Some New Results about Trigonometry in Finite Fields

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Some New Results about Trigonometry in

Finite Fields

Amiri Naser

*

_{, Hasani Fysal}

Department of Mathematics, Payame Noor University, Tehran, Iran

Received 26 December 2015; accepted 11 June 2016; published 14 June 2016

This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/

Abstract

In this paper, we study about trigonometry in finite field, we know that 2∈p, 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 be two fields, we say that F is an extension of K, if K ⊆F or there exists a monomorphism f: K →F. Recall that

[ ]

{

2 ∈

}

0 1 2  , 0

n n i

F x = a +a x+a x + +a x a F n≥ , F[x] is the ring of polynomial over F. If _K_≤e_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 (see [1]-[4]). The algebraic closure of K in F is K, which 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 2∈p, 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

1 2 n

p + p + + p in _p where p p₁, ₂,,pn are prime numbers. To attempt to answer the question, for

which p, 2+ p∈p, we are naturally led to use the formula,

2 1 cos 2

cos .

2

θ

= + Indeed, if π

8

θ

= , we

have cos2π 2 2

8 4

+

= and so cosπ 2 2

8 2

+

= , we can choose θ, a suitable 16th root of unity, such that

*_{Corresponding author.}

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1

2 2

θ θ₊ − ₊

= . The crucial observation is that this formula makes sense any algebraic closure _p of _p

if p≠2.

Let F and K be two fields, we say that F is an extension of K if K⊆F or there exists a monomorphism :

f K→F. Recall that F x

[ ]

=

{

a₀+a x₁ +a x₂ 2++a x an n i∈F n, ≥0

}

, F x

[ ]

is the ring of polynomial over F. If K ≤e 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 K, which is the set of all algebraic ele-ments in F over K.

Definition. Let p be a prime number, p≠2 and let k be an integer such that p k . Then we can define the set

[ ]

( )

1

2

cos k c θ θ 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 di-vide b.

Remark. 1) Recall that θ is a primitive kth root of unity if θ =k 1 but θ ≠n 1, for all 1≤ ≤ −n k 1 (see

[2][4]). The two make the assumption p k because if p k , then there are no primitive kth root of unity in p

 .

2) We can define

[ ]

( )

1

sin

2

k s is the kth root of nit

i u y

θ θ

θ − θ

 

 

= = 



−

, in this set, i is a fixed square root of −1. We know that s

( )

θ

∈_pn. In particular, we have

( )

2 2

1

c

θ

+s

θ

= and θ =c

( )

θ +is

( )

θ .

Theorem 1. If K is a field with 9 elements and if 𝔽𝔽 is a finite extension of K, then the mapping λ: → defined by λ

( )

x =x9 is an automorphism of 𝔽𝔽 which fixes exactly the elements of K.

Proof. It is obviously that λ is onto and one to one (see [5][6]).

Theorem 2. Let θ be a primitive kth root of unity. Then

θ θ

+ −1∈_p if and only if p≡ ±1

(

mod k

)

.

Proof: Assume

θ θ

+ −1∈_p. If θ ∈_p, then p≡1

(

mod k

)

. Since the order of the multiplicative group of p

 is p − 1. If θ ∉_p, then the irreducible polynomial of θ over _p is

(

x−

θ

)

(

x−

θ

−1

)

. Hence θp=θ−1 and so p≡ −1

(

mod k

)

.

Conversely, let p≡ ±1

(

mod k

)

. If p≡1

(

mod k

)

then, since the multiplicative group of _p is cyclic of or-der p− 1, _p contains a primitive kth root of unity. Therefore _p contains all primitive kth root of unity and so

p

θ ∈_ . Hence θ θ+ −1∈_p. If p≡ −1

(

mod k

)

then θp=θ−1_hence

(

_{θ θ}₊ −1

)

p_{= +}_{θ θ}−1_{, so} 1 p

θ θ

₊ − _∈_

. Corollary 3. If p≠ 2 and θ is a primitive kth root of unity, then c

( )

θ ∈_p if and only if p≡ ±1

(

mod k

)

. Remark. We observe that since membership of c

( )

θ in _p depends only on p and k, we have that either

[ ]

cos k ⊂_p or cos

[ ]

k _p= ∅.

