TENSOR PRODUCTS OF POLYNOMIALS

S.P. Glasby

School of Mathematics and Statistics University of Sydney N.S.W. 2006, Australia

1 June 1994

Abstract. This paper defines a binary operation ⊗ on the set R[X]^{∗} of nonzero polynomials over the integral domain R. The set of polynomials with nonzero constant term is a commutative monoid under ⊗ endowed with a natural involution. Let f and g be polynomials over F, the field of fractions of R. We give formulae expressing f ⊗ g as a product of irreducible polynomials over F – a problem which we show is similar to writing the tensor product of simple extensions of F as a direct sum of fields. Finally, we consider questions of unique ⊗-factorization.

1. Introduction

Given two algebras A and B over a field F, the most obvious way to construct an F -algebra of dimension dim_{F} A dim_{F} B is to form the tensor product A ⊗_{F} B. If f and g are two polynomials over an integral domain, there are two natural ways to compute a polynomial of degree deg f deg g. The first involves composing polynomials, while the second it a little known operation which we call the tensor product of polynomials. We start by giving a definition of this binary operation and develop some of its basic properties including finding a connection between tensor product and composition of polynomials. Furthermore, if α, β are elements of an algebraic extension of a field F, we relate the ⊗-product F (α) ⊗_{F} F (β) to the factorization of the ⊗-product m_{α/F} ⊗ m_{β/F} of minimum polynomials. Indeed, F (α) ⊗_{F} F (β) can be shown to be the direct sum of compositums of F (α) and F (β), and to each summand there corresponds a factor of m_{α/F} ⊗ m_{β/F} which is a power of an irreducible polynomial. Finally, we consider ⊗-factorizations and argue that we should only consider ⊗-factorizations of irreducible polynomials, and analogously ⊗-factorizations of simple algebraic extensions of F. We show that there are unique ⊗-factorization theorems (for irreducible polynomials) over some fields, but not for others. This may be seen as an attempt to find a generalized Krull-Schmidt theorem which holds for ⊗-products.

Typeset by AMS-TEX

1991 Mathematics Subject Classification. 12E05, 11T06.

2. Definition of ⊗-product

Let R be an integral domain with field of fractions F. A good example to keep in mind is R = ℤ[X_{1}, …, X_{n}] and F = ℚ(X_{1}, …, X_{n}). Let R[X]^{∗} = R[X]\{0} denote the set of nonzero polynomials. Given f(X) = f_{m}X^{m} + ⋯ + f_{1}X + f_{0} and g(X) = g_{n}X^{n} + ⋯ + g_{1}X + g_{0} in R[X]^{∗} with f_{m} ≠ 0 and g_{n} ≠ 0, let M denote the splitting field of fg and let α_{1}, …, α_{m} be the roots of f, and β_{1}, …, β_{n} the roots of g in M.

Definition. Let f, g ∈ R[X]^{∗} be as above. Define f ⊗g ∈ M[X] to be the following polynomial of degree mn

(f ⊗ g)(X) = f_{m}^{n}g_{n}^{m} ∏_{i=1}^{m} ∏_{j=1}^{n}(X − α_{i}β_{j}).

Note that if g(X) = g_{0} ≠ 0, then f ⊗ g = g_{0}^{m}. Furthermore, if f_{0} ≠ 0, then

(f ⊗ g)(X) = f_{m}^{n} ∏_{i=1}^{m} α_{i}^{n}g(X/α_{i}) = (−1)^{mn}f_{0}^{n} ∏_{i=1}^{m} g(X/α_{i}).

The definition of the tensor product of polynomials (for the case of monic polynomials over a field) occurred in [1] where the term “composed product” and the notation ∘ were used. We prefer to use the notation ⊗ given the connection with the tensor product of fields and matrices mentioned later. The following theorem shows f ⊗ g ∈ R[X]^{∗}.

Theorem 1. Let Y and G be the n × n matrices

Y = ( \begin{matrix}X & 0 & ⋯ & 0 \\ 0 & X & & 0 \\ ⋮ & & ... \\ 0 & 0 & & g_{n}X \end{matrix} ) and G = ( \begin{matrix}0 & 1 & & 0 \\ ⋮ & & ... \\ 0 & 0 & & 1 \\ −g_{0} & −g_{1} & ⋯ & −g_{n−1} \end{matrix} ) .

Then

(f ⊗ g)(X) = det ( \begin{matrix}Y & −G & 0 & ⋯ & 0 \\ 0 & Y & −G & & 0 \\ . & . & . \\ f_{0}G & f_{1}G & f_{2}G & ⋯ & f_{m−1}G + f_{m}Y \end{matrix} )

Hence f ⊗ g ∈ R[X].

Proof. Let

C(h) = ( \begin{matrix}0 & 1 & & 0 \\ ⋮ & & ... & 0 \\ 0 & 0 & & 1 \\ −h_{0} & −h_{1} & ⋯ & −h_{n−1} \end{matrix} )

denote the companion matrix of the monic polynomial h(X) = X^{n} + h_{n−1}X^{n−1} + ⋯ + h_{0} and let f̂ and ĝ denote the monic polynomials f_{m}^{−1}f and g_{n}^{−1}g.

Recall that C(f̂) ⊗ C(ĝ) is the mn × mn matrix which may be identified, using a certain choice of basis, with the n × n block matrix

C(f̂) ⊗ C(ĝ) = (C(f̂)[C(ĝ)]_{ij})

where [C(ĝ)]_{ij} denotes the (i, j)th entry of C(ĝ). By considering Jordan canonical forms, one may show that the eigenvalues of C(f̂) ⊗ C(ĝ) are α_{i}β_{j}, 1 ≤ i ≤ m, 1 ≤ j ≤ n. Therefore

∏_{i,j}(X − α_{i}β_{j}) = det(XI − C(f̂) ⊗ C(ĝ)).

Multiplying this equation by f_{m}^{n}g_{n}^{m} shows

(f ⊗ g)(X) = f_{m}^{n}g_{n}^{m} | \begin{matrix}XI & −Ĝ & 0 & ⋯ & 0 \\ 0 & XI & −Ĝ & & 0 \\ . & . & . \\ f_{0}f_{m}^{−1}Ĝ & f_{1}f_{m}^{−1}Ĝ & f_{2}f_{m}^{−1}Ĝ & ⋯ & f_{m−1}f_{m}^{−1}Ĝ + XI \end{matrix} |

where Ĝ = C(ĝ). The factor f_{m}^{n} may be absorbed into the determinant by multiplying the last n rows by f_{m}. Similarly, the factor g_{n}^{m} may be absorbed by multiplying rows n, 2m, …, mn by g_{n}. This changes XI and Ĝ to Y and G respectively, and completes the proof. □

3. Basic Properties of ⊗-product

The following theorem shows, amongst other things, that R[X]^{∗} with the operation ⊗, is a commutative monoid. Let S denote the submonoid of R[X]^{∗} comprising polynomials with nonzero constant term, then there are two natural self-inverse maps f ↦ f^{r} and f ↦ \overline{f} on S (actually f ↦ f^{r} preserves ⊗-multiplication). Define the reverse polynomial f^{r} of f to be

f^{r}(X) = (−X)^{m}f(X^{−1}) = (−1)^{m}(f_{0}X^{m} + f_{1}X^{m−1} + ⋯ + f_{m}).

and define \overline{f} to be \overline{f}(X) = (−1)^{m}f(−X).

