. In addition to this, we show that $\mathscr O$ is never similar to a positive-definite even unimodular lattice of rank $\leqslant 48$, in particular, $\mathscr O$ is not similar to the Leech lattice. In appendix, we give a general cyclicity criterion for the primary components of the discriminant group of $\mathscr O$.


Introduction
Number fields are natural sources of lattices, i.e., pairs (L, b), where L is a free Z-module of finite rank and b : L×L → Z is a nondegenerate symmetric bilinear form; see [7]. Namely, let K be a number field, n := [K : Q] < ∞, and let O be the ring of integers of K. We fix a field automorphism θ ∈ AutK such that θ 2 = id. (1) Then the map tr K,θ : K × K → Q, tr K,θ (x, y) := Trace K/Q (x·θ(y)) (2) Some remarkable lattices are of the form (I, tr K,θ ). For instance, this is so for some root lattices, the Coxeter-Todd lattice, the Leech lattice, and some others [2], [8]. So, given a lattice (L, b), it arises the problem of finding out whether it is isometric to (I, tr K,θ ) for suitable K, θ, I.
There is a distinguished ideal in the set of all nonzero ideals of O, namely, O itself. This leads to the problem of finding remarkable lattices isometric (or, more generally, similar) to lattices of the form (O, tr K,θ ). This paper is aimed to explore whether (O, tr K,θ ) may be similar to a root lattice, in particular, whether (O, tr K,θ ) itself may be a root lattice. It naturally conjuncts with our previous publication [20]: both papers stem from our wish to explore realizations in number fields of objects associated with root systems.
A nonzero lattice is called a root lattice if it is isometric to orthogonal direct sum of lattices belonging to the union of two infinite series A ℓ (ℓ 1), D ℓ (ℓ 4), and four sporadic lattices Z 1 , E 6 , E 7 , E 8 , whose explicit description is recalled in Appendix 1 below (Section 7). All lattices in this union are indecomposable (i.e., inexpressible as orthogonal direct sums of nonzero summands). By Eichler's theorem [17,Thm. 6.4] decomposition of a root lattice as orthogonal direct sum of indecomposable lattices (called its indecomposable components) is unique.
Given a lattice (L, b), we denote the orthogonal direct sum of s > 0 copies of (L, b) by (L, b) s . For (L, b) = Z 1 , we denote (L, b) s by Z s .
A characterization of root lattices is given by fundamental Witt's theorem: Theorem (E. Witt [23]; see also [16,Thm. 4.10.6]). A lattice (L, b) is a root lattice if and only if the following two conditions hold: (i) the form b is positive-definite; (ii) the Z-module L is generated by the set {x ∈ L | b(x, x) = 1 or 2}.
The lattices (L 1 , b 1 ) and (L 2 , b 2 ) are called similar (equivalently, one of them is called similar to the other) if there are nonzero integers m 1 , m 2 such that the lattices (L 1 , m 1 b 1 ) and (L 2 , m 2 b 2 ) are isometric.
A nonzero lattice (L, b) is called a primitive lattice if the positive integer is 1. For every nonzero lattice (L, b), the lattice (L, b/d (L,b) ) is primitive. Two lattices (L 1 , b 1 ) and (L 2 , b 2 ) are similar if and only if the lattices (L 1 , b 1 /d (L 1 ,b 1 ) ) and (L 2 , b 2 /d (L 2 ,b 2 ) ) are isometric. A root lattice is nonprimitive if and only if it is isometric to A a 1 1 for some a 1 . We first consider a special case of the problem, namely, explore whether (O, tr K,θ ) may be a root lattice. The following examples show that such cases do exist. Example 1. Let n = 1. Then we have K = Q, O = Z, θ = id, and Trace K/Q (x) = x for every x ∈ K. Hence in this case (O, tr K,θ ) is the root lattice Z 1 (which is similar but not isometric to A 1 ).
Example 2. Let n = 2 and let K be a 3rd cyclotomic field: This shows that (O, tr K,θ ) is a root lattice isometric to A 2 ; see [20].
