The norm of a skew polynomial

Let $D$ be a finite-dimensional division algebra over its center and $R=D[t;\sigma,\delta]$ a skew polynomial ring. Under certain assumptions on $\delta$ and $\sigma$, the ring of central quotients $D(t;\sigma,\delta) = \{f/g \,|\, f \in D[t;\sigma,\delta], g \in C(D[t;\sigma,\delta])\}$ of $D[t;\sigma,\delta]$ is a central simple algebra with reduced norm $N$. We calculate the norm $N(f)$ for some skew polynomials $f\in R$ and investigate when and how the reducibility of $N(f)$ reflects the reducibility of $f$.


Introduction
Let A be a central simple algebra, R = A[t; σ, δ] and A(t; σ, , δ) = {f /g | f ∈ A[t; σ, δ], g ∈ C(A[t; σ, δ])} the ring of central quotients of R. Under certain assumptions on δ and σ, A(t; σ, δ) is a central simple algebra with norm N . So far, this norm has been successfully used when investigating factorizations of polynomials f ∈ F q n [t; σ] (cf. [2,3] In particular, let K/F be a finite cyclic Galois extension of degree n with Galois group generated by σ. Then for all f ∈ K[t; σ], we have N (f ) ∈ F [x] where x = t n and f divides N (f ). If f satisfies (f, t) r = 1, and either f is not right-invariant and n is prime, or gcd(m, n) = 1, then f is irreducible in K[t; σ] if and only N (f ) is irreducible as a polynomial in F [x]. This indeed holds more generally for all f ∈ K[t; σ] such that (f, t) r = 1 that are not right-invariant with bound f * of degree mn.
We extend these results to the case that R = D[t; δ], where δ is an algebraic derivation, in particular we look at the case that R = K[t; δ], where K/F is a field extension in characteristic p of degree p e and δ an algebraic derivation with constant field F . Any not right-invariant f ∈ K[t; δ] of degree m and with bound f * of degree mn, is irreducible in K[t; δ] if and only N (f ) is irreducible as a polynomial in some polynomial ring F [x].
Furthermore, in some cases it is still true that a factorization of N (f ) in F [x] provides a factorization of f . This work is part of the second author's PhD thesis [10].

Preliminaries
2.1. Skew polynomial rings. Let A be a unital associative ring, σ a ring endomorphism of A and δ : A → A a left σ-derivation, i.e. an additive map such that δ(ab) = σ(a)δ(b) + δ(a)b for all a, b ∈ S. The skew polynomial ring R = A[t; σ, δ] is the set of skew polynomials g(t) = a 0 + a 1 t + · · · + a n t n with a i ∈ A, with term-wise addition and multiplication defined via ta = σ(a)t + δ(a) for all a ∈ A [7]. Define Fix(σ) = {a ∈ A | σ(a) = a} and Const(δ) = {a ∈ A | δ(a) = 0}. If δ = 0, define A[t; σ] = A[t; σ, 0]. If σ = id, define For f (t) = a 0 + a 1 t + · · · + a n t n ∈ R with a n = 0, we define the d egree of f as deg(f ) = n and deg(0) = −∞. An element f ∈ R is irreducible in R if it is not a unit and it has no proper factors, i.e if there do not exist g, h ∈ R with 1 ≤ deg(g), deg(h) < deg(f ) such that f = gh [6, p. 2 ff.].
Let D be a division algebra. Then a polynomial f (t) ∈ D[t; σ, δ] is bounded if there exists a nonzero polynomial f * ∈ D[t; σ, δ] such that D[t; σ, δ]f * is the largest two-sided ideal of D[t; σ, δ] contained in D[t; σ, δ]f . The polynomial f * is uniquely determined by f up to scalar multiplication by elements of D × . f * is called the bound of f .
In this paper, D will always be a central simple division algebra of degree d over its center C.

The minimal central left multiple of f ∈ D[t; σ].