Lemma 4. Let θ be a primitive kth root of unity in Q, the algebraic closure of the rationales Q. Let

[ ]

R=Z θ , the subring of Q generated by the integers Z andθ, and let P be a prime ideal of R containing of P

R , where (p, k) = 1, where (,) denotes the highest common factor. Let S be the valuation ring of Q

( )

θ con-

taining the ring

{

1

}

, ,

A= xy− x y∈R y∉P , and let M be the maximal ideal of S. Then θ θ= +M is a

primi-tive kthroot of unity in the field of S M .

Proof. The formal derivative kxk−1 of xk−1 is relatively prime to xk−1 and so xk−1 has no repeated

roots in S

M . On the other hand,

(

)

1 0

1 k

k i

i

x − =

∏

₌− x−θ and so, over S

M :

(

)

1 0

1 k

k i

i

x − =

∏

₌− x−θ . It follows

that θ is a primitive root of unity in S M .

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2. Some Properties

Corollary 5. Let (q, 10) = 1. Then 2 2 2 5 1 1

(

2 5

)

2

n q q mod −

+ + + + ∈ ⇔ ≡ ± ⋅ were n is the number

of 2’s occurring under the root signs (excluding the 2 in the denominator!).

Proof. Define 1

5 1 2

r = − , ₁ 10 2 5

2

a = + , and for each n ≥ 2: rn= 2+an−₁, an= 2−an−1. Let 2

n

r n

b

= ,

2

n

a n

d

= . Now b1+id1 is a primitive 5th root of unity viewed as an element of the complex number. Thus b1+id1

is a 5th primitive root of unity in _p provided p≠ 5. Moreover, it is easy to check that

(

b_n+id_n

)

2=b_n−₁±id_n−₁ and so θ = +b_n id_n is a primitive 2n−1⋅5 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: 5 1 1

(

5

)

2 q q mod

− _∈ _{⇔ ≡ ±}

 or 5∈p⇔ p≡ ±1

(

mod5

)

.

Corollary 6. Assume (2, q) = 1. Then 2+ 2+ + 2 ∈_q ⇔q≡ ±1

(

mod2n+2

)

where n is the number of 2’s occurring under root signs.

Proof. Let a₁=0, b₁=2 and for each n ≥ 2 Let a_n= 2+b_n₋₁, b_n= 2−a_n₋₁ where at each stage we make a specific choice of square root.

As before letting

2 2

,

n n

a b n n

r t

= = and we have rn+itn is a primitive

2

2n+ root of unity.

Corollary 7. Let (6, q) = 1. Then ₂ ₂ ₃ ₁

(

₂n 2 ₃

)

q q mod

+

+ + + ∈ ⇔ ≡ ± ⋅ , where n is the number of

2’s under the square root signs.

Proof. Let a₁= 3,b₁=2 and for each n≥ 2 Let an= 2+bn−₁,bn= 2−an−₁. Then with the same nota-tion as above we have r_n+it_n is a primitive 2

2n+ _⋅3_{root of unity.}

Remark. If n = 0 and q = p above we have 3∈p ⇔p≡ ±1

(

mod12

)

which is again a particular case of the quadratic reciprocity Law.

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

(

)

1 17 34 2 17

17 3 17 34 17 2 34 2 17 1 17 .

2 q q mod

a=− + + − + + − − − + ∈ ⇔ ≡ ±

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 _q. Suppose that q≡ ±1

(

mod17

)

, then 17∈q.

Indeed set λ θ θ= + 16+θ4+θ13+θ9+θ8+θ15+θ2 whereθ is a primitive 17th root of unity in _q. Since

(

)

1 17

q≡ ± mod we see λq =λ and λ ∈_q. On the other hand one checks easily that

(

2

λ

+1

)

2=17, hence 17∈q. We climb that also

2 4

λ + is a square in _q. To show this consider α θ θ= + 16+θ4+θ13 and

9 8 15 2

β θ= +θ +θ +θ . Then α β λ+ = . Moreover we have 16

1 1

i i

αβ=

∑

₌θ = − . Thus α α− −1=λ. Since

(

)

1 17

q≡ ± mod we see that both α β ∈, q. Hence

2 ₄

q

λ + ∈_ too. Since 17 1 2

λ

= − or

(

17 1

)

2

λ=− +

we see that 2 17

(

− 17

)

∈_q or 2 17

(

+ 17

)

∈_q. Since 2 17

(

− 17

) (

⋅ 2 17+ 17

)

= ±8 17∈_q.

we see that both element 2 17

(

− 17

)

and 2 17

(

+ 17

)

belong to _q. Combining corollary 8 with the considerations above, we obtain.

