ON THE GRADED ANNIHILATORS OF RIGHT MODULES OVER THE
FROBENIUS SKEW POLYNOMIAL RING

arXiv:1109.4966v2 [] 19 Oct 2012

Ehsan Tavanfar
Dedicated to Professor Hossein Zakeri

Abstract. Let R be a commutative Noetherian ring of prime characteristic and M be an xdivisible right R[x, f ]-module that is Noetherian as R-module. We give an affirmative answer to the
question of Sharp and Yoshino in the case where R is semi-local and prove that the set of graded
annihilators of R[x, f ]-homomorphic images of M is finite. We also give a counterexample in the
general case.

0. Introduction
Throughout this note, R is a commutative Noetherian ring of prime characteristic p > 0 with identity.
Also f : R → R is the Frobenius endomorphism which maps an element r ∈ R to its pth power. Let
x be an indeterminate. By the Frobenius endomorphism we can endow the free R-module generated
by (xi )i∈N0 with a ring structure such that xr = f (r)x = rp x. The result is called the Frobenius
L skew
Rxn .
polynomial ring. We denote it by R[x, f ]. We can consider R[x, f ] as a positively-graded ring
i∈N0

When R is a local ring with maximal ideal m, the top local cohomology module of R with respect to m
has a left R[x, f ]-module structure. In [1] Rodney Y. Sharp has used this structure to study the existence
of the tight closure test element and parameter test element in certain cases. In order to accomplish these
results, he introduced the graded annihilators of left R[x, f ]-modules. The graded annihilator of a left
R[x, f ]-module H is the largest graded two-sided ideal of R[x, f ] contained in annR[x,f ] H. The theory
of graded annihilators is a novel tool to study the test elements in local rings of prime characteristic:
see [1],[2],[3], for example. One of the key points in this theory is the following finiteness condition
([1, Corollary 3.11]): if H is an x-torsion-free left R[x, f ]-module which is either Noetherian or Artinian
as R-module, then the set of graded annihilators of R[x, f ]-submodules of H is finite.
In [4], Sharp and Yoshino have shown that if (R, m) is an F -finite local ring and E(R/m) is the
injective envelope of the residue field of R, then the restriction of the Matlis duality functor (−)∨ :=
HomR (−, E(R/m)) to the left R[x, f ]-modules provides a contravariant functor from the category of left
R[x, f ]-modules to the category of right R[x, f ]-modules. A similar argument holds for the restriction of
the Matlis duality to the right R[x, f ]-modules. Also they proved that if H is a left (right) R[x, f ]-module,
then the natural evaluation map ωH : H → (H ∨ )∨ is a left (right) R[x, f ]-homomorphism. By applying
these interesting results, Sharp and Yoshino deduced [4, Theorem 3.5]: Assume that R is F -finite, local
and complete. Let M be an x-divisible right R[x, f ]-module which is Noetherian as R-module. Then
there are only finitely many graded annihilators of R[x, f ]-homomorphic images of M .
This leads to the following question which is proposed in [4].
Question. Assume that R is a commutative Noetherian ring of prime characteristic p > 0. Let M be
an x-divisible right R[x, f ]-module that is Noetherian as an R-module. Is the set of graded annihilators
of R[x, f ]-homomorphic images of M finite?
2010 Mathematics Subject Classification. 13A35, 13E05, 16S36.
Key words and phrases. Commutative Noetherian ring, prime characteristic, skew polynomial ring.

1

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GRADED ANNIHILATORS

Note that this question can be considered as a dual to the finiteness condition mentioned above. We
give an affirmative answer to this question in the case where R is semi-local. Furthermore, we give a
counterexample in the general case.

1

The Results

Definitions and Remarks 1.1. One can easily verify that an ideal B of R[x, f ] is a two-sided
L graded
bn xn .
ideal if and only if there exists an ascending chain {bn }n∈N0 of ideals of R such that B =
n∈N0

Assume that K is a right R[x, f ]-module.

