Polynomial functions over nite commutative rings

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1 Polynomial functions over nite commutative rings Balázs Bulyovszky a, Gábor Horváth a, a Institute of Mathematics, University of Debrecen, Pf. 400, Debrecen, 4002, Hungary Abstract We prove a necessary and sucient condition for a function being a polynomial function over a nite, commutative, unital ring. Further, we give an algorithm running in quasilinear time that determines whether or not a function given by its function table can be represented by a polynomial, and if the answer is yes then it provides one such polynomial. Keywords: polynomial functions; local rings; interpolation 2010 MSC: 13M10, 13P99, 13H05, 13F10, 11T06 1. Introduction It is well-known that given nitely many pairs (a i, b i ) (0 i n) over a eld, there exists a polynomial p of degree at most n such that p(a i ) = b i for all 0 i n. Several classical interpolation methods exists e.g. by Lagrange, by Newton or by Hermite to name a few. A direct consequence of these results is that an arbitrary function over a nite eld can be represented by a polynomial. These methods, however, do not generalize in a straightforward manner to commutative rings. In fact, not even every function can be represented by a polynomial over a nite commutative ring which is not a eld. The question arises naturally: given a nite ring R and a function f : R R, does there exists a polynomial p R[x] such that p(r) = f(r) for every r R, and if such polynomial exists, then how could one nd such a polynomial? Carlitz [1] gave several necessary and sucient conditions for a function over Z p t being a polynomial function. For example, a function f : Z p t Z p t is a polynomial function if and only if there exists φ 0,..., φ t 1 : Z p t Z p t such that f(r + sp) = φ 0 (r) + (sp)φ 1 (r) + + (sp) t 1 φ t 1 (r) Corresponding author Preprint submitted to Elsevier July 23, 2017

2 holds for every r, s Z p t. Several generalizations of this result have been proved since, e.g by Spira [2] or later by Jiang, Peng, Sun and Zhang [3]. Note, however, that such a condition is not useful from an algorithmic perspective as it does not help nding a polynomial representing the input function f. Guha and Dukkipati [4] gave an algorithmically useful necessary and suf- cient condition for a function f : Z p t Z p t being a polynomial function. Let u 0 : Z p t Z p t be the function dened by { 0, if p r, u 0 (r) = 1, if p r, and let u k : Z p t Z p t (1 k t 1) be u k (r) = { 0, if p r, r k, if p r. Then f can be represented by a polynomial if and only if it is a linear combination of u 0,..., u t 1 and their shifts. Further, they gave an algorithm running in O (p t t + pt 3 ) time nding a polynomial representing f if one exists. Later they generalized their results to functions over Z n [5]. Both papers [4, 5] are based on Carlitz's result [1]. In this paper we generalize the results of Guha and Dukkipati [4, 5] to arbitrary nite, commutative, unital rings. Our proof is direct and is not based on Carlitz's result [1]. Further, we provide an algorithm running in quasilinear time (in the size of the ring) that determines whether or not function (over a nite, commutative, unital ring) given by its function table can be represented by a polynomial, and if yes then computes one such polynomial, as well. As every nite commutative, unital ring is a direct sum of local rings [6, Theorem VI.2], one only needs to consider these problems over nite, commutative, unital, local rings. In Section 2 we recall some basic facts necessary for our work. In particular, in Section 2.1 we summarize the most important properties of local rings, introduce functions u 0,..., u t 1 for local rings and prove that they are indeed polynomial functions. In Section 3 we generalize Guha and Dukkipati's necessary and sucient condition from [4] to arbitrary nite, commutative, unital, local rings by proving the following. Theorem 1. Let R be a nite, commutative, unital, local ring with maximal ideal M. Let t be the smallest positive integer for which M t = { 0 }. Let f : R R be an arbitrary function. Then f is a polynomial function over 2

