Pure and Applied Mathematics Journal
Volume 5, Issue 4, August 2016, Pages: 93-96

Some New Results About Trigonometry in Finite Fields

Habib Hosseini1, *, Naser Amiri2

2Department of Mathematics, Tehran Payame Noor University, Tehran, Iran

(H. Hosseini)
(N. Amiri)

*Corresponding author

Habib Hosseini, Naser Amiri. Some New Results About Trigonometry in Finite Fields. Pure and Applied Mathematics Journal. Vol. 5, No. 4, 2016, pp. 93-96. doi: 10.11648/j.pamj.20160504.11

Received: April 23, 2016; Accepted: May 21, 2016; Published: June 17, 2016

Abstract: In this paper we study about trigonometry in finite field, we know that , the field with p elements, where p is a prime number if and only if p = 8k + 1 or p = 8k−1. Let F and K are two field, we say that F is an extension of K, if K F or there exist a monomorphism f: KF. recall that , F[x] is the ring of polynomial over F. If K  F (means that F is an extension of K) an element u F is algebraic over K if there exists f(x) K[x] such that f(u)=0. The algebraic closure of K in F is , is the set of all algebraic elements in F over K.

Keywords: Trigonometry, Finite Field, Primitive, Root of Unity

Contents

1. Introduction

In this paper we study about trigonometry in finite field, we know that , the field with p elements, where p is a prime number if and only if p = 8k + 1 or p = 8k − 1. More generally, what can be said about  In  where , p are prime numbers. Can we replace square root by cube roots.

In attempting to answer the question, for which p, , we are naturally led to use the formula,  Indeed, if  we have,  and so , we can choose θ, a suitable 16th root of unity, such that . The crucial observation is that this formula makes sense any algebraic closure of  if .

Let F and K are two field, we say that F is an extension of K if K F or there exist a monomorphism. recall that ,  is the ring of polynomial over F. If K  F (means that F is an extension of K) an element u F is algebraic over K if there exists  such that . The algebraic closure of K in F is , is the set of all algebraic elements in F over K.

Definition: Let p be a prime number,  and k an integer such that . Then define the set cos[k] = {c(θ) =  | θ is a primitive kth root of unity}.

Note that symbol | is divisor or divides such that a|b means a divides b and ab means a does not divide b.

Remark:

(1) Recall that θ is a primitive kth root of unity if  but , for all . We have two make the assumption p k because if p | k, then there are no primitive kth root of unity in

(2) We can defined sin[k] = {s(θ) = | θ is the kth root of unity}, in this set i is a fixed square root of -1. We know that s(θ) . In particular we have  and θ = c(θ) + is(θ).

Theorem 1. If K is a field with 9 elements and if is a finite extension of K, then the mapping λ:  → defined by  is an automorphism of which fixes exactly the elements of K.

Proof: It is obviously that λ is onto and one to one.

Theorem 2. Let θ be a primitive kth root of unity. Then  if and only if p ≡ ±1(mod k).

Proof: Assume   . If θ , then p ≡ 1(mod k). Since the order of the multiplicative group of  is p − 1. If θ , then the irreducible polynomial of θ over  is . Hence  and so p ≡ −1(mod k).

Conversely, let p ≡ ±1(mod k). If p ≡ 1(mod k) then, since the multiplicative group of  is cyclic of order p − 1,  contains a primitive kth root of unity. Therefore  contains all primitive kth root of unity and so θ . Hence   . If p ≡ −1(mod k) then  whence , so   .

Corollary 3. If p ≠ 2 and θ is a primitive kth root of unity, then c(θ)  if and only if p ≡ ±1(mod k).

Remark: we observe that since membership of c(θ) in  depends only on p and k We have that either cos[k]  or cos[k] ∩  = Ø.

Lemma 4. Let θ be a primitive kth root of unity in  the algebraic closure of the rationales Q. Let R = Z[θ], the subring of  generated by the integers Z and θ, and let P be a prime ideal of R containing of , where (p, k) = 1, where (,) denote the highest common factor. Let S be the valuation ring of Q(θ) containing the ring A = { }, and let M be the maximal ideal of S. Then = θ + M is a primitive kth root of unity in the field of .

Proof. The formal derivative  of  is relatively prime to and so  has no repeated roots in . On the other hand,  = and so, over :

= It follows that is a primitive root of unity in .

Remark. For the basic properties of valuation rings the reader can consults. In particular, it is worth recalling that each valuation ring is integrally closed in its quotient field K, and so, if , k K, then k A. Moreover, each valuation ring is a local ring which means that for each a A/M, a−1 A/M as well. Expression obtained for the real and imaginary parts of the roots of unity over complex number are meaningful in A/M.

2. Some Properties

Corollary 5. Let (q, 10) = 1. Then  were n is the number of 2’s occurring under the root signs (excluding the 2 in the denominator!).

Proof. Define , and for each n ≥ 2: . Let . Now is a primitive 5th root of unity viewed as an element of the complex number. Thus  is a 5th primitive root of unity in  provided p ≠ 5. Moreover, it is easy to check that  and so  is a primitive  root of 1.