Theorem 2. Let f, g, h ∈ R[X]^{∗}. Then

(a) f ⊗ g = g ⊗ f

(b) (f ⊗ g) ⊗ h = f ⊗ (g ⊗ h)

(c) f ⊗ (gh) = (f ⊗ g)(f ⊗ h)

(d) f ⊗ X^{n} = f_{m}^{n}X^{mn}

(e) f ⊗ (X − β) = f_{m}X^{m} + f_{m−1}βX^{m−1} + ⋯ + f_{0}β^{m} = (−X)^{m}f^{r}(X^{−1}β)

(f) f ⊗ (X − 1) = f

(g) (X + 1) ⊗ (X + 1) = X − 1

(h) \overline{f} = f ⊗ (X + 1), \overline{\overline{f}} = f and \overline{f} ⊗ \overline{g} = f ⊗ g

(i) (f^{r})^{r} = f and f^{r} ⊗ g^{r} = (f ⊗ g)^{r} if f, g ∈ S

(j) (\overline{f})^{r} = (−1)^{m}\overline{f^{r}}.

Proof. The proofs of (a), (d), (f) and (g) are immediate consequences of the definition of ⊗-products. Consider (b). Let h(X) = h_{p}X^{p} + ⋯ + h_{1}X + h_{0} have degree p and roots γ_{1}, …, γ_{p}. Then

(f ⊗ g) ⊗ h = (f_{m}^{n}g_{n}^{m})^{p}h_{p}^{mn} ∏_{i,j,k}(X − (α_{i}β_{j})γ_{k})

and

f ⊗ (g ⊗ h) = f_{m}^{np}(g_{n}^{p}h_{p}^{n})^{m} ∏_{i,j,k}(X − α_{i}(β_{j}γ_{k})).

Therefore ⊗ obeys the associative law. A similar argument proves (c), that ⊗ distributes over polynomial multiplication.

To prove (e), note that f ⊗ (X − β) is a polynomial of degree m with roots α_{1}β, …, α_{m}β and leading coefficient f_{m}. Setting β = −1 in (e) shows f ⊗(X +1) = \overline{f}. By (a), (b), (f) and (g) we have

\overline{\overline{f}} = f ⊗ (X + 1) ⊗ (X + 1) = f ⊗ (X − 1) = f

and

\overline{f} ⊗\overline{g} = f ⊗(X +1)⊗g⊗(X +1) = f ⊗(X +1)⊗(X +1)⊗g = f ⊗(X −1)⊗g = f ⊗g.

Consider the proof of (i). Since f, g ∈ S, so f_{0} ≠ 0 and g_{0} ≠ 0. Thus the roots α_{i} and β_{j} are nonzero, and therefore

f^{r}(X) = (−1)^{m}f_{0} ∏_{i=1}^{m}(X − α_{i}^{−1}) and g^{r}(X) = (−1)^{n}g_{0} ∏_{j=1}^{n}(X − β_{j}^{−1}).

Hence

(f^{r} ⊗ g^{r})(X) = f_{0}^{n}g_{0}^{m} ∏_{i,j}(X − α_{i}^{−1}β_{j}^{−1}).

It follows from the definition of ⊗-product that

(f ⊗ g)(0) = f_{m}^{n}g_{n}^{m}(−1)^{mn}(α_{1} ⋯ α_{m})^{n}(β_{1} ⋯ β_{n})^{m}
= f_{m}^{n}g_{n}^{m}(−1)^{mn}((−1)^{m}f_{0}f_{m}^{−1})^{n}((−1)^{n}g_{0}g_{n}^{−1})^{m}
= (−1)^{mn}f_{0}^{n}g_{0}^{m}.

Thus

(f ⊗ g)^{r}(X) = f_{0}^{n}g_{0}^{m} ∏_{i,j}(X − (α_{i}β_{j})^{−1})
= (f^{r} ⊗ g^{r})(X).

Finally, putting g = X + 1 in (f ⊗ g)^{r} = f^{r} ⊗ g^{r} shows (\overline{f})^{r} = (−1)^{m}\overline{f^{r}}. □

Evaluating tensor products at X = 1 yields connections with classical quantities such as resultants and discriminants. Recall that the resultant Res(f, g) is an element of R that is zero if and only if f and g have common roots in M, and the discriminant D(f) is zero if and only if f has repeated roots. Some well-known properties of Res(f, g) and D(f) such as those below, are proved in [7].

Res(f, g) = det ( \begin{matrix}f_{m} & f_{m−1} & ⋯ & f_{0} \\ & ... & & & ... \\ & & f_{m} & f_{m−1} & ⋯ & f_{0} \\ g_{n} & g_{n−1} & ⋯ & g_{0} \\ & ... & & & ... \\ & & g_{n} & g_{n−1} & ⋯ & g_{0} \end{matrix} )

= f_{m}^{n}g_{n}^{m} ∏_{i,j}(α_{i} − β_{j}) = f_{m}^{n} ∏_{i} g(α_{i}) = (−1)^{mn}g_{n}^{m} ∏_{j} f(β_{j}).

D(f) = f_{m}^{−1} Res(f, f′) = f_{m}^{2m−2} ∏_{i<j}(α_{i} − α_{j})^{2}.

The following lemma relates the determinant of the above (m + n) × (m + n) Sylvester matrix with the mn × mn determinant in Theorem 1.

Lemma 3. Let f, g ∈ R[X]^{∗}. If f_{0} ≠ 0, then Res(f, g) = (f^{r} ⊗ g)(1) and

(1)  (f^{r} ⊗ f)(X) = f_{0}f_{m}(X − 1)^{m}(−1)^{m(m−1)/2}h(X)

where

h(X) = f_{m}^{2m−2} ∏_{i<j}(α_{j}X − α_{i})(α_{j} − α_{i}X) ∈ R[X]

satisfies h(1) = D(f).

Proof.

(f^{r} ⊗ g)(X) = ((−1)^{m}f_{0} )^{n}g_{n}^{m} ∏_{i,j}(X − α_{i}^{−1}β_{j})

= (f_{m}α_{1} ⋯ α_{m})^{n}g_{n}^{m} ∏_{i,j}(X − α_{i}^{−1}β_{j})

= f_{m}^{n}g_{n}^{m} ∏_{i,j}(α_{i}X − β_{j})

When g = f, this expression may be rearranged to give (1). Setting X = 1 shows (f^{r} ⊗ g)(1) = Res(f, g). Clearly,

h(1) = f_{m}^{2m−2} ∏_{i<j}(α_{i} − α_{j})^{2} = D(f).

The proof that h(X) ∈ R[X] involves considering elementary symmetric polynomials in α_{1}, …, α_{m}. □

The assumption above that f_{0} ≠ 0 is not problematic given that Res(Xf, g) = g_{0}Res(f, g). Lemma 3 may be used in conjunction with Theorem 1 and facts such as (fg)^{r} = f^{r}g^{r} and f^{r}(1) = (−1)^{m}f(1) to deduce elementary properties of resultants such as Res(fg, h) = Res(f, h) Res(g, h) and Res(g, f) = (−1)^{mn}Res(f, g).

4. Composites and ⊗-products

Let f, g ∈ R[X]^{∗}. Then f ⊗ g and the composite polynomials f ∘ g and g ∘ f have degree mn. Furthermore, the coefficients of these polynomials bear some similarities to one another:

\begin{matrix}∑_{i=0}^{mn} a_{i}X^{i} & a_{mn} & a_{mn−1} & ⋯ & a_{0} \\ f ⊗ g & f_{m}^{n}g_{n}^{m} & −f_{m−1}f_{m}^{n−1}g_{n−1}g_{n}^{m−1} & ⋯ & (−1)^{mn}f_{0}^{n}g_{0}^{m} \\ f ∘ g & f_{m}g_{n}^{m} & m f_{m}g_{n−1}g_{n}^{m−1} & ⋯ & f(g_{0}) \\ g ∘ f & f_{m}^{n}g_{n} & n g_{n}f_{m−1}f_{m}^{n−1} & ⋯ & g(f_{0}) \end{matrix}

It is a pleasant surprise that Theorem 2(e) generalizes to give a connection between tensor products and composites of polynomials. More importantly, this generalization provides a faster method for computing ⊗-products than Theorem 1.

