INTEGRAL GROUP ALGEBRAS AND CYCLOTOMIC POLYNOMIALS ^{1}

Kaoru MOTOSE

Recently, using cyclotomic polynomials, Z. Marciniak and S. K. Sehgal [3] obtained excellent results about units in integral group rings of cyclic groups. In this paper, we shall give some improvements and alternative proofs of their results.

Let ℤG be the group algebra of a finite abelian group G over the ring ℤ of rational integers. It is well known that the units of finite order in ℤG have the form ±g for some g ∈ G (see [1], p. 262). We study the form of units of infinite order in ℤG where G = ⟨σ⟩.

Let Φ_{m}(x) be cyclotomic polynomial of order m defined inductively by

X^{m} − 1 = ∏_{d|m} Φ_{d}(x).

Z. Marciniak and S.K. Sehgal [3] construct many units of infinite order using cyclotomic polynomials. These units cover the alternative units, the Hoechsmann units [3] and Yamauchi’s results [4].

In this paper, we study the Euclidean algorithm for cyclotomic polynomials in ℤ[x], and we have easy applications to some their results in [3]. The following are well known units. Units in 1, 2 are covered by cyclotomic polynomials.

1. The alternating units:

Φ_{2k}(σ) = 1 − σ + σ^{2} − ⋯ + (−1)^{k}σ^{k}

where k is odd and (2k, |G|) = 1.

2. The Hoechsmann units (the constructible units) (see also K. Yamauchi [4]).

\frac{σ^{kℓ} − 1}{σ^{k} − 1} · \frac{σ − 1}{σ^{ℓ} − 1} = \frac{1 + σ^{k} + σ^{2k} + ⋯ + σ^{(ℓ−1)k}}{1 + σ + σ^{2} + ⋯ + σ^{ℓ−1}}

where k, ℓ ≥ 2, (kℓ, |G|) = 1 and (k, ℓ) = 1.

3. Bass cyclic units,

(1 + σ + ⋯ + σ^{k−1})^{m} − ℓ(1 + σ + ⋯ + σ^{|G|−1})

where k > 1 and k^{m} = 1 + ℓ|G|. Since the group algebra ℤG are isomorphic to ℤ[x]/(x^{n} − 1)ℤ[x], our study on units in

ℤG is equivalent to find polynomials f(x) ∈ ℤ[x] satisfying

f(x)u(x) + (x^{n} − 1)v(x) = 1, where u(x), v(x) ∈ ℤ[x].

For relatively prime polynomials f(x) and g(x) over a field K, it is easy to compute polynomials u(x), v(x) ∈ K[x] by Euclidean algorithm such that

f(x)u(x) + g(x)v(x) = 1.

1The detailed version of this paper will be submitted for publication elsewhere This paper was financially supported by Fund for the Promotion of International Scientific Research B-2, 2004, Aomori, Japan.

However, over ℤ[x], situation is different from this. Of course we can compute u(x), v(x) ∈ ℚ[x] by Euclidean algorithm for relatively prime polynomials f(x), g(x) ∈ ℤ[x]. Thus we have

f(x)u_{0}(x) + g(x)v_{0}(x) = a

where u_{0}(x), v_{0}(x) ∈ ℤ[x] and 0 ≠ a ∈ ℤ.

For example, we obtain for cyclotomic polynomials Φ_{3}(x) = x^{2} + x + 1, Φ_{6}(x) = x^{2} − x + 1,

Φ_{3}(x)(1 − x) + Φ_{6}(x)(x + 1) = 1 − x^{3} + 1 + x^{3} = 2

and we can easily show there is no polynomials u(x), v(x) ∈ ℤ[x] such that

Φ_{3}(x)u(x) + Φ_{6}(x)v(x) = 1.

In fact 1 = Φ_{6}(ω)v(ω) = −2ωv(ω) = −2ω̄v(ω̄) for two roots ω, ω̄ of Φ_{3}(x). We have a contradiction such that 1 = 4 · v(ω)v(ω̄) and v(ω)v(ω̄) is an integer.

Thus it is natural to consider the next problem. For given polynomials f(x), g(x) ∈ ℤ[x], does there exist polynomials u(x), v(x) ∈ ℤ[x] such that

f(x)u(x) + g(x)v(x) = 1 ?

It is easy for f(x) = x and g(x) = x^{n} − 1. But in general, it seems to be difficult for me because the ring ℤ[x] is not Euclidean though it is a unique factorization ring. In this paper, we shall answer to this problem in case f(x) and g(x) are cyclotomic polynomials for units in ℤG.

If m ≠ n, then we have Φ_{m}(x)u(x) + Φ_{n}(x)v(x) = 1 in ℚ[x] since Φ_{m}(x), Φ_{n}(x) are distinct irreducible polynomials in ℚ[x]. Over ℤ[x], we can see the next theorem.