Corollary 9. Suppose that (q, 34) = 1, Then 34−2 17 and 17+3 17− 34−2 17 −2 34+2 17 both belong to _q if q≡ ±1

(

mod17

)

.

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more easily. Indeed, for example, from c1 and c4 in q we deduce that 1 4

1 17 34 2 17

4 q

c +c =− + + − ∈ ,

similarly ₂ ₈ 1 17 34 2 17

4 q

c +c =− + − − ∈ . From this follows that 17 and 34−2 17∈q.

Theorem 10. Suppose (34, q) = 1, Then 2 17

(

+ 17

)

∈_q if and only if q≡ ± ±1, 4

(

mod17

)

.

Proof. If λ= 2 17

(

+ 17

)

∈_q then also 17∈_q and q≡ ± ± ± ±1, 4, 2, 8

(

mod17

)

. Indeed 17∈_q if either q=pr and 17∈q or

r

q= p with r even. In the first case p≡ ± ± ± ±1, 4, 2, 8

(

mod17

)

and therefore r 1, 4, 2, 8

(

17

)

m

p ≡ ± ± ± ± od , 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

α θ θ

1

θ

4

θ

4 q

− −

= + + + ∈_ if

and only if 2 17

(

+ 17

)

∈_q. If q≡ ±1 or q≡ ±4 we see that αq ₌α_and q

α ∈_ . Hence

(

)

1, 4 17

q≡ ± ± mod . So 2 17

(

+ 17

)

∈_q.

We want to prove that 2 17

(

+ 17

)

∈_q then q≡ ± ±1, 4

(

mod17

)

. It is enough to exclude possibilities

(

)

2, 8 17

q≡ ± ± mod . Suppose that q≡ ± ±2, 8

(

mod17

)

, Then α α₌ q₌θ8₊θ9₊θ2₊θ15_{= −}α−1_{. Thus}

(

)

1 17

2

1 17

2 θ α β

− +   = + = 

− +  

iff 17 1

17 1

 =

 

= −

 that this is contradiction.

Corollary 11. Assume (34, q) = 1. If q=pr, then 17+ 17∈q if and only if q≡ ±1

(

mod8

)

and

(

)

1, 4 17

q≡ ± ± mod or q≡ ±3

(

mod8

)

and q≡ ± ±2, 8

(

mod17

)

.

Therefore the inclusion 17+ 17∈q depends only on q(mod 136). we now focus attention on

( )

1

2

s

i

θ θ

θ = − − where θ is a primitive kth root of unity in _q. Note that if p=2,

1 1

2i 2

θ θ₋ − θ θ₊ −

= 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 ∈_q iff one of the following holds : (i) q≡ ±1

(

mod

[ ]

4,k

)

where [,] denotes 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 ∈_q. Then q≡1

(

mod k

)

and set c=c

( )

θ so that θ = c + is. For case (i), Let q

θ ∈_ . Then q≡1

(

mod k

)

and by corollary 3: c∈_q*. Therefore is∈_q and so ∈_q. Hence q≡1

(

mod4

)

and thus q≡1

(

mod

[ ]

4,k

)

.

Case (ii), Let θ ∉_q and c∈_q. Then is∉_q too, and thus i∉_q. Therefore q≡1

(

mod4

)

. On the oth-er hand q≡/1

(

mod k

)

Since θ ∉_q and so q≡ −1

(

mod k

)

, with c∈_q, implies that q≡ −1

(

mod

[ ]

4,k

)

.

Case (iii), θ ∉_q, c∈_q and is belong to _q. In this case i∈_q and so q≡1

(

mod4

)

. Now 2 2 1 c = −s whence q

c

θ = ± . But c∉_q and so q

c = −c Therefore

θ

q =

(

c+is

)

q = − +c is. Hence 1 1 q

θ + _{= −} _{and so}

( )

2 1

1 q

θ + ₌

. Therefore 2q≡ −2

(

mod k

)

. So q≡ ±/ 1

(

mod k

)

and thus k is even and 1 2 k q≡ − mod _

 . There-

fore q≡1

(

mod4

)

, 1 2 k q≡ − mod _

  and q≡ ±/ 1

(

mod k

)

. 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 ⊂_q or sin

[ ]

k _q = ∅.

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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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