Theorem 4. Let Y, G and Ĝ be as in Theorem 1. Then

(f ⊗ g)(X) = det((−Y)^{m}f^{r}(Y^{−1}G))

= g_{n}^{m} det(f_{0}Ĝ^{m} + f_{1}XĜ^{m−1} + ⋯ + f_{m}X^{m}I). (2)

Proof. Note that Y^{−1} is a matrix over F (X), the field of fractions of R[X]. (Since g_{n} ≠ 0, Y is invertible.) By Theorem 1,

(f ⊗ g)(X) = det ( \begin{matrix}Y & −G & 0 & ⋯ & 0 \\ 0 & Y & −G & & 0 \\ . & . & . \\ f_{0}G & f_{1}G & f_{2}G & ⋯ & f_{m−1}G + f_{m}Y \end{matrix} )

Postmultiply column 1 by Y^{−1}G and add the result to column 2. Then postmultiply column 2 by Y^{−1}G and add the result to column 3, etc. Then

(f ⊗ g) = det ( \begin{matrix}Y & 0 & ⋯ & 0 \\ 0 & Y & & 0 \\ . & . & . \\ H_{1} & H_{2} & ⋯ & H_{m} \end{matrix} )

where

H_{i} = Y (f_{i−1}(Y^{−1}G) + ⋯ + f_{0}(Y^{−1}G)^{i}) if 1 ≤ i < m, and
H_{m} = Y (f_{m}I + f_{m−1}(Y^{−1}G) + ⋯ + f_{0}(Y^{−1}G)^{m}) = (−1)^{m}Y f^{r}(Y^{−1}G).

Therefore,

(f ⊗ g)(X) = det(Y^{m−1}H_{m}) = det((−Y)^{m}f^{r}(Y^{−1}G)).

Since Y^{−1}G = X^{−1}Ĝ and X^{−1}I and Ĝ commute, the right-hand side simplifies to g_{n}^{m} det(f_{0}Ĝ^{m} + f_{1}XĜ^{m−1} + ⋯ + f_{m}X^{m}I). □

Now the complexity of multiplying two n × n matrices is (for practical purposes) O(n^{3}), and the complexity of computing an n×n determinant is \frac{1}{3}n^{3}+O(n^{2}). Hence the complexity of computing (2) is (m+ \frac{1}{3})n^{3} +O(n^{2}). By contrast, the complexity of computing f ⊗g via Theorem 1 is \frac{1}{3}(mn)^{3}+O((mn)^{2}). (One practical application of Theorem 4 is to the computation of large degree irreducible polynomials over finite fields, see [3].)

5. Tensor Products of Power Series

The ⊗-product of two monic polynomials is again monic. If f and g are two power series whose constant terms are 1, it is natural to normalize their tensor product so its constant term is also 1. Towards this end define the binary operation ⊙ on polynomials by f ⊙ g = (−1)^{mn}f ⊗ g. If f_{0} = g_{0} = 1 and m, n ≥ k, then the coefficient of X^{k} in f ⊙ g is a polynomial in f_{1}, …, f_{k}, g_{1}, …, g_{k}. The polynomials when k = 0, 1, 2, 3 are

1, −f_{1}g_{1}, f_{1}^{2}g_{2}−2f_{2}g_{2}+f_{2}g_{1}^{2}, and (−f_{1}^{3}−3f_{3}+3f_{1}f_{2})g_{3}−f_{1}f_{2}g_{1}g_{2}+3f_{3}g_{1}g_{2}−f_{3}g_{1}^{3},

respectively. If 1+XR[[X]] denotes the set of power series (with coefficients from R) and constant term 1, then the binary operation ⊙ extends naturally to 1+XR[[X]]. (If f ∈ 1 + XR[[X]], let f̂_{k} denote the polynomial 1 + ∑_{i=1}^{k} f_{i}X^{i}. Then the coefficients of 1, X, …, X^{k} of f ⊙ g are defined to be the same as f̂_{k} ⊙ ĝ_{k}.)

If ⊕ denotes multiplication in the abelian group 1 + XR[[X]], then 1 + XR[[X]] becomes a ring with addition ⊕ and multiplication ⊙. Indeed, 1 + XR[[X]] is an integral domain where zero and one are the power series 1 and 1 − X, respectively (cf. [5]). The polynomials in 1 + XR[[X]] form a subring. We shall show that for certain choices of R this subring is a unique factorization domain.

Given f, g ∈ 1 + XR[[X]], it can be difficult in practice to compute f ⊙ g. It is quite easy to show that

(1 − rX) ⊙ ∑_{i=0}^{∞} f_{i}X^{i} = ∑_{i=0}^{∞} f_{i}(rX)^{i} and (1 − X^{m})^{−1} ⊙ (1 − X^{n})^{−1} = (1 − X^{e})^{d}

where (1 − X^{m})^{−1} denotes the power series ∑_{i=0}^{∞} X^{mi}, e = lcm(m, n) and d = gcd(m, n) (see Section 8). If the characteristic of R is zero and r, ρ ∈ R, one may define (1 − rX)^{ρ} and e^{rX} to be ∑_{i=0}^{∞} \binom{ρ}{i}(rX)^{i} and ∑_{i=0}^{∞}(rX)^{i}/i! where \binom{ρ}{i} = ρ(ρ − 1) ⋯ (ρ − i + 1)/i!. Given r, s, ρ, σ ∈ R one may prove

e^{rX}⊙e^{sX} = e^{−rsX}, (1−rX)^{ρ}⊙(1−sX)^{σ} = (1−rsX)^{ρσ} and (1−rX)^{ρ}⊙e^{sX} = e^{rsρX}.

We shall omit these proofs and return to considering polynomials.

6. Field Extensions

A natural and fundamental question is: How does f ⊗ g factor into irreducibles over F? It follows from the law f ⊗ (gh) = (f ⊗ g)(f ⊗ h), that it is sufficient to answer this question when both f and g are irreducible over F.

Let α be one of α_{1}, …, α_{m} and β one of β_{1}, …, β_{n}. Denote the minimum polynomial of α over F by m_{α/F} or, when there is no confusion over the field, by m_{α}. Henceforth assume that f = m_{α} and g = m_{β} are irreducible over F and monic. In general, m_{α} ⊗ m_{β} is reducible. For example, if α ∉ F, and β = α^{−1}.

If m_{α} ⊗ m_{β} is irreducible, then it equals m_{αβ/F}. Therefore [F (αβ) : F] = mn holds, and one may deduce from the following diagram

• •  • • •  • • FF (α) ∩ F (β)F (β) F (α, β) M F (α)  F (αβ) ≤ m/d ≤ n/d  n/d m/d  d

Figure 1. Subfields of F (α, β) and degrees of field extensions

that F (α) ∩ F (β) = F, F (α, β) = F (αβ), [F (α, β) : F (α)] = n and [F (α, β) : F (β)] = m. The following theorem is now straightforward to prove.

Theorem 5. The following are equivalent

(a) m_{α/F} ⊗ m_{β/F} is irreducible over F,

(b) [F (αβ) : F] = mn,

(c) m_{β/F} is irreducible over F (α),

(d) m_{α/F} is irreducible over F (β).

In general, it is hard to determine whether m_{α} ⊗ m_{β} equals m_{αβ}. However, if [F (α, β) : F] = mn, this happens precisely when F (α, β) = F (αβ).

Theorem 6. If [F (α, β) : F] = mn, then

(m_{α/F} ⊗ m_{β/F})(X) = m_{αβ/F} (X)^{[F (α,β):F (αβ)]}.