Example 3. Let n = 2 and let K be a 4th cyclotomic field: K = Q( √ −1). Let θ be the complex conjugation. Then O = Z + Z √ −1, and formula (4) still holds. This shows that (O, tr K,θ ) is a root lattice isometric to A 2 1 ; see [20]. Our first main result, Theorem 1 below, yields the classification of all pairs K, θ for which (O, tr K,θ ) is a root lattice: Theorem 1. The following properties of a pair K, θ are equivalent: (a) (O, tr K,θ ) is a root lattice; (b) K, θ is one of the following pairs: We then address the general problem of classifying all pairs K, θ such that the lattice (O, tr K,θ ) is similar (but not necessarily isometric) to a root lattice (L, b). It appears that such pairs are far from being exhausted by Examples 1, 2, and 3. We obtain their complete classifications in both "unmixed" cases, namely, when the Z-module L is generated by the set {x ∈ L | b(x, x) = 1} and when it generated by the set {x ∈ L | b(x, x) = 2}. The first case is precisely the one in which (L, b) is isometric to Z n . The second is the one in which every indecomposable component of (L, b) is not isometric to Z 1 ; the latter property, in turn, is equivalent to the evenness of the lattice (L, b). Our next two main results, Theorem 2 and 3 below, yield these classifications. In these theorems, m denotes the unique positive integer such that (such m exists because Trace K/Q : O → Z is a nonzero additive group homomorphism).
Theorem 2. The following properties of a pair K, θ are equivalent: K is a 2 a th cyclotomic field for a positive integer a, and θ is the complex conjugation if a > 1 and θ = id if a = 1. If these properties hold, then n = 2 a−1 and m = n.
In (e), let ζ 2 a ∈ K be a 2 a th primitive root of unity, and let x j := ζ j 2 a . Then the set of all indecomposable components of the root lattice (O, tr K,θ /m) coincides with the set of all its sublattices For every j, the value of tr θ /m at (x j , x j ) is 1. and θ is the complex conjugation. If these properties hold, then n = 2 a 3 b−1 and m = n/2. In (d), let ζ 2 a and ζ 3 b ∈ K be respectively a primitive 2 a th and 3 b th root of unity, and let x i,j := ζ i 2 a ζ j 3 b . Then the set of all indecomposable components of the root lattice (O, tr K,θ /m) coincides with the set of all its sublattices Note that if K is a dth cyclotomic field, and θ is the complex conjugation, then for d = 2 a , the similarity of (O, tr K,θ ) to Z 2 a−1 was observed in [4, Prop. 9.1(ii)], and for d = 2 · 3 b , the similarity of (O, tr K,θ ) to A 3 b−1 2 can be deduced from [4, Prop. 9.1(i)]. Since E 8 is the unique (up to isomorphism) positive-definite even unimodular lattice of rank 8 (see [17, §6]), Theorem 3 solves in the negative for n = 8 the existence problem of a lattice (O, tr K,θ ) similar to a positive-definite even unimodular one. Our last main result, Theorem 4 below, shows that as a matter of fact the following more general statement holds: This paper is organized as follows. Theorems 1, 2, 3, and 4 are proved, respectively, in Sections 3, 4, 5, and 6. Section 2 contains several general auxiliary results used in these proofs and in the proof of Theorem 5 (see below). At the end of this paper two short appendices are placed. Appendix 1 (Section 7) recalls the explicit description of indecomposable root lattices. Appendix 2 (Section 8) contains a general cyclicity criterion for the primary components of the discriminant group of (O, tr K,θ ) (Theorem 5). At the first stage we used another approach to finding a classification of all pairs K, θ such that (O, tr K,θ ) is similar to an indecomposable even root lattice. This approach led us to only a partial answer, see [19,Thms. 4,5] (in contrast, in the present paper we obtain a complete classification, see Theorems 2 and 3). However, within this approach we obtained and applied a general cyclicity criterion, which may be useful for other applications. Therefore, we consider it worthwhile to publish it in Appendix 2 as Theorem 5.
In conclusion, we thank E. Bayer-Fluckiger for her interest in this work and help with references.

Conventions, terminology, and notation
Given a lattice (L, b) of rank r > 0, we canonically embed L in the vector space V = L ⊗ Z Q over Q and extend b to a nondegenerate symmetric bilinear form is called a sublattice of (L, b) and denoted just by M.
