Journal of Mathematics

Volume 2014, Article ID 258187, 7 pages

http://dx.doi.org/10.1155/2014/258187

## Self-Dual Normal Basis of a Galois Ring

Algebra Research Group, Institut Teknologi Bandung, Bandung 40132, Indonesia

Received 12 May 2014; Revised 13 August 2014; Accepted 15 August 2014; Published 1 September 2014

Academic Editor: Fernando Torres

Copyright © 2014 Irwansyah et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

Let and be two Galois rings. In this paper, we show how to construct normal basis in the extension of Galois rings, and we also define weakly self-dual normal basis and self-dual normal basis for over , where is considered as a free module over . Moreover, we explain a way to construct self-dual normal basis using particular system of polynomials. Finally, we show the connection between self-dual normal basis for over and the set of all invertible, circulant, and orthogonal matrices over .

#### 1. Introduction

Normal basis is one important type of basis over Galois fields, because it is computationally manageable. One type of normal basis which has applications in cryptography and coding theory is self-dual normal basis; see [1, 2]. Therefore, some researcher are interested in finding a way to construct self-dual normal basis over Galois fields; see [3, 4].

On the other hand, as a generalization of Galois fields, Galois rings have also several connections with coding theory; see [5, 6]. Normal basis and its variants such as self-dual are also important in Galois ring, especially for computations in codes over this ring [2, 4].

Given two Galois rings and , where is an extension of , a normal basis for over is a basis which consists of all orbit of some from the action of into , where is Galois group of over . As a consequence of the work of Kanzaki [7] and the fact that Galois ring is a local ring [8, Lemma 2], there always exists normal basis for over .

In this paper, we will show some properties of self-dual normal basis of a Galois ring similar to the properties of self-dual normal basis over finite fields. We will also explain how to get normal basis generator and self-dual normal basis generator and show a connection between the set of generators of self-dual normal basis and the set of invertible circulant and orthogonal matrices over . Moreover, we give an application of normal basis, especially self-dual normal basis, in encoding certain cyclic codes over .

#### 2. Normal Basis over Galois Ring

##### 2.1. Some Properties of Normal Basis over Galois Ring

Let be a Galois ring with characteristic and cardinality . If and , then is a subring of . Moreover, if is a basic primitive irreducible polynomial with degree , then is a Galois ring with characteristic and cardinality and contains as a subring. In other words, is an extension of of order . If is a root of and has order , then . Define a map where for every . We can prove that is an automorphism which left fixed. Furthermore, generates the Galois group of order [8]. Some detailed explanations about Galois ring can be found in [8].

We define normal basis of over as follows.

*Definition 1. *A normal basis of over is a basis which consists of the elements , where , for some , where is considered as a free-module over .

If is a normal basis for over , then, for every , we can uniquely write down as follows: for some . Therefore, we can consider as an element in . As an immediate consequence of the above definition, one has the following lemma.

Lemma 2. *Let be a normal basis for over . If with respect to , then .*

*Proof. *Consider ; then , because every element in is fixed by . Thus, we have .

Before we construct a normal basis in Galois ring, we need the following definition and ring homomorphism.

*Definition 3 (see [9]). *Let be a polynomial in with degree , where is Galois field with elements and is a prime power. We say is a Normal Polynomial or N-Polynomial if is irreducible over and its roots is the generator of normal basis for over .

We define a ring homomorphism as in [8] by
where . This ring homomorphism can be extended to the following ring homomorphism:
where . The following theorem gives us a way to construct normal basis for over .

Theorem 4. *Let be the ring of integers modulo and let be a basic primitive polynomial of degree over . If is a generator of normal basis for over , then is also a generator for normal basis for Galois ring over .*

*Proof. *As in [8] the homomorphism (4) induces the following homomorphism:
where the kernel of homomorphism (5) is the ideal . Consequently,
From (6), is a generator of normal basis for over and also for over . Therefore, we can say that if and only if for all .