An R[x, f ]-submodule N of K is said to be a special submodule if N is of the form KB for some
two-sided graded ideal B of R[x, f ].
We denote the annihilator of K by ann(K)R[x,f ] . Note that,
ann(K)R[x,f ] = {α ∈ R[x, f ] : kα = 0 for all k ∈ K},
is a two-sided ideal of R[x, f ]. The graded annihilator of K is defined as follows:
gr-ann(K)R[x,f ] = {

n
P

i=0

ri xi : n ∈ N0 and ri ∈ R, ri xi ∈ ann(K)R[x,f ] for all 0 ≤ i ≤ n}.

In fact, the graded annihilator of K is the largest graded two-sided ideal of R[x, f ] which annihilates
K.
We say that K is x-divisible if K = Kx.
Lemma 1.2. Let K be a right R[x, f ]-module. Suppose that B and B′ are two-sided graded ideals of
R[x, f ] and that N and N ′ are R[x, f ]-submodules of K.
i) Whenever B ⊆ B′ , KB ⊆ KB′ .
ii) Whenever N ⊆ N ′ , gr-ann(K/N )R[x,f ] ⊆ gr-ann(K/N ′ )R[x,f ] .
iii) K(gr-ann(K/N )R[x,f ] ) ⊆ N .
iv) There exists a one-to-one order-preserving correspondence Γ between graded annihilators of R[x, f ]homomorphic images of K and special submodules of K given by
Γ : KB 7−→ gr-ann(K/KB)R[x,f ] .
The inverse map Γ−1 is given by
Γ−1 : gr-ann(K/N )R[x,f ] 7−→ K(gr-ann(M/N )R[x,f ] ).
Proof. Statements (i) and (ii) are obvious. For all α ∈ gr-ann(K/N)R[x,f ] , we have Kα ⊆ N and
this

proves (iii). Now, we prove (iv). By parts (ii) and (iii), we have gr-ann K/ K(gr-ann(K/N )R[x,f ] )
⊆
R[x,f ]

gr-ann(K/N )R[x,f ] . For the reverse inclusion
suppose that rxn ∈ gr-ann(K/N
)R[x,f ] . Then K(rxn ) ⊆



. Therefore
K(gr-ann(K/N )R[x,f ] ). So rxn ∈ gr-ann K/ K(gr-ann(K/N )R[x,f ] )
R[x,f ]



gr-ann K/ K(gr-ann(K/N )R[x,f ] )

R[x,f ]

= gr-ann(K/N )R[x,f ] .

On the other hand, parts (i) and (iii) together with the fact that B ⊆ gr-ann(K/KB)R[x,f ] imply that
KB = K(gr-ann(K/KB)R[x,f ] ).

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Ehsan Tavanfar

Corollary 1.3. Let X be the the set of all special submodules of a right R[x, f ]-module K. Then X
satisfies the ascending chain condition.
Proof. By [1, Lemma 1.4] ([1, Lemma 1.4] is deduced by applying the idea of proof of Yoshino’s [5,
Corollary (2.7)]), R[x, f ] satisfies the ascending chain condition on two-sided graded ideals. Hence the
corollary follows from the previous lemma.
Throughout the rest of this article, we assume that M is an x-divisible right R[x, f ]-module.
Lemma
1.4. Suppose that 0 :R M = b. Then b is a radical ideal of R and gr-ann(M )R[x,f ] = bR[x, f ] =
L
bxn .

n∈N0

Proof. There exists an ascending sequence {an }n∈N0 of ideals of R such that a0 = 0 :R M = b and
L
an xn .
gr-ann(M )R[x,f ] =
n∈N0

√

n

n

Let r ∈ a0 . So we have rp ∈ a0 for some n ∈ N0 , i.e. M rp = 0 for some non-negative integer n. But
n
then, 0 = (M rp )xn = M (xn r) = (M xn )r = M r. It follows that a0 is a radical ideal. Now, it remains
n
to show that an ⊆ a0 for each n ∈ N. Let an ∈ an , i.e. M (an xn ) = 0. Then (M apn −1 )(an xn ) = 0 which
implies that (M xn )an = 0, and hence M an = 0. Thus an ∈ a0 .
Definition 1.5. L
We say that an ideal b of R is a special M -ideal if there exists an R[x, f ]-submodule N
bxn = gr-ann(M/N )R[x,f ] .
of M such that
n∈N0
T
We will denote the set of all special M -ideals by I(M ). Also we will denote I(M ) Spec(R) by
I s (M ).
Corollary 1.6. There exists a one-to-one order-preserving correspondence between I(M ) and the set of
special submodules of M .
Remark 1.7. The following hold:
i) For a family {Ni }i∈I of R[x, f ]-submodules of M we have
that I(M ) is closed under arbitrary intersection.

