Abstract

In the Galois fields , a polynomial basis with a small number of trace-one elements is desirable for its convenience in computing. To find new irreducible polynomials over with this property, we research into the auxiliary polynomial with roots , such that the symmetric polynomials are relative to the symmetric polynomials of . We introduce a new class of polynomials with the number “1” occupying most of the values in its . This indicates that the number “0” occupies most of the values of the traces of the elements . This new class of polynomial gives us an indirect way to find irreducible polynomials having a small number of trace-one elements in their polynomial bases.

1. Introduction

Let denote the finite field with elements. Let be a root of an irreducible polynomial of degree over . The set forms a basis of , called a polynomial basis. An element can be represented as The trace of is defined as which is the linear combination of the traces of elements of the polynomial basis; that is, If the number of basis elements for which is small, the computation of in software and the implementation in hardware are faster. It is benefit for halving a point on an elliptic curve over [1, 2] or generating pseudo random sequences using elliptic curves [3]. Thus it is of interest to find irreducible polynomials having a small number of trace-one elements in their polynomial bases. As the value of equals the symmetric polynomial , the authors in [4] got the conditions for trinomials and pentanomials to have this property using the Newton function for symmetric polynomials [5, 6].

We presented an indirect strategy that is different from the above one, which is inspired by extending the work of [7] in [8]. We still use the Newton function to investigate the polynomial with the following two properties. First, the polynomial should have the factorization with irreducible. Second, let be the roots of , the number of different values of the symmetric polynomials, equaling one should be as many as possible. With these two properties, we can identify as having less trace-one elements in its polynomial basis. A class of polynomials suiting our demands is introduced, and we showed their existence by numerical experiments.

The paper is organized as follows. In Section 2 we give some basic definitions which are needed for our result. The new class of polynomials is introduced in Section 3. Section 4 draws the conclusion of our work.

2. Preliminaries

Let be a polynomial over of degree whose roots are . The elementary symmetric polynomials are the coefficients of :

Another particular kind of symmetric polynomial is defined as where is an integer.

The Newton function is a relation between and the coefficients of the polynomial; it is defined as follows [5, p.12]:

It is widely known that if is irreducible, then the traces of equal the value of .

Let ; the roots of are . We haveSo when more ones appeared in the set of the values of , more zeros appeared in the traces of elements of the polynomial basis of .

3. The Three Piece Polynomials

We first introduce a class of polynomials with roots over fitting the conditions that the value set of has plenty of ones and then search for the polynomials with the decomposition where is irreducible.

3.1. The Three Piece Polynomials

Definition 1. Let be a polynomial over with nonzero constant term. If where and mean consecutive bits of ones and zeros, respectively, we call three-piece polynomial over and denote it by its coefficients as .

Lemma 2. Let be a three-piece polynomial with coefficients over . Let be the roots of ; if , then

Proof. If , then , and the Newton formula tells us that and . Suppose is true for ; then and We have

We mainly divide our theorems into two parts according to the parity of . First, let us consider the condition that is even.

Theorem 3. Let be denoted as above. If both and are even, then the values of functions , all equal one.

Proof. From Lemma 2, we know that for . We have as is even and . Suppose is true for ; then as . Here we come to the next even number with , Suppose is true for ; then as .

In the following proofs, we still focus on the changes of the value in the summations of the symmetric polynomial . To simplify narration, we will divide the summations in into four parts according to the subscript of . That is, the summations with are contained in , the summation with comprises , the summations in are with , and the ending part .

Theorem 4. Let be denoted as above. If is even, is odd, and , then the values of functions , , are

Proof. The first case is contained in Theorem 3.
For the remaining cases, we declare they are periodic. We first show one period in the following. That is, for we haveBy the assumption , we get for and ; these simplify the above equation to .
For the value of , as appears in the of and its value is “0” we have .
We go on our proof by induction. Suppose holds for all satisfying ; only the summation with equals zero in of , so .
For , as goes into the summation in , the value of the summation turns back to and we have . Now, we finished computing the value of in one period in our declaration, which is .
The condition guarantees the subsequent computations can still use for (17), as and still holds. So the above method will work for the next periods, and our declaration holds.

Next we will deal with the situation that is odd, which is more complicated. It has been proved that for in Lemma 2.

Lemma 5. Let be denoted as above. If is odd, then the values of functions , , are

Proof. We follow the idea in the proof of Theorem 4. For , letWe have and it fixes the value of in for to be which makes . If then and the next period starts; otherwise the equation holds within one period.

In the following, we will deal with with ; remember that is odd. We define a comparative sequence here to simplify the narration. Let be the coefficient of a polynomial with and then of , which is denoted by , follows the period shown in Lemma 5. That is,

Theorem 6. Let and be denoted as above. If is odd, is even, and , then the values of functions , have two cases.

Proof. For , we haveFor , we have as enters the summations of , so . By induction, we haveThe value of for is opposite to the corresponding .
We have for as the value of is changed in the of .
As all the summations in and are the same for , we get and our claim follows.

Using the same method, we can also prove the following theorem.

Theorem 7. Let and be denoted as above. If is odd, is odd, and , then the values of functions , have two cases.

That is, the changed value appears at the position where .

Corollary 8. Let and be denoted as above. If and , then the values of function , are

Proof. As all the coefficients , equal zero, the first case is straightforward. For , we have and the corollary follows.

We have discussed the four cases concerning the three-piece polynomials over . The condition guarantees that the above cases cover all the possible kinds of three-piece polynomials. The number “1” does occupy a great proportion of the values of , as the examples shown in Table 1.

3.2. Experiment Results

Next, we search for the three-piece polynomial whose decomposition is with irreducible. As there exists a trinomial or pentanomial basis for having at most four elements of trace one for each [4], the polynomial with less number of trace-one elements that appeared in its polynomial basis is acceptable.

With the formula , we choose the number of trace-one elements to be at most four, and the result showed that the three-piece polynomials fitting these requirements exist for a great number of , and we guess they exist for . The proof for this conjecture is difficult as it remains open to prove the existence of irreducible polynomials when almost all coefficients of them are prescribed [9].

If we want to get a polynomial basis with less than two trace-one elements from a three-piece polynomial, then almost every odd degree will upset us. These two results are illustrated in Tables 2 and 3.