Isometric embeddings of finite fields

Finite Fields and Their Applications 25 (2014) 134–145

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Finite Fields and Their Applications www.elsevier.com/locate/ffa

Isometric embeddings of finite fields Yasushi Mizusawa a,∗ , Shinya Nishikawa b a b

Department of Mathematics, Nagoya Institute of Technology, Gokiso, Showa, Nagoya 466-8555, Japan Department of Computer Science and Engineering, Nagoya Institute of Technology, Gokiso, Showa, Nagoya 466-8555, Japan

a r t i c l e

i n f o

Article history: Received 12 February 2013 Received in revised form 10 June 2013 Accepted 10 September 2013 Available online 2 October 2013 Communicated by L. Storme MSC: primary 11T55 secondary 11R18

a b s t r a c t By regarding a finite field as a vector space over the prime field with a basis consisting of powers of an element, a Hamming distance is defined on the finite field with respect to the power basis. We consider the existence of isometric homomorphisms between such finite fields, and characterize the isometric embeddings for even characteristic by arithmetical conditions. Moreover, a canonical Hamming metric is defined in a certain infinite dimensional algebraic extension of a finite field. © 2013 Elsevier Inc. All rights reserved.

Keywords: Finite field Hamming distance Cyclotomic field

1. Introduction The finite field F pn of order pn , where p is a prime number, is identified with the residue ring F p [x]/( f (x)) of the polynomial ring F p [x] modulo an irreducible polynomial f (x) of degree n. The Hamming weight   f on the F p -vector space F pn with respect to the power basis {xi mod f (x) | 0  i  n − 1} is defined as

    c 0 + c 1 x + · · · + cn−1 xn−1 mod f (x) = {i | c i = 0} f where | | denotes the cardinality of the set. Then (F pn , d f ) is a metric space with the Hamming distance d f (c, c ) = c − c  f (c, c ∈ F pn ). While the Hamming distance induces a discrete topology

*

Corresponding author. E-mail address: [email protected] (Y. Mizusawa).

1071-5797/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.ffa.2013.09.003

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on F pn , the definition of this metric d f depends on the choice of f (x). The abilities of error-correcting codes over F p are estimated by the minimum Hamming distance (cf. e.g. [3]). An infinite dimensional algebraic extension F over F pn is also an F p -vector space. Hence we can define a Hamming distance on F if we choose a basis. However, the choice of the basis is not canonical. Then a simple question arises: How can we define a canonical Hamming distance on F ? Toward the answer of this question, for an algebraic extension F pmn /F pn , it is expected that there are irreducible polynomials f (x) and g (x) such that the embedding ε : (F pn , d f ) → (F pmn , d g ) is isometric, i.e., the injective ring homomorphism ε : F pn → F pmn satisfies c f = ε (c) g for any c ∈ F pn . If ε is isometric, then ε (x mod f (x)) = cxk mod g (x) with some 0 < k < mn = deg g (x) and 0 = c ∈ F p . Then moreover k  m, since f (cxk ) ≡ 0 (mod g (x)). In this paper, we focus on the case p = 2. By using the feature of even characteristic, we determine when the embedding ε is isometric. The main theorem is the following. Theorem 1.1. Let f (x) ∈ F2 [x] be an irreducible polynomial of degree n  2, and r the minimal positive integer satisfying xr ≡ 1 (mod f (x)). Then, for a pair (k, m) of positive integers such that k < mn, the following two conditions are equivalent: (i) There is an irreducible polynomial g (x) ∈ F2 [x] of degree mn such that

ε : (F2n , d f ) → (F2mn , d g ) : x mod f (x) → xk mod g (x) is an isometric embedding. (ii) k is odd, (n − 1)k < mn and the order of 2 mod kr ∈ (Z/kr Z)× is mn. Moreover, if these equivalent conditions hold, g (x) is uniquely determined as the irreducible polynomial of degree mn dividing f (xk ). Remark 1.1. Theorem 1.1 implies that k  m if the condition (ii) holds. The order of 2 mod kr is always divisible by n for any k (cf. Lemma 2.1). The condition (ii) states that the quotient is just m. In some cases where p is odd, there exists a nontrivial isometry ε : (F pn , d f ) (F pn , d g ) (cf. Section 5.1). On the other hand, Theorem 1.1 induces the nonexistence of nontrivial isometry for the characteristic p = 2. Corollary 1.1. If ε : (F2n , d f ) → (F2n , d g ) is an isometric isomorphism, then f (x) = g (x) and ε = id. Proof. If the condition (i) holds for m = 1, then k = 1 by the condition (ii). Since g (x) divides f (x), we have f (x) = g (x) and ε = id. 2 We give a proof of Theorem 1.1 in Section 2. In Section 3, we represent the condition (ii) in terms of prime factorizations of k and m (cf. Theorem 3.1). In Section 4, we give an inductive system of isometric embeddings. The inductive system induces a canonical Hamming metric in some infinite dimensional algebraic extensions F /F2n . In Section 5, we give more explicit examples. 2. Proof of Theorem 1.1 2.1. Preliminaries For irreducible polynomials f (x), g (x) ∈ F2 [x] with deg f (x) = n and deg g (x) = mn, we denote x mod f (x) ∈ F2 [x]/( f (x)) and x mod g (x) ∈ F2 [x]/( g (x)) by x ∈ F2n and x ∈ F2mn respectively. Let