3 R if and only if f can be written as a linear combination of the shifts of u 0,..., u t 1, where u 0 is the characteristic function of M, and u k (x) = x k u 0 (x) (1 k t 1). Let f : R R be an arbitrary function given by its function table. That is, f is given as the set of pairs (r, f(r)) for all r R, and the size of f is O ( R ). In Section 4 we provide an algorithm running in quasilinear time in R determining whether or not f is a polynomial function, and if yes then computing a polynomial representing f. Theorem 2. Let R be a nite, commutative, unital, local ring with maximal ideal M. Let t be the smallest positive integer for which M t = { 0 }. Let f : R R be an arbitrary function given by its function table. Then there exists an algorithm that decides whether or not f is a polynomial function, and if yes, then gives a polynomial that represents f, and the running time of this algorithm is O ( R t), if M is a principal ideal, and R/M t, T O ( R t 2 ), if M is a principal ideal, and R/M < t, O ( R t 2 log 3 M ), if M is not a principal ideal. Here and throughout the paper by log we mean base 2 logarithm. The running time of our algorithm is similar to that of Guha and Dukkipati [4, 5] for Z p t, p t. We need the notion of Galois rings in our algorithm, therefore we recall their main properties in Section Preliminaries Let R be a nite, commutative, unital ring. A polynomial p R[x] naturally induces a function p f : R R by substitution. A function f : R R is a polynomial function if there exists a polynomial p f R[x] such that p f (r) = f(r) for every r R. Every nite commutative, unital ring is a direct sum of local rings [6, Theorem VI.2]. Therefore, to understand polynomial functions over an arbitrary nite, commutative, unital ring, it is enough to consider local rings in the following Local rings A ring is local if it has a unique maximal ideal. We summarize some of the most important properties of local rings by [6, Chapter V]. Let R be a nite, commutative, unital, local ring with maximal ideal M. Let t denote the smallest positive integer for which M t = { 0 }. Note, that the quotient R/M is a eld, and for the set of invertible elements we have R = R \ M. 3

4 Further, if M = (m) is a principal ideal, then every r R can be written in the form sm i for some s R and 0 i t, and then all ideals of R are principal ideals generated by m i for some 0 i t. Let r R and f : R R be an arbitrary function. Let the shift of f by r be the function f r : R R, f r (x) = f(x r). Note that if f, g : R R are polynomial functions, then f + g, r f and f r are polynomial functions, as well (for every r R). Let u 1,..., u k : R R be arbitrary functions. Let u 1,..., u k denote the set of functions that can be written as a linear combination of shifts of u 1,..., u k with coecients from R. For every k { 0,..., t 1 } let u k : R R be the function dened as { 0, if x / M, u 0 (x) = (1) 1, if x M, and u k (x) = { 0, if x / M, x k, if x M. (2) Lemma 3. The functions u 0,..., u t 1 dened by (1) and (2) are polynomial functions over R. Proof. It is enough to prove that u 0 is a polynomial function, since u k (x) = x k u 0 (x). Recall that R = R \ M, and t was the smallest positive integer for which M t = { 0 }. Let d be the smallest multiple of exp R, which is at least t. That is, d t is the smallest positive integer, such that r d = 1 for every r R. Let p u0 (x) = 1 x d. We prove that p u0 (r) = u 0 (r) for every r R. Indeed, if r M, then d t implies r d = 0, hence p u0 (r) = 1 r d = 1 0 = 1. On the other hand, if r R \ M, then r is invertible, thus r d = 1, and p u0 (r) = 1 r d = 1 1 = 0. 4

5 2.2. Galois-rings In order to explain our algorithm in detail, we will need the notion of Galois-rings. Galois-rings have an important role in the theory of nite commutative rings [7, 8, 9]. Now, we summarize some of their properties by [6, Chapters XVIXVII]. Let h e (x) Z[x] be a monic polynomial of degree e, which is irreducible over Z p. Let c be positive integer. Then the Galois-ring GR(p c, e) is dened as the quotient Z[x]/(p c, h e (x)). Note, that the numbers p, c, e determine the Galois-ring GR(p c, e) up to isomorphism, that is GR(p c, e) is independent from the choice of h e. The Galois-ring GR(p c, e) is a nite, commutative, unital ring of size p ce and of characteristic p c. In particular, if c = 1, then GR(p, e) is isomorphic to the eld F p e, and if e = 1 (i.e. the degree of h e is 1), then GR(p c, 1) is isomorphic to Z p c. The Galois-ring GR(p c, e) is a local ring, its unique maximal ideal is (p), and (p) is its Jacobson-radical, as well. Furthermore, for every ideal I GR (p c, e) there exists 0 i c, such that I = (p i ). That is, every ideal of the Galois-ring is a principal ideal. Further, for every 1 i c the quotient GR(p c, e)/(p i ) is isomorphic to the Galois-ring GR(p i, e). In particular, the quotient by the Jacobson-radical is isomorphic to the eld F p e. Let R be a nite, commutative, unital, local ring with maximal ideal M and of characteristic p c, and assume R/M = p e. Then there exists a subring R 0 R, such that R 0 GR(p c, e). Consider R as an R 0 -module. Since every ideal of R 0 is principal, there exist elements b 1,..., b d R, such that R = R 0 b 1 R 0 b d as R 0 -modules [10]. That is, every r R can be written in the form d r = r i b i, i=1 where r i R 0, and r = 0 if and only if r i b i = 0 for every 1 i d. That is, r = 0 if and only if r i Ann(b i ), where Ann(b i ) = { r R 0 rb i = 0 }. Further, Ann(b i ) R 0 implies Ann(b i ) = (p c i ) for some 1 c i c. In particular, r i b i = 0 if and only if p c c i r i = 0 in R 0. Finally, b 1,..., b d can be chosen such that b 1 = 1, b 2,..., b d M, and Ann(b i ) R 0 (1 j d). In particular, Ann(b 1 ) = { 0 }, and R = d R 0 / Ann(b i ) R 0 2 d 1 R/M 2 d 1, i=1 thus d 1 log M. 5

