Some New Results About Trigonometry in Finite Fields
Habib Hosseini^{1,}^{ *}, Naser Amiri^{2}
^{1}Department of Mathematics, Firoozabad Branch, Islamic Azad University, Firoozabad, Iran
^{2}Department of Mathematics, Tehran Payame Noor University, Tehran, Iran
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To cite this article:
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: K → F. 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
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 a ∤ b 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 17^{th} 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.
References