A=







α ∈ F2n  α  f = 1 ,





B = β ∈ F2mn  β g = 1



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be the sets of weight 1 elements. As a feature of characteristic 2, we use the fact that A = {1, x, . . . , xn−1 } ⊂ x and B = {1, x, . . . , xmn−1 } ⊂ x , where x (resp. x ) is the cyclic subgroup of the × multiplicative group F× 2n (resp. F2mn ) generated by x (resp. x). As in the assumption of Theorem 1.1, we put r = | x |. Lemma 2.1. r is odd, and the order of 2 mod r ∈ (Z/r Z)× is n. n Proof. Since x2 −1 = 1, r divides 2n − 1. Hence r is odd and 2n ≡ 1 (mod r ). If 2d ≡ 1 (mod r ) for

some d < n, we have x2 −1 = 1, i.e., x ∈ F2d . This implies that there is some h(x) ∈ F2 [x] such that deg h(x)  d < n and h(x) = 0, i.e., f (x) divides h(x). This is a contradiction. Therefore | 2 mod r | = n. 2 d

Let Φd (x) ∈ Z[x] be the dth cyclotomic √ polynomial (cf. [1,4] for the properties), and put Φ d (x) = Φd (x) mod 2 ∈ F2 [x]. Let ζd = cos 2dπ + −1 sin 2dπ be a primitive dth root of unity, and Q(ζd ) the

dth cyclotomic field (cf. e.g. [4]). 2.2. Proof of (i) ⇒ (ii)

Suppose that the condition (i) holds. Since ε (x) = xk may not be the lowest power of x in ε( x ) = xk , we put  = min{  1 | x ∈ xk }  k. Our goal is to prove (ii) with the equality k = . By using the minimality of , we show that the pair (, m) satisfies the inequality in the condition (ii) as follows. Lemma 2.2. We have (n − 1) < mn  n and ε ( A ) = {1, x , x2 , . . . , x(n−1) }. Proof. First, we check that x = xk . Since x ∈ xk , we have x ⊂ xk . The integer k can be written in the form k = a + b with some 0  a ∈ Z and 0  b < . Since xb = xk (x )−a ∈ xk , the minimality of  yields that b = 0, i.e., xk ∈ x . Hence x = xk = ε ( x ). Since ε is isometric and injective, ε ( A ) = B ∩ ε (F2n ). Since A ⊂ x , we have ε ( A ) ⊂ ε ( x ). Therefore ε ( A ) = ε ( A ) ∩ ε ( x ) = B ∩ ε ( x ) = B ∩ x . We claim that

 



 

B ∩ x = xu  0  u 



mn − 1





where [ mn−1 ] is the largest integer smaller than or equal to mn−1 . If 0  u  [ mn−1 ], then 0  u  mn − 1, and hence xu ∈ B. Therefore the right-hand side is a subset of B ∩ x . Conversely, we take an element β ∈ B ∩ x arbitrarily. Then β = x j with some 0  j < mn, and j is written in the form j = u  + v with some 0  u ∈ Z and 0  v < . Since x v = β(x )−u ∈ x = xk , we have v = 0 by the minimality of . Then u  = j  mn − 1, and hence β = xu with 0  u  [ mn−1 ]. Thus we obtain the claim. Since [ mn−1 ]  mn − 1, the elements xu (0  u  [ mn−1 ]) are distinct. Therefore n = | A | =

|ε ( A )| = | B ∩ x | = [ mn−1 ] + 1. This yields that ε ( A ) = B ∩ x = {xu | 0  u  n − 1}. Moreover, we have (n − 1) = [ mn−1 ]  mn − 1 and mn − 1 = ( mn−1 ) < ([ mn−1 ] + 1) = n. Thus we obtain the inequalities (n − 1) < mn  n. The proof of Lemma 2.2 is completed. 2 As a feature of characteristic 2, f (x) is never of the form xn − a (a ∈ F p ). In particular, xn − xi ≡ 0 (mod f (x)) for any 0  i  n − 1, and hence xn  f > 1. By using this fact, we obtain the following key lemma. Lemma 2.3. ε (xn ) = xn .

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Proof. Since 1 = x ∈ ε ( A ) by Lemma 2.2, there is some i such that ε (xi ) = x and 1  i  n − 1. Since 1  n − i  n − 1, we have 1 = xn−i ∈ A. By Lemma 2.2, ε (xn−i ) = xu with some u such that 1  u  n − 1. Then ε (xn ) = ε (xi )ε (xn−i ) = x(u +1) and 2  u + 1  n. Since x(u +1)  g = ε (xn ) g = xn  f > 1, / B, in particular x(u +1) ∈ / ε ( A ). By Lemma 2.2, u + 1 must be n. 2 we have x(u +1) ∈ Since A is an F2 -basis of F2n ,

n−1 i =0

xi =



xn − 1

ε(x) = ε n−1 i =0

xi

α ∈ A α = 0. By Lemmas 2.2 and 2.3, we have



xn − 1 + 1 = n−1 + 1 = x . i i =0

x

Therefore  = k. In particular, (n − 1)k < mn. Lemma 2.4. | x | = kr.  Proof. Put k = | x / xk |  k = . Then xk ∈ xk and hence k = k by the minimality of  = k. Since | xk | = | x | = r, we obtain the claim. 2