T

(0 :R (M/Ni )) = 0 :R M/(

T

i∈I

i∈I


Ni ) . So

ii) Special submodules of L
M are x-divisible. To see why this is the case, consider a special submodule
bn xn is a two-sided graded ideal of R[x, f ] (Hence {bn }n∈N0 is an ascending
M B of M where B =
n∈N0

chain of ideals of R). Let mbn xn ∈ M B (m ∈ M, bn ∈ bn , n ∈ N0 ). Since M is x-divisible, we
have m = m′ x for some m′ ∈ M . Hence mbn xn = (m′ x)bn xn = m′ (xbn xn ) = (m′ bpn xn )x ∈ (M B)x.
Therefore, the fact that each element of M B can be written as a sum of the elements of the form
mbn xn , implies that M B ⊆ (M B)x.

iii) Assume that a, b are ideals of R. Then (M (aR[x, f ]))(bR[x, f ]) = M ((ab)R[x, f ]).
Theorem 1.8. Assume that N is a special submodule of M and that b is the corresponding special√M ideal to N (Note that
T b =T0 :R (M/N ), N = M (gr-ann(M/N )R[x,f ] ) = M (bR[x, f ]) and that b = b).
Suppose that b = p1 . . . pn is a minimal primary decomposition of b with n ≥ 2 and that
T U and VTare
pi
pi , c =
two non-empty disjoint subsets of {1, . . . , n} such that their union is {1, . . . , n}. Let a =
i∈U

i∈V

and L = M (aR[x, f ]). In addition, assume that b′ is an ideal of R such that N ⊆ M (b′ R[x, f ]). Then
i) gr-ann(M (b′ R[x, f ])/N )R[x,f ] = (b :R b′ )R[x, f ].
ii) gr-ann(L/N )R[x,f ] = cR[x, f ]. In particular, c ∈ I(L).
iii) The corresponding M -special ideal to L is a. In particular, a ∈ I(M ).

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GRADED ANNIHILATORS

Proof.

i) Since M (b′ R[x, f ]) is an x-divisible right R[x, f ]-module, we only need to prove that
0 :R M (b′ R[x, f ])/N = b :R b′ .

It is easy to verify that b :R b′ ⊆ 0 :R M (b′ R[x, f ])/N . For the reverse inclusion, let m ∈ M ,
r ∈ 0 :R M (b′ R[x, f ])/N and t ∈ 0 :R M/M (b′ R[x, f ]). Then mt ∈ M (b′ R[x, f ]) and (mt)r ∈ N . So
tr ∈ 0 :R M/N . This implies that (0 :R M/M (b′ R[x, f ]))r ⊆ 0 :R M/N . But,
b′ ⊆ 0 :R M/M (b′ R[x, f ]).
Hence b′ r ⊆ 0 :R M/N = b, i.e. r ∈ (b :R b′ ).
ii) By [1, Lemma 3.5(ii)], b :R a = c.
iii) Suppose that gr-ann(M/L)R[x,f ] = dR[x, f ]. If we prove that d ⊆ a, thenTwe are
T done. Note that
d(0 :R L/N ) ⊆ 0 :R M/N = b and 0 :R L/N = c. Thus dc ⊆ b. Since p1 . . . pn is the minimal
primary
T decomposition of b, we have c * pi for each i ∈ U . But dc ⊆ b ⊆ pi for each i ∈ U . Therefore
pi = a.
d⊆
i∈U

