Polynomials Inducing the Zero Function on Local Rings

For a Noetherian local ring (R, m) having a finite residue field of cardinality q, we study the connections between the ideal Z(R) of R[x], which is the set of polynomials that vanish on R, and the ideal Z(m), the polynomials that vanish on m, using what we call pi-polynomials: polynomials of the form p(x) = \prod_{i = 1}^{q} (x - c_i), where c1, ..., cq is a set of representatives of the residue classes of m. When R is Henselian we prove that p(R) = m and show that a generating set for Z(R) may be obtained from a generating set for Z(m) by composing with p(x). When m is principal and has index of nilpotency e, we prove that if e \leq q then Z(m) = (x, m)^e, and if e = q + 1 then Z(m) = (x, m)^e + (x^q - m^{q - 1} x). When R is finite, we prove that Z(R) = \cap_{i = 1}^{q} Z(c_i + m) is a minimal primary decomposition. We determine when Z(R) is nonzero, regular, or principal, respectively, and do the same for Z(m). We prove that when R is complete, repeated application of p(x) + x to elements of R will produce a sequence converging to the roots of p(x). We show that Z(R) is the intersection of the principal ideals generated by the pi-polynomials.


Introduction
One of the surprising facts about finite rings is that a polynomial can be nonzero and yet induce the zero function. An interesting first example is provided by Fermat's Little Theorem: If p is prime, then the nonzero polynomial x p ¡x induces the zero function on Z p . Using this we can build obvious examples, such as ppx p ¡ xq and px p ¡xq 2 on Z p 2 , and more surprising examples, such as px p ¡xq p ¡p p¡1 px p ¡xq on Z p p 1 ; see Corollary 4.5 for more information on these examples. For a ring R, it's easy to see that the set of polynomials in Rrxs that induce the zero function on R is an ideal of Rrxs; we call this the null ideal of R, denoted N pRq. The null ideal has been studied, often with particular focus on the rings R Z p n , for its connection with integer-valued polynomials and functions induced by polynomials [3,5,10,13] and coding theory [6].
For most of our paper, pR, mq is a Noetherian local ring; we will see that N pRq is nonzero only when the residue field R R{m is finite, so we focus most of our attention on this case and let q § § R § § . Many authors have studied the problem of finding a generating set for N pRq, most often in the case R Z p n [1,2,3,7,9,10,11]. In this paper, we argue that in many cases, focus should be shifted from N pRq to the simpler ideal N pmq, which is the set of polynomials that induce the zero function on m. The connection between the two is the ideal N pR, mq, the set of polynomials that take elements of R into m; it's easy to show that N pR, mq px q ¡ x, mq. Certainly polynomials in N pR, mq can be composed with those in N pmq to obtain polynomials in N pRq; visually, N pmq ¥ N pR, mq N pRq. One of the main themes of this paper is to show that in some sense the opposite is true: A generating set for N pRq can be obtained by composing generators of N pR, mq with generators of N pmq (see Theorem 4.2). Example 1.1. Let R Z 9 , so that m p3q and R ! Z 3 . By direct computation or by Theorem 4.4, N pmq px, mq 2 . In Lemma 2.5 we easily find that with πpxq x 3 ¡ x, N pR, mq pπpxq, mq px 3 ¡ x, 3q; from this, according to Theorem 4.2, we deduce N pRq N pmq ¥ N pR, mq px 3 ¡ x, 3q 2 . This makes it clear that the simpler ideal N pmq controls the structure of the generating set of the more complicated ideal N pRq.
The polynomial x q ¡ x has played an important role in the research on null ideals, due primarily to the fact that the image of x q ¡ x generates N R¨and, to a lesser extent, the fact that when R is finite, x q ¡ x maps R surjectively onto m; see A. Bandini's paper [1] for applications of surjectivity when R Z p n . We generalize the surjectivity result in Corollary 2.11 and apply it in Theorem 3.3 and Proposition 4.1. A secondary theme of this paper is that there is actually a class of polynomials with these properties that can play the role of x q ¡ x; we call these π-polynomials, defined to be polynomials of the form πpxq ± q i1 px ¡ c i q where c 1 , . . . , c q is any set of representatives of the residue classes of m. As we will see, x q ¡ x is a π-polynomial when R is Henselian (which holds true in the common case where R is finite). If R is complete, we also provide a computational way of obtaining the factorization of any π-polynomial, such as x q ¡ x itself. In the case of x q ¡ x, this method is as simple as choosing any element of R and repeatedly taking the qth power; the results converge to a root of x q ¡ x. (See Theorems 2.10,

5.2.)