By Lemma 2.4, we have xkr = 1. Since kr divides 2mn − 1, k is odd. Since xkr − 1 =



d|kr

Φ d (x),

g (x) divides Φ d (x) for some factor d of kr. Then xd = 1. By Lemma 2.4, d must be kr. Therefore g (x) is an irreducible factor of Φ kr (x). Let P be a prime ideal of Z[ζkr ] lying over 2, and e the order kr i mod P)) of 2 mod kr ∈ (Z/kr Z)× . Then F2 ⊂ Z[ζkr ]/P F2e and Φ kr (x) = i =1, gcd(i ,kr )=1 (x − (ζkr i in F2e [x]. Hence there is some 1  i < kr such that gcd(i , kr ) = 1 and g (ζkr mod P) = 0. Since the i kernel of the surjective homomorphism F2 [x] → Z[ζkr ]/P : h(x) → h(ζkr mod P) is ( g (x)), we have F2mn F2 [x]/( g (x)) Z[ζkr ]/P F2e , i.e., e = mn. Thus the condition (ii) holds.

2.3. Proof of (ii) ⇒ (i)



Since f (x) divides xr − 1 = d|r Φ d (x) and | x | = r, f (x) is an irreducible factor of Φ r (x). Let P be a prime ideal of Z[ζkr ] lying over 2, and put p = P ∩ Z[ζr ] which is a prime ideal of Z[ζr ]. ⊂ Z[ζr ]/p ⊂ Z[ζkr ]/P F2mn and Z[ζr ]/p F2n by the condition (ii) and Lemma 2.1. Since Then F2 Φ r (x) = ri=1, gcd(i ,r )=1 (x − (ζri mod p)) in F2n [x], we can choose 1  i < r such that gcd(i , r ) = 1 and i mod P is a common root of f (xk ) and Φ kr (x). However, f (ζri mod p) = 0. If gcd(i , k) = 1, then ζkr i is not necessarily satisfying gcd(i , k) = 1, so that we need the following lemma.

Lemma 2.5. gcd( f (xk ), Φ kr (x)) = 1. Proof. For 1  a < r such that gcd(a, r ) = 1, we denote by σa the element of the Galois group Gal(Q(ζr )/Q) such that σa (ζr ) = ζra . The elements σj ∈ Galσ(Q(ζr )/Q) can bej extended to the ring isomorphisms σ : Q(ζr )[x] Q(ζr )[x] : h(x) = j c j x → h (x) = j σ (c j )x . Let K = {ξ ∈ Q(ζr ) | σ2 (ξ ) = ξ } be the decomposition field of the prime number 2 in the abelian extension Q(ζr )/Q, and put O K = K ∩ Z[ζr ] the ring of algebraic integers in K . Then O K /(p ∩ O K ) F2 . Put F (x) = xn +

n−1 j =0

c jxj =

n−1

j =0 (x

j

− σ2 (ζri )) ∈ O K [x]. Since deg F (x) = n and F (ζri ) = 0, we have f (x) = −1

−1

F (x) mod p ∩ O K . Since F σi ((ζkr )k ) = σi−1 ( F (ζri )) = 0, we have O K [x]  gcd( F σi (xk ), Φkr (x)) = 1. Then O K [x]  gcd( F (xk ), Φkr (x)) = Φ kr (x)) = 1. 2

−1

σi (gcd( F σi (xk ), Φkr (x))) = 1, and therefore F2 [x]  gcd( f (xk ),

By Lemma 2.5, there is an irreducible polynomial g (x) dividing gcd( f (xk ), Φ kr (x)). There is j some 1  j < kr such that gcd( j , kr ) = 1 and g (ζkr mod P) = 0. Since F2 [x]/( g (x)) Z[ζkr ]/P :

h(x) mod g (x) → h(ζkr mod P) is an isomorphism, we have deg g (x) = mn. Let ( f  (x)) be the kerj

nel of the homomorphism F2 [x] → F2 [x]/( g (x)) : x → xk . Since f (x) ∈ ( f  (x)) and f (x) is irreducible,

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f  (x) = f (x). Thus we obtain the injective homomorphism ε : F2n = F2 [x]/( f (x)) → F2 [x]/( g (x)) = F2mn : x → xk . By the condition (n − 1)k < mn, ε ( A ) ⊂ B, i.e., ε is isometric. Since deg f (xk ) = nk < mn + k < 2mn, g (x) is the unique irreducible factor of f (xk ) of degree mn. This completes the proof of Theorem 1.1. 3. A condition equivalent to (ii) 3.1. An equivalent condition In this section, we give an equivalent condition of (ii) in Theorem 1.1 in terms of prime factorizations of k and m. By using this condition, it becomes easy to find many pairs (k, m) satisfying (ii). For an odd integer r > 1 and the order n of 2 mod r ∈ (Z/r Z)× , we put