6 3. The necessary and sucient condition We prove Theorem 1 in this section. Proof of Theorem 1. First, let f u 0,..., u t 1. By Lemma 3 the functions u 0,..., u t 1 can be represented by a polynomial. This implies that any shift and any linear combination of these shifts can be represented by a polynomial. Thus, the function f can be represented by a polynomial, as well. For the other direction, let f be a polynomial function. That is, there exists a polynomial p f R[x], such that p f (r) = f(r) for every r R. Let q = R/M, and let s 1,..., s q be an arbitrary representation system for the cosets of M. For every 1 j q let v j (x) = p f (x) u 0 (x s j ). Note that the functions v j (x) are polynomials as well, since p f (x) and u 0 (x) are polynomials. Then { f(r), if r s j + M, v j (r) = 0, if r / s j + M, and f = v v q. Therefore, it is enough to prove v j u 0,..., u t 1 for every j { 1,..., q }. For every 1 j q let w j (x) := v j (x + s j ). (3) Since v j is a polynomial, w j (x) is a polynomial as well, hence w j (x) = n a k x k for some a k R. We prove that for every r R w j (r) = a k u k (r). First, let r M. Recall, that M t = { 0 }. Now, u k (r) = r k for 0 k t 1, and u k (r) = 0 if k t. Now, we have w j (r) = n a k r k = a k r k = a k u k (r). 6

7 Now, let r / M. Then u k (r) = 0 for every 0 k t 1, that is This implies By (3) we have w j (r) = v j (r + s j ) = 0 = a k u k (r). w j (x) = a k u k (x). v j (x) = a k u k (x s j ), hence v j u 0, u 1,..., u t 1. Therefore, we have f = v v q u 0, u 1,..., u t 1, as well. 4. The algorithm Let R be a nite, commutative, unital, local ring with maximal ideal M, such that R/M = q = p e. Let s 1,..., s q be an arbitrary representation system for the cosets of M, but choose s q = 0 for convenience. Let t be the smallest positive integer for which M t = { 0 }. We assume M 2, otherwise R is a eld, in which case interpolation methods are well-known. Note, that log M M 1. We assume that the ring elements and the ideal structure of R are given in advance, and are not part of the input. Namely, R, M, q, t, s 1,... s q are already known. Further, we assume that every ring operation (including addition, multiplication and nding the inverse if one exists) takes one time-step. Let u 0 be the characteristic function of M dened by (1), and for every 1 k t 1 let u k be dened as u k (x) = x k u 0 (x). Let f : R R be a function given by its function table. That is, f is given as the set of pairs (r, f(r)) for all r R. Note, that only f is the input of our algorithm, the ring R and its elements, etc. are not. With the notations from Theorem 1 we have f(x) = q j=1 v j(x), where { f(x), if x s j M, v j (x) = 0, if x s j / M, for every j {1,..., q}. Let us assume that f is a polynomial function. By Theorem 1 every v j is a polynomial function, that is v j (x) = α j,k u k (x s j ) (4) 7