Corollary 1.9. I(M )\{R} is precisely the set of all finite intersections of members of I s (M ).
Theorem 1.10. Suppose that M is Noetherian as R-module. Then
I s (M ) = {p ∈ Spec(R) : p ∈ AssR (M/M (pR[x, f ]))}.
In particular, I s (M ) ⊆ SuppR (M ).
Proof. Note that p ∈ AssR (M/M (pR[x, f ])) if and only if p = 0 :R M/M (pR[x, f ]), and the latter
is equivalent to that p ∈ I s (M ). On the other hand, if p ∈ AssR (M/M (pR[x, f ])), then we have
p = 0 :R (m + M (pR[x, f ])) for some m ∈ M . Hence
p ⊆ 0 :R (m + M p) ⊆ 0 :R (m + M (pR[x, f ])) = p.
Therefore p ∈ AssR (M/M p). Thus I s (M ) ⊆ {p ∈ Spec(R) : p ∈ AssR (M/M p)} ⊆ SuppR (M ).
Theorem 1.11. Suppose that L := M (aR[x, f ]) for some ideal a ∈ I(M )\{0 :R M }. Then
I s (L) = {p ∈ I s (M ) : a * p}.
Proof. Let p ∈ I s (L), i.e. p = 0 :R L/L(pR[x, f ]). Let b ∈ 0 :R M/M (pR[x, f ]) and m(axi ) ∈ L
(m ∈ M, a ∈ a and i ∈ N0 ). By 1.7(iii), we have
i

(maxi )b = (mbp )axi ∈ (M (pR[x, f ]))(aR[x, f ]) = (M (aR[x, f ]))(pR[x, f ]) = L(pR[x, f ]).
It follows that 0 :R M/M (pR[x, f ]) ⊆ 0 :R L/L(pR[x, f ]) = p. Thus, 0 :R M/M (pR[x, f ]) = p, i.e.
p ∈ I s (M ).
We have L(pR[x, f ]) = M ((ap)R[x, f ]) = M (cR[x, f ]) for some radical ideal c such that
c = 0 :R M/M ((ap)R[x, f ]).
So if a ⊆ p, then a2 ⊆ ap ⊆ c ⊆ a, hence c = a and L(pR[x, f ]) = L which contradicts the choice of p. It
follows that p ∈ {p′ ∈ I s (M ) : a * p′ }.
Conversely, let p ∈ {p′ ∈ I s (M ) : a * p′ }. Note that, in particular, p :R (a + p) = p. Hence by
Theorem 1.8(i), p = 0 :R M ((a + p)R[x, f ])/M (pR[x, f ]).
For each r ∈ 0 :R L/L(pR[x, f ]) and m((a + p)xi ) ∈ M ((a + p)R[x, f ]) (m ∈ M, a ∈ a, p ∈ p and
i ∈ N0 ) we have
(m((a + p)xi ))r = (m(axi ))r + (m(pxi ))r ∈ M (pR[x, f ]).

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Ehsan Tavanfar

It follows that 0 :R L/L(pR[x, f ]) ⊆ 0 :R M ((a + p)R[x, f ])/M (pR[x, f ]) = p. Thus p ∈ I s (L).
Theorem 1.12. Assume that R is a semi-local ring and that M is Noetherian as R-module. Then there
are only finitely many graded annihilators of R[x, f ]-homomorphic images of M .
Proof. Firstly, we claim that the set of maximal member of I s (M ) is finite. Suppose the contrary, and
let {pi }i∈N be a countable (infinite) subset of the maximal members of I s (M ). By our hypothesis,
M ((

n
T

pi) + pn+1 )R[x, f ]) = M

i=1

for all n ≥ 1. So using the exact sequence

 n
T
T
M (piR[x, f ])
M (pn+1 R[x, f ]) −→
0 −→ M/
i=1


n
n
L


T
T
M/M (pn+1 R[x, f ]) −→ M/
M (pi R[x, f ]) + M (pn+1 R[x, f ]) −→ 0
M/( M (pi R[x, f ]))
i=1

i=1

and by induction on n, we deduce that M/

n
T

i=1

n
 L
M/M (piR[x, f ]) for each n ∈ N.
M (pi R[x, f ]) ∼
=
i=1

We denote the jacobson radical of R by a. Furthermore, if an R-module L is Noetherian, then we use
ν(L) to denote the length of the R-module L/aL. By the previous paragraph, there exists an epimorphism
n
n