Proof. Let f = m_{α/F} and g = m_{β/F}. Then f_{m} = g_{n} = 1, so f = f̂, g = ĝ and G = Ĝ. Now 1, β, …, β^{n−1} is a basis for F (β) over F. Since [F (α, β) : F] = mn, it follows from Figure 1 that 1, α, …, α^{m−1} is a basis for F (α, β) over F (β). Hence

1, β, …, β^{n−1}, α, αβ, …, αβ^{n−1}, …, α^{m−1}, α^{m−1}β, …, α^{m−1}β^{n−1}

is a basis for F (α, β) over F. Relative to this basis, multiplication by αβ has matrix

( \begin{matrix}0 & G & 0 & … & 0 \\ 0 & 0 & G & & 0 \\ . & . & . \\ −f_{0}G & −f_{1}G & −f_{2}G & ⋯ & −f_{m−1}G \end{matrix} )

where G = C(g). The above matrix equals C(f) ⊗ C(g), and its characteristic polynomial equals m_{αβ/F} (X)^{[F (α,β):F (αβ)]}. Since m_{αβ/F} is monic, Theorem 1 gives the desired result. □

Corollary 7. If [F (α, β) : F] = mn and m_{α/F} ⊗m_{β/F} is square-free, then m_{α/F} ⊗ m_{β/F} is irreducible over F.

The hypothesis [F (α, β) : F] = mn holds automatically in the case when m and n are coprime. The following theorem has as a corollary, the well-known and important fact that if F is a finite field and gcd(m, n) = 1, then m_{α/F} ⊗ m_{β/F} is irreducible over F of degree mn (see [3]).

Theorem 8. Let d = gcd(m, n) and e = lcm(m, n). Suppose that M:F is Galois where Γ(M : F) is cyclic and generated by η. Then

m_{α/F} ⊗ m_{β/F} = h_{0}h_{1} ⋯ h_{d−1}

where h_{k} is a polynomial over F of degree e. Furthermore, h_{k}(X) = p_{k}(X)^{e_{k}} where p_{k}(X) is irreducible of degree e_{k}, and e_{k} divides d. (In particular, when m_{α/F} ⊗ m_{β/F} is square-free, the polynomials h_{k}(X) are irreducible and pairwise coprime.)

Proof. Set α_{i} = αη^{i} and β_{i} = βη^{i}. Then α_{i}, i = 0, …, m − 1, are distinct and α_{m} = α_{0}, and similarly β_{i}, i = 0, …, n − 1, are distinct and β_{n} = β_{0}. Therefore m_{α/F}, m_{β/F} and m_{α/F} ⊗ m_{β/F} equal

∏_{i=0}^{m−1}(X − α_{i}), ∏_{i=0}^{n−1}(X − β_{i}) and ∏_{i=0}^{m−1} ∏_{j=0}^{n−1}(X − α_{i}β_{j})

respectively. Define h_{k} to be the polynomial h_{k}(X) = ∏_{i=0}^{e−1}(X − α_{i}β_{i+k}) where the subscripts for α (respectively β) are read modulo m (respectively n). Since (α_{e−1}β_{e−1+k})η = α_{e}β_{e+k} = α_{0}β_{k}, it follows that h_{k}(X) ∈ F [X]. In addition, h_{k} = h_{k+d}. Note that if M, N ∈ ℤ satisfy Mm + Nn = d, then

α_{Mm+i}β_{Mm+i+k} = α_{Mm+i}β_{d−Nn+i+k} = α_{i}β_{i+k+d}.

Furthermore, m_{α/F} ⊗ m_{β/F} = h_{0}h_{1} ⋯ h_{d−1}. (Note that h_{0}h_{1} ⋯ h_{d−1} has de = mn roots, counting multiplicities. Also, α_{i}β_{j} is a root of h_{j−i}, and if j − i ≡ k (mod d) where 0 ≤ k < d, then h_{j−i} = h_{k}.)

In general, the roots α_{i}β_{j}, 1 ≤ i ≤ m, 1 ≤ j ≤ n, are not all distinct (i.e. m_{α/F} ⊗ m_{β/F} is not square-free). Let E = e_{k} be the smallest positive integer such that α_{E}β_{E+k} = α_{0}β_{k}. It follows that α_{i+E}β_{i+E+k} = α_{i}β_{i+k} holds for all i, and α_{i+f}β_{i+f+k} ≠ α_{i}β_{i+k} for 1 ≤ f < E. Thus the roots of h_{k} fall into orbits of length e_{k} and so h_{k}(X) = p_{k}(X)^{e_{k}} where p_{k}(X) = ∏_{i=0}^{e_{k}−1}(X − α_{i}β_{i+k}) is the minimum polynomial of α_{0}β_{k}. Thus p_{k}(X) is irreducible over F.

We now show that if d = 1, then m_{α/F} ⊗ m_{β/F} is irreducible over F. It suffices to prove that the α_{i}β_{j}, 1 ≤ i ≤ m, 1 ≤ j ≤ n, are all distinct. Suppose that two are equal, say α_{i}β_{j} = α_{k}β_{ℓ}. Thus α_{i}α_{k}^{−1} = β_{ℓ}β_{j}^{−1} = λ lies in the fixed field of η^{m} and of η^{n}. Hence λ lies in the fixed subfield of η, i.e. in F. Thus α_{i} = λα_{k} or α_{0} = λα_{k−i}, and similarly β_{0} = λβ_{j−ℓ}. Thus α_{0} = λ^{m}α_{m(k−i)} = λ^{m}α_{0} and β_{0} = λ^{n}β_{0}. Therefore, λ^{m} = λ^{n} = 1 and so λ = 1. Because the roots of m_{α} and m_{β} are distinct, this shows that i ≡ k (mod m) and j ≡ ℓ (mod n), as desired.

Now let f_{k}(X) = ∏_{i=0}^{m/d−1}(X − α_{di+k}) and g_{k}(X) = ∏_{i=0}^{n/d−1}(X − β_{di+k}). Then f_{k} and g_{k} are irreducible over the fixed subfield K of η^{d}. Furthermore,

m_{α/F} = f_{0}f_{1} ⋯ f_{d−1} and m_{β/F} = g_{0}g_{1} ⋯ g_{d−1}

and f_{d} = f_{0} and g_{d} = g_{0}. Since deg f_{i} = m/d and deg g_{j} = n/d are coprime, it follows from the previous paragraph that the f_{i} ⊗ g_{j} are irreducible over K of degree (m/d)(n/d) = e/d and m_{α/F} ⊗m_{β/F} = ∏_{i,j} f_{i} ⊗g_{j} has d^{2} irreducible factors over K. Suppose that p_{k} factors as a product of r_{k} irreducible factors over K. Then deg p_{k} = r_{k}e/d. However, e = e_{k} deg p_{k} so e_{k}r_{k} = d. □

The polynomials h_{0}, h_{1}, …, h_{d−1} in Theorem 8 need not be distinct. For example, if α = β, then e = d = m and h_{k} = h_{−k}. Furthermore, the exponents e_{k} may be greater than 1. For example, if α = β and m is even, then

h_{m/2}(X) = ( ∏_{i=0}^{m/2−1}(X − α_{i}α_{i+m/2}))^{2} .

Also, if α = β and xη = x^{2} for all x ∈ M, then h_{0}(X) = m_{α/F} (X), so e_{0} = 1.

7. Tensor Products of Fields

We relate the problem of factoring m_{α/F} ⊗ m_{β/F} into irreducible factors over F to the problem of writing F (α) ⊗ F (β) as as a direct sum of fields.

Let m_{α/F} (X) = ∏_{i=1}^{m}(X −α_{i}) and m_{β/F} (X) = ∏_{i=1}^{n}(X −β_{i}) be separable polynomials over F. (Note that an irreducible polynomial is automatically separable if the characteristic of F does not divide its degree, or if F is finite.) Let M be the splitting field of the separable polynomial m_{α}m_{β}. Then M : F is a Galois extension. Let A and B denote the diagonal matrices A = diag(α_{1}, …, α_{m}) and B = diag(β_{1}, …, β_{n}) over M. Consider the F -algebra monomorphism

ϕ: F (α) ⊗_{F} F (β) ⟶ M_{mn}(M) defined by ∑_{i,j} a_{i,j}α^{i} ⊗ β^{j} ↦ ∑_{i,j} a_{i,j}A^{i} ⊗ B^{j}

where a_{i,j} ∈ F.