8 for some coecients α j,k. This implies f(x) = q α j,k u k (x s j ). j=1 Our aim is to determine the coecients α j,k (1 j q, 0 k t 1). The pseudocode for the main algorithm can be found as Algorithm 4.4 in Section 4.4. The algorithm consists of a main loop for j running from 1 to q (lines 434 of Algorithm 4.4), and each iteration determines the coecients α j,k (0 k t 1) and thus v j by (4). We explain the iteration for j = q, that is we detail the case for s q = 0. For every 1 j q 1 the coecients α j,k are determined similarly after an appropriate shift by s j (cf. lines 6 and 32 of Algorithm 4.4). First, for every r R let us substitute x = r into (4) for j = q to obtain v q (r) = α q,k u k (r). (5) Note, that if r / M, then u k (r) = 0 for every k {0,..., t 1}, that is v q (r) = 0. Thus, to determine all α q,k, one only needs to consider the equations in (5) for r M. From now on for easier notation we use α k instead of α q,k. Now, the M -many equations (5) take the form v q (r) = α k u k (r) = α k r k (r M). (6) Let A R M t be the matrix, whose rows are indexed by the elements of M, and the kth element of row r is r k 1. Let α denote the column vector (α 0,..., α t 1 ) T, and let the column vector v R M contain the element v q (r) in row r. (The matrix A, the vector v and the independent variables α are dened in lines 511 of Algorithm 4.4.) Now, the system of equations (6) is equivalent to the system of linear equations Aα = v. (7) Note, however, that if R is not a eld, then it is not straightforward whether or not Gaussian elimination can be used to solve such a system. We separate three cases to determine the vector α, depending on whether or not M is a principal ideal and q t or q < t. 8

9 4.1. M is a principal ideal and q t Let M = (m) for some m M. Again, we consider m as given in advance and not part of the input. Let us write every r R in the form r = sm i such that the power i is maximal. Then s R \ M = R. See Algorithm 4.1 in Section 4.4 for the pseudocode. Now, s 1,..., s q was a representation system of R/M. Consider the rst t of these elements: s 1,..., s t. Now, s i s j / M if 1 i j t. Let r 1 = s 1 m,..., r t = s t m, and consider the t t submatrix of A created by the rows corresponding to r 1,..., r t. Note, that this matrix is a Vandermonde-matrix, let us denote this matrix by V (r 1,..., r t ). We prove that (7) has a solution if and only if its subsystem corresponding to V (r 1,..., r t ) has a solution, and if that (arbitrary) solution satises (7). Now, V (r 1,..., r t ) = V (s 1,..., s t ) D, where V (s 1,..., s t ) denotes the Vandermonde-matrix corresponding to s 1,..., s t, and m D = 0 0 m m t 1 Now, det(v (s 1,..., s t )) = i<j (s i s j ) R = R \ M, since for every 1 i j t we have s i s j / M, and hence s i s j is invertible. This implies that the matrix V (s 1,..., s t ) is invertible, let L denote its inverse. One can determine L by applying GaussJordan elimination on V (s 1,..., s t ) in O (t 3 ) time. Note, that LV (r 1,..., r t ) = LV (s 1,..., s t ) D = D. That is, performing the same row operations on V (r 1,..., r t ), one obtains D as a result. That is, the GaussJordan elimination on V (r 1,..., r t ) results in the system m m m t 1 α 0 α 1 α 2. α t 1 = for b 0, b 1, b 2,..., b t 1 R, where b = (b 0, b 1, b 2,... b t 1 ) T = Lv. The system (8) is solvable if and only if b k is divisible by m k for every 0 k t 1, but the solution of (8) is not necessarily unique. Note, 9 b 0 b 1 b 2. b t 1 (8)