Now, we need to show that is also a generator for normal basis of over . Since is already linearly independent over , if we want to show that is linearly independent over , we only need to prove that the the following equation over
is only satisfied by for all . Let , where for all . Consider
where is the rest terms. By (7), we have is or and from the first paragraph, we can say that . Therefore, using the fact that the term with power of is more than or is equal to , we have will not be the case. Hence, , which means will be equal to or . Again, from the first paragraph, we have . Consequently, we have , and because is linearly independent over , we have is the only solution. Continuing this process for the terms which contain , , and so on, we have , for all , is the only solution. Hence , for all , is the only solution for (7) as we hope. Because Galois ring has invariant basis number by [10, Lemma 2.6], so must be a basis.

Let be a basic primitive polymomial of degree over and is a root of . Based on [8, Theorem 14.8 part (ii)], we have where . We call this representation as -adic representation of the Galois ring. The next theorem gives a way to construct normal basis for extension of Galois rings in general.

Theorem 5. *Let and be two Galois rings, where , and is a basic primitive polynomial of degree . If is a generator of normal basis for over , then is also a generator for normal basis of Galois ring over .*

*Proof. *Let be a primitive polynomial of degree over . Since, , we can take a basic primitive polynomial over . Let be a root of , so is a primitive element of . Therefore, by -adic representation we have

Since , Galois field is a subfield of Galois field . We have
Let . In order to be a basis, we only need to show that is linearly independent over . Now, consider
where for all . Using -adic representation of and the methods in Theorem 4, we will have for all is the only solution and is a normal basis for over .

Let be a normal basis of over . Let where
for some . Let ; based on [10, Lemma 2.6], Galois ring has invariant basis number. Therefore, the transition matrix between two basis is an invertible square matrix. We have the following characterization.

Proposition 6. * is a normal basis of over if and only if is an invertible circulant matrix.*

*Proof. *The proof is similar to the proof of Theorem 1.6 in [9].

If is a root of a basic irreducible polynomial in , then as in [8] we have . Now, let , where and for some , for all . Then we have the following characterization.

Proposition 7. *The set forms a normal basis for over if and only if is an invertible matrix and for all .*

Now, let be a normal basis for over . As before, we can write down uniquely and for some , where , with respect to normal basis . We are also able to represent , with respect to the above normal basis. Therefore, we would like to know how to write every coordinate of in terms of and . Let
for some ; then, we will have
where is an -by- matrix over . Then, we would like to know how many nonzero entries in every matrix in the collection , which called* complexity* of normal basis .

*Definition 8. *Let be the number of nonzero entries in the matrix . Then, is called the complexity of the matrix . Furthermore, if , then is called the complexity of normal basis .

Using the generalization of homomorphism in (4) and [8, Lemma 14.31], we have that is the Frobenius automorphism in finite field over . In fact, is isomorphic as a group to via the above homomorphism; see [8, Theorem 14.32]. Now, consider (15)
which implies
Then we will have the following result.

Proposition 9. *If is the complexity for , then .*

*Proof. *Note that, by Theorems 4 and 5, is also a generator of normal basis for over . Therefore, is a normal basis for over . We will consider first. Let be the number of nonzero entries in and let be the multiplication table for . By Theorem 5.1 in [9], we have the number of nonzero entries in ; that is, , satisfies which implies . Therefore, we have . Now, by (18), we have the conclusion for all .

From the above result, we have an immediate consequence regarding the complexity of normal basis in Galois rings as follows.

Corollary 10. *For any normal basis of over , one has .*

##### 2.2. Self-Dual Normal Basis

A dual basis of a basis is defined using a map called trace. We define trace over as follows: for every . More properties of trace function can be found in [8]. The definition of dual basis is as follows.

*Definition 11. *Let and be two bases of over as a free-module. We say that is a dual of if
for all , where is Kronecker’s delta function.

One has the following properties for dual basis which are similar to the properties in [11].

Lemma 12. *If and are two bases which are dual to each other, then one has the following.*(1)*For every one has
*(2)*, for all .*

*Lemma 13. There always exists a unique dual basis for any basis of over .*

*Proof. *This lemma follows from the fact that is a surjective map. The rest of the proof is similar to the proof of Theorem 1.1 in [9].

