On the group of unit-valued polynomial functions

Let $R$ be a finite commutative ring with $1\ne 0$. The set $\mathcal{F}(R)$ of polynomial functions on $R$ is a finite commutative ring with pointwise operations. Its group of units $\mathcal{F}(R)^\times$ is just the set of all unit-valued polynomial functions, that is the set of polynomial functions which map $R$ into its group of units. We show that $\mathcal{P}_R(R[x]/(x^2))$ the group of polynomial permutations on the ring $R[x]/(x^2)$, consisting of permutations represented by polynomials over $R$, is embedded in a semidirect product of $\mathcal{F}(R)^\times$ by $\mathcal{P}(R)$ the group of polynomial permutations on $R$. In particular, when $R=\mathbb{F}_q$, we prove that $\mathcal{P}_{\mathbb{F}_q}(\mathbb{F}_q[x]/(x^2))\cong \mathcal{P}(\mathbb{F}_q) \ltimes_\theta \mathcal{F}(\mathbb{F}_q)^\times$. Furthermore, we count unit-valued polynomial functions $\pmod{p^n}$ and obtain canonical representations for these functions.


Introduction
Throughout this paper R is a finite commutative ring with unity 1 = 0. We denote by R × the group of units of R. A function F : R −→ R is called a polynomial function on R if there exists a polynomial f ∈ R[x] such that F (r) = f (r) for each r ∈ R. In this case, we say that f induces (represents) F or F is induced (represented) by f . If F is a bijection, we say that F is a polynomial permutation on R and f is a permutation polynomial on R (or f permutes R). When F is the constant zero, f is called a null polynomial on R or shortly, null on R. The set of all null polynomials is an ideal of R[x], which we denote by N R .
It is evident that the set F(R) of all polynomial functions on R is a monoid with composition of functions. Its group of invertible elements P(R) consists of polynomial permutations on R, and it is called the group of polynomial permutations on R. Further, F(R) is a ring with addition and multiplications defined pointwise.
In this paper we are interested in the group of units of the pointwise ring structure on F(R), which we denote by F(R) × . We show that there is a relation between the group F(R) × and the group of polynomial permutations on R[x]/(x 2 ), which is represented by polynomials with coefficients in R only. Moreover, when R = Z p n the ring of integers modulo p n we find the order of F(Z p n ) × and give canonical representations for its elements.