10 however, that if any solution of (8) satises (7), then every solution satises (7). Indeed, let α 0,..., α t 1 be an arbitrary solution of (8), and let r M be arbitrary. Write r = sm i for some i 1. Then multiplying the rth row of A by α results in α k r k = α k s k m ik = α k m k s k m (i 1)k = b k s k m (i 1)k, which is independent of the choice of α as a solution of (8). Thus, Algorithm 4.1 runs as follows. In lines 37 it obtains the system (8) in O (t 3 ) time if L is not computed, yet. Then in lines 814 it solves (8) in O (t) time. Finally, in lines 1519 it checks in O ( M t) time whether or not this solution of (8) satises the system (7). That is the system (7) can be solved in O (t 3 + M t) time. If a solution is found, then v q can immediately be obtained from it as a linear combination of u k s (line 32 of Algorithm 4.4), and the polynomial representation of v q can be obtained by Lemma 3 (line 35 of Algorithm 4.4). Repeating the algorithm for every coset s j + M and function v j gives a polynomial representation of f if one exists. Note that one does not need to recompute the steps of the GaussJordanelimination on every coset s j + M if one chooses the rows corresponding to the elements r 1 +s j,... r t +s j. Indeed, in that case GaussJordan-elimination would result in the same row operations, that is the matrix L is independent of j. Thus, only b needs to be recomputed (line 7 in Algorithm 4.1), which only needs an additional O (t 2 ) O ( M t) time for each j. Further, t 1 + log M 2 M 2 R, thus t 3 2 R t. Therefore, the time requirement of the entire algorithm in this case is O ( t 3 + ( M t) R/M ) O ( t 3 + R t ) O ( R t) M is a principal ideal (and q < t) See Algorithm 4.2 in Section 4.4 for the pseudocode. Recall, that M being principal yields that every ideal of R is principal (see Section 2.1). Then the matrix of the system (7) can be written in Hermite normal form [11]. A matrix is in Hermite normal form, if every row's rst occurring non-zero element is strictly right from that of the previous row. If R is a commutative unital ring, and every ideal of R is a principal ideal, then any matrix over R can be written in Hermite normal form using only invertible operations [11]. Note, however, that the algorithm described in [11] is more general than what we need. Therefore, Algorithm 4.2 explains a slightly more ecient way of writing the matrix A (in fact, any matrix over R) into Hermite normal form using modied Gauss-elimination. Again, let M = (m) for some m M, 10

11 and write every element of A in the form r = sm i, where s R (line 2 of Algorithm 4.2). Further, assume that A has t A columns and M A rows (if A is the matrix in (7), then t A = t and M A = M ). Now we explain lines 613 of Algorithm 4.2. Let a 1 = u 1 m k 1 (u 1 R ) be an element of A with smallest power of m. Move a 1 to the top left corner by swapping (possibly) two rows and two columns. From here on, every row operation is applied to the vector on the right-hand side of (7), and together with a column-swap we swap the appropriate coordinates of the vector α, as well. Now, multiply the rst row with u 1 1, and then subtract the appropriate multiple of the rst row from every other row such that the rst element of every other row becomes 0. Let A (1) denote the matrix obtained this way. By the choice of a 1, every element of A (1) is divisible by the top left element m k 1. The matrix A (1) can be obtained in O ( M t) time. This is the rst iteration of lines 613 of Algorithm 4.2. Assume that after the lth step we have a matrix A (l). Let a (l) j,k denote the k-th element of the j-th row in the matrix A (l). Assume further, that the top left l l-block in A (l) is an upper-triangular matrix with increasing powers of m in the diagonal, and a (l) j,k is divisible by a(l) i,i (1 i l) if j i and k i, and a (l) j,k = 0 if k l < j. We explain step l + 1. Let a l+1 = u l+1 m k l+1 (ul+1 R ) be an element among a (l) j,k (j, k > l) with smallest power of m. Replace a l+1 by a (l) l+1,l+1 using a row-swap and a column-swap. Multiply row l+1 with u 1 l+1, and then subtract the appropriate multiple of row l + 1 from every other row below row l + 1 such that column l + 1 below row l + 1 becomes 0. Let A (l+1) denote the matrix obtained this way. Now, the top left (l + 1) (l + 1)-block in A (l+1) is an upper-triangular matrix with increasing powers of m in the diagonal, and a (l+1) j,k a (l+1) i,i is divisible by (1 i l + 1) if j i and k i, and a (l+1) j,k = 0 if k l + 1 < j. Now, A R M A t A, thus repeating these steps min { ta, M A } times results in a matrix (denoted by A (t A) ), having increasing powers of m in its diagonal and every element under the diagonal is 0. Further, every element right or below a diagonal element is divisible by that diagonal element. The matrix A (t A) can be determined in O ( M A t A min { t A, M A }) steps from A. Hence, in O ( M A t A min { t A, M A }) time one can write the system (7) in Hermite- 11