P
L
ν M/M (pi R[x, f ]) for all n ∈ N. Thus
M/M (pi R[x, f ]) for each n ∈ N. Therefore ν(M ) ≥
M→
i=1 
i=1
there must exist a natural number i such that ν M/M (pi R[x, f ]) = 0, hence M/M (pi R[x, f ]) = 0 (by
Nakayama’s lemma). This is a contradiction. Therefore the set of maximal members of I s (M ) is finite.
By Corollary 1.9, we only need to prove that I s (M ) is finite. Suppose the contrary and let dim(R) = t.
By the previous paragraph I s (M ) has only finitely many maximal members. If t = 0, then each member
of I s (M ) is maximal which is a contradiction. So t > 0. Let M0 = M and a1 be the intersection of the
maximal members of I s (M ). By Remark 1.7(i), a1 ∈ I(M ). Note that 0 :R M 6= a1 , otherwise I s (M ) is
finite (Assume that q ∈ I s (M ) and that {pi }ni=1 are maximal members of I s (M ). If 0 :R M = a1 , then
n
T
pi . Hence q = pi for some i by the maximality).
we have q ⊇ 0 :R M = a1 =
i=1

Let M1 = M (a1 R[x, f ]). By Theorem 1.11

I s (M1 ) = I s (M )\{maximal members of I s (M )}
and I s (M1 ) is infinite. Assume that, by induction, we have constructed an x-divisible R[x, f ]-submodule
Mt of M such that
I s (Mt ) = I s (Mt−1 )\{maximal members of I s (Mt−1 )}
and I s (Mt ) is infinite. As before, let at+1 be the intersection of maximal members of I s (Mt ) and
Mt+1 = Mt (at+1 R[x, f ]). Note that again at+1 6= 0 :R Mt . Therefore I s (Mt+1 ) is infinite (In particular,
it is non-empty). Let pt+1 ∈ I s (Mt+1 ). There exists a maximal member of I s (Mt ), say pt such that
pt+1 ( pt . Proceeding in this way we can make an ascending chain of prime ideals of R of length t + 1.
This is a contradiction.
Corollary 1.13. Suppose that M is Noetherian as R-module and that
A := {m ∈ M ax(R) : there exists a maximal member p ∈ I s (M ) such that p ⊆ m}
is a finite set. Then the set of graded annihilators of R[x, f ]-homomorphic images of M is finite.
In particular, if the set of maximal members of I s (M ) is finite, then there are only finitely many graded
annihilators of R[x, f ]-homomorphic images of M .

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GRADED ANNIHILATORS

Proof. Suppose the contrary and assume that I s (M ) is not finite. By our hypothesis, there exists a
maximal ideal m of R such that there are infinitely many members of I s (M ) contained in m. By
msp−1 x
for all
[4, Lemma 3.3], Mm has an x-divisible right Rm [x, f ]-module structure such that m
s x =
s
m ∈ M and s ∈ R\m. Since (M (qR[x, f ]))m = Mm ((qRm )Rm [x, f ]), by applying Theorem 1.10, we
deduce that,
s
IR
(Mm ) = {qRm : q ∈ I s (M ), q ⊆ m}.
m

s
But then IR
(Mm ) is infinite which contradicts the previous theorem.
m
Similarly, the second assertion can be proved by localization at maximal members of I s (M ).

The following example shows that Theorem 1.12 does not hold without the assumption for the ring
to be semi-local.
Counterexample 1.14. Consider the polynomial ring
A := K[θ1 , θ2 , . . . , θn , . . .]
over a perfect field K of prime characteristic p > 0, in a countably infinite set of indeterminates. Let
pi := (θi ) for each i ∈ N. Thus {pS
i }i∈N is a family of prime ideals. Therefore we can consider the
multiplicative closed subset S := A\( pi ) of A. It is easy to verify that R := S −1 A is a Noetherian ring
i∈N

of dimension 1. We are going to prove that R has an x-divisible right R[x, f ]-module structure and that
S −1 pi ∈ I s (R) for each i ∈ N.
The Frobenius map can be used to endow A with a left A-module structure, denoted by F (A), such that
b.a = bp a for all a, b ∈ A. We define the map h : F (A) → A by
h(