Let σ be an automorphism of D of finite order n modulo inner automorphisms, that means that σ n = i u for some inner automorphism i u (z) = uzu −1 . Then the order of σ| C is n. W.l.o.g., we choose u ∈ Fix(σ). Let R = D[t; σ] and define F = C ∩ Fix(σ). Then R has center [6, Theorem 1. 1.22]. All polynomials f ∈ R are bounded.
If the greatest common right divisor of f and t (denoted (f, t) r ) is one, then f * ∈ C(R) [3, Lemma 2.11]). From now on we assume that (f, t) r = 1. For any f ∈ R = D[t; σ] with a bound in C(R), we can define the minimal central left multiple of f in R to be the unique (i) If (f, t) r = 1, then the minimal central left multiple of f exists and is unique. It is equal to the bound of f up to a scalar multiple from D.
Proof. (cf. [8]) (i) Let f * be a bound of f . Then f * is unique up to scalar multiplication by elements in D × and Rf * is the largest two-sided ideal of R contained in the left ideal Rf . In particular, this implies f * = gf for some g ∈ R.
The assumption that (f, t) r = 1 implies that f * ∈ C(R) [3,Lemma 2.11]) thus f * is the unique minimal central left multiple of f , up to some scalar.
on the right, this contradicts the minimality of h. Moreover, as f is irreducible the greatest common right divisor of f and h 1 is 1. As R is a right Euclidean domain, there exist p, q ∈ R such that pf + qh 1 = 1. Multiplying everything on the right by h 2 (t) =ĥ 2 (u −1 t n ), we obtain pf h 2 + qh = h 2 . As f is a right divisor of h by definition, h = rf for some r ∈ R. Noting that h 2 (t) ∈ C(R), this yields Thus h 2 is a central left multiple of f of degree strictly less than h contradicting the mini- for some p(t) ∈ R and so comparing the irreducible factors of f and h and employing [6,Theorem 1.2.9], we see that f = f 1 · · · f r for irreducible f i ∈ R such that f i ∼ f j for all i, j and r ≤ k.
The above results apply in particular to the special case that D is a finite field extension and σ ∈ Aut(K) has order n. Then R = K[t; σ] has center where C is a field of characteristic p (we allow D = C). Moreover, we assume that δ is a derivation of D, such that δ| C is algebraic with minimum polynomial are of the form f (t) = uc(t) with u ∈ D and c(t) ∈ C(R) [6, Theorem 1.1.32]. All polynomials f ∈ R are bounded.
For every f ∈ R = D[t; σ] we define the minimal central left multiple of f in R to be the unique polynomial of minimal degree h ∈ C(R) = F [x] such that h = gf for some g ∈ R, and such that h(t) =ĥ(g(t) − d 0 ) for some monicĥ(x) ∈ F [x]: (i) The minimal central left multiple of f exists and is unique. It is equal to f * up to a scalar multiple in D × .
Proof. (cf. [8]) (i) Let f * be a bound of f . Then f * is unique up to scalar multiplication by elements in D × and Rf * is the largest two-sided ideal of R contained in the left ideal Rf . Since f * is two-sided, we know that f * (t) = dc(t) for some c(t) ∈ C(R) and d ∈ D × . So assume w.l.o.g. that f * ∈ C(R). The rest of the proof is identical to the one of Lemma 1. The proofs of (ii), (iii) are identical to the one of Lemma 1 (ii), (iii).
2.4. The algebra (A(x), σ, ux). Let C/F be a finite cyclic field extension of degree n with Gal(C/F ) = σ . Let A be a central simple algebra of degree d with center C and suppose that σ extends to a C-algebra automorphism of A that we also will call σ. Then there exists an element u ∈ A × such that σ n = i u and σ(u) = u. These two relations determine u up to multiplication with elements from F × [9, Lemma 19.7] . Let R = A[t; σ] and Note that when regarding A(t; σ) as an F (x)-algebra, the choice of u is lost: x depends on u, and different choices of u lead to different actions of F (x) on A(t; σ). Here and in the following we thus assume that u is fixed and x = u −1 t n .