For every s ∈ Z, we put where e 1 , . . . , e r is a basis of L over Z (the right-hand side of (7) does not depend on the choice of basis). L * is the dual of L with respect to b, i.e., The discriminant group of (L, b) is the (finite Abelian) group L * /L. discr K/Q := discr(O, tr K,id ) is the discriminant of K/Q. Trace K/Q (x) and Norm K/Q (x) are respectively the trace and norm over Q of an element x ∈ K.
c and d are respectively the codifferent and different of K/Q, i.e., (c is a fractional ideal of K, O ⊂ c, and d is an ideal of O) [12], [18]. If a and b are nonzero ideals of O and a is prime, then ord a b is the highest nonnegative integer t such that a t ⊇ b.
µ K is the (finite cyclic) multiplicative group of all roots of unity in K. ϕ is Euler's totient function.

Generalities
In this section, several auxiliary results are collected. We do not claim priority for all of them and give a reference always when we know it. At the same time, we prove all the statements, wanting to make this paper reasonably self-contained, and because our proofs, while being rather short, provide somewhat more information than can be found in the literature. Lemma 1. The following properties of a prime p are equivalent: Proof. (i)⇒(ii) is clear. We shall prove the inclusion which readily implies (ii)⇒(i). The set of all field embeddings K ֒→ C contains exactly n elements σ 1 , . . . , σ n , and for every x ∈ K, we have (see, e.g., [11,Chap. 12]). From (11) we deduce the existence of a symmetric polynomial f = f (t 1 , . . . , t n ) in variables t 1 , . . . , t n with integer coefficients such that Let s i = s i (t 1 , . . . , t n ) be the elementary symmetric polynomial in t 1 , . . . , t n of degree i. Then f may be represented a polynomial with integer coefficients in s 1 , . . . , s n (see, e.g., [15, The latter inclusion and (12) yield (10). Proof. This follows from (5) and Lemma 1 with p = 2.
Lemma 2. Let (L, b) be a nonzero lattice of rank r.
(i) If e 1 , . . . , e r is a basis of the Z-module L and s 1 , . . . , s r are the invariant factors of the matrix (b(e i , e j )) (see [9, (16.6)]), then the group r i=1 Z/s i Z is isomorphic to the discriminant group of (L, b). In particular, s 1 · · · s r = |discr(L, b)| is the order of the latter group.
(iii 2 ) This follows from (iii 1 ) and (3) because (iv) It follows from θ(O) = O and formulas (2), (8), (9) that c is the dual of O with respect to tr K,θ . Therefore, the discriminant group of (O, tr K,θ ) is c/O. By [9, (4.15), p. 85], the latter group is isomorphic to cd/Od. Since cd = O and Od = d, this proves the first claim.
By (i), the order of the discriminant group of (O, tr K,id ) is |disc K/Q|. Since, by the first claim, this group is isomorphic to the discriminant group of (O, tr K,θ ), this proves the second claim.
(v) The first claim follows from (3). Let (L, b) = (O, tr θ ). By (iii 1 ), the lattice (O, tr θ /m) is even. By (13) this implies that n/m is an even integer; whence (v 1 ). If, moreover, θ = id, then m is even by Corollary 2; whence (v 2 ) because of the second formula in (v 1 ).  (i) tr K,θ is a definite bilinear form; (ii) tr K,θ is a positive-definite bilinear form; (iii) either K is a totally real field and θ = id, or K is a CM-field and θ is the complex conjugation.
Note that if K is a CM-field and θ is the complex conjugation, then n is even, [K θ : Q] = n/2, and (discr K θ /Q) 2 divides discr K/Q; Corollary 3. For a pair K, θ, if the lattice (O, tr K,θ ) is similar to a root lattice, then either K is a totally real field and θ = id, or K is a CM-field and θ is the complex conjugation. Proof. The equalities (see, e.g., [11,Chap. 12]) imply that Trace K/Q (x) and Norm K/Q (x) are positive integers. Combining (17) with the classic inequality of arithmetic and geometric means (see, e.g., [21, Sect. 2]), we then obtain the inequalities This proves (i). If x = 1, then the equality in (16) holds by (13). Conversely, assume that the equality in (16) holds for some x. In view of (18) we then conclude that Norm K/Q (x) = 1 (19) and the equalities in (18) hold too. The classic inequality of arithmetic and geometric means tells us that the latter happens if and only if σ 1 (x) = . . . = σ n (x), i.e., x is a positive integer. From (19) and (17) we then obtain x n = 1, hence x = 1. This proves (ii).