Recall that

A^{k} ⊗ B^{ℓ} = diag(α_{1}^{k}β_{1}^{ℓ}, …, α_{m}^{k}β_{1}^{ℓ}, …, α_{1}^{k}β_{n}^{ℓ}, …, α_{m}^{k}β_{n}^{ℓ}).

Let ϕ_{i}: F (α) ⊗_{F} F (β) ⟶ M map x to the ith diagonal entry of xϕ. Then im ϕ_{(j−1)m+i} = F (α_{i}, β_{j}). If σ ∈ Γ(M : F), then σ permutes the roots of m_{α} and m_{β}, and so permutes the fields F (α_{i}, β_{j}). Now Γ(M : F) acts transitively on the roots of m_{β}, and so by renumbering α_{1}, …, α_{m} if necessary, we may assume that the orbit representatives of the fields F (α_{i}, β_{j}), 1 ≤ i ≤ m, 1 ≤ j ≤ n, under Γ(M : F) are F (α_{1}, β), …, F (α_{s}, β). Thus s ≤ m (and similarly s ≤ n).

Let E_{(j−1)m+i} denote the field F (α_{i}, β_{j}). Suppose that β_{1} = β and consider the F -algebra homomorphism Φ : F (α) ⊗_{F} F (β) ⟶ ⊕_{i=1}^{s}E_{i} defined by xΦ = (xϕ_{1}, …, xϕ_{s}). If x ∈ ker Φ, then x ∈ ker ϕ_{i} for 1 ≤ i ≤ s and so x ∈ ker ϕ_{i} for 1 ≤ i ≤ mn. Thus x ∈ ker ϕ = {0} and so Φ is injective. Now im Φ is an F -subalgebra of ⊕_{i=1}^{s}E_{i} and so it is semisimple. By the Wedderburn-Artin theorem im Φ is isomorphic to a direct sum of some of the fields E_{i}, 1 ≤ i ≤ s. Now im Φ projects onto each summand E_{i} and so if im Φ were a proper F -subalgebra of ⊕_{i=1}^{s}E_{i}, then it follows that there is an F -isomorphism E_{i} ⟶ E_{j} where i ≠ j and 1 ≤ i, j ≤ s. Since M : F is normal, such an F -isomorphism can be extended to an F -automorphism of M mapping E_{i} to E_{j}. This is a contradiction as E_{1}, …, E_{s}, come from different orbits under Γ(M : F). Hence Φ is an F -isomorphism.

Consider the characteristic polynomial of the transformation induced by multiplication by (α ⊗ β)Φ on ⊕_{i=1}^{s}E_{i}. By Theorem 6 it equals m_{α_{i}β}(X)^{e_{i}} where e_{i} = [F (α_{i}, β), F (α_{i}β)]. Therefore m_{α} ⊗ m_{β} = ∏_{i=1}^{s} m_{α_{i}β}(X)^{e_{i}}, and we have proved

Theorem 9. Let m_{α/F} and m_{β/F} be separable polynomials and let M be the splitting field of m_{α/F} m_{β/F}. Let F (α_{1}, β), …, F (α_{s}, β) be the representatives of different orbits of {F (α_{i}, β_{j}) | 1 ≤ i ≤ m, 1 ≤ j ≤ n} under Γ(M : F). Then

F (α) ⊗_{F} F (β) ≅ ⊕_{i=1}^{s} F (α_{i}, β) and m_{α} ⊗ m_{β} = ∏_{i=1}^{s} m_{α_{i}β}(X)^{e_{i}}

where e_{i} = [F (α_{i}, β), F (α_{i}β)].

The polynomials m_{α_{i}β}(X), 1 ≤ i ≤ s, in Theorem 9 need not be distinct as the following example shows.

Example. Suppose that q is a prime-power and that p is a prime such that ord_{q}(p) = p−1 (i.e. p has order p−1 modulo q). Then Φ_{p}(X) = X^{p−1} +⋯+X +1 is irreducible over F = 𝔽_{q} and 𝔽_{q^{p}} = F (α) where α is a root of Φ_{p}(X). Thus m_{α/F} = Φ_{p} and

F (α) ⊗_{F} F (α) ≅ ⨁_{i=1}^{p−1} F (α^{i}, α) = ⨁_{i=1}^{p−1} F (α).

Furthermore, e_{i} = [F (α^{i}, α) : F (α^{i+1})] equals 1 if i ≠ p − 1 and p − 1 otherwise, and m_{α^{i+1}/F} (X) = m_{α/F} (X) if i ≠ p − 1 and X − 1 otherwise.

Let A be an F -algebra with 1, and let B and C be F -subalgebras (with the same 1) of A. Suppose that B and C generate A as an F -algebra, dim_{F} A = dim_{F} B dim_{F} C and bc = cb for all b ∈ B and c ∈ C. Then the map B ⊗_{F} C ⟶ A defined by b ⊗ c ↦ bc is an F -algebra isomorphism. In this case A is called an internal tensor product of B and C, and one can show that B ∩ C equals F · 1.

Theorem 10. Let D = F (α) ∩ F (β) and suppose that the extensions F (α) : D and D : F are both Galois. Then F (α) ⊗_{F} F (β) is isomorphic to a direct sum of [D : F] copies of the field F (α, β).

Proof. Note that F (α), F (β) and F (α, β) are equal to D(α), D(β) and D(α, β) respectively. Since D(α) : D is Galois, so is D(α, β) : D(β) and

Γ(D(α, β) : D(β)) ≅ Γ(D(α) : D).

Thus [D(α, β) : D(β)] = [D(α) : D] and so [D(α, β) : D] = [D(α) : D][D(β) : D]. By the above remarks D(α, β) is an internal ⊗-product D(α) ⊗_{D} D(β).

Since D(α) : D and D : F are Galois, so is D(α) : F. Let σ ∈ Γ(D(α) : F) and consider the F -algebra homomorphism

ϕ_{σ}: F (α) ⊗_{F} F (β) ⟶ D(ασ, β) defined by ∑ a_{i,j}α^{i} ⊗ β^{j} ↦ ∑ a_{i,j}(ασ)^{i}β^{j}

where a_{i,j} ∈ F. Clearly im ϕ_{σ} = D(ασ, β) = D(α, β) as ασ ∈ D(α). Therefore ker ϕ_{σ} is a maximal ideal of F (α) ⊗_{F} F (β). By Theorem 9, F (α) ⊗_{F} F (β) is a direct sum of s fields and so has precisely s maximal ideals. We shall prove that each maximal ideal has the form ker ϕ_{σ} for some σ ∈ Γ(D(α) : F).

Let τ be an element of the subgroup Γ(D(α) : D) of Γ(D(α) : F). Let τ′ ∈ Γ(D(α, β) : D(β)) be the extension of τ that fixes β. Then ϕ_{σ} ∘ τ′ = ϕ_{στ} and so ker ϕ_{στ} = ker ϕ_{σ}. Suppose now that σ, τ ∈ Γ(D(α) : F) lie in different left cosets of Γ(D(α) : D). Then σ^{−1}τ ∉ Γ(D(α) : D) and so there exists a δ ∈ D such that δσ^{−1}τ ≠ δ. Hence 1⊗δ−δσ^{−1}⊗1 lies in ker ϕ_{σ} but not in ker ϕ_{τ}. Thus the number, s, of maximal ideals is at least [D : F] (as each coset of Γ(D(α) : D) in Γ(D(α) : F) gives rise to a different maximal ideal of the form ker ϕ_{σ}). We have thus shown that ⊕_{i=1}^{[D:F]}D(α, β) is a quotient of F (α) ⊗_{F} F (β). By counting dimensions, we see that F (α) ⊗_{F} F (β) ≅ ⊕_{i=1}^{[D:F]}D(α, β). Note that

dim_{F} ⊕_{i=1}^{[D:F]} D(α, β) = [D : F][D(α, β) : F]
= [D : F][D(α) : D][D(β) : D][D : F]
= [D(α) : F][D(β) : F]
= dim_{F} F (α) ⊗_{F} F (β).