Preliminaries
In this section, we introduce the concepts and notations used frequently in the paper. Throughout this paper whenever A is a ring and f ∈ A[x], let [f ] A denote the polynomial function on A represented by f . That is, if F is the function induced by f on R, then [f ] A = F .
is called a unit-valued polynomial if f (r) ∈ R × for each r ∈ R [7]. In this case [f ] R , the function induced by f on R, is called a unit-valued polynomial function.
Throughout this paper for every f ∈ R[x], let f ′ denote its first formal derivative. Unit-valued polynomials and unit-valued polynomial functions have been appeared in the literature within studying other structures. For example, Loper [7] has used unit-valued polynomials for examining non-D-rings. One other example is seen in the criteria of permutation polynomials on finite local rings. To illustrate this we recall the well known fact. Indeed, the second condition of the previous fact requires that f ′ to be a unit-valued polynomial on R (equivalentlyf ′ is a unit-valued polynomial on R/M ), that is [f ′ ] R is a unit-valued polynomial function.
To prove our first fact about unit-valued polynomial functions we need the following lemma, which is a special case of a general property proved in [6]. Lemma 2.3. Let R be a finite commutative ring with 1 = 0. Then every regular element (i.e., element which is not a zero divisor) is invertible.
Proof. Let a ∈ R be a regular element. Define a function φ a : R −→ R by φ a (r) = ar for each r ∈ R. We claim that φ a is injective, and hence it is surjective since R is finite. Because, if there exist r 1 , r 2 ∈ R such that r 1 = r 2 and φ a (r 1 ) = ar 1 = ar 2 = φ a (r 2 ), then a(r 1 − r 2 ) = 0. But this implies that a is a zero divisor, which is a contradiction. Thus φ a is bijective, and there exists a unique element r such that 1 = φ a (r) = ar.
From now on, let " · " denote the pointwise multiplication of functions.
Fact 2.4. Let F(R) be the set of polynomial functions on R. Then F(R) is a finite commutative ring with nonzero unity, where addition and multiplication are defined pointwise. Moreover, F(R) × is an abelian group and; Proof. We leave to the reader t check that F(R) forms a finite commutative ring under pointwise operations with with unity is the constant 1 on R, which we denote by 1 F (R) .
Moreover, since F(R) is commutative ring, it follows that F(R) × is an abelian group. Now, it is easy to see that every unit-valued polynomial function is regular, and hence it is invertible by Lemma 2.3. Thus F(R) × contains every unit-valued polynomial functions.
For the other inclusion, let F ∈ F(R) × . Then there exists F −1 ∈ F(R) × such that F · F −1 = 1 F (R) . Hence F (r)F −1 (r) = 1 for each r ∈ R, whence F (r) ∈ R × for each r ∈ R. Therefore F is a unit-valued polynomial function by Definition 2.1. This completes the proof. Remark 2.5. When R is an infinite commutative ring, still true that F(R) is a commutative ring (infinite) and every element of F(R) × is a unit-valued polynomial function. However, F(R) × can be properly contained in the set of all unit-valued polynomial functions.
The following example illustrates the previous remark.
Then the polynomial f = 1 + 2x is a unit-valued polynomial on R, and F = [f ] R is a unit-valued polynomial function. We claim that F has no inverse in F(R). Assume, on the contrary, that F is invertible, that is there exists Further, h has infinite roots in R since h(r) = F (r)F 1 (r) − 1 = 0 for every r ∈ R, which contradicts the fundamental theorem of algebra. The following fact about the polynomials of R[α] can be proved easily.
Fact 2.8. Let R be a commutative ring, and a, b ∈ R.
Recall from the introduction that P(R[α]) denotes the group of polynomial permutations on R[α]. It is apparent that P(R[α]) is a finite set, being a subset of a finite set (i.e., a subset of the set of polynomial functions on R[α]).
From now on, let " • " denote the composition of functions (polynomials) and let id R denote the identity function on R.
. Then their composition g • f induces a function on R, which is the composition of the functions induced by f and g on R. Similarly, f + g and f g induce two functions on R, namely the pointwise addition and multiplication respectively of the functions induced by f and g. In terms of our notations this equivalent to the following: The above equalities appear periodically in our arguments in the next sections.