The ideal N pRq for R Z p n was studied as early as 1929 by L. E. Dickson [2,Theorem 27]; in that work, the polynomials in N pRq were referred to as residual polynomials. Dickson found a generating set for N pZ p n q when n ¤ p. In our notation, he found N pRq pπpxq, mq n , where πpxq x p ¡ x and m pR. We generalize and recover this work as another application of our Theorem 4.2: We show in Theorem 4.4 and its corollary that if pR, mq is an Artinian local ring with a principal maximal ideal having index of nilpotency e ¤ q, then N pmq px, mq e , and thus N pRq pπpxq, mq e for any π-polynomial. When e ¡ q, the situation is more complicated, but we take care of the case e q 1; the result is related to results on N pRq for specific rings R ([1, Theorem 2.1] and [7, Theorem II]).
As further indication of the importance of π-polynomials and N pmq, we provide two additional results. Under suitable conditions, we prove in Proposition 2.7 that N pRq is the intersection of the principal ideals generated by the π-polynomials, and in Proposition 2.9 we provide a minimal primary decomposition of N pRq as the intersection of the ideals N pc i mq, where c 1 , . . . , c q is a set of representatives of the residue classes of m. Since generators for N pc i mq may be obtained from generators for N pmq by composition with x ¡ c i , this shows that a primary decomposition for N pRq may be obtained from knowing only a generating set for N pmq.
This result on primary decomposition is a generalization of results from the paper [10] of G. Peruginelli, which was concerned with the ring R Z p n .
The remaining theme of our paper is provided in Theorem 3.3, Theorem 3.4, and Corollary 3.5, where we identify conditions under which N pRq is nonzero, principal, and regular, and the same for N pmq; these results explain why we often focus our attention on finite rings. The results generalize, have some overlap with, and were inspired by R. Gilmer's paper [4].

Null ideals and π-polynomials
We begin with a precise definition of the null ideal of a ring, and we define a class of polynomials that plays an important role in the study of null ideals.
Definition 2.1. Let R be a commutative ring with identity, let S be a subset of R, and let J be an ideal of R. The set N pS, Jq of polynomials in Rrxs which map S into J is easily seen to be an ideal of Rrxs. When the ideal J is omitted, it is assumed to be zero. The focus of this paper is on N pRq, which we call the null ideal of R.
Definition 2.2. Suppose the local ring pR, mq has a finite residue field and let f pxq Rrxs. If f pxq ± q i1 px¡c i q for some set of representatives c 1 , . . . , c q of the residue classes of m, then we call the polynomial f pxq a π-polynomial for R. Example 2.3. We mentioned in the introduction that for R Z 9 , N pmq px, 3q 2 . However, for R Z 8 , N pmq px, 2q 3 . In fact, according to Theorem 4.4 and a few brief calculations, N pmq px, 2q 3 px 2 ¡ 2xq px 2 ¡ 2x, 4xq. We may then use Theorem 4.2 to compose with the π-polynomial πpxq x 2 ¡ x and conclude that N pRq ppx 2 ¡ xq 2 ¡ 2px 2 ¡ xq, 4px 2 ¡ xqq.
The following basic result is a generalized factor theorem that will be useful in a few proofs. This result appeared in Gilmer's proof of Theorem 4 in [4], albeit restricted to units rather than regular elements. We omit the trivial proof of this result.
Lemma 2.4. Let R be a commutative ring with identity. If f pxq Rrxs is a polynomial with roots c 1 , c 2 , . . . , c n such that each difference c i ¡ c j (i $ j) is a regular element, then px ¡ c 1 qpx ¡ c 2 q ¤ ¤ ¤ px ¡ c n q divides f pxq in Rrxs.