This concludes the proof. □

An immediate corollary of Theorem 10 is that 𝔽_{q^{m}} ⊗_{𝔽_{q}} 𝔽_{q^{n}} ≅ 𝔽_{q^{e}} ⊕ ⋯ ⊕ 𝔽_{q^{e}} where e = lcm(m, n) and there are d = gcd(m, n) summands.

8. Some Examples

It is of interest to know which binomials in F [X] are irreducible (c.f. [2]). This question can be reduced to binomials of prime-power degree using tensor products as the following argument shows.

Let X^{m}−a, X^{n}−b be binomials over F (so a and b are nonzero). If d = gcd(m, n), then it is straightforward to prove

(3)  (X^{m} − a) ⊗ (X^{n} − b) = (X^{mn/d} − a^{n/d}b^{m/d})^{d}.

Now consider the binomial X^{r} − c ∈ F [X]. Suppose that r is not a prime power. Let r = mn where gcd(m, n) = 1 and m, n > 1. There exist integers M and N such that

Mm + Nn = 1.

Then c = c^{Mm+Nn} = a^{n}b^{m} where a = c^{N} and b = c^{M}. Therefore,

(X^{m} − a) ⊗ (X^{n} − b) = X^{r} − c.

Thus a necessary condition for X^{r} − c to be irreducible over F is that both X^{m} − a and X^{n} − b are irreducible over F. We show that this condition is also sufficient.

Suppose gcd(m, n) = 1 and α and β are roots of X^{m} −a and X^{n} −b respectively. Since (αβ)^{Mm} = α^{Mm}β^{1−Nn} = a^{M}b^{−N}β, we see β ∈ F (αβ). Similarly, α ∈ F (αβ) so F (α, β) = F (αβ). Hence [F (αβ) : F] = mn and by Theorem 5, (X^{m} − a) ⊗ (X^{n} − b) is irreducible over F.

The next question to ask is: Which binomials X^{p^{m}} − a, where p is a prime, are irreducible over F? This is answered in [6, VIII, Theorem 16]. If p is an odd prime or p^{m} = 2, then X^{p^{m}} − a is irreducible if and only if a ∉ F^{p} (i.e. a is not a pth power). If 4 divides p^{m}, then X^{2^{m}} − a is irreducible if and only if a ∉ −4F^{4} (i.e. a ≠ −4b^{4} for all b ∈ F). Hence, one can completely classify the irreducible binomials. (With more work one can factor binomials into irreducible factors.)

Putting a = b = 1 in (3) shows (X^{m} − 1) ⊗ (X^{n} − 1) = (X^{e} − 1)^{d} where e = lcm(m, n). Thus the tensor product Φ_{m} ⊗ Φ_{n} of two cyclotomic polynomials is a product of certain Φ_{r} where r divides e. This factorization may be determined from the following facts

(4)  Φ_{m} ⊗ Φ_{n} = Φ_{mn}  if  gcd(m, n) = 1,

and if p is a prime and m ≥ n, then

(5)  Φ_{p^{m}} ⊗ Φ_{p^{n}} = \{ \begin{matrix}(X^{p^{n}} − 1)^{ϕ(p^{n})}/Φ_{p^{n}}(X)^{p^{n−1}} & if m = n, \\ Φ_{p^{m}}(X)^{ϕ(p^{n})} & if m > n. \end{matrix}

(Here ϕ denotes Euler’s totient function.) The proof of (4) is straightforward, and (5) follows from the fact that the if α has order p^{m} and β has order p^{n} where m > n, then αβ has order p^{m}.

By using (4) and (5) and the factorization X^{p^{n}} − 1 = ∏_{i=0}^{n} Φ_{p^{i}}(X), we may obtain factorizations of the form Φ_{m} ⊗ Φ_{n} = ∏ Φ_{r}^{e_{r}}. For example,

Φ_{60} ⊗ Φ_{40} = (Φ_{4} ⊗ Φ_{3} ⊗ Φ_{5}) ⊗ (Φ_{8} ⊗ Φ_{5})
= (Φ_{8} ⊗ Φ_{4}) ⊗ Φ_{3} ⊗ (Φ_{5} ⊗ Φ_{5})
= Φ_{8}^{2} ⊗ Φ_{3} ⊗ Φ_{1}^{4}Φ_{5}^{3}
= (Φ_{8} ⊗ Φ_{3} ⊗ Φ_{1})^{8}(Φ_{8} ⊗ Φ_{3} ⊗ Φ_{5})^{6}
= Φ_{24}^{8}Φ_{120}^{6}

Taking the degree of both sides gives identities such as

ϕ(60)ϕ(40) = 8ϕ(24) + 6ϕ(120).

9. Uniqueness of ⊗-factorizations

Let U(R) denote the group of units of R, and U(R[X]^{∗}) the group of ⊗-units in R[X]^{∗}. We call f ∈ R[X]^{∗} a ⊗-unit if there exists a g ∈ R[X]^{∗} such that f ⊗ g = X − 1.

Lemma 10. Every f ∈ U(R[X]^{∗}) may be uniquely written as f = α(X −β) where α, β ∈ U(R). Indeed, the map

U(R[X]^{∗}) ⟶ U(R) × U(R) defined by f ⟼ (α, β)

is a isomorphism.

The proof of this lemma is omitted as it is straightforward. We call f ∈ R[X]^{∗} ⊗-irreducible if f is not a ⊗-unit and f = g ⊗ h implies that g or h is a ⊗-unit. We call f, g ∈ R[X]^{∗} ⊗-associates if there is an h ∈ U(R[X]^{∗}) such that f = g ⊗ h. There are two obvious questions to ask about ⊗-factorizations:

(F1) Can every f ∈ R[X]^{∗} which is not a ⊗-unit be factored as a ⊗-product of

⊗-irreducibles? (F2) Are such factorizations unique? (That is, are the ⊗-irreducibles in one

⊗-factorization, up to order, ⊗-associates of the other?) Now R[X]^{∗} does not satisfy (F1) as 1 ⊗ 1 = 1, X ⊗ X = X and 1 and X are not

⊗-units. For this reason we consider the following submonoid of R[X]^{∗}

T = {f ∈ R[X]^{∗} | deg f ≥ 1 and f_{0} ≠ 0}.

We show in Theorem 12 that T has property (F1) if R is a unique factorization domain.

Lemma 11. (a) If f ∈ R[X]^{∗} is monic and f = g ⊗ h holds in R[X]^{∗}, then there exist monic ĝ, ĥ ∈ R[X]^{∗} such that f = ĝ ⊗ ĥ.

(b) ^{−} and ^{r} respect ⊗-units, and ⊗-irreducibles with nonzero constant terms.

(c) If m is prime and f_{m}, f_{0} ∈ U(R), then f is ⊗-irreducible.

(d) If m is prime, f_{m} is irreducible in R and f_{0} ∈ U(R), then f is ⊗-irreducible.

Proof. (a) The leading term of f = g ⊗ h is 1 = g_{n}^{p}h_{p}^{n}. Thus g_{n}, h_{p} ∈ U(R), so setting ĝ = g_{n}^{−1}g and ĥ = h_{p}^{−1}h shows ĝ ⊗ ĥ = (g_{n}^{−p}h_{p}^{−n})(g ⊗ h) = f.

(b) This follows from Theorem 2(h) and (i).