Theorem 1. Assume n > m ≥ 1. Then we have

(1) If m is not a devisor of n, then there exist u(x), v(x) ∈ ℤ[x] such that

Φ_{m}(x)u(x) + Φ_{n}(x)v(x) = 1.

(2) If m is a divisor of n, then we set n = mk and k_{0} is the product of all distinct prime divisors k. There exist u(x), v(x) ∈ ℤ[x] such that

Φ_{m}(x)u(x) + Φ_{n}(x)v(x) = Φ_{k_{0}}(1).

Proof. (1) If we set n = mq + r, 0 ≤ r < m, then we have easily

x^{n} − 1 = (x^{m} − 1) · (\frac{x^{mq} − 1}{x^{m} − 1} · x^{r}) + x^{r} − 1.

Hence, we can use Euclidean algorithm in ℤ[x] and so

(x^{n} − 1)s(x) + (x^{m} − 1)t(x) = x^{d} − 1, for some s(x), t(x) ∈ ℤ[x]

where d = (n, m). Thus we have

\frac{x^{n} − 1}{x^{d} − 1}s(x) + \frac{x^{m} − 1}{x^{d} − 1}t(x) = 1.

Therefore, we obtain the next equation excluding cases m|n

Φ_{n}(x)u(x) + Φ_{m}(x)v(x) = 1 for some u(x), v(x) ∈ ℤ[x].

(2) Since x − 1 divides Φ_{k_{0}}(x) − Φ_{k_{0}}(1) in ℤ[x], we have x^{hm} − 1 and so Φ_{m}(x) divides Φ_{k_{0}}(x^{hm}) − Φ_{k_{0}}(1) where h = \frac{k}{k_{0}} . Let n_{0} be the product of all distinct prime divisors n. We set n_{0} = ℓk_{0} and

u(x) = \frac{Φ_{k_{0}}(1) − Φ_{k_{0}}(x^{hm})}{Φ_{m}(x)} and v(x) = ∏_{d|ℓ, d<ℓ} Φ_{k_{0}d}(x^{\frac{n}{n_{0}}} ).

Then u(x) and v(x) ∈ ℤ[x]. Noting \frac{n}{n_{0}} ℓ = \frac{k}{k_{0}} m = hm and (ℓ, k_{0}) = 1, we have

Φ_{m}(x)u(x) + Φ_{n}(x)v(x) = Φ_{m}(x)u(x) + Φ_{n_{0}}(x^{\frac{n}{n_{0}}} ) ∏_{d|ℓ, d<ℓ} Φ_{k_{0}d}(x^{\frac{n}{n_{0}}} )
= Φ_{k_{0}}(1) − Φ_{k_{0}}(x^{hm}) + Φ_{k_{0}}((x^{\frac{n}{n_{0}}} )^{ℓ})
= Φ_{k_{0}}(1).

Let m be a natural number and let q be a power of a prime with (q, m) = 1. Then we can see from Theorem 1 (2) that there exist u(x), v(x) ∈ ℤ[x] such that

Φ_{m}(x)u(x) + Φ_{mq}(x)v(x) = p.

However, the next proposition shows that p is the smallest positive integer satisfying the above equation.

Proposition 1. There exist no s(x), t(x) ∈ ℤ[x] such that

Φ_{m}(x)s(x) + Φ_{mq}(x)t(x) = 1

for a natural number m and a power q of a prime p with (q, m) = 1. Proof. Let Δ be the set of roots of Φ_{m}(x). Using ∏_{d|m} Φ_{dq}(x) = Φ_{q}(x^{m}) , we have the next

∏_{d|m} Φ_{dq}(η) = Φ_{q}(η^{m}) = Φ_{q}(1) = p

where η ∈ Δ. Thus

p^{|Δ|} = ∏_{η∈Δ} ∏_{d|m} Φ_{dq}(η) = ∏_{d|m} ∏_{η∈Δ} Φ_{dq}(η).

We set a_{d} = ∏_{η∈Δ} Φ_{dq}(η). Then a_{d} is an integer because a_{d} is a symmetric polynomial in ℤ[η ∈ Δ] and so a_{d} ∈ ℤ[coefficients of Φ_{m}(x)]. Hence we have from the above equation.

p^{|Δ|} = ∏_{d|m} |a_{d}| and |a_{d}| = p^{α(d)}

where α(d) is a nonnegative integer. Therefore we have

φ(m) = |Δ| = ∑_{d|m} α(d).

Using Möbius inversion formula, we obtain

α(m) = ∑_{d|m} φ(d)μ(\frac{m}{d}).

–30–

For a prime r,

α(r^{e}) = φ(r^{e}) − φ(r^{e−1}) = \{ \begin{matrix}r^{e−2}(r − 1)^{2} & for e ≥ 2, \\ r − 2 & for e = 1. \end{matrix}

Since φ(i) is multiplicative, α(i) is also multiplicative. Thus if α(i) = 0, then i = 2j and j is odd.