The embedding of the group
In this section we show that the group P R (R[α]), which consists of permutations represented by polynomials from R[x], is embedded in a semidirect product of the group of unit-valued polynomial functions on R, F(R) × by the group of polynomial permutations on R, P(R), via a homomorphism θ defined in Lemma 3.2 below.
Lemma 3.1. Let F, F 1 ∈ F(R) × and G ∈ P(R). Then the following hold: (2) Put F 2 = F · F 1 . So if r ∈ R then, by the pointwise multiplication on F(R) × , (3) By (2), An expert reader should notice that Lemma 3.1 defines a group action of G on F(R) × in which every element of G induces a homomorphism on F(R) × , and what is coming now is a consequence of that. However, we do not refer to this action explicitly to avoid recalling additional materials. In fact, our arguments are elementary and depend on direct calculations.
Proof. In view of Lemma 3.1-(2) we need only to show that θ G is a bijection. Let F ∈ F(R) × . Then F • G −1 ∈ F(R) × by Lemma 3.1, and we have that This shows that θ is a surjection, and hence it is a bijection since F(R) × is finite. Therefore Moreover, if θ : P(R) −→ Aut(F(R) × ) is given by (G)θ = θ G , then for every G 1 , G 2 ∈ P(R) and any F ∈ F(R) × , we have Notation and Remark 3.3. Let H be the set of all pairs (G, F ), where G ∈ P(R) and F ∈ F(R) × . We put In the following proposition we use the homomorphism θ of Lemma 3.2 to define a multiplication on H. Such a multiplication allows us to view H as the semidirect of F(R) × by P(R).
Proposition 3.4. Assume the above notation. Define a multiplication on H by Then H is a group containing P(R), F(R) × as subgroups. Furthermore, the following hold: The proof of Proposition 3.4 depends essentially on Lemma 3.2, and it is just the justifications of the semidirect product properties (see for example [3,5]).
Remark 3.5. In fact, the subgroups P(R) and F(R) × are isomorphic to P(R) and F(R) × respectively. In this case H is called the (external) semidirect product of F(R) × by P(R), and it is denoted by P(R) ⋉ θ F(R) × (see for example [3,5]). For this reason, we simply use P(R) ⋉ θ F(R) × to mention the semidirect product constructed in Proposition 3.4.
Our next aim is to show that the group P R (R[α]) defined in Definition 2.10 is embedded in P(R) ⋉ θ F(R) × , i.e., P(R) ⋉ θ F(R) × contains an isomorphic copy of P R (R[α]). To do so let us first prove the following lemma, which is a special case of [1, Theorem 2.8]. Proof. (⇒) Let c ∈ R. Then c ∈ R[α]. Since g permutes R[α], there exist a, b ∈ R such that g(a + bα) = c. Thus g(a) + bg ′ (a)α = c by Fact 2.8. So g(a) = c, and therefore g is onto on the ring R, and hence a permutation polynomial on R.
Suppose that g is not a unit-valued polynomial. Then there exists a ∈ R such that g ′ (a) is a zero divisor of R. For, if 0 = b ∈ R such that bg ′ (a) = 0, then by Fact 2.8, g(a + bα) = g(a) + bg ′ (a)α = g(a).
Then we have g(a) = g(c) and bg ′ (a) = dg ′ (c). Hence a = c since g permutes R. Then, since g ′ (a) is a unit of R, b = d follows.
where [f ] R and [f ′ ] R denote the polynomial functions induced by f and f ′ on R respectively. This shows that the pair ( Proof. The first statement follows by applying Fact 2.8 and the chain rule to the polynomial f • g. The last statement follows from the first part and Remark 3.7. Recall from Definition 2.10 and Fact 2.4 the definitions of the groups P R (R[α]) and F(R) × respectively. Proposition 3.9. Let R be a finite commutative ring. The group of polynomial permutations

Proof. Let F ∈ P R (R[α]). Then by Definition 2.10, F is induced by a polynomial f ∈ R[x].
Hence f permutes R and f ′ is a unit-valued polynomial by Lemma 3.6. Thus [f ] R ∈ P(R) and Lemma 3.8. Therefore, by the operation of P(R) ⋉ θ F(R) × defined in Proposition 3.4, . Thus φ is a homomorphism. Now suppose that F = F 1 . Then by Remark 3.7,

The pointwise stabilizer group of R and the group F(R) ×
In this section, we show that the group of unit-valued polynomial functions contains an isomorphic copy of the pointwise stabilizer group of R (defined below). In particular, when R = F q the finite field of q elements, we prove that F(F q ) × is isomorphic to this group. We employ this result in the end of this section to prove that P Now we recall the definition of the pointwise stabilizer group of R from [1]. It is evident that Stab α (R) is closed under composition, and hence it is a subgroup of P(R[α]) since it is a non-empty finite set. We call this group the pointwise stabilizer group of R.
Recall from the introduction that the ideal N R is consisting of all null polynomials on R. We need the following proposition from [1]. However, we prove it as the proof does not depend on extra materials.
Proof. Obviously, , and so r = F (r) = h 0 (r) + h 1 (r) α for every r ∈ R. It follows that h 1 (r) = 0 for every r ∈ R, i.e., h 1 is null on R. Hence h 1 α is null on R[α] by Fact 2.9. Thus represents F . Such a polynomial is called the Lagrange's polynomial and this method of construction is called Lagrange's interpolation. Therefore every function on a finite field is a polynomial function, and hence |F(F q )| = q q . In particular, every permutation (bijection) on F q is a polynomial permutation, and so |P(F q )| = q!. Further, every unit-valued function is a unit-valued polynomial function, and thus |F(F q ) × | = (q − 1) q since F × q = F q \ {0}. Moreover, it is obvious that Lagrange's interpolation assigns to every function on F q a unique polynomial of degree at most q − 1. Hence every polynomial of degree at most q − 1 is the Lagrange's polynomial of a function on F q since the number of these polynomials is q q , which is the number of functions on F q .

