International Journal of Algebra, Vol. 8, 2014, no. 14, 687 - 697
HIKARI Ltd, www.m-hikari.com
http://dx.doi.org/10.12988/ija.2014.4887
Irreducibility of Polynomials and the Resultant
Boukari Dahani
Université de Ouagadougou 03 B.P. 7021
Ouagadougou, Burkina Faso
Gérard Kientega
Département de mathématiques et d’informatique
Université de Ouagadougou 03 B.P. 7021
Ouagadougou, Burkina Faso
c 2014 Boukari Dahani and Gérard Kientega. This is an open access article
Copyright distributed under the Creative Commons Attribution License, which permits unrestricted
use, distribution, and reproduction in any medium, provided the original work is properly
cited.
Abstract
Let f and g be polynomials of degrees p and q respectively over a
noetherian ring A (with p > q). If the polynomial g is monic then we
know whether g divides f or not. Otherwise, we have no means to solve
this problem. In this paper, our purpose is to give conditions on the
coefficients of f and g for divisibility of f by g. Using the result, we will
obtain conditions of irreducibility of any polynomial. This result can
be extended to polynomials over a commutative ring. We follow ideas
of S. S. Woo.
Mathematics Subject Classification: 11R09, 12D05, 13B25
Keywords: resultant, Fitting invariant, irreducible polynomial
1
Introduction
Given f and g two monic polynomials over a noetherian ring, S.S.Woo found in
[4] a condition of divisibility of f by g by looking at the resultant matrix. Thus
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Boukari Dahani and Gérard Kientega
she got a criterion of irreducibility for a monic polynomial and a generalization
of the Eisenstein’s irreducibility criterion. In this paper we will generalize her
results to not necessarily monic polynomials over a noetherian ring. In section
2, we give some preliminary results that will be needed in the others parts of the
paper. In section 3, for a given polynomial f we give criteria of irreducibility
and a condition using diophantine equations for the reducibility of f . In section
4, we generalize Eisenstein’s irreducibility criterion. Throughout all rings are
commutative with a unit element denoted by 1.
2
Basic results
In this section we generalize some results obtained in the case of monic polynomials [4]. We first define some notions to make things clear. In this part
the ring A is not nesessarily noetherian.
A zero divisor in A is a nonzero element a ∈ A such that there exists a
nonzero element b ∈ A with ab = 0.
An element c of A is nilpotent if there exists an integer n > 0 such that
n
c = 0; the smallest such integer is the nilpotency of c.
Throughout for f (X) = ap X p + · · · + a0 , g(X) = bq X q + · · · + b0 two
polynomials over A with respective degrees p and q, we set Y = ap bq X,
p
q q−1
F(Y) = ap−1
p bq f (X), G(Y) = ap bq g(X). Observe that F and G are monic
polynomials in Y. We also suppose that ap and bq are not zero divisors.
Given a strictly positive integer n, Sn will denote the A-submodule of A[Y]
consisting of the polynomials of degree < n. Since ap and bq are not zero