Convention. Throughout this paper, we let R be a local ring with maximal ideal m; we do not automatically assume that R is Noetherian. Unless otherwise specified, the residue field R{m will be denoted by R. The image in R of an element r R will be denoted by r. If the residue field is finite, c 1 , . . . , c q will denote a set of representatives of the residue classes of m and πpxq will denote the π-polynomial πpxq ± q i1 px ¡ c i q.
The following lemma may be viewed as a generalization of Fermat's Little Theorem; it is well-known, at least in special cases, as remarked by D. J. Lewis in [7].
A simple but important consequence of this lemma is that πpRq m. Later in Corollary 2.11 we will show that if R is Henselian, then πpRq m.
Lemma 2.5. If pR, mq is a local ring with finite residue field of cardinality q and πpxq is any π-polynomial, then N pR, mq pπpxq, mq.
Proof. Let f pxq N pR, mq; then f pxq N R¨. By Lemma 2.4, f pxq is in the ideal generated by πpxq in Rrxs; pull this back to Rrxs to get f pxq pπpxq, mq.
For the opposite containment, certainly the constant polynomials in m are in N pR, mq. Now suppose πpxq ± q i1 px ¡ c i q. Since any element in R is congruent modulo m to one of the c i , the polynomial πpxq is in N pR, mq, as desired.
Example 2.6. Let R Z p5q (Z localized at the prime ideal p5q), so that m p5q, R Z 5 , and q 5. The polynomial x q ¡ x is not a π-polynomial since it doesn't factor completely over R Q: x 5 ¡ x xpx ¡ 1qpx 1qpx 2 1q. However, as we shall see in Theorem 2.10, this polynomial is a π-polynomial for the Henselian rinĝ R of 5-adic integers. In this case, this is due to the existence of a square root of -1 inR.
By Lagrange's Theorem applied to the group of units of R, it is true that x 5 ¡ x N pR, mq. According to the lemma, we expect x 5 ¡ x pπpxq, mq for any πpolynomial. In fact, if for example we let πpxq px¡2qpx 1qxpx ¡1qpx 2q, then In the next result we show that the polynomials in the null ideal are precisely those polynomials that are multiples of each π-polynomial. Proposition 2.7. Let pR, mq be a local ring with finite residue field R of cardinality q. The null ideal of R is the intersection of the principal ideals generated by the π-polynomials. That is, N pRq pπpxqq, where the intersection is taken over all π-polynomials πpxq for R.
Proof. The fact that any polynomial f pxq N pRq is a multiple of any π-polynomial πpxq follows immediately from Lemma 2.4. This shows N pRq pπpxqq.
Let f pxq pπpxqq and let r R; we show f prq 0. We may extend r to a set r, c 2 , . . . , c q of representatives of the residue classes of m, and thus the polynomial In the following result we give a minimal primary decomposition of N pRq if R is finite. First, we give a simple example: Example 2.8. For R Z 9 , as mentioned in the introduction, we have N pmq px, 3q 2 px 2 , 3xq, so the proposition below gives the following minimal primary decomposition of N pRq: where the ideals on the right are primary for the maximal ideals px, 3q, px ¡ 1, 3q, and px ¡ 2, 3q, respectively. Proposition 2.9. Let pR, mq be a finite local ring with residue field R of cardinality q. Let c 1 , . . . , c q be a set of representatives of the residue classes of m. Then Proof. For the minimality of the decomposition, let j be an integer between 1 and q; we show that N pRq i$j N pc i mq. Let hpxq ± i$j px ¡ c i q e ; then hpxq i$j N pc i mq since m e 0. To see that hpxq does not induce the zero function, note that hpc j q ± i$j pc j ¡c i q e is a product of units, and is thus nonzero.
The proofs of the remaining assertions are straightforward and thus omitted.
Next we provide an equivalent way to view π-polynomials, provided the ring is Henselian; of course, this holds for the finite local rings in which we are mainly interested. For an example where the two conditions below are not equivalent, see Example 2.6. This theorem is also needed in our proof that π-polynomials map R surjectively onto m. (A Henselian local ring is a local ring satisfying Hensel's Lemma.) Theorem 2.10. Let R be a Henselian local ring with finite residue field R of cardinality q. For any polynomial ppxq Rrxs, the following statements are equivalent: (i) The polynomial ppxq is a π-polynomial.