The inequality in Lemma 5(ii) below can be found in [3,Cor. 4].
Lemma 5. Suppose a pair K, θ shares one of the following properies: (TR) K is a totally real field, θ = id; (CM) K is a CM-field, θ is the complex conjugation. Then, for every nonzero element x ∈ O, the following holds: (i) tr K,θ (x, x) is a positive integer; Proof. First, we note that if σ : K ֒→ C is a field embedding, then σ(x · θ(x)) is a positive real number. Indeed, if (TR) holds, then whence the claim in this case. If (CM) holds, then σ(K) is stable with respect to the complex conjugation of C and σ(θ(x)) = σ(x) (see, e.g., [22, p. 38]). Therefore, which proves the claim in this case.

In view of this, (i) and the inequality in (ii) follow from (2) and Lemma 4(i). By Lemma 4(ii), the equality in (ii) holds if and only if
x · θ(x) = 1.
Assume that (22) holds. Then, in the notation of Lemma 4, every complex number σ i (x) has the modulus 1. Since x is an algebraic integer, by Kroneker's theorem [13] (see also, e.g., [22, Lem. 1.6]) this implies that x is a root of unity. Conversely, if x is a root of unity, then, in the above notation, σ(x) is a root of unity too, hence the right-hand sides of the equalities in (20), (21) are equal to 1. Whence, (22) holds. This completes the proof of (ii). Suppose that (O, tr K,θ /m)[s] = ∅, i.e., there is a ∈ O such that tr K,θ (a, a)/m = s. This and (ii) yield (a).
Since, by Lemma 2(iii 2 ), n/m is an integer, it follows from (a) that m = n if s = 1. This proves (b).
Suppose, moreover, that (O, tr K,θ /m) is even. Then n/m is even by Lemma 2(v 1 ). This and (a) then imply that m = n/2 if s = 2. This proves (c). Finally, if s = 4, then (a) implies that n/m = 2 or 4. This proves (d).
Lemma 6 (cf. [4, Prop. 9.1(ii)]). Fix a positive integer a. Let K be a 2 a th cyclotomic field, let θ be the complex conjugation if a > 1 and θ = id if a = 1, and let ζ ∈ K be a 2 a th primitive root of unity. Then the following holds: (i) n = 2 a−1 ; (ii) {ζ j | 0 j 2 a−1 − 1} is a basis of the Z-module O; (iii) for every elements ζ i and ζ j of this basis, (iv) m = n.
Lemma 7 (cf. [4, Prop. 9.1(i)]). Fix a positive integer b. Let K be a 3 b th cyclotomic field, let θ be the complex conjugation, and let ζ ∈ K be a 3 b th primitive root of unity. Then the following holds: (iii) for every two elements ζ i and ζ j of this basis, (iv) m = n/2.
Proof. The same argument as in the proof of Lemma 6 yields (i), (ii), and (iv).
(iii) Formula (23) still holds and the same argument as in the proof of Lemma 6(iii) yields the validity of (iii) for i = j. Assume now that i = j. Then ζ i−j is a primitive 3 s th root of unity for some positive integer s b. The degree of its minimal polynomial over Q is Since the 3rd cyclotomic polynomial is t 2 + t + 1, this minimal polynomial is t 2·3 s−1 + t 3 s−1 + 1 (see [15, Chap. VIII, §3]). Given that 2 · 3 s−1 − 3 s−1 = 1 if and only if s = 1, from this we infer that Given a qth cyclotomic field K and a positive integer r dividing q, we denote by K r and O r respectively the unique rth cyclotomic subfield of K and its ring of integers. They are Aut K-stable; for α ∈ Aut K, we denote the restriction α| Kr still by α. If α is the complex conjugation of K, this restriction is the complex conjugation of K r . Lemma 8. Let i and j be coprime positive integers. Let K be a ij-th cyclotomic field and let θ be the complex conjugation. Then the following holds: (i) the natural Q-algebra homomorphism is an isomorphism; (ii) for every x ∈ K i and y ∈ K j , the following equality holds: (iii) the restriction of homomorphism (28) to O i ⊗ Z O j is a ring isomorphism with O, which is also a lattice isometry Proof. Let ζ i ∈ K i and ζ j ∈ K j be respectively a primitive ith and a primitive jth root of unity. Then ζ i ζ j is a primitive ijth root of unity, hence K = Q(ζ i ζ j ). This shows that K is the compositum of K i and K j ; whence (28) is a surjective homomorphism. The coprimeness of i and j implies that [K : Q] = ϕ(ij) = ϕ(i)ϕ(j) = [K i : Q][K j : Q]; hence (28) is an injective homomorphism. This proves (i).