divisors then B = (Y n−1 , · · · , 1) is a basis of Sn .
Choosing bases B1 (respectively B2 ) of Sq × Sp (respectively Sp+q ) by
B1 = {(Y q−1 , 0), (Y q−2 , 0), · · · ,(Y, 0), (1, 0),(0, Y p−1 ), · · · , (0, Y), (0, 1)}
and B2 = {Y p+q−1 , · · · , 1} we observe that the resultant matrix of F and G
which are elements of A[Y] that we denote by R(F, G), is the matrix of the
A-linear map ϕ: Sq × Sp → Sp+q , (u, v) 7→ uF + vG with respect to B1 and
B2 .
Proposition 2.1 The A − − module A[Y ]/(F ) is free of rank p.
Proof. Indeed, let H be an element of A[Y]. Since F is a monic polynomial of
A[Y], then there exist Q and R two polynomials of A[Y] such that
H = Q·F + R with R = 0 or deg R ≤ p − 1. Then in A[Y]/(F), we have
H = R and B = ((Y )p−1 , (Y )p−2 , · · · , 1 ) generates A[Y]/(F). If
p−1
αp−1 (Y )p−1 + · · · + α0 1 = 0 then αp−1 app−1 bp−1
+ · · · + α0 = 0 and
q X
k−1 k−1
αk−1 ap bq = 0 for all k ∈ {p, p − 1, · · · , 1}. Thence αk−1 = 0 for k =
1, · · · , p.
Lemma 2.2 A[Y ] = Sp+q + (F ).
689
Irreducibility of polynomials and the resultant
Proof. Since (F) and Sp+q are submodules of A[Y] then
(F) + Sp+q is a submodule of A[Y]. Let h be an element of A[Y]. Then there
exist two polynomials Q and R of A[Y] such that h = QF + R with
R = 0 or deg R < p. Then h ∈ (F) + Sp+q and A[Y] ⊂ (F) + Sp+q .
Lemma 2.3 Sp+q = Sp ⊕ F Sq .
Proof. Let W be an element of Sp ∩ FSq . If W is a nonzero polynomial then
there exists a polynomial u of degree at least one such that
W = F · u and we have deg W = deg F + deg u; so W = 0. Moreover, it is
obvious that Sp + F Sq ⊂ Sp+q and if u is an element of Sp+q , then
u = Q · F + R with R = 0 or deg R < deg F. Thus, Sp+q ⊂ FSq + Sp and
Sp+q ⊂ FSq + Sp .
Proposition 2.4 We have F Sq = Sp+q ∩ (F ).
Proof. Let w0 be an element of FSq . Then there exists an element w1 of Sq
such that w0 = Fw1 . Since F is a monic polynomial, then we have deg w0 =
deg F + deg w1 . Thus, deg w0 < p + q and w0 is an element of Sp+q . Then
FSq ⊂ Sp+q ∩ (F).
Now let h0 be an element of Sp+q ∩ (F). Then there exists an element h1
of A[Y] such that h0 = h1 F and deg h0 < p + q. We have
deg h0 = deg h1 + deg F. Then deg h1 < q and h0 is an element of FSq . Thus,
Sp+q ∩ (F) ⊂ FSq and FSq = Sp+q ∩ (F). By the second isomorphism theorem, Lemma 2.2 and Proposition 2.4 we have an isomorphism δ: Sp+q /F Sq ∼
=
A[Y]/(F).
Define the map ϕ’: Sp+q → Sp+q by ϕ’(uF + v) = uF + vG; then
ϕ0 |F Sq = id.
Now define the map θ: Sq × Sp → Sp+q by θ(u, v) = uF + v and the map ψ:
A[Y]/(F) → A[Y]/(F), h 7→ h · G. Then we have the following result.
Theorem 2.5 The following diagram where cl is the canonical surjection
is commutative.
Sq × Sp
θ
ϕ0
ϕ
Sp+q
/ Sp+q
id
cl /
Sp+q /F Sq
ϕ0
/ Sp+q
cl /
δ
/
A[Y ]/(F )
ψ
Sp+q /F Sq o
δ −1
A[Y ]/(F )
Proof. Consider the diagram:
Sq × Sp
/
θ
ϕ0
ϕ
Sp+q
Sp+q
id
/
Sp+q .
(1)
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Boukari Dahani and Gérard Kientega
Let u be an element of Sq and v an element of Sp . We have
ϕ0 o θ(u, v) = ϕ0 (uF + v) = uF + vG and id o ϕ(u, v) = ϕ(u, v) = uF +
