Abstract
Swan’s theorem determines the parity of the number of irreducible factors of a binary trinomial. In this work, we study the parity of the number of irreducible factors for a special binary pentanomial with even degree , where , and exactly one of , and is odd. This kind of irreducible pentanomials can be used for a fast implementation of trace and square root computations in finite fields of characteristic 2.
1. Introduction
Irreducible polynomials of low weight over a finite field are frequently used in many applications such as coding theory and cryptography due to efficient arithmetic implementation in an extension field and, thus, it is important to determine the irreducibility of such polynomials. The weight of a polynomial means the number of its nonzero coefficients.
Characterization of the parity of the number of irreducible factors of a given polynomial is of significance in this context. If a polynomial has an even number of irreducible factors, then it is reducible and, thus, the study on the parity of this number can give a necessary condition for irreducibility. Swan [1] gives the first result determining the parity of the number of irreducible factors of trinomials over . Vishne [2] extends Swan’s theorem to trinomials over an even-dimensional extension of . Many Swan-like results focus on determining the reducibility of higher weight polynomials over ; see for example [3, 4]. Some researchers obtain the results on the reducibility of polynomials over a finite field of odd characteristic. We refer to [5, 6].
On the other hand, Ahmadi and Menezes [7] estimate the number of trace-one elements on the trinomial and pentanomial bases for a fast and low-cost implementation of trace computation. They also present a table of irreducible pentanomials whose corresponding polynomial bases have exactly one trace-one element. Each pentanomial of even degree in this table is of the form , where , and exactly one of , and is odd. In this work, we characterize the parity of the number of irreducible factors of this pentanomial. We describe some preliminary results related to Swan-like results in Section 2 and determine the reducibility of the pentanomial mentioned above in Section 3.
2. Preliminaries
In this section, we recall Swan’s theorem determining the parity of the number of irreducible factors of a polynomial over and some results about the discriminant and the resultant of polynomials.
Let be a field and let , where are the roots of in an extension of . The discriminant of is defined by From the definition, it is clear that has a repeated root if and only if . Since is a symmetric function with respect to the roots of , it is an element of .
The following theorem, due to Swan, relates the parity of the number of irreducible factors of a polynomial over with its discriminant.
Theorem 1 (see [1, 8]). Suppose that the polynomial of degree has no repeated roots and let be a number of irreducible factors of over . Let be any monic lift of to the integers. Then, or , and if and only if .
Let , where are the roots of in an extension of . The resultant of and is defined by It is well known that where denotes the derivative of with respect to . An alternate formula for the discriminant of a monic polynomial is see [9].
Let then, for all , the coefficients of are the elementary symmetric polynomials of . Since each , it follows that for every symmetric polynomial .
The following natation will be used throughout the paper. For all integers and , let We denote simply as and put . Then, the following lemma holds.
Lemma 2 (see [10, 11]). (1) . (2) . (3) .
The following formula, called Newton’s identity, is often used for computation of the discriminant.
Theorem 3 (see [12]). Let and be as above. Then, for every , where .
The reciprocal polynomial of with over a finite field is defined by See Lidl and Niederreiter [12] for more details.
3. Main Results
In this section, we characterize the parity of the number of irreducible factors for the pentanomial where is even; and exactly one of , and is odd. For our purpose, we use Swan’s theorem and Newton’s identity. In [10, 11], Newton’s identity has also been used to solve similar problems where it is enough to determine the power sums with indices , but, for (10), one should calculate much more negative indexed power sums. We return this calculation to one of positive indexed power sums by using reciprocals.
It is clear that (10) has no repeated roots because its derivative has a unique root . Let be the monic lift of in (10) to the integers and let denote the roots of in some extension of the rational numbers. The derivative of is Note that . Our work is divided into three cases according to which one of , and is odd.
Case 1 ( is odd). We can write the resultant of and as Since and are even, we have Using Lemma 2 and the fact that the square of every odd integer is congruent to 1 modulo 8, we get Newton’s identity shows that if , then and Therefore, The indices of terms in the above equation have a relation Since , we determine all for by applying Newton’s identity to get Table 1.
Note that does only cover the case of ; that is, . Since , , and are all even, we have Therefore, With reference to Table 1, we can determine all unknown terms in the above equation. We consider two subcases.
Subcase 1 ( is divisible by ). Then, we see easily that hence, the value of modulo depends on a pair . Let .