On the other hand, it follows from the assumption that Φ_{mq}(η)t(η) = 1 for η ∈ Δ and so a_{m} = ∏_{η∈Δ} Φ_{mq}(η) = ±1. Thus |a_{m}| = 1, and so α(m) = 0. This implies m = 2ℓ, ℓ is odd, and q > 2. Hence we have a contradiction for ℓ ≥ 3 by above arguments

1 = Φ_{2ℓ}(−x)s(−x) + Φ_{2ℓq}(−x)t(−x) = Φ_{ℓ}(x)s(−x) + Φ_{ℓq}(x)t(−x).

We have also a contradiction for ℓ = 1 by Φ_{2}(−1) = 0

1 = Φ_{2}(−1)s(−1) + Φ_{2q}(−1)t(−1) = pt(−1).

Remark 1. It follows from Φ_{m}(x^{p^{s}}) = Φ_{mp^{s}}(x)Φ_{m}(x^{p^{s−1}}) for (p, m) = 1 that

Φ_{mp^{s}}(x) ≡ Φ_{m}(x)^{p^{s−1}(p−1)} or Φ_{m}(x)^{p^{s}} mod p.

We can see from Theorem 1 and the above that the ideal of ℤ[x] generated by Φ_{m}(x), Φ_{n}(x) (m < n) can be calculated as follows:

(Φ_{m}(x), Φ_{n}(x)) = \{ \begin{matrix}(p, Φ_{m}(x)) & if m|n and \frac{n}{m} is a power of a prime p, \\ ℤ[x] & otherwise. \end{matrix}

The first part is an alternative proof of Proposition 1.

In the remainder of this paper, we consider our problem about x^{n} − 1 and Φ_{m}(x).

Theorem 2. Let m_{0} be the product of distinct prime divisors of m. If m_{0} is not a divisor of n, then there exist u(x), v(x) ∈ ℤ[x] such that

(x^{n} − 1)u(x) + Φ_{m}(x)v(x) = ∏_{d|(m_{0},n)} Φ_{\frac{m_{0}}{d}} (1).

Proof. We may assume that m = m_{0} from

Φ_{m}(x) = Φ_{m_{0}}(x^{\frac{m}{m_{0}}} ) and (x^{\frac{m}{m_{0}}} )^{n} − 1 = (x^{n} − 1) · \frac{(x^{n})^{\frac{m}{m_{0}}} − 1}{x^{n} − 1}.

We assume d is a divisor of n. If d is not a divisor of m, there exist u_{d}(x), v_{d}(x) ∈ ℤ[x] from Theorem 1 (1) such that

Φ_{d}(x)u_{d}(x) + Φ_{m}(x)v_{d}(x) = 1.

If d is a divisor of m, there exist u_{d}(x), v_{d}(x) ∈ ℤ[x] from Theorem 1 (2) such that

Φ_{d}(x)u_{d}(x) + Φ_{m}(x)v_{d}(x) = Φ\frac{m}{d} (1).

Thus we have from x^{n} − 1 = ∏_{d|n} Φ_{d}(x),

(x^{n} − 1)u(x) + Φ_{m}(x)v(x) = ∏_{d|(m,n)} Φ\frac{m}{d} (1).

Theorem 3 (Marciniak and Sehgal [3]). Let m_{0} be the product of distinct prime divisors of m. If t = \frac{m_{0}}{(n,m_{0})} > 1 is not a prime, there exist integral polynomials u(x), v(x) ∈ ℤ[x] such that

Φ_{m}(x)u(x) + (x^{n} − 1)v(x) = 1.

Proof. We may assume m = m_{0} from the same reason in Theorem 2. If t is not a prime, we have Φ\frac{m}{d} (1) = 1 for all d|(m, n) because \frac{m}{d} = \frac{m}{(m,d)} is not a prime since t = \frac{m}{(m,n)} is a divisor of \frac{m}{(m,d)} = \frac{m}{d}.

Remark 2. If t is a prime p, then we have

Φ_{m}(x)u(x) + (x^{n} − 1)v(x) = Φ_{t}(1) = p.

References

[1] C. W. Curtis and I. Reiner, Representation theory of finite groups and associative algebras, 1962, Interscience, New York.

[2] R. Lidl and H. Niederreiter, Finite fields, 1983, Addison Wesley.

[3] Z. Marciniak and S. K. Sehgal, Generic units in abelian group rings, J. Group Theory, to appear.

[4] K. Yamauchi, The construction of units of infinite order in the character ring of a finite group, Yokohama Math. J. 51 (2005), 89-97.

K. Motose Department of Mathematical System Science Faculty of Science and Technology Hirosaki University Hirosaki 036-8561, Japan

E-mail address: skm@cc.hirosaki-u.ac.jp