*The following lemma is similar to [9, Corollary 1.4].*

*Lemma 14. If is a normal basis of over and is a dual basis for , then is also a normal basis.*

*Self-dual normal basis is a normal basis which is equal to its unique dual basis. Bagio et al. [12] show that there always exist self-dual normal basis in an extension of a local ring of odd order. Therefore, there always exist self-dual normal bases of over when is odd, since Galois rings are local rings. Now, we will give a way to obtain self-dual normal basis generator of over whenever is odd, where is the Galois group of over .*

*Let ; then, we can show that is an algebra over . For , we define conjugate of as . The following theorem gives the generator of self-dual normal basis for over .*

*Theorem 15.
(1) Let be a normal basis generator for in over . If satisfies
then has an inverse in .*

(2) The mapping gives a one-to-one correspondence between the set of solutions for (22) and the set of elements in which generate self-dual normal basis.

*Proof. *The proof of this theorem is similar to the proof of [4, Theorem 2.5].

*The following proposition gives us a way to find which satisfies (22).*

*Proposition 16. If , then
for all .*

*Therefore, based on the previous proposition, which satisfies (22) are the solutions of the following system of polymomials:
*

*3. Connection between Self-Dual Normal Basis and Orthogonal Circulant Matrices*

*In this section we will show that the generators of self-dual normal basis for over are closely related to invertible, circulant, and orthogonal matrix over Galois ring . Therefore, the enumeration of self-dual normal basis for over is just the enumeration of the elements of the set of all invertible, circulant, and orthogonal matrices over with suitable size. The following proposition gives the easiest way to enumerate the solutions of system (24).*

*Proposition 17. Solutions of the system (24) has a one-to-one correspondence with the solutions of the following system:
*

*Proof. *Let be a solution of (24). By Proposition 16, we will have the corresponding element , and is a generator of self-dual normal basis by Theorem 15. Then, we can define
where is related to a solution of (25) by . Consider
Consequently, the images of are related to the solutions of (24). Moreover, since is nonzero, we have is a one-to-one correspondence. Furthermore, if is an element in which is related to a solution of (24), then satisfies
Therefore, is an element of which is related to a solution of (25). Consequently, given which is related to a solution of (24), then we can choose such that . In other words, is a bijection.

*Let be the set of all -by- invertible, circulant, and orthogonal matrices over . The following proposition gives the connection between the number of solutions of system (25) and .*

*Proposition 18. The number of solutions of system (25) is .*

*Proof. *The element as in (25) satisfies
By using the map , the last equation becomes . Therefore, . Since every solutions of (25) is uniquely related to which satisfies (29), we have that the number of solutions of system (25) is .

*The following theorem is the corollary of two previous propositions.*

*Theorem 19. The number of self-dual normal basis of over is .*

*4. Encoding Cyclic Codes over *

*4. Encoding Cyclic Codes over*

*4.1. Cyclic Codes over , Discrete Fourier Transform, and Galois Rings*

*4.1. Cyclic Codes over , Discrete Fourier Transform, and Galois Rings*

*Definition 20. *A cyclic code over , where is a prime, of length is an ideal of residue class polynomial ring .

*In [13], cyclic codes over were characterized using the following discrete Fourier transform (DFT).*

*Definition 21. *Let and . The DFT of is defined as
where is a primitive th root of unity in the Galois ring and is the least integer such that divides .

*Note that we can calculate the inverse discrete Fourier transform by the following formula:
Moreover, the vector is called the transform vector or spectrum of , and the components , , are called the DFT coefficients or spectral components of . Note that all the -tuples of which are DFT vectors of some -tuple over satisfy
for all , where is the generalized Frobenius automorphism on the Galois ring ; see [13].*

*As we can see, the DFT maps to a subset of . Sundar Rajan and Siddiqi [13] show that
where is the number of conjugacy classes for the integer and are the exponents of the conjugacy classses, for all . Therefore, by convolution property of DFT, we have
Hence, a cyclic code with length over consists of the inverse DFT coefficients of all vectors of the subring of which is isomorphic to whose specified spectral components are the elements of ideal , , for . Consequently, if is a cyclic codes over , then we have
Therefore, decoding of cyclic codes over can be accomplished by calculating inverse DFT in the Galois ring .*

