MATEMATIQKI VESNIK

60 (2008), 3–8

UDK 517.984

originalni nauqni rad research paper

THE RESULTANT OF NON-COMMUTATIVE POLYNOMIALS

Aleksandra Lj. Erić

Abstract. Let R = K[x; σ] be a skew polynomial ring over a division ring K. Necessary and sufficient condition under which common right factor of two skew polynomials exists is established. It is shown that the existence of common factor depends on the value of non-commutative (Dieudonné) determinant built on coefficients of polynomials and their σ^{l}-images.

1. Introduction

The main purpose of this paper is to extend the well known criterion for existence of common factor of two polynomials. Our basic tool is the notion of resultant of polynomials in non-commutative sense which we will introduce in Section 2. Several examples are discussed in Section 3.

In Ore extension R = K[x; σ], one can “right divide” a polynomial f by another polynomial h ≠ 0 via an Euclidean algorithm: f = gh + r where either r = 0 or deg r < deg h. From this, it follows easily that R is a PLID (principal left ideal domain). However, it is also widely known that, if σ(K) ≠ K (i.e. when σ fails to be an automorphism of K), then left division does not work, and R is not a PRID (principal right ideal domain).

Let us recall some basic facts. The first is the Remainder Theorem [2]: f(x) = q(x)(x − a) + f(a) where q(x) is uniquely determined by f and by a. It follows that f is divisible by x − a iff f(a) = 0. In this case, we say that a is a right root of f. The second is the Product Formula [2] (Theorem 8.6.4) for evaluating f = gh at any a ∈ K:

f(a) = \{ \begin{matrix}0 & if h(a) = 0 \\ g(a^{h(a)})h(a) & if h(a) ≠ 0 \end{matrix}

where, for any c ∈ K^{∗}, a^{c} denotes a^{c} = σ(c)ac^{−1} + δ(c)c^{−1}, which is called the (σ, δ)-conjugate of a by c. We will denote σ(a) by a^{σ}.

The third is the Evaluating formula: for f(x) = ∑_{i} a_{i}x^{i} we have

f(a) = ∑_{i} a_{i}N_{i}(a) (a ∈ K)

AMS Subject Classification: 12E15

Keywords and phrases: Polynomial ring, skew polynomial, resultant.

where N_{0}(a) = 1 and inductively

N_{n+1}(a) = σ(N_{n}(a))a + δ(N_{n}(a)) (n ∈ N_{0}).

In the case when δ = 0, N_{n}(a) = σ^{n−1}(a)σ^{n−2}(a) ⋯ σ(a)a.

2. The resultant of non-commutative polynomials

First let us recall the well known criterion for existence of a common factor of two ordinary polynomials in the ring F [x], where F is a field:

Let f, g ∈ F [x], F be a field and deg f = m, deg g = n. Polynomials f, g have a common right factor iff there exists polynomials c, d ∈ F [x] such that:

cf = dg

where deg c < deg g and deg d < deg f.

In this section, we are going to extend this criterion to the skew polynomials with δ = 0. Although the ring R := K[x; σ, δ] is non-commutative, it is still a UFD (unique factorization domain) in the sense of [3] (page 28). Let us mention some more advanced facts about the skew polynomials.

Proposition 2.1. ([6], Theorem 2.2) (1) Let f ∈ K[x, σ]. Then R/Rf is a left K-vector space and dim R/Rf = deg f.

(2) Let f, g ∈ K[x, σ]. Then Rf/Rg is a left K-vector space and dim Rf/Rg = deg g − deg f.

Proposition 2.2. ([6], Theorem 2.2) Let E_{1} and E_{2} be submoduls of an R-modul E, when R is a ring. Then the sequence

0 ⟶ E_{1} ∩ E_{2} ⟶ E_{1} + E_{2} ⟶ (E_{1} + E_{2})/E_{1} ⊕ (E_{1} + E_{2})/E_{2} ⟶ 0

is exact.

Proposition 2.3. ([6], Theorem 2.2) Let f, g ∈ R = K[x; σ] and deg f > 0, deg g > 0. If k, h ∈ R so that Rf ∩ Rg = Rh and Rf + Rg = Rh, then

deg f + deg g = deg h + deg k.

Here is the main result of this section.