t
P

i=0

α

αi,ni

ki θ1 i,1 . . . θni

)=

t
P

i=0
αi,1 ,...,αi,ni are
multiple of p

1/p αi,1 /p
θ1

ki

αi,ni /p

. . . θni

,

for all t ∈ N0 , n0 , . . . , nt ∈ N0 , k0 , . . . , kt ∈ K and α0,1 , . . . , α0,n1 , . . . , αt,1 , . . . , αt,nt ∈ N0 . It is straightforward to check that h is an A-homomorphism which splits the exact sequence
f

p

0 → A → F (A) →
p

F (A)
A

→ 0.

Therefore there exists an A -epimorphism π : A → A such that π(a) = h(a)p for all a ∈ A. Hence by
1
[4, Example 1.9(ii)], A is an x-divisible right A[x, f ]-module such that ax = π(a) p = h(a) for all a ∈ A.
So By [4, Lemma 3.3], R has a natural x-divisible right R[x, f ]-module structure.
It is easy to verify that pi = A(pi A[x, f ]) for each i ∈ N0 . So pi ∈ AssA (A/A(pi A[x, f ])). Thus

S −1 pi ∈ AssR R/S −1 (A(pi A[x, f ])) .

But S −1 (A(pi A[x, f ])) = S −1 A(S −1 pi (S −1 A[x, f ])) = R(S −1 pi R[x, f ]). Hence
S −1 pi ∈ AssR (R/R(S −1 pi R[x, f ])).

Therefore S −1 pi ∈ I s (R) by Theorem 1.10. This implies that I s (R) is infinite.
(The statements of [4, Example 1.9(ii)] and [4, Lemma 3.3] still hold if we drop the assumption that
R is Noetherian.)
Theorem 1.15. Assume that M is Artinian as R-module. Then the set of graded annihilators of R[x, f ]homomorphic images of M is finite.
Proof. Suppose, to the contrary, that I s (M ) is infinite. Let p be a maximal member of I s (M ) and
M1 = M (pR[x, f ]). Note that M1 ( M . By Theorem 1.11, I s (M1 ) is infinite. Let M0 = M and t ≥ 1.
Assume, inductively, that we have constructed an x-divisible R[x, f ]-submodule Mt of M such that
Mt ( Mt−1 and I s (Mt ) is infinite. Let pt be a maximal member of I s (Mt ) and Mt+1 = Mt (pt R[x, f ])
again. So Mt+1 ( Mt and I s (Mt+1 ) is infinite. Hence by induction, we have constructed a strictly
descending chain of R-submodules of M which is a contradiction.

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Ehsan Tavanfar

Acknowledgements. I would like to thank and express my deepest gratitude to my MSc supervisor,
Professor Hossein Zakeri, for his valuable comments, patience and encouragement during preparation of
this paper and my thesis.
References
[1] Rodney Y. Sharp, Graded annihilators of modules over the Frobenius skew polynomial ring, and tight closure, Transactions Amer. Math. Soc. 359 (2007), 4237-4258.
[2]
[3]

, Graded annihilators and tight closure test ideals, J. of Algebra 322 (2009), 3410-3426.
, An excellent F-pure ring of prime characteristic has a big tight closure test element, Transactions Amer. Math.
Soc. 362 (2010), 5455-5481.

[4] Rodney Y. Sharp and Y. Yoshino, Right and left modules over the Frobenius skew polynomial ring in the F-finite case,
Mathematical Proceedings of the Cambridge Philosophical Society 150 (2011), 419-438.
[5] Y. Yoshino, Skew-polynomial rings of Frobenius type and the theory of tight closure, Communications in Algebra 22
(1994), 2473-2502.

Faculty of Mathematical Sciences and Computer, Tarbiat Moallem University, 599 Taleghani Avenue, Tehran 15618, IRAN.
School of Mathematics, Institute for Research in Fundamental Sciences (IPM), P.O. Box: 193955746, Tehran, IRAN.
E-mail address: tavanfar@gmail.com