P r = p: prime factor of r  v p (r ) − v p 2 p −1 − 1  v p  2 mod r / p v p (r )  , and put

 q > n, (Z/qZ)× = 2 mod q ,  . gcd(q, r ) = gcd(q − 1, n) = 1



Q r = q: prime number 

The following theorem and Lemma 2.1 characterize all the pairs (k, m) satisfying the condition (ii) of Theorem 1.1. We denote by ϕ ( z) = |(Z/ zZ)× | the Euler function. Theorem 3.1. Let r > 1 be an odd integer, and n the order of 2 mod r ∈ (Z/r Z)× . Then the pair (k, m) of positive integers satisfies the condition (ii) of Theorem 1.1 if and only if

k = q eq



p e p = 1



and m = ϕ qeq

p∈ P r



p e p = 1

p∈ P r

with some q ∈ Q r ∪ {1}, eq  0 and e p  0 such that (a) eq  1 if q = 1 and 2q−1 ≡ 1 (mod q2 ), (b) e p = 0 if eq  1 and 1 = q ≡ 1 (mod p ). Remark 3.1. P r ∩ Q r = ∅. If n is even, Q r = ∅. The prime number p such that 2 p −1 ≡ 1 (mod p 2 ), i.e., v p (2 p −1 − 1)  2, is called Wieferich prime. (cf. [4, Exercise 2.4] etc.) The smallest Wieferich prime is 1093. If the largest prime factor p of r is not a Wieferich prime, we have v p (r ) − v p (2 p −1 − 1) = v p (r ) − 1  0 = v p (ϕ (r / p v p (r ) ))  v p (| 2 mod r / p v p (r ) |) and hence p ∈ P r . Then, in particular, P r = ∅. Even if r has no Wieferich prime as a factor, P r does not necessarily contain all prime factors of r. For example, if r = 21 and p = 3, we have v p (r ) − v p (2 p −1 − 1) = 0 < 1 = v p (| 2 mod 7 |) and hence 3∈ / Pr . Example 3.1. For r = 7, we have n = | 2 mod 7 | = 3. Since p = 7 is not a Wieferich prime, P r = {7} (cf. Remark 3.1). The prime numbers q < 100 with a primitive root 2 are

3, 5, 11, 13, 19, 29, 37, 53, 59, 61, 67, 83. By picking q satisfying q > 3 and gcd(q, 7) = gcd(q − 1, 3) = 1, we have

{q ∈ Q 7 | q < 100} = {5, 11, 29, 53, 59, 83}.

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Put q = 5 and q = 29 particularly. If (k, m) is either ( p , p ) = (7, 7), (q, q − 1) = (5, 4) or (q , q − 1) = (29, 28), then the factorizations of k and m satisfy the conditions (a) and (b). Moreover, since q ≡ 1 (mod p ), (a) and (b) are also satisfied for (k, m) = (qp , (q − 1) p ) = (35, 28). On the other hand, since q ≡ 1 (mod p ), the condition (b) does not hold for (k, m) = (q p , (q − 1) p ) = (203, 196). Actually, | 2 mod kr | = | 2 mod 203 · 7 | = 84 = 84 · 7 = 196 · 3 = mn, and hence the condition (ii) is not satisfied. This example implies that the pair (qp , (q − 1) p ) does not necessarily satisfy (ii) even if both ( p , p ) and (q, q − 1) satisfy (ii). 3.2. Only-if part of the proof of Theorem 3.1 Assume that (k, m) satisfies the condition (ii). Let k be the maximal factor of k which is prime to r, and put r  = k/k . Let M be the kernel of the surjective homomorphism (Z/kr Z)× → (Z/r Z)× : x mod kr → x mod r. Then | M | = r  ϕ (k ). By the assumptions, | 2 mod kr | = mn and | 2 mod r | = n.  ep In particular, 2n mod kr ∈ M. The following lemma implies that k and r  will be qeq and p∈ P r p respectively. Lemma 3.1. m = r  ϕ (k ). Proof. Since the order of 2n mod kr ∈ M is m, m is a factor of | M | = r  ϕ (k ). Suppose that m = r  ϕ (k ). r  ϕ (k )

Then m  2  2k . By the condition (ii), we have (n − 1)k < nm  n 2k and hence n < 2. This is a contradiction. Therefore m = r  ϕ (k ). 2 Let K be the decomposition field of the prime number 2 in Q(ζkr )/Q. By Lemma 3.1, we have













Gal Q(ζkr )/Q(ζr ) M = 2n mod kr ⊂ 2 mod kr Gal Q(ζkr )/ K . This implies that K ⊂ Q(ζr ). Then Gal(Q(ζr )/ K ) 2 mod r . Since Q(ζk ) ∩ Q(ζrr  ) = Q, we have













Gal Q(ζkr )/ K Gal Q(ζrr  )/ K × Gal Q(ζk )/Q . The cyclicity of this group yields the cyclicity of Gal(Q(ζk )/Q) (Z/k Z)× . Hence k = qeq with some prime number q and an integer eq  0. The following lemma yields that q ∈ Q r and the condition (a). Lemma 3.2. If k = 1, then q ∈ Q r and gcd(q − 1, r  ) = 1. Moreover, eq  1 if 2q−1 ≡ 1 (mod q2 ). Proof. Suppose that k =  1. Then eq  1 and we have

ϕ (k ) = (q − 1)qeq −1 . By the condition (ii) and Lemma 3.1,

  (n − 1)r  qeq = (n − 1)k < mn = r  ϕ k n = r  (q − 1)qeq −1n, i.e., (n − 1)q < (q − 1)n. Therefore q > n. On the other hand, since Q(ζk ) ∩ K = Q, the prime number 2 does not split in Q(ζk )/Q. Therefore (Z/k Z)× = 2 mod k and (Z/qZ)× = 2 mod q . If eq  2, q must satisfy 2q−1 ≡ 1 (mod q2 ). Moreover, the cyclicity of Gal(Q(ζkr )/ K ) also induces that



 



1 = gcd Q(ζrr  ) : K , Q(ζk ) : Q Therefore, gcd(q − 1, n) = gcd(q − 1, r  ) = 1.