(ii) The polynomial ppxq is monic and maps to x q ¡ x in Rrxs.
Proof. Let ppxq be any π-polynomial. Since R is a field with q elements, by Lagrange's theorem on the group of units of R, x q ¡ x induces the zero function on R. By Lemma 2.4, ppxq divides x q ¡ x in Rrxs. Since these are monic polynomials of the same degree, they are equal.
For the converse, suppose ppxq is any monic polynomial with ppxq x q ¡ x. Let R td 1 , . . . , d q u; as discussed in the previous paragraph, in Rrxs. By Hensel's Lemma, this factorization of ppxq can be pulled back to a factorization in Rrxs: There exist c i in R with c i d i such that ppxq ± q i1 px¡c i q.
In the following corollary, we improve upon part of Lemma 2.5 by showing that the induced function π : R Ñ m is actually surjective when R is Henselian. This generalizes Lemma 1.3 of [1], where A. Bandini proved that, for any prime p, πpRq m in case R Z p n and πpxq x p ¡ x. We use this corollary in our Theorem 3.3, where we characterize finite rings with principal null ideals, expanding upon Gilmer [4]. It is used again in Proposition 4.1, which is fundamental for our Theorem 4.2, which shows that generators for N pRq may be obtained by composing generators for N pmq with a π-polynomial.
Corollary 2.11. If pR, mq is a Henselian local ring with finite residue field R of cardinality q, then πpRq πpcq m for any π-polynomial πpxq and any coset c of m.
Proof. We show that πpcq πpRq m πpcq. The first containment is clear since c R, and we saw the second containment in Lemma 2.5. For the final containment, let m m. By Theorem 2.10, the polynomial πpxq ¡ m is still a π-polynomial, and thus it factors over R: This shows that for each i, πpc i q m. Since c 1 , c 2 , . . . , c q is a set of representatives of the residue classes of m, one of them, say c j , is in c. Thus m πpc j q πpcq, as desired.
3. When N pRq and N pmq are nonzero, regular, or principal In the upcoming Theorems 3.3 and 3.4 we will use the following result from B. R. McDonald. McDonald states and proves the theorem for any finite local ring pR, mq, but the theorem and proof still hold when R is just Artinian. The notation McDonald uses is different from ours but the part we will use is that over an Artinian local ring, any regular polynomial is an associate of a monic polynomial.
McDonald writes µf where we would write f , the image of f in Rrxs. Then there is a monic polynomial f ¦ with µf µf ¦ and, for an element a in R, f paq 0 if and only if f ¦ paq 0. Furthermore, there is a unit v in Rrxs with vf f ¦ .
One of the motivations for the current paper is Theorem 4 from [4], which states that if pR, mq is a zero-dimensional local ring, then N pRq is principal if and only if either R is infinite (when N pRq 0) or R is a finite field (when N pRq is generated by It should be noted that in Gilmer's result, local rings are assumed to be Noetherian. This is clear for several reasons, including his reference to Zariski-Samuel [12], and due to the following, which would be a counterexample to Gilmer's Theorem 4 if the local ring were not assumed to be Noetherian. We claim that N pRq is principal even though R is not infinite and R is not a finite field.
Since R has characteristic 2, x 2 N pmq; thus f Rrxs is in N pmq if and only if its linear term, say f 1 x, is in N pmq, and this is true if and only if f 1 m 0. However, the annihilator of m is 0, so N pmq px 2 q. One may now prove directly that N pRq ppx 2 ¡ xq 2 q, but it's easier to note that R is Henselian and apply our main theorem, Theorem 4.2.
Theorem 3.3. Let pR, mq be a finite local ring and let πpxq be any π-polynomial for R. The following statements are equivalent: (1) R is a field.