12 normal form and obtain the equivalent system m k 1 a (t A) 1,2 a (t A) 1,3... a (t A) 1,t A 1 0 m k 2 a (t A) 2,3... a (t A) 2,t A m k 3... a (t A) 3,t A m kt A α 0 α 1 α 2. α ta 1 = b 1 b 2 b 3. b t b t+1. b M (9) for some appropriate b 1,..., b M R. (Note, that the coordinates of the vector α may be permuted, but that can be handled by simply re-indexing them.) Now, (9) (and thus (7)) has no solutions if b i 0 for any i > t A (lines 2123 of Algorithm 4.2). From now on we assume b i = 0 for every i > t A. Further, for every 1 i t A the element a (t A) i,j is divisible by m k i. Hence, if b i is not divisible by m k i for some 1 i t A, then (9) (and thus (7)) has no solutions (line 25 of Algorithm 4.2). Now we provide a solution of (9) (and thus of (7)) if b i is divisible by m k i for every 1 i t A. We solve (9) from bottom to top. Assume that for some 1 i t A 1 we have managed to nd α i+1,..., α ta 1 satisfying all but the rst i rows of (9). Now, the ith row is m k i α i + t A j=i+1 a (t A) i,j α j = b i. (10) Now, a (t A) i,j is divisible by m k i for every j > i. Hence, if b i is divisible by m k i, as well, then one can nd α i satisfying (10) (line 28 of Algorithm 4.2). Running i from t to 1 one can obtain a solution of (9), and thus, one of (7) (lines 2430 of Algorithm 4.2). Therefore, a linear system is solvable in O ( M A t A min { t A, M A }) time, if the matrix A has t A -many columns and M A -many rows. In particular, the system (7) is solvable in O ( M t 2 ) time, giving v q as a linear combination of u k s (line 32 of Algorithm 4.4), if possible. From this one can easily obtain the polynomial representation of v q using e.g. Lemma 3 (line 35 of Algorithm 4.4). Repeating Algorithm 4.2 for every coset s j + M and function v j, one obtains a polynomial representation of f = v v q, if one exists. The time needed for the entire algorithm in this case is O ( M t 2 R/M ) = O ( R t 2). 12

13 4.3. M is not a principal ideal See Algorithm 4.3 in Section 4.4 for the pseudocode. Let the characteristic of R be p c, and recall that R/M = p e. Then there exists a subring R 0 R such that R 0 GR(p c, e). Consider R as an R 0 -module. Since R 0 is isomorphic to a Galois-ring, every ideal of R 0 is a principal ideal. By [10] R is the direct sum of cyclic R 0 -modules. That is, there exist elements b 1,..., b d R (for some d) such that R = R 0 b 1 R 0 b d as a module. In fact, these elements can be chosen such that b 1 = 1, b 2,..., b d M. Now, every r R can be written in the form d r = r i b i, (11) i=1 where r i R 0. Further, r = 0 if and only if r i b i = 0 for every 1 i d, that is if r i Ann(b i ) R 0. In particular, Ann(b i ) = (p c i ) for some 1 c i c, and r i b i = 0 if and only if p c c i r i = 0 in R 0. Note, that one may run r i over R 0 / Ann(b i ) for all 1 i d and compute the right-hand side of (11), thus obtaining the decomposition of every r R in O ( R d) time (lines 37 of Algorithm 4.3). Storing these computations in an appropriate lookup table, we may assume that nding the form (11) of an arbitrary r R takes O (d) time. Consider system (7). Let us rewrite A and v using (11) as d A = A i b i, and (12) v = i=1 d v k b k, (13) k=1 where the elements of the matrices A i and of the vectors v k are all in R 0. Write α into the formal sum β 1 b β d b d where every β j is considered to be a vector of variables over R 0. Then system (7) can be rewritten as ( d ) d β j b j = v k b k. (14) i=1 A i b i) ( d j=1 Expanding the product of two sums on the left-hand side of (14), we obtain summands in the form of A i b i b j β j (1 i, j d). Again, by (11), b i b j can be written in the form d k=1 r i,j,kb k for some r i,j,k R 0 (1 i, j, k d). Hence, the system (14) can be rewritten as ( d d ) d A i r i,j,k β j v k b k = 0. k=1 i=1 j=1 13 k=1