   = gcd nr  , ϕ k .

2

 = qeq = 1 and q ≡ Let p be an arbitrary prime factor of r, and put e p = v p (r  ) = v p (k)  0. If k 1 (mod p ), then e p = 0 by Lemma 3.2. By the following lemma, we have r  = p ∈ P r p e p . Then the condition (b) also holds.

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Lemma 3.3. If p divides r  , then p ∈ P r . Proof. Put N = v p (r )  1, v = v p (| 2 mod r / p N |)  v p (n) and w = v p (2 p −1 − 1). By considering the isomorphism

 ×   ×    (Z/rp Z)× Z/ p N +1 Z × Z/ r / p N Z : 2 mod rp → 2 mod p N +1 , 2 mod r / p N , we have v p (| 2 mod rp |) = max{ v p (| 2 mod p N +1 |), v }. Suppose that v p (| 2 mod p N +1 |)  v. Then v p (| 2 mod rp |) = v  v p (n) = v p (| 2 mod r |). This implies that any prime ideal lying over 2 splits in the extension Q(ζrp )/Q(ζr ) of degree p. However, K ⊂ Q(ζrp ) ⊂ Q(ζkr ). This is a contradiction. Therefore v < v p (| 2 mod p N +1 |) = max{0, N + 1 − w }. This yields that v < N + 1 − w and hence p ∈ Pr . 2 Thus we obtain the claim of only-if part. 3.3. If part of the proof of Theorem 3.1 First, we prove the claim in the case qeq = 1. Lemma 3.4. Let p 1 , . . . , p s be distinct elements of P r . If k = m = the condition (ii).

s

i =1

e

p i i with e i  1, then (k, m) satisfies

Proof. Since r is odd, then k is also odd. Since k = m, we have (n − 1)k < mn. We show that | 2 mod kr | = mn by the induction on s. Suppose that s = 1. Put p = p 1 ∈ P r , e = e 1  1, N = v p (r )  1, v = v p (| 2 mod r / p N |)  v p (n) and w = v p (2 p −1 − 1). Then k = m = p e and N − w  v. Since N  w, we have v p (| 2 mod p N +e |) = N + e − w and v p (| 2 mod p N |) = N − w. By considering the isomorphisms

 ×   ×    (Z/kr Z)× Z/ p N +e Z × Z/ r / p N Z : 2 mod kr → 2 mod p N +e , 2 mod r / p N and

 ×   ×    (Z/r Z)× Z/ p N Z × Z/ r / p N Z : 2 mod r → 2 mod p N , 2 mod r / p N , we have v p (| 2 mod kr |) = max{ N + e − w , v } = N + e − w and v p (| 2 mod r |) = max{ N − w , v } = N − w. This implies that any prime ideal lying over 2 does not split in the cyclic extension Q(ζkr )/Q(ζr ) of degree k = p e . Therefore | 2 mod kr | = k| 2 mod = kn = mn. s−r | 1 e Assume that s  2 and | 2 mod k0 r | = m0 n for k0 = m0 = i =1 p i i . Put p = p s and e = e s . Then | 2 mod p e r | = p e n by the above arguments. Hence any prime ideal lying over 2 does not split in the extensions Q(ζk0 r )/Q(ζr ) of degree m0 and Q(ζ pe r )/Q(ζr ) of degree p e . Since gcd(m0 , p e ) = 1, the prime ideals lying over 2 also do not split in the extension Q(ζkr )/Q(ζr ) of degree m. Therefore | 2 mod kr | = m| 2 mod r | = mn. 2 The following lemma shows the claim in the case where e p = 0 for all p ∈ P r . Lemma 3.5. For q ∈ Q r and eq  1, assume that 2q−1 ≡ 1 (mod q2 ) if eq > 1 (i.e., the condition (a)). Then (k, m) = (qeq , ϕ (qeq )) satisfies the condition (ii).

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Proof. By the assumption, (Z/kZ)× = 2 mod k . Since q > n > 1, k is odd and 1 −

1 < 1 − 1q . Then n (n − 1)q < (q − 1)n and hence (n − 1)k = (n − 1)qeq < nϕ (qeq ) = mn. Since gcd(q, r ) = 1, we obtain the

isomorphism

(Z/kr Z)× (Z/kZ)× × (Z/r Z)× : 2 mod kr → (2 mod k, 2 mod r ). Since gcd(q − 1, n) = gcd(q, n) = 1, we have | 2 mod kr | = lcm(| 2 mod k |, | 2 mod r |) = lcm(ϕ (k), n) = ϕ (k)n = mn. 2