Theorem 2.4. Let f, g ∈ R = K[x; σ] and deg f = n, deg g = m. Polynomials f and g have a common right (nonunit) factor if and only if there exist polynomials c, d ∈ R such that cf = dg and deg c < m and deg d < n.

Proof. Since R is a PLID, we have Rf ∩ Rg = Rh, and Rf + Rg = Rk. From h ∈ Rf, it follows that h = cf for some c ∈ R, and similarly h ∈ Rg implies that h = dg for some d ∈ R.

Assume that polynomials f and g have common right (nonunit) factor k_{1}. Then k_{1} is a right factor of k and deg k ≥ deg k_{1} > 0. According to the Proposition 2.3., we have

deg k + deg h = deg f + deg g.  (1)

It follows that deg h < deg f + deg g which means that deg c < deg g, and deg d < deg f.

Conversely, suppose that there exist c, d ∈ R, deg c < m, deg d < n such that cf = dg = h_{1}. Then h_{1} ∈ Rh, so deg h ≤ deg h_{1} < m + n ≤ deg f + deg g. Thus it follows from equation (1) that deg k > 0. It means that k is nonunit and also a right factor of the polynomials f, g (according to the fact Rf + Rg = Rk).

Now, we are going to introduce a notion of resultant in non-commutative setting. If

f = ∑_{i=0}^{m} a_{i}x^{i}, g = ∑_{i=0}^{n} b_{i}x^{i}

and

c = ∑_{i=0}^{n−1} c_{i}x^{i}, d = ∑_{i=0}^{m−1} d_{i}x^{i}

are the polynomials from Theorem 2.4., then

cf = ∑_{k} ∑_{i=0}^{k} c_{i}a_{k−i}^{σ^{i}}x^{k}, dg = ∑_{k} ∑_{i=0}^{k} d_{i}b_{k−i}^{σ^{i}}x^{k}.

So the equality cf = dg means ∑_{i=0}^{k} c_{i}a_{k−i}^{σ^{i}} = ∑_{i=0}^{k} d_{i}b_{k−i}^{σ^{i}} for all 0 ≤ k < m + n.

For example, for m = n = 2 we get the system of linear equations with unknowns c_{0}, c_{1}, −d_{0}, −d_{1}:

c_{0}a_{0}  − d_{0}b_{0}  = 0
c_{0}a_{1} + c_{1}a_{0}^{σ} − d_{0}b_{1} − d_{1}b_{0}^{σ} = 0
c_{0}a_{2} + c_{1}a_{1}^{σ} − d_{0}b_{2} − d_{1}b_{1}^{σ} = 0
c_{1}a_{2}^{σ}  − d_{1}b_{2}^{σ} = 0

The determinant of this system is

| \begin{matrix}a_{0} & a_{1} & a_{2} & 0 \\ 0 & a_{0}^{σ} & a_{1}^{σ} & a_{2}^{σ} \\ b_{0} & b_{1} & b_{2} & 0 \\ 0 & b_{0}^{σ} & b_{1}^{σ} & b_{2}^{σ} \end{matrix} | .

In general, we get a system of m + n linear equations with m + n unknowns c_{0}, …, c_{m−1}, −d_{0}, …, −d_{n−1} with coefficients in division ring K ([4]). The determinant of the system is

| \begin{matrix}a_{0} & a_{1} & a_{2} & … & a_{n} & 0 & … & 0 \\ 0 & a_{0}^{σ} & a_{1}^{σ} & … & a_{n−1}^{σ} & a_{n}^{σ} & … & 0 \\ ⋮ & ⋮ & ... & ⋮ & ⋮ & ⋮ & ... \\ 0 & 0 & 0 & … & a_{0}^{σ^{n−1}} & a_{1}^{σ^{n−1}} & … & a_{n}^{σ^{n−1}} \\ b_{0} & b_{1} & b_{2} & … & b_{m} & 0 & … & 0 \\ 0 & b_{0}^{σ} & b_{1}^{σ} & … & b_{m−1}^{σ} & b_{m}^{σ} & … & 0 \\ ⋮ & ⋮ & ... & ⋮ & ⋮ & ⋮ & ... \\ 0 & 0 & 0 & … & b_{0}^{σ^{m−1}} & b_{1}^{σ^{m−1}} & … & b_{m}^{σ^{m−1}} \end{matrix} | .