The algebra A(t; σ) has center F (x) and degA(t; σ) = dn. The reduced norm N of (A(x), σ, ux) is a nondegenerate form of degree dn over F (x). In particular, if A is a division algebra then A(t; σ) is a division algebra [4, Theorem 2.2.].
If A contains a strictly maximal subfield that is Galois over F with Galois group G, then A(t; σ) is a crossed product with group G. If A is a symbol algebra over a global field F , then A(t; σ) is a crossed product. If A is a p-algebra over a global field F , then A(t; σ) is a cyclic crossed product [4, Corollary 2.4.].
is a subring of D(t; δ). Let δ denote the extension of δ to D(x) such that δ = id t | D(x) . Then δ| C(x) is algebraic with minimum polynomial g(t), and D(t; δ) is the central simple F (x)-algebra i.e. a generalized differential algebra of degree p e d [6, p. 23]. Let N be its reduced norm. Then N has degree p e d. Moreover, (D(x), δ, d 0 + x) contains D(x) as the centralizer of C(x) [5, Theorem 3.1], and is free of rank p e as a left D(x)-module.
2.6. The algebra (K(x), δ, x). Let K be a field extension of characteristic p and δ a derivation on K that is algebraic with F = Const(δ). Let g(t) = t p e + c 1 t p e−1 + · · · + c e t be the minimum polynomial of δ. Then K/F is a purely inseparable extension of exponent one, and K p ⊂ F . More precisely, is algebraic with minimum polynomial g(t). The algebra K(t; δ) ∼ = (K(x), δ, x) is a generalized differential algebra of degree p e over F (x) [6, p. 23]. Let N be its reduced norm. N has degree p e .
3. Using the norm of (K(x), σ, x) We start with the special case that A = C in 2.4 (with some small changes in our notation): Let K/F be a cyclic field extension of degree n with Gal(K/F ) = σ , R = K[t; σ] and x = t n . Let N be the reduced norm of the cyclic algebra (K(x), σ, x) over F (x) (cf. also [6, Proposition 1.4.6]). We have σ| K = σ, and N is a nondegenerate form of degree n.
Caruso and Le Borgne [2] use N (f ) successfully to factorize skew polynomials f ∈ F q [t, σ] over finite fields. For certain f ∈ K[t, σ], the norm N of (K(x), σ, x) can also be used to obtain a reducibility criterium: . Moreover, we have: This is the generalized and corrected version of [6, Proposition 1.7.1 (ii)], which stated (−1) mn N K/F (a m )x m for the leading term, and also required m < n. Furthermore, our proof fixes a small mistake in the proof of [2, Lemma 2.1.15].
Proof. Write f (t) = a 0 + a 1 t + · · · + a m t m as . We can use verbatim the same proof as given in [2, Lemma 2.1.15] to obtain the matrix in M n (K[x]) representing the left multiplication ρ(f (t)) with respect to the basis 1, t, . . . , t n−1 : we have Thus N (f (t)), which is the determinant of this matrix, has as constant term the constant There are unique integers k, r, 0 ≤ r ≤ n − 1, such that we can write m as m = kn + r. In the sum giving the determinant of this matrix, the term of highest degree is ) has as highest term the highest term of this sum. The highest term is thus given by Remark 4. In [3, Theorem 2.9, Corollary 2.12], it was proved that deg(f * ) ≤ n · deg(f ). We recover this result as a byproduct of the above observation that , which immediately yields the assertion.
Theorem 6. Let f (t) ∈ R be a polynomial of degree m such that (f, t) r = 1. Suppose that deg(ĥ) = m (this is always the case if, for instance, either n is prime or gcd(m, n) = 1).
Since we assume (f, t) r = 1, so we know that f * ∈ C(R) and therefore f * equals h up to some invertible factor in F . Thus Corollary 7. Let f ∈ R be a monic polynomial of degree. Suppose that (f, t) r = 1 and that . Then f has exactly l! irreducible decompositions corresponding to each possible ordering of the factors of N (f ).