14 Thus, (7) is solvable if and only if d d ( ) i=1 j=1 Ai r i,j,k β j v k bk = 0 is solvable for every 1 k d, that is if d d i=1 j=1 A ir i,j,k β j v k Ann(b k ) for every 1 k d for some β j from R 0. Since Ann(b k ) = (p c k ), the system (7) is solvable if and only if ( d ) d p c ck A i r i,j,k β j v k = 0 (15) i=1 j=1 has a solution in R 0 for every 1 k d. That is, we reduced solving (7) over R to solving a system of equations over R 0 having a base matrix of size M d td. This system is computed in lines 1520 of Algorithm 4.3. Finally, R 0 GR(p c, e) with maximal ideal (p), thus we are able to solve system (15) over R 0 by Algorithm 4.2 in Section 4.2 in O ( ( M d) (td) 2) = O ( M t 2 d 3 ) time. Now, we compute the time requirement of the algorithm in the case when M is not a principal ideal. Computing the form (11) for every r R takes O ( R d) time, but this needs to be computed only once (lines 37 of Algorithm 4.3). Rewriting every b i b j as d k=1 r i,j,kb k takes O(d 3 ) time (lines 814 of Algorithm 4.3). Obtaining (15) takes O ( M td 3 ) time (lines 1520 of Algorithm 4.3), solving (15) takes O ( M t 2 d 3 ) time (lines 2126 of Algorithm 4.3). In particular, the system (7) is solvable in O ( M t 2 d 3 ) time, giving v q as a linear combination of u k s (line 32 of Algorithm 4.4), if possible. From this one can easily obtain the polynomial representation of v q using e.g. Lemma 3 (line 35 of Algorithm 4.4). Repeating Algorithm 4.3 for every coset s j + M and function v j, one obtains a polynomial representation of f = v 1 + +v q, if one exists. Thus, the time requirement of the algorithm is O ( R d + R/M M t 2 d 3) = O ( R t 2 d 3). Now, b 2,..., b d M yields d 1 + log M 2 log M. Therefore, the time requirement of the algorithm is 4.4. Pseudocode of the algorithm O ( R t 2 d 3) O ( R t 2 log 3 M ). Finally, we list the pseudocode for the main algorithm (as Algorithm 4.4) and for the three cases (as Algorithms 4.1, 4.2, and 4.3 for Section 4.1, 4.2, and 4.3, respectively). 14

15 Algorithm 4.1 Pseudo code for SolveByInverse (Section 4.1) Input: 1: R, M, t, q, s 1,..., s q, m // given in advance 2: A, v, (optional: L) // input Output: α solution of Aα = v, if exists, and false, otherwise 3: if L is not dened then 4: V V (s 1,..., s t ) // Vandermonde-matrix 5: L V 1 // by GaussJordan elimination 6: end if 7: b Lv 8: for k = 0 to t 1 do 9: if m k b k then 10: α k one instance of b k /m k 11: else 12: return false and exit 13: end if 14: end for 15: for all r M do 16: if t 1 α kr k v(r) then 17: return false and exit 18: end if 19: end for 20: return (α 0,..., α t 1 ) T 15

16 Algorithm 4.2 Pseudo code for SolveByHermite (Section 4.2) Input: 1: A, v, R, M, m // input Output: α solution of Aα = v, if exists, and false, otherwise 2: write every element of A in the form sm i, where s R 3: A (0) A, and v (0) v 4: t A number of columns in A // A has t columns in (7) 5: M A number of rows in A // A has M rows in (7) 6: for l = 1 to min { t A, M A } do // modied Gauss-elimination 7: nd a l = u l m k l with smallest m-power in A (l 1) 8: switch a l 's row by the lth row of A (l 1) and switch the corresponding elements in v (l 1) 9: switch a l 's column by the lth column of A (l 1) and switch the corresponding variables in α 10: multiply row l of A (l 1) and of v (l 1) by u 1 l 11: for k = l + 1 to m A do 12: subtract appropriate multiple of lth row from row k such that its lth element becomes 0 (apply to both A (l 1) and v (l 1) ) 13: end for 14: A (l) matrix obtained from A (l 1) 15: k l power of m in a (l) l,l 16: v (l) vector obtained from v (l 1) 17: end for 18: A (ta) last A (l) // A (ta) is in Hermite normal form, see (9) 19: b last v (l) 20: for i = M A downto 1 do 21: if i > t A and b i 0 then 22: return false and exit 23: end if 24: if i t A then 25: if m k i b i then 26: return false and exit 27: else 28: α i b i /m k i t A j=i+1 α j a (t A) i,j /m k i 29: end if 30: end if 31: end for 32: return (α 0,..., α ta 1) T 16