This is an (m + n) determinant over division ring where m and n are degrees of polynomials f and g. It is also called the Dieudonné determinant ([5]) and it takes values in the factor group K^{∗}/[K^{∗}, K^{∗}] ∪ {0} where [K^{∗}, K^{∗}] is the commutator of the multiplicative group K^{∗} = K \{0}. We will denote this determinant by R(f, g), and we will call it the resultant of polynomials f and g.

If the system has a nontrivial solution, determinant must be zero. Conversely, if determinant is zero, then system has a nontrivial solution. So, there exist polynomials c, d ∈ R such that cf = dg and deg c < deg g, deg d < deg f.

Theorem 2.5. Polynomials f, g from R = K[x, σ] have a common (nonunit) right factor iff R(f, g) = 0.

Corollary 1. Let f, g ∈ R = K[x; σ], f(x) = a_{n}x^{n} + ⋯ + a_{0}, g(x) = x − d. Then

R(f, g) = | \begin{matrix}a_{0} & a_{1} & a_{2} & … & a_{n−1} & a_{n} \\ −d & 1 & 0 & … & 0 & 0 \\ 0 & −d^{σ} & 1 & … & 0 & 0 \\ ⋮ & ⋮ & ... & ⋮ & ⋮ & ⋮ \\ 0 & 0 & 0 & … & −d^{σ^{n−1}} & 1 \end{matrix} | .

We switch the first and the last row and then move columns cyclically. We get

R(f, g) = | \begin{matrix}1 & −d^{σ^{n−1}} & 0 & … & 0 & 0 \\ 0 & 1 & −d^{σ^{n−2}} & … & 0 & 0 \\ 0 & 0 & 1 & … & −d^{σ^{n−3}} & … \\ ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ \\ a_{n} & a_{n−1} & a_{n−2} & … & a_{1} & a_{0} \end{matrix} | .

We continue the procedure until we get a determinant which differs from unit in the last place ((n + 1, n + 1) entry). The last place will be

a_{0} + … + a_{n−1}d^{σ^{n−2}} ⋯ d^{σ}d + a_{n}d^{σ^{n−1}} ⋯ d^{σ}d

and that is f(d). So R(f, g) = 0 iff f(d) = 0. This means that f is divisible by x − d iff f(d) = 0.

Example 1. Let R = D[x] where D is the ring of real quaternions. First, we consider a pair of polynomials

f(x) = x^{2} − (i + j)x − ij = (x − j)(x − i),
g(x) = x^{2} − jx + (1 − ij) = (x + i − j)(x − i).

R(f, g) = | \begin{matrix}−ij & −(i + j) & 1 & 0 \\ 0 & −ij & −(i + j) & 1 \\ 1 − ij & −j & 1 & 0 \\ 0 & 1 − ij & −j & 1 \end{matrix} | .

This is a Dieudonné determinant. We get a determinant which differs from unit in the last place by left multiplying its rows. The last place is R(f, g) = 0, so the polynomials f, g have a common right factor.

Next, let us consider

f(x) = (x − j)^{2} = x^{2} − 2jx − 1,
g(x) = (x + i − j)(x − i) = x^{2} − jx + 1 − ij.

We get a determinant which differs from unit in the last place by left multiplying its rows. The last place is R(f, g) = −2 + 4i + [D^{∗}, D^{∗}] ≠ 0, so the polynomials f, g have no common right factor.

Example 2. Let R = K[x], where K is the field of fractions ([2]) of the skew polynomial ring C[t;¯ ], i.e. K = C(t;¯ ). For the pair of polynomials

f(x) = x^{2} + t^{−1},
g(x) = (t − 1)^{−1}x^{3} + (t^{2} − 1)(t − i)x^{2} + (t^{2} − t)^{−1}x + (t^{3} − t)^{−1}(t − i),

R(f, g) is

| \begin{matrix}t^{−1} & 0 & 1 & 0 & 0 \\ 0 & t^{−1} & 0 & 1 & 0 \\ 0 & 0 & t^{−1} & 0 & 1 \\ (t^{3} − t)^{−1}(t − i) & (t^{2} − t)^{−1} & (t^{2} − 1)(t − i) & (t − 1)^{−1} & 0 \\ 0 & (t^{3} − t)^{−1}(t − i) & (t^{2} − t)^{−1} & (t^{2} − 1)(t − i) & (t − 1)^{−1} \end{matrix} | .