Since N (f ) = af * with a ∈ F × because of our assumption that deg(ĥ) = m and knowing that deg(N (f )) = m by Theorem 3, we conclude that there is a "rough factorization" of f given by f = g 1 · · · g l where each g i has minimal central left multipleĥ i [3,Proposition 5.2].
Thus N (f ) = N (g 1 ) · · · N (g l ) =ĥ 1 · · ·ĥ l in the commutative polynomial ring Thus there is a permutation π such that N (g π(i) ) =ĥ i for all i. Sinceĥ i is the minimal central left multiple of g i and thus divides N (g i ), we can conclude that π = id. Moreover, determine the degrees of the g i . Since (f, t) r = 1, also (g i , t) r = 1 for all i. By Lemma 5, we conclude that the g i must be irreducible for all i. Denote them by f i to conform with our previous notation.
(ii) Assume that N (f ) is the product of l distinct irreducible factors. Then f is a product of l irreducible factors by (i), containing exactly one each whose reduced norm isĥ i for each irreducibleĥ i in F [x]. Therefore f has exactly l! different irreducible decompositions corresponding to each possible ordering of the distinct factorsĥ i of N (f ).

Using the norm of the twisted function field A(t; σ)
Let A be a central simple algebra of degree d with center C and R = A[t; σ], where σ is an automorphism of A of finite order n modulo inner automorphisms, i.e. σ n = i u , such that σ| C ∈ Aut F (C) has order n. Let N be the reduced norm of (A(x), σ, ut). t) is a subalgebra of (A(x), σ, ut) of degree n over F (x). Note that σ| C ∈ Aut F (C) by our global assumption on σ at the beginning of this section. In particular, this means σ| C(x) ∈ Aut(C(x)). Thus [9,Proposition. p.304]. This yields the assertions: The proof works similarly as the one of [6, Proposition 1.7.1]: , the set {1, t, . . . , t n−1 } also generates A(t; σ) over A(x). Furthermore, we have I t | K(x) = σ, where σ denotes the extension of σ to K(x) fixing x, and A(x) ⊂ C A(t;σ) (K(x)). [4, Lemma 1.27] therefore shows that {1, t, . . . , t n−1 } is free over A(x), thus and t n = ux, every f ∈ R ⊂ (A(x), σ, ut) can be written as a linear combination of 1, t, . . . , t n−1 with coefficients in A[x]. We therefore obtain a representation ρ of (A(x), σ, ut) by matrices in M n (A(x)) by writing , σ, ut) and 0 ≤ i, j ≤ n−1. Hence the matrix ρ(f (t)) has entries in A[x] for every f ∈ R. Since A has a subfield E of degree d, we can regard A as a left module over E. Let {v 1 , . . . , v d } be a basis for A over E(x). Then {v 1 , . . . , v d , v 1 t, . . . , v d t, . . . v d t n−1 } is a basis of (A(x), σ, ut) = A(t; σ) as a left module over E and we now analogously obtain a representation ρ of (A(x), σ, ut) by matrices in M dn (E(x)) with respect to that basis.
For f (t) ∈ R, the matrix ρ(f (t)) has entries in E[x], therefore it follows that (ii) Similarly as in (i), it can be shown that all the coefficients of the characteristic polynomial of ρ(f (t)) are contained in F [x] (cf. also [9,Proposition,p. 295]) and thus f (t) ♯ ∈ R by [6, (1.6.12 Proof. Write m = kn + r for some 0 ≤ r < n. Substituting t n = ux, we obtain f (t) = Computing the left regular representation of ρ : [6, Proposition 1.6.9]. Moreover, it follows that Comparing the above equation with the expressions for P i (x), it follows that This means the bottom left r × r minor of ρ(f (t)) has elements of degree at most k + 1 in lower triangular entries (including the diagonal which attains this maximum degree) and the top right n − r × n − r minor of ρ(f (t)) has elements of degree at most k − 1 in the upper triangular entries (excluding the diagonal which has elements of exactly degree k). Every other element of ρ(f (t)) has degree at most k.