Assume that k = qeq p ∈ P r p e p and m = ϕ (qeq ) p ∈ P r p e p (q ∈ Q r ∪ {1}, eq  0, e p  0) satisfy the conditions (a), (b). Put k = qeq , m = ϕ (qeq ), and put k0 = m0 = k/k . Then k = k k0 and m = m m0 . If k = 1 or k0 = 1, then (k, m) satisfies the condition (ii) by Lemma 3.4 and Lemma 3.5. Suppose that k = 1 and k0 = 1. Then (k , m ) and (k0 , m0 ) satisfy the condition (ii) by Lemma 3.4 and Lemma 3.5. Clearly, k is odd. Since (n − 1)k < nm , we have (n − 1)k = (n − 1)k m0 < nm m0 = mn. Moreover, any prime ideal lying over 2 does not split in the extensions Q(ζk r )/Q(ζr ) of degree m and Q(ζk0 r )/Q(ζr )  of degree m0 . Since q ≡ 1 (mod p ) if e p  1 by the condition (b), we have gcd(m0 , m ) = gcd( p ∈ P r p e p , ϕ (qeq )) = 1. Hence the prime ideals lying over 2 also do not split in the extension Q(ζkr )/Q(ζr ) of degree m. Therefore | 2 mod kr | = m| 2 mod r | = mn. Thus the proof of Theorem 3.1 is completed. 4. Sequences of isometric embeddings As a consequence of Theorem 1.1 and Theorem 3.1, we obtain the following proposition which gives a sequence of isometric algebraic extensions over (F2n , d f ). This sequence induces an infinite dimensional algebraic extension F /F2n with  a canonical Hamming distance. For an odd integer r > 1, put M r = { p ∈ P r p e p | e p  0}. Proposition 4.1. Let f (x) ∈ F2 [x] be an irreducible polynomial of degree n  2, and r = | x mod f (x) |. Then, for any m, m ∈ M r such that m divides m ,









εm,m : (F2mn , d f (xm ) ) → (F2m n , d f (xm ) ) : x mod f xm → xm /m mod f xm





is an isometric embedding. Proof. Assume that m, m ∈ M r and m /m ∈ M r . If m = 1, the pair (m, m) satisfies the condition (ii) by Theorem 3.1. Hence f (xm ) has an irreducible factor of degree mn = deg f (xm ) by Theorem 1.1. Then f (xm ) is irreducible. Moreover, ε1,m : (F2n , d f ) → (F2mn , d f (xm ) ) : x mod f (x) → xm mod f (xm ) is an isometric embedding. In particular, we have r ∗ = | x mod f (xm ) | = mr. Put n∗ = deg f (xm ) = mn and k∗ = m∗ = m /m. Since P r ∗ = P r , the pair (k∗ , m∗ ) satisfies that k∗ is odd, (n∗ − 1)k∗ < m∗ n∗ and | 2 mod k∗ r ∗ | = m∗n∗ by Theorem 3.1. By Theorem 1.1, there is an irreducible polynomial g (x) ∈ F2 [x] ∗ of degree m∗ n∗ = m n such that ε : (F2n∗ , d f (xm ) ) → (F2m∗ n∗ , d g ) : x mod f (xm ) → xk mod g (x) is an 

isometric embedding. Since g (x) = f (xm ), we have

εm,m = ε . 2

For (F2n , d f ) and r = | x mod f (x) |, if P r = ∅ then we obtain a system of isometric embeddings F nm with respect to {εm,m }. Then F is {εm,m } by Proposition 4.1. Put the inductive limit F = lim − → 2 the composition of all F2nm (m ∈ M r ), and Gal( F /F2n ) p ∈ P r Z p where Z p denotes the additive group of the ring of p-adicintegers (cf. e.g. [4]). The system {εm,m } induces a Hamming weight   F = lim − →   f (xm ) on the ( p ∈ P r Z p )-extension F /F2n such that α  F = α  f (xm ) if α ∈ F2mn . This definition of α  F does not depend on the choice of m. Example 4.1. We consider (F22 , d f ). Then f (x) = Φ 3 (x) = x2 + x + 1, r = | x mod f (x) | = 3, P r = {3} and Q r = ∅. For any m = 3i ∈ M r , f (xm ) = Φ 3i+1 (x) is an irreducible polynomial of even degree n∗ =

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2 · 3i , and r ∗ = | x mod f (xm ) | = 3i +1 . Hence P r ∗ = {3} and Q r ∗ = ∅. By Theorem 1.1, Theorem 3.1 and Proposition 4.1, there exists uniquely a tower

F22 ⊂ F22·3 ⊂ F22·32 ⊂ · · · ⊂ F22·3i ⊂ · · ·

dΦ 3

dΦ

32

dΦ

of isometric embeddings over F22 . Then F = weight   F = lim − →  Φ i+1 .

dΦ

33

 i

3 i +1

F22·3i is the Z3 -extension of F22 with a Hamming

3

5. Examples 5.1. Isometries for odd characteristics Only in this subsection, we consider the case where the characteristic p = |F p | is odd. Then we can find many examples of nontrivial isometric isomorphisms ε : (F pn , d f ) (F pn , d g ) as follows. We denote by A ε the presentation matrix such that (1, ε (x), . . . , ε (xn−1 )) = (1, x, . . . , xn−1 ) A ε where x = x mod f (x) and x = x mod g (x). −n f (cx) = f (x), then For an irreducible polynomial f (x) ∈ F p [x] of degree n and c ∈ F× p , if g (x) = c

ε : (F pn , d f ) (F pn , d g ) : x → cx is an isometric isomorphism with a diagonal presentation matrix A ε = diag(1, c , . . . , cn−1 ). Suppose that f (x) is an irreducible polynomial of the form f (x) = xn − a. Then, since x p − a = (x − a) p , p does not divide n. For each i ∈ {1, 2, . . . , n − 1}, there exists uniquely j ∈ {1, 2, . . . , n − 1} such that ip ≡ j (mod n). Since xip ≡ a(ip − j )/n x j (mod f (x)), the Frobenius automorphism

ι : (F pn , d f ) (F pn , d f ) : x → x p is isometric, and hence nal.