This is a Dieudonné determinant and R(f, g) = 0, so the polynomials f and g have a common right factor. But polynomial f is irreducible, so f is a factor of the polynomial g. Really,

g(x) = ((t − 1)^{−1}x + (t^{2} − 1)(t + i))(x^{2} + t^{−1}).

Note that (t^{2} − 1)^{−1}(t + i)t^{−1} = (t^{3} − t)^{−1}(t − i).

Let R = K[x; 1; δ] where δ is a σ-derivation. R is a PLID and it has the algorithm of dividing. Thus we can apply the same argument as before.

For f, g ∈ R, where deg f = deg g = 2, R(f, g) is

| \begin{matrix}a_{0} & a_{1} & a_{2} & 0 \\ a_{0}^{δ} & a_{1}^{δ} + a_{0} & a_{2}^{δ} + a_{1} & a_{2} \\ b_{0} & b_{1} & b_{2} & 0 \\ b_{0}^{δ} & b_{1}^{δ} + b_{0} & b_{2}^{δ} + b_{1} & b_{2} \end{matrix} | .

In general, if f(x) = ∑_{0}^{n} a_{i}x^{i} and g(x) = ∑_{0}^{m} b_{i}x^{i}, than R(f, g) is a determinant of dimension (m + n).

The first row is (a_{0}, …,a_{n},0,...0), and the (i + 1)-th row is A_{i+1,j} (1 < j ≤ m − 1), where

A_{i+1,j} = ∑_{0}^{j−1} \binom{i}{i − k}a_{j−1−k}^{δ^{i−k}}

where \binom{i}{−l} = 0 for l > 0 and a_{i} = 0 for i > n and the m + 1-th row is (b_{0}, …,b_{m},0, …,0).

A_{i+1+m,j} = ∑_{0}^{j−1} \binom{i}{i − k}b_{j−1−k}^{δ^{i−k}}

where \binom{i}{−l} = 0 for l > 0 and b_{i} = 0 for i > m and 0 < i ≤ m − 1.

Example 3. Let K = R(t), R = K[x; 1;′] where ′ is the standard derivation. Then

xr(t) = r(t)x + r′(t)

for r(t) ∈ R(t). For the polynomials

f(x) = (x − \frac{1}{t})(x − t) = x^{2} − (t + \frac{1}{t}) and g(t) = x(x − t) = x^{2} − tx − 1,

R(f, g) = | \begin{matrix}0 & −(t + \frac{1}{t}) & 1 & 0 \\ 0 & −(1 − \frac{1}{t^{2}} ) & −(t + \frac{1}{t}) & 1 \\ −1 & −t & 1 & 0 \\ 0 & −2 & −t & 1 \end{matrix} | = 0,

so polynomials f, g have a common factor.

On the other hand, for

f(x) = (x − t)x = x^{2} − tx, g(x) = x(x − t) = x^{2} − tx − 1,

R(f, g) = −1, so the polynomials have no common factors.

REFERENCES

[1] O. Ore, Theory of non-commutative polynomials, Ann. Math. 34 (1933), 480–508.

[2] P. M. Cohn, Free Rings and Their Relations, 2nd Ed., London Math. Soc. Monograph No. 19, Academic Press, London, 1985.

[3] P. M. Cohn, Skew Fields. Theory of General Division Rings, Encyclopedia in Math., Vol. 57, Cambridge Univ. Press, Cambridge, 1995.

[4] E. Artin, Geometric Algebra. Interscience Publishers Ltd., London, 1957.

[5] J. Dieudonné, Les déterminants sur un corps non commutatif, Bull. Soc. Math. France 71 (1943), 27–45.

[6] J. B. Castillon, These de doctorat d’etat sciences mathematiques a la faculte des science de Montpellier, 1971.

(received 20.08.2006, in revised form 16.11.2007)

Faculty of Civil Engineering, University of Belgrade, Bulevar Kralja Aleksandra 73, 11000 Beograd, Serbia

E-mail: eric@grf.bg.ac.yu