As D has a subfield of degree d, there exists a left regular representation ω : D → M d (E) which extends to D[x] by setting ω(x) = xId. The d × d block matrices representing Q i,j (x) are inserted for every entry Q i,j (x) in ρ(f (t)) to obtain a representation for As ω is additive and ω(x) is a diagonal matrix, ω(σ j (g(x))) = ω(σ j (g k )u k )(xId) k + · · · + ω(σ j (g 0 )) for any polynomial g(x) = k i=0 g i x i ∈ D[x]. As we are computing the determinant only to find the degree of N (f (t)), it is sufficient to only consider the term of highest degree in Q i,j (x) and ignore all terms of lower degree. We truncate Q i,j (x) at the highest term and apply ω to all the entries of the matrix. To determine the term of highest degree, we expand the determinant along the columns and consider only the portion of the determinant expansion which yields the maximum possible degree. As ω(a m u k ) = ω(a m )ω(u) k is invertible, there are no zero columns in ω(a m u k ) so it is always possible to find an expansion of the matrix yielding the highest degree. Hence the degree of N (f (t)) is at most dr(k + 1) + d(n − r)k = d(kn + r) = dm. We wish to show the coefficient of x dm in N (f (t)) is non-zero: following a suitable expansion of ω • ρ(f (t)) the coefficient of x dm is equal to ±det(ω(a m u k ))det(ω(σ(a m u k )) · · · det(ω(σ n−r−1 (a m u k ))det(ω(σ n−r (a m u k+1 )) · · · det(ω(σ n−1 (a m u k+1 )).
As σ is an automorphism, det(ω(σ i (a m u k )) = 0 by our assumption on ω(a m u k ). Thus the coefficient of x dm is non-zero and deg(N (f (t)) = dm.
Truncate the polynomials in the matrix at their highest terms and apply ω : for some a (i,j) k,l ∈ E for i, j ∈ {1, 2} and k, l ∈ {1, 2, 3}. Then the determinant of the above matrix is equal to Comparing this to the matrix in terms of ω, this is equal to
From now on, we assume that A = (E/C, γ, a) is a cyclic algebra over C of degree d, Then σ| E has order n. Write m = kn + r for some 0 ≤ r < n.

given by the entries
We obtain the constant term of N (f (t)) by substituting x = 0. Thus the constant term equals a 0 γ(a 0 ) . . . γ d−1 (a 0 )), since γ commutes with σ. As a 0 ∈ E, this is equal to F (a 0 ). Similarly, the leading term of N (f (t)) is given by the leading term of (P r (x)) . . . σ n−1 (P r (x))u r x r ], which is given by since u ∈ E. As σ and γ commute and a m ∈ E, we can express this as As d(kn + r) = dm, this implies the assertion.
If (f, t) r = 1 then f * ∈ C(R), that means f * is up to some scalar α ∈ D × equal to the minimal central left multiple h of f , where h(t) =ĥ(u −1 t n ) withĥ(x) ∈ F [x]. Analogously as Lemma 5 we obtain: Theorem 14. Let A = (E/C, γ, a) be a cyclic algebra over C of degree d such that σ| E ∈ Aut(E), u ∈ E, and γ • σ = σ • γ. Let f ∈ R be a polynomial such that (f, t) r = 1. Suppose that deg(ĥ) = dm.
Proof. N (f ) is a two-sided multiple of f in R, therefore the bound f * of f must divide N (f ) in R. Since (f, t) r = 1, we know that f * ∈ C(R) and therefore f * equals h up to some invertible factor in F . Thus h(t) =ĥ(u −1 t n ) must divide N (f ) in R. Write N (f ) = g(t)h(t) for some g ∈ R. By Theorem 12 we have deg(N (f )) = dmn in R. Comparing degrees in R we obtain degN (f ) = deg(g(t))+dmn = dmn, which implies deg(g) = 0, i.e. g(t) = a ∈ A × . This gives N (f ) = ah(t) = aĥ(u −1 t n ).