Remark. If in corollary 5 we take n = 0, q = p, we obtain a special case of the quadratic reciprocity law, namely:     p ≡ ±1(mod 5)  or     p ≡ ±1(mod 5)

Corrollary 6. Assume (2, q) = 1. Then     q ≡ ±1(mod  where n is the number of 2’s occurring under root signs.

Proof. Let  and for each n ≥ 2 Let  where at each stage we make a specific choice of square root.

As before letting  and we have  is a primitive  root of unity.

Corollary 7: Let (6, q) = 1. Then     q ≡ ±1(mod , where n is the number of 2’s under the square root signs.

Proof. Let and for each n ≥ 2 Let . Then with the same notation as above we have  is a primitive  root of unity.

Remark. If n = 0 and q = p above we have     p ≡ ±1(mod  which is again a particular case of the quadratic reciprocity Law.

Corollary 8: Let (q, 34) = 1. Then

The Formula in corollary 8 is quite complicated and one is naturally interested to know whether already some subformula of this formula is an element of . Suppose that q ≡ ±1(mod 17), then .

Indeed set λ =  where θ is a primitive 17th root of unity in . Since q ≡ ±1(mod 17) we see  = λ and λ . On the other hand one checks easily that , hence . We climb that also is a square in . To show this consider α =  and β = . Then α + β = λ. Moreover we have = -1. Thus . Since q ≡ ±1(mod 17) we see that both α, β . Hence  too. Since  or  we see that  or . Since we see that both element  belong to. Combining corollary 8 with the considerations above we obtain.

Corollary 9: Suppose that (q, 34) = 1, Then  and  both belong to  if q ≡ ±1(mod 17).

Remark. One could use the formula given in the table at the end of this note to deduce corollary 9, more easily. Indeed, for example, from  and  in  we deduce that

, similarly . From this follows that  and .

Theorem 10: Suppose (34, q) = 1, Then  if and only if q ≡ ±1, ±4(mod 17).

Proof. if  then also  and q ≡ ±1, ±4, ±2, ±8(mod 17). Indeed  if either  and  or  with r even. In the first case p ≡ ±1, ±4, ±2, ±8(mod 17) and therefore ≡ ±1, ±4, ±2, ±8( mod 17), too. On the other hand p, when r is even, is congruent to one of the elements ±1, ±4, ±2, ±8. On the other hand, in the notation as above, we have  if and only if . If q ≡ ±1 or q ≡ ±4 we see that  α and α . Hence q ≡ ±1, ±4(mod 17). So .

We want to prove that  then q ≡ ±1, ±4(mod 17). It is enough to exclude possibilities q ≡ ±2, ±8(mod 17). Suppose that q ≡ ±2, ±8(mod 17), Then

. Thus  iff  that this is contradiction.

Corollary 11: Assume (34, q) = 1. If , then  if and only if q ≡ ±1(mod 8) and q ≡ ±1, ±4(mod 17) or q ≡ ±3(mod 8) and q ≡ ±2, ±8(mod 17).

Therefore the inclusion  depends only on q(mod 136). we now focus attention on s(θ) =  where θ is a primitive kth root of unity in . Note that if  = which has been dealt with is lemma 4 from now on we assume 2 q.

Definition: Let sin[k] = {s(θ)|θ is a primitive kth root of unity}, we shall abbreviate s(θ) to s. The reader should beware that ‘ is ’ is not necessarily the third person singular of the present tense of the verb to be!

Theorem 12: Let θ be a primitive kth root of unity. Then s(θ)  iff one of the following holds:

(i) q ≡ ±1(mod [4, k]) where [,] denote the least common multiple.

(ii) k has the form 8m + 4 and q ≡ 4m + 1(mod k)

(iii) k has the form 8m + 4 and q ≡ 4m + 3(mod k)

Proof. Assume s = s(θ) . Then q ≡ 1(mod k) and set c = c(θ) so that θ = c + is. For case (i), Let . Then q ≡ 1(mod k) and by corollary 3: . Therefore is  and so . Hence q ≡ 1(mod 4) and thus q ≡ 1(mod [4, k]).

Case(ii), Let  and . Then  too, and thus. Therefore q ≡ −1(mod 4). On the other hand  Since  and so q ≡ −1(mod k), with , implies that  q ≡ −1(mod [4, k]).

Case (iii), ,  and is belong to. In this case  and so q ≡ 1(mod 4). Now  whence . But  and so  Therefore  Hence  and so . Therefore 2q ≡ −2(mod k). So  and thus k is even and. Therefore q ≡ 1(mod 4),  and . It is easily seen that these three condition are equivalent to k = 8m + 4 and q ≡ 4m + 1(mod k) for some m.

Corollary 13: For any k, either sin[k]  or sin[k] ∩ = .

Proof. As s(θ) depends only on q and k and not particular primitive root chosen. Finally, we determine how many distinct values of c(θ) and s(θ) there are as θ varies over the primitive kth root of unity

3. Conclusion

We conclude that in the field of real numbers trigonometric ratios are defined as defined in finite fields. As well as relations between trigonometric ratios hold in the field of real numbers, finite fields are also established under the circumstances.

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