ε ◦ ι is also isometric. If p ≡ 1 (mod n), A ι and A ε◦ι = A ι A ε are not diago-

Example 5.1. Let a be a generator of the cyclic group F× p. −2 2 (1) For f (x) = (x + 1)2 − a and  c = −1, we have g (x) = c f (cx) = (x − 1) − a = f (x). Then

isometric and A ε =

1 0 0 −1

ε is

. If p = 5 and a = 2 ∈ F5 , then f (x) = x2 + 2x + 4 and g (x) = x2 − 2x + 4.

2 ( p −1)/2 = −1, (2) Assume that p ≡ 1 (mod 4), and put f (x) = x2 − a and c = a( p −1)/4 . Since  c = a 10 − 2 2 . Moreover, since we have g (x) = c f (cx) = x + a = f (x). Then ε is isometric and A ε =

xp = ( x2 )( p −1)/2 x = a( p −1)/2 x = −x, the Frobenius automorphism 1 0 . If p = 5 and a = 2, then c = 2. 0 −1

0c

ι : x → −x is isometric and A ι =

(3) Assume that p ≡ 2 (mod 3). For f (x) = x3 − a and c = −1, we have g (x) = c −3 f (cx) = x3 + a = f (x). Then ε is isometric and

 Aε =

1 0 0

0 0 −1 0 0 1

 .

Y. Mizusawa, S. Nishikawa / Finite Fields and Their Applications 25 (2014) 134–145

143

Moreover, since x p = (x3 )( p −2)/3 x2 = a( p −2)/3 x2 and x2p = a(2p −4)/3 x4 = a(2p −1)/3 x, the Frobenius automorphism ι : x → a( p −2)/3 x2 , x2 → a(2p −1)/3 x is isometric and A ι is not diagonal. If p = 5 and a = 2, then

 Aι =

1 0 0 0 0 a( p −2)/3

0

a(2p −1)/3



 =

0

1 0 0 0 0 3 0 2 0



 and

A ε ◦ι =

1 0 0 0 0 2 0 2 0

 .

We obtain an isometric isomorphism

ε ◦ ι : (F53 , d f ) (F53 , d g ) : x → 2x2 where f (x) = x3 − 2, g (x) = x3 + 2 ∈ F5 [x]. 5.2. Embeddings of F8 We choose an irreducible polynomial f (x) = x3 + x + 1 ∈ F2 [x] of degree n = 3, and consider (F8 , d f ). Since f (x) is a primitive polynomial, we have r = |F× 8 | = 7. By Theorem 3.1 and Example 3.1, we obtain the pairs (5, 4), (7, 7), (29, 28) and (35, 28) as examples of (k, m) satisfying (ii). By using the PARI/GP calculator [2], we obtain the following examples of isometric embeddings ε . Put (k, m) = (5, 4). Then (n − 1)k = 10 < 12 = mn and | 2 mod kr | = | 2 mod 35 | = 12 = mn, and hence the condition (ii) is certainly satisfied. By Theorem 1.1, we obtain an isometric embedding

ε : (F8 , d f ) → (F84 , d g ) : x mod f (x) → x5 mod g (x) where g (x) = x12 + x11 + x10 + x8 + x5 + x4 + x3 + x2 + 1 = f (x5 )/(x3 + x2 + 1). The presentation matrix A ε such that (1, ε (x), ε (x2 )) = (1, x, . . . , x11 ) A ε where x = x mod f (x) and x = x mod g (x) are the transpose of



1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0

 .

For (k, m) = (7, 7), since | 2 mod kr | = | 2 mod 72 | = 21 = mn, the condition (ii) is certainly satisfied. By Theorem 1.1, we obtain an isometric embedding

ε : (F8 , d f ) → (F87 , d g ) : x mod f (x) → x7 mod g (x) where g (x) = f (x7 ) = x21 + x7 + 1. Put m = 28, k = 29 and k = 35. Then mn = 84, k r = 203 and kr = 245. Since (n − 1)k = 58 < 84, (n − 1)k = 70 < 84 and | 2 mod 203 | = | 2 mod 245 | = 84, the pairs (k , m) = (29, 28) and (k, m) = (35, 28) certainly satisfy (ii). For (k , m) = (29, 28), we obtain an isometric embedding

ε : (F8 , d f ) → (F828 , d g  ) : x mod f (x) → x29 mod g  (x) where g  (x) = x84 + x82 + x81 + x80 + x77 + x75 + x74 + x73 + x70 + x68 + x67 + x66 + x63 + x61 + x60 + x59 + x56 + x54 + x53 + x52 + x49 + x47 + x46 + x45 + x42 + x40 + x39 + x38 + x35 + x33 + x32 + x31 + x28 + x25 + x23 + x22 + x21 + x18 + x16 + x15 + x14 + x11 + x9 + x8 + x7 + x4 + x2 + x + 1 = f (x29 )/(x3 + x + 1). On the other hand, for (k, m) = (35, 28), we obtain

ε : (F8 , d f ) → (F828 , d g ) : x mod f (x) → x35 mod g (x)

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Y. Mizusawa, S. Nishikawa / Finite Fields and Their Applications 25 (2014) 134–145