*4.2. Encoding Algorithm*

*4.2. Encoding Algorithm*

*As in the previous subsection, we know that every cyclic code can be associated with an ideal , in . Hence, if we work in , then we can encode a cyclic code via inverse discrete Fourier transform. Therefore, our main purpose in this part is to formulate an algorithm for calculating inverse discrete Fourier transform. Before we explain the encoding algorithm, we will need the following materials first.*

*The following lemma is a direct consequence from the results in the last subsection.*

*Lemma 22. The inverse DFT of a vector is a vector with components in if and only if
for all , and
for all , where is the exponent of the conjugacy class which contains .*

*Therefore, the DFT spectrum is determined by one spectral coefficient for each conjugacy class. Our approach on calculating inverse DFT is similar to Fumy’s approach in [2] over finite fields, but now we apply it over Galois rings.*

*Let be an ideal in . From the previous subsection, we can say that is a cyclic code in spectral domain from conjugacy classes of . Then must be an ideal in , where is a least common multiple of .*

*Let be a DFT spectrum of a vector . We can associate and with polynomials
respectively. Let be a normal basis of over , where . We can write every spectral coefficient as follows:
where . Define
From Lemma 22, we will have . Therefore, can be obtained from . The following theorem is similar to [2, Theorem 1].*

*Theorem 23. Let be a basis of over which is dual to , and let
where , for all . Then .*

*The following corollary is an immediate consequence of the above theorem.*

*Corollary 24. , for and .*

*Recall that if is a monic basic irreducible polynomial over with minimal degree, where is a root of , then with basis . Therefore, we have the following lemma.*

*Lemma 25. Let be a monic basic irreducible polynomial with minimal degree where is a root of . Then, for any polynomial over ,
and is given in basis by the coefficients of .*

*Hence, we have the following algorithm to calculate inverse DFT over due to Fumy [2].(1)Represent the coefficients of DFT spectrum with respect to a normal basis of over and generate the polynomial .(2)Reduce modulo the monic basic irreducible minimal polynomial of the th root of unity over .(3)Transform resulting residues into the normal basis dual to which yields the coefficients of polynomial according to Corollary 24.*

*Recall that, when is an odd integer, the Galois ring has self-dual normal basis over . Therefore, we can make the previous algorithm complete by using only one normal basis as follows.(1)Given normal basis , solve (25) to obtain a generator for self-dual normal basis for over .(2)Represent the coefficients of DFT spectrum with respect to a normal basis of over and generate the polynomial .(3)Reduce modulo the monic basic irreducible minimal polynomial of the th root of unity over .(4)Transform resulting residues into basis which yields the coefficients of polynomial according to Corollary 24.Note that the algorithm can only calculate some which are in the scope of Corollary 24, and for the rest of , we can calculate using the formula of inverse discrete Fourier transform.*

*Example 26. *We would like to encode cyclic codes of length over . So, we let , , and . Then we will have , and the conjugacy classes are and . Therefore, and
We can show that , where is a root of , is a self-dual normal basis for over . Also, is the 3rd root of unity in . First, if we choose ideal , then every , and we can write down for some . Let be the codeword which we would like to find and ; then, and , because (from conjugacy property). Also, , because which implies and . By Corollary 24, we have and . We can calculate using inverse DFT formula, and we have . Consequently, we have and . Hence, our calculation gives a cyclic code .

*Second, if we choose ideal , then for all , where , because the ideal represents the conjugacy class and the ideal represents the conjugacy class . Since , we have and for and . Consequently, and . Therefore, by similar way as before, we have , , and . Hence, we have cyclic code .*

*5. Conclusion*

*5. Conclusion*

*In this paper we show that the generator of normal basis over Galois rings can be constructed using the generator of normal basis over Galois fields, and weakly self-dual normal basis over Galois rings has similar properties to the one over finite fields. We also can have self-dual normal basis generator for over simply by solving a particular system of polynomials, and the number of self-dual normal basis for over is . As an application, we may use normal basis and self-dual normal basis to encode certain cyclic codes over .*

*Conflict of Interests*

*Conflict of Interests*

*The authors declare that there is no conflict of interests regarding the publication of this paper.*

*Acknowledgment*

*Acknowledgment*

*This research is supported by Hibah Riset dan Inovasi KK ITB 2013 and BPKLN-DIKTI through Program Doktor Unggulan FMIPA-ITB Batch III-B.*

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