Proof. p1q ñ p2q If R is a field, then Rrxs is a principal ideal domain, so N pRq is principal. p2q ñ p3q Assume N pRq is principal. Since πpRq m and m e 0 for some e ¥ 1, N pRq contains regular polynomials, such as πpxq e ; thus, the generator of N pRq must be regular. According to Theorem 3.1, we may assume the generator is monic: N pRq pfpxqq for some monic polynomial in Rrxs. Let r be a nonzero element in the annihilator of m, so that, by Lemma 2.5, rπpxq is a polynomial of degree q in N pRq pfpxqq. This forces f pxq to have degree at most q, and since we know that all polynomials in N pRq are multiples of πpxq (Proposition 2.7), the degree of f pxq is exactly q. Since f pxq is a monic multiple of πpxq with the same degree, f pxq πpxq. p3q ñ p4q If N pRq pπpxqq, then according to Corollary 2.11, 0 πpRq m. From this we easily conclude that N pmq pxq, and thus N pmq is principal. p4q ñ p5q If N pmq is principal, we use an argument similar to the part where we assumed N pRq is principal. Since N pmq contains x e for some e ¥ 1, N pmq contains regular elements, so the generator of N pmq is regular, and we may assume it is monic: N pmq pfpxqq for some monic f pxq Rrxs. Let r be a nonzero element in the annihilator of m, so that rx N pmq. This forces f pxq to have degree at most 1. Since f pxq is monic and f p0q 0, f pxq x, as desired. p5q ñ p1q If N pmq pxq, then m 0, so R is a field.
In the next proposition it becomes clear how the conditions of being Artinian or having a finite residue field affect N pRq and N pmq. We will use the concept of the embedding dimension of R, denoted edim R; this is the minimal number of generators of the maximal ideal. Recall that depth R ¤ dim R ¤ edim R.  In order to make the similarity with the other parts more clear, parts (4), (5), and (6) of the previous proposition were not stated as concisely as possible. Before we present the proof, we state a corollary to clarify those three parts; the proof of the corollary is straightforward and is thus omitted. In the development of this paper, we were particularly interested in the monic polynomials in N pRq, so this corollary explains why we were mainly focused on finite rings. If R does not have depth 0, then R contains a regular element t. Let gpxq g 0 g 1 x ¤ ¤ ¤ g n x n N pmq. Since gptq gpt 2 q gpt 3 q ¤ ¤ ¤ gpt n 1 q 0, there is a matrix equation .

The determinant of this Vandermonde matrix is
Since each t j¡i ¡1 is a unit, this determinant is an associate of a power of t; namely t k where k npn 1qpn 2q{6. After multiplying both sides of the matrix equation by the adjugate and dividing by the unit, we find that t k g i 0 for each i. Since t k is regular, the polynomial gpxq is zero, as desired.
(2) Assume R tc 1 , . . . , c q u and R has depth 0. Since R is Noetherian with depth 0, there is a nonzero element m that annihilates m. By Lemma 2.5, the polynomial πpxq ± q i1 px ¡ c i q is in N pR, mq, so the polynomial mπpxq is a nonzero element of N pRq.
For the converse, if R is not finite, then there is an infinite sequence of elements tc n u n¥1 such that no two come from the same residue class of m; thus each difference c j ¡ c i (j $ i) is a unit. On the other hand, if R does not have depth 0, then there is a regular element t m. Consider the sequence tt n u n¥1 ; for i ¡ j we have t i ¡t j t j pt i¡j ¡1q. Since t j is regular and t i¡j ¡1 is a unit, each difference t i ¡t j (i $ j) is regular.
In either case, we may apply Lemma 2.4 to conclude that any nonzero polynomial in the null ideal has arbitrarily high degree. This is a contradiction, so the null ideal contains only the zero polynomial, as desired.
(3) If R has dimension 0, then m e 0 for some e ¥ 1. Thus x e N pmq.
Conversely, if R has positive dimension, then for any minimal prime ideal p of R, R{p is an integral domain of positive dimension; in particular, it has positive depth. According to Part (1), N pm{pq is the zero ideal of R{p. The image of any f pxq N pmq in pR{pqrxs is in N pm{pq, and thus f pxq prxs. Since the coefficients of f pxq are in every minimal prime, they are all nilpotent, so f pxq is not regular.