Proof. By Lagrange's interpolation (Remark 4.4), there exist two polynomials
is a null polynomial on F q . The last part follows by Lemma 3.6 and Definition 2.10.
Remark 4.6. Let F, F 1 ∈ Stab α (R), with F = F 1 . Then, by Proposition 4.3, F and F 1 are induced by x + f (x) and x + g(x) respectively for some f, g ∈ N R . By Definition 4.1, Throughout for any set A let |A| denote the number of elements in A.
Theorem 4.7. The pointwise stabilizer group of R is embedded in the group of unit-valued polynomial functions F(R) × . In particular, if R = F q is the finite field of q elements, then By Remark 3.7, φ is well defined; and it is injective by Remark 4.6. Now if F 1 ∈ Stab α (R), then F 1 is induced by x + g(x) for some g ∈ N R . By routine calculations and Fact 2.8, one shows easily that Thus φ is a homomorphism. Therefore Stab α (R) is embedded in F(R) × . For the case R = F q , we need only to show that φ is surjective. Let F ∈ F(F q ) × . Then there  In Proposition 3.9 we have proved for any finite ring R that the group P R (R[α]) is embedded in P(R) ⋉ θ F(R) × . In the following theorem we show that, for a finite field F q , Theorem 4.9. Let F q be the finite field of q elements, and let θ be the homomorphism defined in Lemma 3.2
The following lemma from [4] gives the cardinality of ker φ n of the epimorphism φ n mentioned in Remark 5.2.
Proof. Consider the epimorphism φ n given in Remark 5.2. We have, by Lemma 5.1, Proof. Assume to the contrary that [f ] p k = [g] p k . Then f (a) ≡ g(a) (mod p k ) for every a ∈ Z, and hence f (a) ≡ g(a) (mod p n ) for every a ∈ Z since n ≤ k. But this means that [f ] p n = [g] p n which is a contradiction. We need the following fact from [4] in which we use our notations.  Proof. We have Thus every polynomial of the form (5.1) represents an element of F(Z p n ) × by Lemma 5.1.
Fix an integer 2 ≤ k ≤ n. Then, by Lemma 5.4, the sum i+vp(j!)=k−1 a kij p i (x) j is a member of ker φ k , and hence can be chosen in | ker φ k | ways. Therefore the total number of polynomials of the form (5.1) is (p − 1) p n k=2 | ker φ k | = |F(Z p n ) × | by Theorem 5.6. So to complete the proof we need only to show that every non equal two polynomials of the form (5.1) induce two different functions on Z p n . Now let f, g ∈ Z[x] of the form (5.1) with f = g. Then, for simplicity, we write First, we notice that if r = s, then by Remark 5.8, [f ] p = [l r ] p = [l s ] p = [g] p . Thus [f ] p n = [g] p n by Lemma 5.7. So we may assume that r = s, and hence n k=2 f k (x) = n k=2 g k (x). Since n k=2 f k (x) = n k=2 g k (x), there exists 2 ≤ k 0 ≤ n such that f k 0 = g k 0 and we choose k 0 to be minimal with this property. Now f k 0 , g k 0 are two different polynomials of the form (5.2) i+vp(j!)=k 0 −1 a k 0 ij p i (x) j , where i, j ≥ 0; 0 ≤ a k 0 ij ≤ p − 1.