vG. Then ϕ0 o θ = id o ϕ and diagram ( 1) is commutative. Now consider the
diagram:
cl
Sp+q
ϕ0
/
(2)
ϕ0
cl /
Sp+q
Sp+q /F Sq
Sp+q /F Sq .
Let w be an element of Sq . Since ϕ0 |FSq = id, we have
cl o ϕ0 (Fw) = cl(Fw) = 0 because Fw is an element of FSq = ker cl. Then, FSq
⊂ ker (cl o ϕ0 ). By the first isomorphism theorem, there exists one and only
one morphism ϕ0 : Sp+q /F Sq → Sp+q /F Sq such that the following diagram is
commutative:
ϕ0
Sp+q
cl /
/ Sp+q
4
Sp+q /F Sq .
cl
Sp+q /F Sq
Then diagram ( 2) is commutative. To finish the proof, let us consider the
following diagram:
Sp+q /F Sq
ϕ0
Sp+q /F Sq o
δ
/
A[Y ]/(F )
ψ
δ −1
(3)
A[Y ]/(F ).
Let h be an element of Sp+q . Set h = cl(h). We have ϕ0 o cl(h) = cl o ϕ0 (h).
Then ϕ0 (h) = ϕ0 (h) = ϕ0 (uF + v) with h = uF + v, where u and v are some
uniquely defined elements of Sq and Sp respectively. (The unicity is due to the
equality Sp+q = Sp ⊕ F Sq .) Then we have: ϕ0 (h) = uF + vG + FSq = vG +
FSq because uF belongs to FSq . Moreover, since δ(cl(h)) = h + (F) and
GuF ∈ (F ) we have:
δ −1 o ψ o δ(h) = δ −1 o ψ(h + (F)) = δ −1 (GuF + Gv + (F)) =
δ −1 (Gv + (F)) = Gv + FSq ( because deg Gv ≤ deg G + deg v < q + p).
Hence, diagram ( 3) is commutative.
Proposition 2.6 Let M be the cokernel of
ψ : A[Y ]/(F ) → A[Y ]/(F ), h 7→ h · G.
If G is a factor of F then M ∼
= A[Y]/(G).
Irreducibility of polynomials and the resultant
691
Proof. Indeed, since M is the cokernel of ψ we have
∼
M =(A[Y]/(F))/Im ψ. Since F is a multiple of G, we have (F) ⊂ (G). Let t be
an element of Im ψ. There exists an element y of A[Y]/(F) such that ψ(y) =
t. We have ψ(y) = yG = t. Since y·G is an element of (G), t is an element of
(G)/(F). Thus, Im ψ ⊂ (G)/(F). Now let u be an element of (G)/(F). Then u
is an element of (G), and there exists an element k of A[Y] such that
u = kG. Now, ψ( k) = kG = u and ψ( k) is an element of Im ψ. Then we
have (G)/(F) ⊂ Im ψ and Im ψ = (G)/(F). The first isomorphism theorem
gives A[Y]/(G) ∼
= M.
Corollary 2.7 Let M be the cokernel of
ψ : A[Y ]/(F ) → A[Y ]/(F ), h 7→ h · G.
If G is a factor of F then M is also the cokernel of
ϕ : Sq × Sp → Sp+q , (u, v) 7→ uF + vG
Proof. Let Gv0 be an element of GSp . We have ϕ(0, v) = Gv0 . Then
GSp ⊆ Im ϕ. Let h be an element of Im ϕ. Then there exists an element u0 of
Sq and an element v 0 of Sp such that u0 F + v 0 G = h. Now u0 F is an element of
GSp because (F) ⊆ (G) and v 0 G is an element of GSp . Thus, h is an element
of GSp and Im ϕ = GSp . By the isomorphism
Sp+q /gSp ∼
= A[Y]/(G) and the previous proposition, M ∼
= A[Y]/(G). Then
∼
Sp+q /Im ϕ = M.
Lemma 2.8 Let f (X) = ap X p + · · · + a0 , g(X) = bq X q + · · · + b0 be two
polynomials over A with respective degrees p and q (p > q). Then g is a factor
of f if and only if F = GH and app−q−1 bqp−q+1 divides H.
Proof. If f(X) = g(X)·h(X) then
p
p−1−q p−q+1 q q−1
ap−1
bq
ap bq g(X)h(X). Setting F(Y) = app−1−q bp−q+1
h(X),
p bq f (X) = ap
q
p−q−1
p−q+1
we have F(Y) = G(Y)·H(Y) and ap
· bq
divides H.
Conversely, suppose that F(Y) = G(Y)·H(Y) and app−q−1 bqp−q+1 is a factor