We can therefore again determine the similarity classes of irreducible polynomials appearing in a factorization of certain f , and show that any order is possible for the appearance of these similarity classes: Corollary 15. Let A = (E/C, γ, a) be a cyclic algebra over C of degree d such that σ| E ∈ Aut(E), u ∈ E, and γ •σ = σ•γ. Let f ∈ A[t; σ] be a monic polynomial, such that (f, t) r = 1 and that deg(ĥ) = dm. Theorem 16. Let A = (E/C, γ, a) be a cyclic algebra over C of degree d, and f (t) = a 0 + a 1 t + · · · + a m t m ∈ C[t; σ] ⊂ R be a polynomial of degree m, such that (f, t) r = 1 and deg(ĥ) = dm. Then f is reducible in R and has at least d irreducible factors.
Proof. By Proposition 8, N (f (t)) = (N C/F (a 0 ) + · · · + (−1) m(n−1) N C/F (a m )x m ) d is clearly reducible in F [x]. Since deg(ĥ) = dm, f is reducible in R. The at least d irreducible divisorsĝ 1 , . . . ,ĝ l of N (f (t)) correspond to a decomposition f = f 1 · · · f l of f into irreducible factors, such that N (f i ) =ĝ j for some suitable j, for all i, 1 ≤ i ≤ l. The degree of f i in R equals the degree ofĝ j in F [x] for a suitable j.
Note that if σ is an inner automorphism (i.e. n = 1 and so R = D[t] by a change of indeterminants), d is prime and f not right-invariant, then any f of degree m will satisfy deg(ĥ) = dm.
More generally, we also have by [9, Proposition, p. 304]: Proposition 17. Let A = (E/C, γ, a) be a cyclic algebra. Let B be a central simple algebra over K(x) that is a subalgebra of (A(x), σ, ut) and assume that K(x)/F (x) is a finite field extension. Then every f (t) ∈ A[t; σ] ∩ B with deg(h) = dmn is reducible.
Proof. Let K(x)/F (x) be a finite field extension of degree c and B be a central simple algebra f (t))) e for a suitable integer e with d = bce [9,Proposition,p. 304]. Thus N (f ) is reducible and hence so is f , since we assumed that deg(h) = dmn.

5.
The norm condition for differential polynomials in characteristic p 5.1. The case that R = K[t; δ]. We use the notation from 2.6: let F be a field of characteristic p and K/F be a field extension which is purely inseparable of exponent one. Let δ be a derivation on K that is algebraic with Const(δ) = F , and g(t) = t p e +c 1 t p e−1 +· · ·+c e t ∈ F [t] the minimum polynomial of δ. Let R = K[t; δ] and C(R) = F [x], where x = g(t). Let N be the norm of (K(x), δ, x).
Part (ii) follows similarly to Theorem 3. This generalizes and refines [6, p. 31], which states that for f (t) = a 0 + a 1 t + · · · + a m t m ∈ R we have N (f (t)) = ±a p e m x m + . . . , if m < p e , where the omitted terms are of lower degree than m in x (no proof was given).
To find the constant term of N (f ) in Theorem 18 is difficult, let us look at one example: Example 19. Let p e = 5, f (t) = t 4 + a for some a ∈ K × , and g(t) = t 5 + t. Computing . Setting x = 0 and taking the determinant of ρ(f (t)) shows that the constant term of N (f (t)) is It is possible to compute special cases though: Proposition 20. For f (t) = g(t) + a for some a ∈ K, Proof. Following the proof of Theorem 18, we substitute Computing the left regular representation ρ : K[t; σ] → M p e (K[x]), it follows that ρ(f (t)) is a lower triangular matrix where each diagonal entry is equal to x + a. As the determinant of a triangular matrix is the product of its diagonal entries, the result follows. Since . Thus deg(f * ) ≤ p e · deg(f ), because the degree of N (f ) as a polynomial in R is m · p e by Theorem 18 (ii). In particular, if deg(f * ) = m in F [x], comparing degrees in F [x] shows that N (f ) is the bound of f . The bound f * is up to some scalar α ∈ K × equal to the minimal central left multiple h of f , where as before we write h(t) =ĥ(g(t)) withĥ(x) ∈ F [x] monic.