Theorem 4. Suppose that is odd and is divisible by 4. Then, the pentanomial in (10) has an even number of irreducible factors over if and only if one of the following conditions hold.
Consider(1) :(a);(b) and ;(c) and ;(d) and .(2) :(a) and ;(b) and ;(c) and .
Proof. If , then and, therefore, we have If , then and . So if , that is, , then And if , then (21) holds again. Similarly, if , then and . Thus, if , that is, , then (23) holds and if , then (21) holds. If , then and or can be nonzero only when it is equal to either or . Analyzing the possible cases shows that implies (23) and implies (21). Now, applying Swan’s theorem completes the proof.
Subcase 2 ( is not divisible by ). Then, and we can write
where
It is clear that if , then
and if , then
Now determine and . First, assume that ; that is, . Since and , we have
And then because . Next, assume that ; that is, . Clearly, and . Since , if , that is, , then and if , then . Therefore, we get
and also , similarly. When and , a similar consideration shows
Summarizing the above discussion and applying Swan’s theorem, we have the following theorem.
Theorem 5. Suppose that is odd and . Then, the pentanomial in (10) has an even number of irreducible factors over if and only if one of the following conditions hold.
Consider(1) or :(a);(b) and ;(c) and either or ;(d) and either or ;(2) or :(a) and ;(b) and either or ;(c) and .
Case 2 ( is odd). Similarly, we have From Newton’s identity, we see easily that , , , and are nonzero for and if is even with , then is even. To calculate for negative indices, we observe a monic lift of the reciprocal polynomial of to the integers. Denote the th power sum of the roots of in some extension of the rational numbers by . Then, clearly for every positive integer . We can apply Newton’s identity to to see that is equal to for odd and is even for even . From the above discussion, we have
First consider the case when .
Theorem 6. Suppose that is odd and . Then, the pentanomial in (10) has an even number of irreducible factors over if and only if one of the following conditions hold.
Consider(1):(a) and either or ;(b) and either or ;(c) and either or ;(2):(a), and ;(b), and either or ;(c), and either or ;(d), and either or .
Proof. We determine the unknown terms in (35). Clearly, from . Let again It is easy to see that We also see that is equal to if and equal to , otherwise, since . And, by Newton’s identity, we have With reference to the nonzero coefficients of , we obtain that if and otherwise. Now we can determine modulo . If and , then and thus . Clearly and ; hence . Consideration for the other cases is similar so we describe only the results: if , then if , , then and if , then Next, we compute modulo . If we denote the coefficient of in by , then Since is odd and is even , and is even. It follows, therefore, that . Repeating this process, we get , where . Thus, we obtain Now, Swan’s theorem is used to complete the proof.
The remaining cases when or gives the following theorems, whose proofs follow a similar way and, hence, are omitted.
Theorem 7. Suppose that is odd and . Then, the pentanomial in (10) has an even number of irreducible factors over if and only if one of the following conditions hold.(1), and either or ;(2), and either or .
Theorem 8. Suppose that is odd and . Then, the pentanomial in (10) has an even number of irreducible factors over if and only if one of the following conditions hold.
Consider(1), , and either or ;(2), , and either or ;(3), , and either or ;(4), and either or ;(5), , and either or .
Case 3 ( is odd). Analogously, to Case 2, we can write the resultant of and its derivative as follows: Straightforward calculations show that and and are even. In this case, we have also that is equal to for odd and is even for even . It follows, therefore, that Now, let
We present the following result for the reducibility of in (10) depending on the value of modulo .
Theorem 9. Suppose that is odd. Then, the pentanomial in (10) has an even number of irreducible factors over if and only if one of the following conditions hold.(1) and ;(2), , and either or ;(3), , and either or .
Proof. First, we compute in (45). By Newton’s identity, we get Since , we have and ; hence is equal to if and equal to otherwise. A simple calculation shows that if , then . If , then and, thus, we have Now applying Swan’s theorem completes the proof.
4. Conclusion
We have determined the parity of the number of irreducible factors of a pentanomial (10) under the condition that and exactly one of , and is odd. Our discussion is based on Swan’s theorem. If is odd, we obtained only a result which depends on modulo instead of exponents of the terms of a given pentanomial. In this case, a complete characterization of the reducibility of the given pentanomial seems to be more difficult.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Acknowledgment
The authors would like to thank the anonymous referees for their useful comments and suggestions.