of H. Then there exists a polynomial h(X) over A[X], such that
H(Y) = ap−q−1
bp−q+1
h(X). Therefore, we have
p
q
p−1 p
q q−1
ap bq f(X) = ap bq g(X)ap−q−1
bp−q+1
h(X). Since ap and bq are not zero divisors,
p
q
then f(X) = g(X)h(X).
3
Some irreducibility criteria
For an A–module M and an integer i, we denote by F itti (M) the i-th Fitting
invariant of M. [1]
692
Boukari Dahani and Gérard Kientega
Recall that for f (X) = ap X p + · · · + a0 , g(X) = bq X q + · · · + b0 two
polynomials over A with respective degrees p and q, we set Y = ap bq X,
p−k−1 p−k
q q−1
p
bq for
F(Y) = ap−1
p bq f (X), G(Y) = ap bq g(X). Also we set Ak = ak ap
j q−j−1
for j = 0, 1, · · · , q − 1 will denote
k = 0, 1, · · · , p − 1, and Bj = bj ap bq
the respective coefficients of the polynomials F and G. Recall that also
Ap = Bq = 1.
In this section we suppose that the ring A is noetherian.
Theorem 3.1 Let f (X) = ap X p + · · · + a0 , g(X) = bq X q + · · · + b0 be
two polynomials over A with respective degrees p and q (p > q) which leading
coefficients are not zero divisors. Let M be the cokernel of the map:
ψ : A[Y ]/(F ) −→ A[Y ]/(F ), H 7→ H · G.
Then the following propositions are equivalent:
1. G(Y ) is a factor of F (Y ).
2. M is free of rank q.
3. M is projective of constant rank q.
4. The minors of ψ of size p − q generates A and the minors of size bigger
than p − q of ψ vanish.
5. The minors of R(F, G) of size p generate A and the minors of
R(F, G) of size bigger than p vanish.
Proof.1) ⇒ 2). By Proposition 2.6 we have M ∼
= A[Y]/(G). Then M is free of
rank q.
2) ⇒ 3). Obvious.
3) ⇒ 4). Since A is noetherian M is an A – module with finite presentation.
Thus if M is projective of rank q then F ittq (M ) = A and F ittq−1 (M) = 0.
4) ⇐⇒ 5). Consider the following commutative diagram:
Sq × Sp
/
θ
Sp+q
ϕ0
ϕ
id
∼
=
Sp+q
M
0
/
cl /
ϕ0
Sp+q
cl /
∼
=
/
M
Sp+q /F Sq
/ A[Y ]/(F )
ψ
Sp+q /F Sq o
/
δ
M
δ −1
id
A[Y ]/(F ).
/
M
0
Irreducibility of polynomials and the resultant
693
By Corollary 2.7, the first and the last vertical rows of the diagram are finite
free presentations of the same module M. Therefore by Fitting’s Lemma,
F ittq (R(F, G)) = A and F ittq−1 (R(F, G)) = 0 if and only if F ittq (ψ) = A and
F ittq−1 (ψ) = 0.
5) ⇒ 1). Indeed, consider the following determinant of size p + 1 that is
zero:
0
B
0
.
.
.
.
.
.
.
.
.
.
.
.
0
q
..
..
..
..
.
.
Bq
.
.
.
.
.
.
0
..
..
..
..
..
..
..
...
Ap
.
.
.
..
..
.
... ...
Dc (f , g) = A
B0
.
.
p−1
..
..
..
...
.
.
.
0
B1 . . . . . . . . .
.
.
.
.
.
.
.
.
0
.
B0 . . . . . . . . . . . .
Aq
0
.
.
.
0
B
.
.
.
.
.
.
.
.
.
B
0
q
F (Y ) Y p−1 G(Y ) · . . . . . . . . . . . . Y G(Y ) G(Y ) Expanding this determinant according to the first column we obtain the result.
The previous theorem gives a criterion to know whether a polynomial g which
p
leading coefficient bq is not a zero divisor and divides ap−1
p bq f and a method to
p−1 p
obtain the factorization ap bq f = gh.
Example 3.2 Let f (X) = 5X 3 − 4X 2 − 1 and g(X) = 2X − 2 be two
polynomials over Z. We have F (Y ) = Y 3 − 8Y 2 − 200 and G(Y ) = Y − 10.