17 Algorithm 4.3 Pseudo code for SolveByModule (Section 4.3) Input: 1: R, M, t, q, s 1,..., s q, R 0, b 1,..., b d, p, c, c 1,..., c d // given in advance 2: A, v (optional: LookupT able) // input Output: α solution of Aα = v, if exists, and false, otherwise 3: if LookupT able is not dened then 4: for all r R do // by iterating over cosets of R/ Ann(b i ) 5: compute r = d i=1 r ib i form and store the values in LookupT able 6: end for 7: end if 8: for i = 1 to d do 9: for j = 1 to d do 10: for k = 1 to d do 11: r i,j,k from LookupT able (b i b j ) 12: end for 13: end for 14: end for 15: for k = 1 to d do 16: Matrix k p c c k d i=1 d j=1 r i,j,ka i 17: v k p c c k vk 18: end for 19: Matrix ( Matrix T 1,..., Matrix T d ) T 20: vector ( v T 1,..., v T d 21: if SolveByHermite(Matrix, vector, R 0, (p), p) = false then 22: return false and exit 23: else ( ) 24: β T 1,..., β T T d SolveByHermite(Matrix, vector, R0, (p), p) 25: return d j=1 β jb j 26: end if ) T 17

18 Algorithm 4.4 Pseudo code for the main algorithm FindPolynomial Input: 1: R, M, t, q, s 1,..., s q, m, R 0, b 1,..., b d, p, c, c 1,..., c d // given in advance 2: f given as (r, f(r)) for all r R // input Output: a polynomial representing f if exists, and false otherwise 3: p f (x) 0 4: for j = 1 to q do // run through all cosets of M 5: for all r M do // dene vector and matrix of (7) 6: v(r) f(r + s j ) 7: for k = 1 to t do 8: a r,k r k 1 9: end for 10: end for 11: A ((a r,k )) 12: if M = (m) then 13: if q t then // Section : if L is already computed from a previous SolveByInverse run then 15: α SolveByInverse (A, v, L) 16: else 17: α SolveByInverse (A, v) 18: end if 19: else // Section : α SolveByHermite (A, v, R, M, m) 21: end if 22: else // Section : if LookupT able is already computed from a previous SolveByModule run then 24: α SolveByModule (A, v, LookupT able) 25: else 26: α SolveByModule (A, v) 27: end if 28: end if 29: if α = false then 30: return false and exit 31: else 32: p f (x) p f (x) + t 1 α ku k (x s j ) 33: end if 34: end for 35: p f (x) replace each u k by its representing polynomial from Lemma 3 36: return p f (x) 18

19 5. Acknowledgements We thank the anonymous referee for their suggestions, which signicantly improved the presentation of the paper. The research was supported by the Hungarian National Research, Development and Innovation Oce (NKFIH) grant no. K and grant no. FK [1] L. Carlitz, Functions and polynomials (mod p n ), Acta Arith. 9 (1964) [2] R. Spira, Polynomial interpolation over commutative rings, The American Mathematical Monthly 75 (6) (1968) [3] J. J. Jiang, G. H. Peng, Q. Sun, Q. F. Zhang, On polynomial functions over nite commutative rings, Acta Math. Sin. (Engl. Ser.) 22 (4) (2006) doi: /s URL [4] A. Guha, A. Dukkipati, An algorithmic characterization of polynomial functions over Z p n, Algorithmica 71 (2015) [5] A. Guha, A. Dukkipati, A faster algorithm for testing polynomial representability of functions over nite integer rings, Theoretical Computer Science 579 (2015) [6] B. R. MacDonald, Finite rings with identity, M. Dekker, [7] R. Raghavendran, Finite associative rings, Compositio Math. 21 (2) (1969) [8] R. S. Wilson, On the structure of nite rings, Compositio Math. 26 (1) (1973) [9] R. S. Wilson, On the structure of nite rings II, Pacic Journal of Mathematics 51 (1) (1974) [10] G. Köthe, Verallgemeinerte Abelsche Gruppen mit hyperkomplexem Operatorenring, Math. Z. 39 (1) (1935) doi: /bf URL [11] J. L. Hafner, K. S. McCurley, Asymptotically fast triangularization of matrices over rings, SIAM J. Comput. 20 (6) (1991) doi: / URL 19