(4) Suppose R is finite and dim R 0. Let e be such that m e 0 and let R tc 1 , . . . , c q u. By Lemma 2.5, the polynomial πpxq ± q i1 px¡c i q is in N pR, mq, so the regular polynomial πpxq e is in N pRq.
Conversely, suppose either R is infinite or dim R ¥ 1. If R is infinite, then N pRq does not contain regular polynomials since it is the zero ideal, by Part (2). If dim R ¥ 1, the proof continues as in the "conversely" part of the proof of (3), replacing N pm{pq with N pR{pq. (5) If edim R 0, then R is a field, so m 0 and N pmq pxq. Conversely, assume N pmq pfpxqq for some regular polynomial f pxq Rrxs. By Part (3), dim R 0, so m e 0 for some e ¥ 1. Due to this and the result from McDonald (our Theorem 3.1), we may assume f pxq is monic. By Part (1), depth R 0, so there is an element m m with mx N pmq. This shows that the monic polynomial f pxq must have degree 1; the only choice is f pxq x. From this we see that m 0 so R is a field.
(6) If edim R 0 and R is finite, then R is a finite field, so N pRq pπpxqq for any π-polynomial, by Lemma 2.5.
Conversely, assume N pRq pfpxqq for some regular polynomial f pxq. From Part (4) we know that dim R 0 and R is finite; this implies that R is finite. By Theorem 3.3, R is a finite field, so edim R 0.
In the following example we illustrate the use of this theorem and contrast the behavior of the null ideals over rings with finite and infinite residue fields.
Example 3.6. The ring R Z 2 S, T {pS 2 , ST q is a complete Noetherian local ring with depth zero, dimension one, and a finite residue field with q 2 elements; let s and t be the images of S and T in R. According to Theorem 3.4, N pmq is nonzero but does not contain regular polynomials, and the same goes for N pRq. We argue that N pmq psxq and conclude that N pRq pspx 2 ¡ xqq by applying Theorem 4.2. Certainly N pmq psxq, since the annihilator of the maximal ideal of R is sR. For the opposite containment, letR R{sR ! Z 2 T , a local Noetherian ring of positive depth. If f pxq N pmq, thenf N pmq, wherem andf are the images of m and f inR andRrxs. By Theorem 3.4 (1), N pmq 0, so we conclude that f psq. Since f p0q 0, f psq pxq psxq, as desired.
If we switch to an infinite coefficient ring and residue field, say, Q instead of Z 2 , we still have N pmq psxq (nonzero but containing no regular polynomials).

Obtaining N pRq from N pmq; applications
The following proposition is the key to our main result, Theorem 4.2.
Proposition 4.1. Let pR, mq be a Henselian local ring with finite residue field R and let πpxq be an arbitrary π-polynomial. Any f pxq N pRq may be written in the form f pxq p 0 pπpxqq xp 1 pπpxqq x 2 p 2 pπpxqq ¤ ¤ ¤ x q¡1 p q¡1 pπpxqq with each p i pxq N pmq.
Proof. Let f pxq N pRq. In the polynomial ring Rrx, ys Rrysrxs, f may be divided by the monic polynomial πpxq¡y, so that f pxq Qpx, yqpπpxq¡yq Gpx, yq for some Qpx, yq, Gpx, yq Rrx, ys with Gpx, yq 0 or the degree of Gpx, yq with respect to x is less than q. Now set y πpxq to obtain f pxq p 0 pπpxqq xp 1 pπpxqq x 2 p 2 pπpxqq ¤ ¤ ¤ x q¡1 p q¡1 pπpxqq where the polynomials p i pyq Rrys are the coefficients of the powers of x in Gpx, yq.
It remains to see that each p i pxq N pmq. Let m m. Since (according to Corollary 2.11) π maps each coset of m onto m, there exists a set c 1 , c 2 , . . . , c q of representatives of the residue classes of m, with πpc i q m for each c i ; each difference c i ¡ c j (i $ j) is a unit. Since f pxq N pRq, we may evaluate f pxq at c i for each i from 1 to q to obtain In matrix form, this system becomes .

The matrix is a Vandermonde matrix, and its determinant is
which is a unit since it is a product of units. Thus, the matrix is invertible, so each p i pmq 0, as desired.