By definition of the trace of an element of a number field, the lefthand side of (29) is the trace of the Q-linear transformation of K given by multiplication by xy. In view of (i), it is equal to the trace of the Q-linear transformation of Q(ζ i ) ⊗ Q Q(ζ j ) given by multiplication by x ⊗ y. Hence (see [6, Chap. VII, §5, no. 6]) it is equal to the product of traces of the Q-linear transformations of Q(ζ i ) and Q(ζ j ) given by multiplications by respectively x and y. By the mentioned definition, the latter product is equal to the right-hand side of (29). This proves (ii).
In In the next lemma, we use that if K is totally real, then discr K/Q is positive (see, e.g., [12, Prop. 1.2a]). Lemma 9. Let K be a totally real number field of degree n > 1 over Q.
(i) If n 24, then n discr K/Q > n.
Lemma 10. Let K be a CM-field of degree n > 2 over Q. If n 48, then n |discr K/Q| > n/2.
The claim then readily follows from the lower bounds on the right-hand side of (30), which is obtained by applying Lemma 9(i) to the totally real number field K θ of degree n/2.
Proof of Theorem 1. In view of Examples 1, 2, 3, the "if" part is clear.
To prove the "only if" one, suppose that (O, tr K,θ ) is a root lattice. Then by Corollary 3, either K is a totally real field and θ = id, or K is a CM-field and θ is the complex conjugation. There are two possibilities, which we will consider separately: (a) the lattice (O, tr K,θ ) is primitive or, equivalently, at least one of its indecomposable components is not isometric to A 1 . (b) the lattice (O, tr K,θ ) is nonprimitive or, equivalently, it is isometric to A a 1 1 for some a 1 .
(a)⇒(e) Suppose that (O, tr K,θ ) is similar to Z n . Then, by Corollary 3, either K is a totally real field and θ = id, or K is a CM-field and θ is the complex conjugation. Since (O, tr K,θ /m) and Z n are isometric, and the Z-module Z n is generated by Z n [1], we deduce from Lemma Suppose that the order of the cyclic group µ K is divisible by an odd prime p. Then there is x ∈ µ K which is a pth primitive root of unity. The minimal polynomial of x over Q is t p−1 + t p−2 + · · · + 1, hence (38) However, in view of (5) and (36), the integer Trace K/Q (−x) is divisible by n. This contradics (39). The obtained contradiction proves that there a positive integer a such that µ K is a cyclic group of order 2 a . In view of (37), this implies that K is a 2 a th cyclotomic field. In particular, it is a CM-field, and therefore, θ is the complex conjugation. This proves implication (a)⇒(e).
(e)⇒(c) Let K be a 2 a th cyclotomic field for a positive integer a and let θ be the complex conjugation. By Lemma 6 (whose notation we use), n = 2 a−1 , m = n, and (O, tr K,θ /m) is the orthogonal direct sum of the sublattices Zζ j , 0 j 2 a−1 − 1, each of which is isometric to Z 1 . This proves implication (e)⇒(c).
The last statement of Theorem 2 follows from Lemma 6(iii).
and (40) implies by Lemma 5(ii) that O is the Z-linear span of µ K . Hence K is a cyclotomic field and θ is the complex conjugation.