1
1
0
0
 −8 −10 1

0
.
R(F, G) = 
 0

0 −10 1
−200 0
0 −10
satisfy the fifth condition of theorem 3.1. Then G(Y) is a factor of F(Y) and
F(Y) = G(Y)·H(Y) with H(Y) = Y 2 + 2Y + 20 = 100X 2 + 20X + 20 =
20(5X 2 + X + 1). Then setting h(X) = 5X 2 + X + 1, we have
10f(X) = g(X)·h(X).
The previous Theorem and the Lemma 2.8 allow us to obtain new results
on divisibility of polynomials.
Corollary 3.3 Let f (X) = ap X p + · · · + a0 , g(X) = bq X q + · · · + b0 be two
polynomials over A with respective degrees p and q. g is a factor of f if and
only if the minors of R(F, G) of size bigger than p vanish and app−q−1 bqp−q+1
divides H, where H is a polynomial defined by F = GH.
694
Boukari Dahani and Gérard Kientega
Example 3.4 Let f (X) = 6X 3 + 3X 2 − 2X − 1 and g(X) = 3X 2 − 1 be
two polynomials over Z. We have F (Y ) = Y 3 + 9Y 2 − 108Y − 972 and
G(Y ) = Y 2 − 108.



R(F, G) = 


1
0
1
0
0
9
1
0
1
0
−108
9
−108
0
1
−972 −108
0
−108
0
0
−972
0
0
−108



.


satisfy the fifth condition of theorem 3.1. Then G(Y) is a factor of F (Y ) and
F (Y ) = G(Y ) · H(Y )withH(Y ) = Y + 9 = 32 (2X + 1). Then
setting h(X) = 2X + 1 we have f (X) = g(X) · h(X).
The previous corollary gives also an irreducibility criterion for a polynomial
over A, which leading coefficient is not a zero divisor in A. Thus, let
gt (X) = tq X q + tq−1 X q−1 + · · · + t0 be a generic polynomial over A which
p
leading coefficient tq is not a zero divisor. Set Ft (Y ) = ap−1
p tq f (X) and
Gt (Y ) = aqp tq−1
q g(X).
p
, there exists a
2
nonzero minor of size > p in R(Ft , Gt ) or if Ft = Gt Ht then ap−q−1
tp−q+1
p
q
divides Ht .
Corollary 3.5 f is irreducible if and only if for all q ≤
Now, we suppose that A is a noetherian domain and we will adapt the
notations of [5, p.106]. Then we obtain conditions for f to be reducible.
Corollary 3.6 Let f (X) = ap X p + · · · + a0 and g(X) = bq X q + · · · + b0 be
two polynomials over A with respective degrees p and q (p > q), which leading
coefficients are not zero divisors. g divides f if and only if
(B0 , B1 , · · · , Bq−1 ) is a solution of the system of equations:

(F,q)

w1 (T0 , · · · , Tq−1 ) = 0


 (F,q)
w2 (T0 , · · · , Tq−1 ) = 0

·····················


 (F,q)
wq (T0 , · · · , Tq−1 ) = 0
divides H where H is defined by F = GH.
and ap−q−1
bp−q+1
p
q
Proof. We apply the ideas of S. S. Woo [5, p.105 − 106] to F and G. As a
consequence of the above Corollary, we have the following result:
695
Irreducibility of polynomials and the resultant
Corollary 3.7 Let f (X) = ap X p +· · ·+a0 be polynomial over A with degree
p, which leading coefficient is not a zero divisor. f is irreducible if and only if
p
for all q ≤ the system of equations:
2

(F ,q)