An application of the previous result, we come to our main theorem, which states that generators for N pRq can be obtained by composing generators for N pmq with any π-polynomial, which we could roughly describe by writing N pRq N pmq ¥ N pR, mq, if one keeps in mind Lemma 2.5 which states that N pR, mq pπpxq, mq.
An obvious consequence of the next theorem is that the minimal number of generators of N pRq is less than or equal to the minimal number of generators of N pmq.
Proof. Since πpRq m, certainly N pRq pF 1 pπpxqq, . . . , F n pπpxqqq. Now let f pxq N pRq. Use Proposition 4.1 to write f pxq p 0 pπpxqq xp 1 pπpxqq x 2 p 2 pπpxqq ¤ ¤ ¤ x q¡1 p q¡1 pπpxqq with each p i pxq N pmq. Since each p i pxq is an Rrxs-linear combination of F 1 pxq, . . . , F n pxq, each p i pπpxqq is an Rrxs-linear (actually Rrπpxqslinear) combination of F 1 pπpxqq, . . . , F n pπpxqq. Since f pxq is an Rrxs-linear combination of the p i pπpxqq, the proof is complete.
Remark 4.3. The equality N pRq N pmq ¥ N pR, mq should not be taken too literally. Certainly polynomials in N pmq composed with polynomials in N pR, mq are in N pRq, but it's not true that every polynomial in N pRq can be obtained in that way. For example, xpx 2 ¡ xq N pZ 2 q, but since its degree is not even, it does not equal f px 2 ¡ xq for any polynomial f pxq.
As mentioned in the introduction, the theorem below is a version of results of  Proof. We prove the first result using induction on e. The base case e 1 is clear, since then R is a field, m 0, and N pmq pxq. Assume the result is true for rings whose maximal ideal has index of nilpotency e¡1 ¤ q; we prove the result for a ring whose maximal ideal has index of nilpotency e ¤ q. The containment is clear. Let f pxq N pmq; then f pxq N m{m e¡1¨. By induction, f pxq px, mq e¡1 , and thus f pxq px, mq e¡1 . We have f pxq °e ¡1 k0 x k m e¡1¡k f k pxq for some f k pxq Rrxs; it remains to see that each f k pxq px, mq, i.e. that f k p0q m.
For each r R, prmq k m e¡1¡k f k prmq m e¡1 e¡1 k0 r k f k prmq.
Since the annihilator of m e¡1 is m,°e ¡1 k0 r k f k prmq m, and thus°e ¡1 k0 r k f k p0q m.
This shows that°e ¡1 k0 f k p0qx k N R¨. Since by Lemma 2.5 N R¨ px q ¡xq, this polynomial with degree less than q must be the zero polynomial. Therefore each f k p0q m, as desired.
For the second result, the only part of the containment N pmq px, mq e px q ¡ m q¡1 xq that is not clear is x q ¡ m q¡1 x N pmq; for this, take any rm m and compute prmq q ¡ m q¡1 prmq m q pr q ¡ rq m q 1 0. For the opposite containment, assume f pxq N pmq and reduce module m e¡1 as above to obtain a similar expression for f pxq, and again deduce that°e ¡1 k0 f k p0qx k N R¨ px q ¡xq. Since e ¡ 1 q, we must have°e ¡1 k0 f k p0qx k upx q ¡ xq for some unit u R; thus each f k p0q 0 except for f q p0q u and f 1 p0q ¡u . Define polynomials g k pxq identical to f k pxq except for g q pxq f q pxq ¡ u and g 1 pxq f 1 pxq u. Now the constant term of each g k pxq is in m and we have x k m e¡1¡k g k pxq so that f pxq px, mq e px q ¡ m q¡1 xq, as desired.
The following corollary follows immediately from the theorem and Theorem 4.2.
Corollary 4.5. Let pR, mq be a finite local ring with principal maximal ideal m pmq; set q |R{m|. Suppose e is the index of nilpotency of m, and let πpxq be any π-polynomial. If e ¤ q then N pRq pπpxq, mq e ; if e q 1 then N pRq pπpxq, mq e pπpxq q ¡ m q¡1 πpxqq.