(c) Suppose f = g ⊗ h. Then f_{m} = g_{n}^{p}h_{p}^{n} and f_{0} = (−1)^{np}g_{0}^{p}h_{0}^{n}. Since f_{m}, f_{0} ∈ U(R), so g_{n}, h_{p}, g_{0}, h_{0} ∈ U(R). Since m = np is prime, we may assume without loss of generality that n = 1. Therefore, g ∈ U(R[X]^{∗}) by Lemma 10, and hence f is ⊗-irreducible.

(d) This proof is similar to that of (c) and hence is omitted. □

Theorem 12. If R is a unique factorization domain, then every element of T, which is not a unit, may be ⊗-factored as a product of ⊗-irreducibles.

Proof. Let f ∈ T and suppose f_{m} = εp_{1} ⋯ p_{r} and f_{0} = ηq_{1} ⋯ q_{s} where ε, η ∈ U(R) and the p_{i} and q_{j} are irreducible in R. Because R is a unique factorization domain, the quantity d(f) = m + r + s is an invariant of f. Furthermore, d(f) = 1 if and only if m = 1 and r = s = 0. That is, if and only if f ∈ U(R[X]^{∗}). In addition, if f = g ⊗ h, then d(f) ≥ d(g) with equality if and only if h ∈ U(R[X]^{∗}).

Our proof uses induction on d(f). It is vacuously true if d(f) = 1. Suppose now d(f) > 1. If f is ⊗-irreducible, then the result is true. If not, then there exist g, h ∈ R[X]^{∗} such that f = g ⊗ h where d(g) < d(f) and d(h) < d(f). The proof now follows from the inductive hypothesis. □

There seems to be little connection between irreducible elements of R[X] and ⊗-irreducible elements of R[X]^{∗}. For example, if m is prime (X − 1)^{m} is reducible in R[X] but is ⊗-irreducible by Lemma 11 (c). Conversely, if m is odd (X^{m} − 2) ⊗ (X^{2} + 1) = X^{2m} + 4 is irreducible in ℤ[X], but is clearly ⊗-reducible in ℤ[X]^{∗}.

Unfortunately Theorem 12 can not be strengthened to conclude that T has unique ⊗-factorization. The following example shows unique ⊗-factorization fails whenever U(R) ≠ {1}.

Example. Let γ ∈ U(R) have prime order p, or let γ have infinite order and let p be an arbitrary prime. Let q be a prime different to p and consider the ⊗-factorizations

∏_{i=0}^{p−1}(X − γ^{i}) ⊗ ∏_{j=0}^{q−1}(X − γ^{pj}) = ∏_{i=0}^{q−1}(X − γ^{i}) ⊗ ∏_{j=0}^{p−1}(X − γ^{qj}) = ∏_{i=0}^{pq−1}(X − γ^{i}).

By Lemma 11 (c) the ⊗-factors of degree p and q are ⊗-irreducible. Since p ≠ q, the polynomials of degree p and q are not ⊗-associates. Suppose the two polynomials of degree q are ⊗-associates. Then there is some ε(X − η) ∈ U(R[X]^{∗}) such that

ε(X − η) ⊗ ∏_{i=0}^{q−1}(X − γ^{i}) = ∏_{i=0}^{q−1}(X − γ^{pi}).

Thus η = γ^{pi} and ηγ = γ^{pj} where 0 ≤ i, j < q and i ≠ j. Therefore γ^{p(j−i)} = γ. This is impossible when γ has infinite order, and when γ has order p.

Because of this example and the identity (fg) ⊗ h = (f ⊗ h)(g ⊗ h), it seems unproductive to consider unique ⊗-factorization of reducible polynomials. Henceforth consider the ⊗-factorization of irreducibles. Now if m_{γ/F} = f ⊗ g, then both f and g are irreducible. Hence, we may write m_{γ/F} = m_{α/F} ⊗m_{β/F} where γ = αβ. If m_{α/F} has degree m and m_{β/F} has degree n, then m_{γ/F} has degree mn and F (γ) is an internal tensor product F (α) ⊗_{F} F (β) because F (α)F (β) = F (γ) and [F (γ) : F] = mn.

We shall investigate unique ⊗-factorization for finite fields. Although Theorems 13 and 14 were first proved by Brawley and Carlitz [1], their proofs were somewhat long and indirect. We offer shorter elementary proofs.

Theorem 13. Let F = 𝔽_{q}. If m_{γ/F} = m_{α/F} ⊗ m_{β/F} is ⊗-reducible over F where m = deg m_{α/F} and n = deg m_{β/F} are > 1, then deg m_{γ/F} = mn, gcd(m, n) = 1 and the order of γ divides (q^{m} − 1)(q^{n} − 1)/(q − 1). Conversely, if deg m_{γ/F} = mn where gcd(m, n) = 1, and the order of γ divides (q^{m} − 1)(q^{n} − 1)/(q − 1), then there exist α, β ∈ F (γ) such that m_{γ/F} = m_{α/F} ⊗ m_{β/F} and deg m_{α/F} = m and deg m_{β/F} = n.

Proof. Suppose that m_{γ} = m_{α} ⊗m_{β} is ⊗-reducible where γ = αβ and m = deg m_{α}, n = deg m_{β} are > 1. Then deg m_{γ} = mn where m = deg m_{α} and n = deg m_{β}. The lattice of subfields of F (γ) is isomorphic to the lattice of divisors of mn and hence gcd(m, n) = 1. Furthermore, it follows from γ = αβ that the order of γ divides lcm(q^{m} − 1, q^{n} − 1) = (q^{m} − 1)(q^{n} − 1)/(q − 1).

Conversely, suppose that deg m_{γ} = mn where gcd(m, n) = 1, and the order of γ divides (q^{m}−1)(q^{n}−1)/(q−1). Let Q_{r} be an abbreviation for (q^{r} −1)/(q−1). Then gcd(Q_{r}, Q_{s}) = Q_{gcd(r,s)}. Thus there exist M, N ∈ ℤ such that MQ_{m} + NQ_{n} = 1. Set α = γ^{NQ_{n}} and β = γ^{MQ_{m}}. Then αβ = γ^{MQ_{m}+NQ_{n}} = γ and

α^{q^{m}−1} = γ^{N(q^{m}−1)(q^{n}−1)/(q−1)} = 1,

so α ∈ 𝔽_{q^{m}}. Similarly, β ∈ 𝔽_{q^{n}}. Now deg m_{α} divides m, deg m_{β} divides n and m_{α} ⊗ m_{β} divides m_{γ}. Since m_{γ} is irreducible over F, m_{γ} = m_{α} ⊗ m_{β} and so deg m_{α} = m and deg m_{β} = n. □

When cancellation laws hold, it is standard to prove that unique factorization holds by showing that the notions of “prime” and “irreducible” coincide. In general, cancellation laws do not hold. If F = ℚ, and f ⊗ g = f ⊗ h is irreducible, then it does not imply that g and h are ⊗-associate. For example, the factorizations e^{2πi/8} = (1 + i)\frac{1}{\sqrt{2}} = (1 − i)\frac{i}{\sqrt{2}} give rise to the ⊗-factorizations

X^{4} + 1 = (X^{2} − 2X + 2) ⊗ (X^{2} − 1/2) = (X^{2} − 2X + 2) ⊗ (X^{2} + 1/2)

over ℚ. However, X^{2} − 1/2 and X^{2} + 1/2 are not ⊗-associate over ℚ. (Indeed, unique ⊗-factorization fails over ℚ!) As a consequence of Theorem 14, the above cancellation law holds for finite fields, however, we can not use this fact to prove

Theorem 14. Every irreducible polynomial over 𝔽_{q} can be uniquely ⊗-factored into ⊗-irreducibles.