w1 t (T0 , · · · , Tq−1 ) = 0


 (Ft ,q)
w2
(T0 , · · · , Tq−1 ) = 0

·
·
·
·
·
·
···············


 (Ft ,q)
wq
(T0 , · · · , Tq−1 ) = 0
has no solution or if Ft = Gt Ht then ap−q−1
bp−q+1
does not divide Ht .
p
q
4
Generalized Eisenstein criterion
In this section we will generalize the Eisenstein’s irreducibility criterion. Given
f(X) = ap X p + · · · + a0 a polynomial over a ring B and ϕ: B −→ C a ring
p
X
p
homomorphism, ϕ(f (X)) = ϕ(ap )X + · · · + ϕ(a0 ) =
then ϕ(ai )X i is a
i=0
polynomial over C.
We still suppose that A is noetherian.
Lemma
P 4.1 Let ϕ: B −→ C be a ring homomorphism and let
f(X) = pi=0 ai X i be a polynomial over B of degree p. If ϕ(ap ) is a nonzero
element of A and not a zero divisor in C and ϕ(f (X)) is irreducible over C
then f (X) is irreducible.
This lemma is in fact, a generalization of those of S. Lang [3, p.185] and S. S.
Woo [6, p.502].
P
Theorem 4.2 Let f(X) = pi=0 ai X i be a polynomial over a ring B and let
ϕ: B −→ A be a ring homomorphism such that:
1. ϕ(ap ) is a nonzero element of A and not a zero divisor in A.
2. ϕ(ai ) = 0 for i = 0, · · · , p − 1.
3. If a0 = b0 c0 then either ϕ(b0 ) or ϕ(c0 ) is of of nilpotency > p.
Then f(X) is irreducible over A.
P
Proof. Suppose that f(X) = g(X)·h(X) with g(X) = qj=0 bj X j and
P
h(X) = rk=0 ck X k . Then a0 = b0 · c0 and by the condition 3 of theorem,
either ϕ(b0 ) or ϕ(c0 ) is of nilpotency > p. We may suppose that ϕ(b0 ) is of of
nilpotency > p. (Otherwise reverse the role of g and h.) Since
696
Boukari Dahani and Gérard Kientega
ϕ(f(X)) = ϕ(g(X))· ϕ(h(X)), then setting Z = ϕ(ap ) · ϕ(bq ) · X,
P(Z) = ϕ(ap )p−1 ϕ(bq )p ϕ(f (X)), Q(Z) = ϕ(ap )q ϕ(bq )q−1 ϕ(g(X)), we have P(Z)
= Q(Z)·S(Z) for some polynomial S of A[Z]. Then, the minors of
R(P, Q) of size bigger than p vanish. Now, setting Cj = ϕ(ap )q ϕ(bq )q−1 ϕ(bj )
for j = 0, · · · q − 1 the coefficients of Q(Z) and using the condition 2,
R(P, Q) is of the form:
1 0 ... 0
1
0
......
0
..
. . ..
..
. . Cq−1
.
1
.
0 1
.
..
..
.. 0 . . . 0
.
.
Cq−1
0
. . .
.
.
.
.. 1
..
..
..
.. ..
1
.
...
R(P, Q) = 0
0 C0
C1
Cq−1
.
.
...
0 0 . . . ..
.
0
C
.
0
. .
..
..
..
..
.. . . . . . ...
.
.
.
.
0 ... 0 0
0
.
.
.
0
C
0
Then we have R(P, Q) = C0p = ϕ(ap )pq · ϕ(bq )p(q−1) ϕ(b0 )p which is nonzero
because ϕ(b0 ) is of nilpotency > p and ϕ(ap ) and ϕ(bq ) are nonzero elements
of A and not zero divisors in A.
The following result may be viewed as a generalization of Eisenstein criterion.
In fact, it is a result of S.S. Woo. [6, p. 505].
P
Corollary 4.3 Let ϕ: B −→ A be a ring homomorphism, f(X) = pi=0 ai X i
a polynomial over B of degree p > 0 and k an integer such that 0 ≤ k < p. If:
1. ϕ(ap ) is a nonzero element of A and not a zero divisor in A.
2. ϕ(ai ) = 0 for all i ∈ { 0, 1, · · · , k }.
Then f(X) has no factor g(X) of degree ≥ p − k with ϕ(g(0)) of nilpotency >
k + 1.
Proof. Suppose that f (X) = g(X) · h(X) with g(X) =
q
X
bj X j ,
j=0
h(X) =
r
X
ck X k such that q ≥ p − k and the nilpotency d of ϕ(g(0)) > k +
k=0
1. Then we have F(Y) = G(Y)H(Y). Now, since F is monic then ϕ(Ap ) = 1
and ϕ (Ai ) = ϕ(ai ap−i−1
bp−i
p
q ) = 0 for all i ∈ { 0, 1, · · · , k } because ϕ(ai ) =
0. By [6, p.505], F has no factor of degree ≥ p − k with
ϕ (G(0)) of index of nilpotence > k + 1. Now, we have
Irreducibility of polynomials and the resultant
697
d
ϕ (G(0))d = ϕ (b0 aqp bq−1
q ) = 0. Then ϕ(B0 ) is of nilpotency d.
In the following result we show that the condition 2) of the above corollary
can be replaced by a weaker one.
Corollary 4.4 Let ϕ: B −→ A be a ring homomorphism and let
p
X
ai X i be a polynomial over B of degree ≥ 1 and k an integer such
f (X) =
i=1
that 0 ≤ k < p. If:
1. ϕ(ap ) is a nonzero element of A and not a zero divisor in A.
2. ϕ(ai ) is nilpotent for i = 0, · · · , p − 1.
Then f (X) has no factor g(X) over B of degree ≥ p − k with g(0) a
nonnilpotent element.
Proof. Assuming that f (X) has a factor g(X) over B of degree ≥ p − k with
g(0) a nonnilpotent element, we obtain F = GH for some polynomial H of
A[Y ] and F satisfy to the hypotheses of Theorem 4.6 of [6, p.507] and G(0) is
nonnilpotent, which is impossible.
References
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New York, Berlin, Springer-Verlag, 1995.
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Springer-Verlag, 1999.
[3] S. Lang, Algebra, Springer GTM 211, 2002.
[4] S.S.Woo, Dividing polynomials using the resultant matrix, Comm. Algebra
35, no. 11, (2007), 3263-3272 .
[5] S.S.Woo, Irreducibility of polynomials and diophantine equations, Math.
Soc. 47, no. 1, (2010), 101-112.
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(2008), 499-509.
[7] V.V. Prasolov, Polynomials, Springer-Verlag, 2010.
Received: August 8, 2014