Factoring π-polynomials
The following lemma is the heart of a more constructive approach (Theorem 5.2) to the converse part of the proof of Theorem 2.10, which gave two equivalent conditions for π-polynomials. Note that according to this lemma, if m e 0, then for any r R, the sequence tp n prqu stabilizes at n e ¡ 1.
Lemma 5.1. Let pR, mq be a Noetherian local ring with finite residue field R of cardinality q, and let ppxq be any polynomial mapping to x q ¡ x in Rrxs. Let p 0 pxq x and p n pxq ppp n¡1 pxqq p n¡1 pxq, so that p n pxq denotes the function obtained by successively applying the function ppxq x, n times. For every n ¥ 1, p n pxq ¡ p n¡1 pxq N pR, m n q .
Proof. We use induction on n. For the base case (n 1) just note that p 1 pxq ¡ p 0 pxq pppxq xq ¡ x ppxq N R, m 1¨b y Lemma 2.5. Now assume the induction hypothesis: For some n ¥ 1, p n pxq ¡ p n¡1 pxq N pR, m n q. We show that p n 1 pxq¡p n pxq N R, m n 1¨. Since ppxq is a polynomial mapping to x q ¡ x modulo mrxs, there is some mpxq mrxs such that ppxq x q ¡ x ¡ mpxq; thus p n pxq may also be viewed as applying x q ¡ mpxq, n times. For simplicity of notation, set c p n pxq, a p n¡1 pxq, and b c ¡ a, so that c ppaq a a q ¡ mpaq. We have p n 1 pxq ¡ p n pxq ppp n pxqq p n pxq ¡ p n pxq ppcq c q ¡ c ¡ mpcq pa bq q ¡ c ¡ mpcq since a q ¡ c mpaq. By induction, b N pR, m n q, and since q m, we see that the first term is in N R, m n 1¨. Since b N pR, m n q and n ¥ 1, b 2 N R, m n 1¨, which takes care of all but the last two terms: mpaq ¡ mpcq. Now mpaq ¡ mpcq °d eg mpxq i1 m i pa i ¡ c i q, where m i is the coefficient of x i in mpxq, and is thus in m. Since ¡b a ¡c is a factor of a i ¡c i for all positive integers i and ¡b N pR, m n q, it follows that mpaq ¡ mpcq N R, m n 1¨, as desired.
The following theorem provides, in particular, a more constructive approach to the proof of the result in Theorem 2.10 which states that any monic polynomial ppxq mapping to x q ¡ x is actually a π-polynomial. In a ring with m e 0, it allows discovery of the roots by successively applying the function ppxq x (e ¡ 1 times) to representatives of the residue classes of m. When ppxq x q ¡ x, this amounts to successively taking qth powers. In the case of a finite ring, the resulting roots of x q ¡ x are called Teichmüller elements in Jian Jun Jiang's paper [5]. Theorem 5.2. Let pR, mq be a complete Noetherian local ring with finite residue field R tc 1 , . . . , c q u. Let ppxq be any polynomial mapping to x q ¡ x in Rrxs and let p n pxq be the function obtained by applying ppxq x successively, n times. The limit lim nÑV p n pc i q exists. Set d i lim nÑV p n pc i q; then d i is a root of ppxq and d i c i . If ppxq is monic, then there is a factorization ppxq px ¡ d 1 qpx ¡ d 2 q ¤ ¤ ¤ px ¡ d q q, and thus ppxq is a π-polynomial.
Proof. The limit exists since, by Lemma 5.1, the sequence tp n pc i qu n¥1 is a Cauchy sequence. For any r R, pprq r and r are in the same coset of m, since pprq r q ¡ r ¡ mprq m. We may apply this fact successively, beginning with r c i , to see that each p n pc i q is congruent to c i modulo m. Since m is closed under the m-adic topology, we conclude that d i c i .
To see that ppd i q 0, use the Cauchy sequence mentioned above and the fact that polynomials are continuous under the m-adic topology: ppd i q pplim p n pc i qq lim ppp n pc i qq limpp n 1 pc i q ¡ p n pc i qq 0.
If ppxq is monic then it must have degree q; an application of Lemma 2.4 completes the proof.