There is an odd prime p dividing the order of µ K : otherwise, this order is 2 a for some a, hence m = n by Theorem 2, which contradicts (40). Thus, we can find an element x ∈ µ K which is a pth primitive root of unity. The same argument as in the above proof of implication (a)⇒(e) of Theorem 2 shows that (39) holds. In view of (5) and (40), the integer Trace K/Q (−x) is divisible by n/2. This and (39) then yield p − 1 = 2, i.e., p = 3. Thereby we proved that K is a 2 a 3 b th cyclotomic field for some a 0, b > 0. Since every 3 b th cyclotomic field is simultaneously a 2 1 3 b th cyclotomic field, we may assume that a > 0. This completes the proof of implication (a)⇒(d).
(d)⇒(c) Let K be a 2 a 3 b th cyclotomic field for some a > 0, b > 0. We use the notation introduced in the paragraph immediately preceding By Lemma 8(i) we have n = n i n j . Hence by Lemma 8(iii) there is a lattice isometry In view of (41) and Lemmas 6, 7, the lattices (O i , tr K i ,θ /n i ) and O j , tr K j ,θ /(n j /2) are isometric to respectively Z n i and A n j /2 2 . Since the lattices Z 1 ⊗ Z A 2 and A 2 are isometric, the existence of isomorphism (42) then implies that the lattice O, tr K,θ /(n/2) is isometric to A n/2 2 . This completes the proof of implication (d)⇒(c).
Proof of Theorem 4. Arguing on the contrary, assume that (O, tr K,θ ) is similar to a positive-definite even unimodular lattice (L, b) of rank n, where 8 n 48.
7. Appendix 1: The Z n , A n , D n , and E n lattices To be self-contained, here we briefly describe the lattices Z n , A n , D n , and E n as they play a key role in this paper. For details and a discussion of their properties see [5], [8], [16].
Let R m be the m-dimensional coordinate real vector space of rows endowed with the standard Euclidean structure Denote by e j the row (0, . . . , 0, 1, 0, . . . , 0), where the number of 0's to the left of 1 is j − 1.
If L is the Z-linear span of a set of linearly independent elements of R m such that b(L × L) ⊆ Z, where b is the restriction of map (47) to L × L, then (L, b) is called a lattice in R m and denoted just by L.
With these notation and conventions, we have: Z n is the lattice {(x 1 , . . . , x n ) | x j ∈ Z for all j} in R n . A n is the lattice {(x 1 , . . . , x n+1 ) ∈ Z n+1 | n+1 j=1 x j = 0} in R n+1 . D n is the lattice {(x 1 , . . . , x n ) ∈ Z n | n j=1 x j ≡ 0 mod 2} in R n , n 4. E 8 is the lattice D 8 D 8 + 1 2 (e 1 + · · · + e 8 ) in R 8 . E 7 is the orthogonal in E 8 of the sublattice Z(e 7 + e 8 ). E 6 is the orthogonal in E 8 of the sublattice Z(e 7 + e 8 ) + Z(e 6 + e 8 ). Each of these lattices except A 1 is primitive. Each of them except Z n for every n is even.

Appendix 2: Cyclicity criterion
Theorem 5 (cyclicity criterion). Let n > 1 and let p be a prime ramified in K. Then the following properties of O are equivalent: (i) The p-primary component of the additive group of O/d is cyclic.
(ii) The following conditions hold: (ii 1 ) in O, there is exactly one ramified prime ideal p which lies over p; (ii 2 ) p is odd, ord p pO = 2, and O/p is the field of p elements.
Proof. Let r 1 , . . . , r m be all pairwise distinct prime ideals of O ramified in K/Q. For every r i , there is a prime integer r i and a positive integer f i such that By the Chinese remainder theorem (see, e.g., [11,Props. 12.3.1]), decomposition (50) yields the following ring isomorphism: As every r i is prime, (52), (48) imply that (i) is satisfied if and only if the following two conditions (a) and (b) hold: (a) there is exactly one i such that r i = p, (b) the additive group of O/r d i i is cyclic. Clearly (a) is equivalent to (ii 1 ). We shall now show that (b) is equivalent to (ii 2 ) with p = r i and p = r i ; this will complete the proof.
Next, assume that (W) holds. Then r e i i ⊇ r d i i ; whence there is a ring epimorphism This completes the proof of (b)⇒(ii 2 ) with p = r i and p = r i . The converse implication is immediate.