The proof is identical to the one of Lemma 13.
Theorem 22. Let f ∈ R be a polynomial of degree m. Suppose that deg(ĥ) = m.
The proof is identical to the one of Theorem 14, using Theorem 18, Proposition 2 and Lemma 21.
Corollary 23. Let f ∈ R be a monic polynomial of degree m. Suppose that deg(ĥ) = m.
. Then f has exactly l! irreducible decompositions corresponding to each possible ordering of the factors of N (f ).
The proof is identical to the one of Corollary 7, using Theorem 18 and Lemma 21.

5.2.
The case that R = D[t; δ]. Let C be a field of characteristic p and D a central division algebra over C of degree d. Let δ be a derivation of D, such that δ| C is algebraic with minimum polynomial g(t) = t p e + c 1 t p e−1 + · · · + c e t ∈ F [t] of degree p e , where F = Const(δ) ∩ C. Write g(t) = t p e + g 0 (t). Then g(δ) = id d0 is an inner derivation of D and w.l.o.g. we choose d 0 ∈ Const(δ) so that δ(d 0 ) = 0. R = D[t; δ] has center C( Let N be the reduced norm of D(t; δ) ∼ = (D(x), δ, d 0 + x) of degree p e d [6, p. 23].
is well-defined: since δ| C is an algebraic derivation on C by our assumption, we can canonically extend it to get the algebraic derivation δ| C(x) : C(x) −→ C(x). ρ ij (f (t))t j for all f ∈ R ⊂ (A(x), δ, d 0 + x) and 0 ≤ i ≤ n − 1, where ρ ij (f (t)) is the (i, j) th entry of ρ(f (t)). The matrix ρ(f (t)) then has entries in D[x] for every f ∈ R.
Let {v 1 , . . . , v d } be a canonical basis of D as a left E-module. Then the set {v 1 , . . . , v d , v 1 t, . . . , v d t, . . . v d t p e −1 } is a basis of (D(x), δ, d 0 + x) as a left module over E(x). We can therefore analogously obtain a representation ρ of (D(x), δ, d 0 +x) by matrices in M dp e (E(x)) with respect to that basis. For f (t) ∈ R, the matrix ρ(f (t)) has entries in E[x], therefore it follows that N (f (t)) = det(ρ(f (t))) ∈ E[x] ∩ F (x) = F In particular, N (f ) has degree dm.
The proof follows analogously to the proof of Theorem 10.
Corollary 27. Let D = (E, δ 0 , a) be a differential algebra, δ| E be a derivation on E, and let f ∈ D[t; δ] have coefficients in E and be monic. Then N (f (t)) = ±x dm + . . . .
Proof. Through direct computations of the left regular representation of D, we see that for each a ∈ E, ω(a) is a lower triangular matrix with each entry on the lead diagonal equal to a. Hence the result follows analogously to Theorem 26.
Since Theorem 29. Let f ∈ R be a polynomial of degree m. Suppose that deg(ĥ) = dm.
Corollary 30. Let f ∈ R be a monic polynomial of degree m. Suppose that deg(ĥ) = dm. (i) If N (f (t)) =ĥ 1 · · ·ĥ l such thatĥ i ∈ F [x] is irreducible, 1 ≤ i ≤ l, then there exists a decomposition f = f 1 · · · f l of f into irreducible factors, such that N (f i ) =ĥ i for all i, 1 ≤ i ≤ l. The degree of f i in R equals the degree ofĥ i in F [x] for all i.
(ii) Assume that N (f ) is the product of l distinct irreducible factorsĥ 1 · · ·ĥ l in F [x]. Then f has exactly l! irreducible decompositions corresponding to each possible ordering of the factors of N (f ).
The proofs of the above results are identical to the ones of Lemmata 13, 21, Theorem 14 and Corollary 23.