Table 1 For even n. f (x)

r

x2 + x + 1

3

x4 + x + 1

15

x4 + x3 + x2 + x + 1

5

k=m

g (x)

3, 9, 27, 81

f (xk )

3, 5, 9, 15, 25, 27, 45, 75, 81 5, 25

x6 + x + 1

63

3, 7, 9, 21, 27, 49, 63, 81

x6 + x4 + x2 + x + 1

21

7, 49

x6 + x3 + 1

9

3, 9, 27, 81

x +x +x +x+1

255

x8 + x7 + x3 + x + 1

85

5, 17, 25, 85

x8 + x4 + x3 + x + 1

51

3, 9, 17, 27, 51, 81

x8 + x5 + x4 + x3 + 1

17

17

8

5

3

3, 5, 9, 15, 17, 25, 27, 45, 51, 75, 81, 85

Table 2 For odd n. f (x) x +x+1 3

x5 + x2 + 1

x +x+1 7

r

(k, m) 7

31

127

(5, 4) (7, 7) (11, 10) (25, 20) (29, 28) (35, 28) (49, 49) (53, 52) (59, 58) (77, 70) (83, 82) (13, 12) (19, 18) (29, 28) (31, 31) (37, 36) (53, 52) (59, 58) (67, 66) (83, 82) (11, 10) (13, 12) (19, 18) (37, 36) (53, 52) (59, 58) (61, 60) (67, 66) (83, 82)

g (x) f (xk )/(x3 + x2 + 1) f (xk ) f (xk )/(x3 + x + 1) f (xk )/(x15 + x10 + 1) f (xk )/(x3 + x + 1) f (xk )/(x21 + x14 + 1) f (xk ) f (xk )/(x3 + x + 1) f (xk )/(x3 + x2 + 1) f (xk )/(x21 + x7 + 1) f (xk )/(x3 + x2 + 1) f (xk )/(x5 + x4 + x3 + x2 + 1) f (xk )/(x5 + x4 + x2 + x + 1) f (xk )/(x5 + x3 + 1) f (xk ) f (xk )/(x5 + x4 + x3 + x + 1) f (xk )/(x5 + x4 + x3 + x2 + 1) f (xk )/(x5 + x4 + x2 + x + 1) f (xk )/(x5 + x3 + x2 + x + 1) f (xk )/(x5 + x4 + x3 + x2 + 1) f (xk )/(x7 + x6 + x5 + x2 + 1) f (xk )/(x7 + x3 + 1) f (xk )/(x7 + x6 + x5 + x4 + 1) f (xk )/(x7 + x6 + x4 + x2 + 1) f (xk )/(x7 + x5 + x3 + x + 1) f (xk )/(x7 + x6 + x5 + x4 + x3 + x2 + 1) f (xk )/(x7 + x6 + x5 + x3 + x2 + x + 1) f (xk )/(x7 + x4 + x3 + x2 + 1) f (xk )/(x7 + x5 + x2 + x + 1)

where g (x) = x84 + x77 + x70 + x56 + x35 + x28 + x21 + x14 + 1 = f (x35 )/(x21 + x14 + 1). Thus (F8 , d f ) is embedded into F828 in two ways.

Y. Mizusawa, S. Nishikawa / Finite Fields and Their Applications 25 (2014) 134–145

145

5.3. Embeddings of F2364 Put r = 1093. Then xr − 1 = (x − 1) f 1 (x) f 2 (x) f 3 (x) with three irreducible polynomials f i (x) ∈ F2 [x] of degree n = 364, and | x mod f i (x) | = r and | 2 mod r | = n (cf. [2]). Since n is even, Q r = ∅. Moreover, the prime number p = r = 1093 satisfies that









v p (r ) − v p 2 p −1 − 1 = 1 − 2 < 0 = v p (1) = v p  2 mod r / p v p (r )  , and hence P r = ∅. Therefore, by Theorem 1.1 and Theorem 3.1, (F2n , d f i ) cannot be embedded into any (F2mn , d g ) with m > 1. 5.4. Tables For each factor r of 2n − 1 such that r = | x mod f (x) | for some irreducible polynomial f (x) ∈ F2 [x] of degree n, we choose one of such f (x). In the ranges 2  n  8 and 1 < k < 100, by using the PARI/GP calculator [2], we give tables of (k, m) and g (x) which give isometric embeddings

ε : (F2n , d f ) → (F2mn , d g ) : x mod f (x) → xk mod g (x) (Table 1 for even n, and Table 2 for odd n). Note that k = m if n is even, and that g (x) = f (xk ) if k = m. Acknowledgments The authors thank Professor Masakazu Yamagishi for helpful discussions. This work was partially supported by KAKENHI (23540014); Grant-in-Aid for Scientific Research (C). The authors thank the referees for valuable comments and suggestions for the improvement of this paper. References [1] [2] [3] [4]

A. Fröhlich, M.J. Taylor, Algebraic Number Theory, Cambridge Stud. Adv. Math., vol. 27, Cambridge University Press, 1993. The Pari Group, PARI/GP Version 2.5.2, Bordeaux, 2008. J.H. van Lint, Introduction to Coding Theory, third ed., Grad. Texts in Math., vol. 86, Springer-Verlag, 1999. L.C. Washington, Introduction to Cyclotomic Fields, second ed., Grad. Texts in Math., vol. 83, Springer-Verlag, 1997.