Proof. As we are interested only in ⊗-factorizations of irreducible polynomials, we define a polynomial in R[X]^{∗} to be ⊗-prime if it is not a ⊗-unit, and if it is a ⊗-factor of f ⊗ g where f ⊗ g is irreducible, then it is a ⊗-factor of f or g. If f is ⊗-prime, then it is certainly ⊗-irreducible. (Let R = 𝔽_{q}. If f = g ⊗ h, then g = f ⊗g_{1} or h = h_{1}⊗f. Thus f = f ⊗g_{1}⊗h or f = g⊗h_{1}⊗f so g or h is a ⊗-unit.) Unique ⊗-factorization of irreducibles will follow once we prove that an irreducible ⊗-irreducible polynomial is ⊗-prime. (Suppose that f_{1} ⊗ ⋯ ⊗ f_{r} = g_{1} ⊗ ⋯ ⊗ g_{s} are two ⊗-factorizations of an irreducible polynomial h into ⊗-irreducibles. Then f_{1} is a ⊗-associate of some g_{i}. Similarly, f_{2} is a ⊗-associate of some g_{j}. If i = j, then f_{1} ⊗ f_{1} is a ⊗-factor of h. This is impossible as f_{1} ⊗ f_{1} is reducible. Thus i ≠ j and we may show that r = s and each f_{i} is a ⊗-associate of a unique g_{j}.)

Suppose that m_{α} is ⊗-irreducible and m_{α} ⊗ m_{β} = m_{γ} ⊗ m_{δ} = m_{ε} is irreducible where αβ = γδ. Let a, b, c, d denote the degrees of m_{α}, m_{β}, m_{γ}, m_{δ} respectively. Then by Theorem 13, ab = cd and gcd(a, b) = gcd(c, d) = 1. Assume, without loss of generality, that gcd(a, c) > 1. We consider two cases: when gcd(a, c) = a and gcd(a, c) < a.

Case (i) gcd(a, c) = a. First suppose that a = c. Then b = d and λ = β/δ = γ/α ∈ 𝔽_{q^{a}} ∩ 𝔽_{q^{b}} = 𝔽_{q}. Thus γ = αλ, so m_{γ} = m_{α} ⊗ m_{λ} = m_{α} ⊗ (X − λ) and m_{α} is a ⊗-factor of m_{γ} as desired.

Assume now that c = ar where r > 1. Since c = gcd(ab, c) = gcd(a, c) gcd(b, c), it follows that r = gcd(b, c) and gcd(a, r) = 1. Thus there exist C, D ∈ ℤ such that CQ_{c} + DQ_{d} = 1. By the above comments, we may assume, without loss of generality, that

(6)  γ = ε^{DQ_{d}} and δ = ε^{CQ_{c}}.

The order of ε divides e = lcm(q^{a} − 1, q^{b} − 1) and f = lcm(q^{c} − 1, q^{d} − 1) and hence divides gcd(e, f). However,

gcd(e, f) = (q − 1) gcd(e/(q − 1), f/(q − 1)) = (q − 1) gcd(Q_{a}Q_{b}, Q_{c}Q_{d})
= (q − 1) gcd(Q_{a}, Q_{c}) gcd(Q_{b}, Q_{c}) gcd(Q_{a}, Q_{d}) gcd(Q_{b}, Q_{d})
= (q − 1)Q_{a}Q_{r}Q_{d} = (q^{a} − 1)(q^{r} − 1)(q^{d} − 1)/(q − 1)^{2}.

It follows from (6) that γ^{(q^{a}−1)(q^{r}−1)/(q−1)} = ε^{D(q^{a}−1)(q^{r}−1)(q^{d}−1)/(q−1)^{2}} = 1. Hence by Theorem 13, m_{γ} = m_{ρ} ⊗ m_{σ} where deg m_{ρ} = r and deg m_{σ} = a. Since m_{α} and m_{σ} have the same degree, m_{α} is a ⊗-factor of m_{σ} and hence of m_{γ}, as desired.

Case (ii) gcd(a, c) < a. This case never arises. If it did, an argument similar to the previous case would show that m_{α} is ⊗-reducible – a contradiction. □

While it is true that ⊗-factorizations of m_{γ/F} give rise to ⊗-factorizations of F (γ), the converse is false.

Theorem 15. The cyclotomic polynomial Φ_{3^{k}} is ⊗-irreducible over ℚ for all k ≥ 1.

Proof. Let ζ_{n} = e^{2πi/n} be a primitive nth root of 1. Then ℚ(ζ_{n}) : ℚ is Galois with Galois group isomorphic to the group U(ℤ_{n}) of order ϕ(n). Let n = 3^{k} and ζ = ζ_{n}. Then U(ℤ_{n}) ≅ C_{2} × C_{3^{k−1}} is isomorphic to a direct product of cyclic groups of order 2 and 3^{k−1}. The only possible factorization of ϕ(n) = 2 · 3^{k−1} as a product with coprime factors is, 2 times 3^{k−1}. Hence the only possible ⊗-factorization of ℚ(ζ) is, ℚ(ω) ⊗ ℚ(θ) where ω = ζ^{n/3} and θ = ζ + ζ^{−1}. If Φ_{n} were ⊗-reducible, then there exist α, β ∈ ℚ(ζ) such that ζ = αβ, ℚ(α) = ℚ(ω) and ℚ(β) = ℚ(θ). Replacing α, β by αμ^{−1}, βμ where μ ∈ ℚ if necessary, we may assume that α may be written in the form α = λ + ω where λ ∈ ℚ.

Now (αβ)^{n} = ζ^{n} = 1 and so α^{n} = β^{−n} ∈ ℚ(α) ∩ ℚ(β) = ℚ. Consider α^{n} = ∑_{i=0}^{n} \binom{n}{i}λ^{i}ω^{n−i}. If i ≡ 1 (mod 3), then ω^{n−i} = ω^{2} = −1 − ω and if i ≡ 2 (mod 3), then ω^{n−i} = ω. Writing (λ + ω)^{n} in the form σ + τω where σ, τ ∈ ℚ shows

τ = ∑_{i≡1 (mod 3)} −\binom{n}{i}λ^{i} + ∑_{i≡2 (mod 3)} \binom{n}{i}λ^{i} = ∑_{i≡1 (mod 3)} −\binom{n}{i}λ^{i} + \binom{n}{i + 1}λ^{i+1}.

Since when 0 < i < n, the largest power of 3 dividing n − i is the largest power of 3 dividing i, it follows that n divides \binom{n}{i} if n ≡ 1, 2 (mod 3). Hence the above polynomial in λ has the form τ = nλa(λ) where

a(λ) = −1 + a_{1}λ − a_{3}λ^{3} + a_{4}λ^{4} − ⋯ + λ^{n−1} and a_{i} = \frac{1}{n}\binom{n}{i + 1}

are positive integers. Now the only possible rational roots of a(λ) are ±1, because a(λ) is monic and a(0) = −1. It is clear that a(−1) < 0. Also, a(1) = 0 because \binom{n}{i} = \binom{n}{n−i} and, if i ≡ 1 (mod 3), then n − i ≡ 2 (mod 3). Since λ ∈ ℚ satisfies nλa(λ) = 0, it follows that λ = 0, 1. Hence α = λ + ω = ω, −ω^{2} and β = ζα^{−1} = ζ^{1−n/3}, −ζ^{1−2n/3}. This is a contradiction, as then ℚ(β) = ℚ(ζ) ≠ ℚ(θ). Hence Φ_{3^{k}} is ⊗-irreducible over ℚ for all k ≥ 1. □

It is shown in [4] that, if n is odd, the ⊗-factorization of Φ_{n}(X) over ℚ into ⊗-irreducibles is the same as the factorization of n into prime-powers.

Acknowledgements

I am most grateful to L.G. Kovács for comments and discussions which, amongst other things, lead to an improved version of Theorem 8. I would also like to thank A.M. Nelson for comments relating to Theorem 10.

References

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