Received 26 December 2015; accepted 11 June 2016; published 14 June 2016
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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
are prime numbers. To attempt 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 be two fields, we say that F is an extension of K if
or there exists a monomorphism
. Recall that
,
is the ring of polynomial over F. If
(means that F is an extension of K), an element
is algebraic over K if there exists
such that
. The algebraic closure of K in F is
, which is the set of all algebraic elements in F over K.
Definition. Let p be a prime number,
and let k be an integer such that
. Then we can define the set
.
Note that symbol “|” is divisor or divides such that
means a divides b and
means a does not divide b.
Remark. 1) Recall that θ is a primitive kth root of unity if
but
, for all
(see [2] [4] ). The two make the assumption
because if
, then there are no primitive kth root of unity in
.
2) We can define
, in this set, i is a fixed square root of −1. We know that
. In particular, we have
and
.
Theorem 1. If K is a field with 9 elements and if 𝔽 is a finite extension of K, then the mapping
defined by
is an automorphism of 𝔽 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
if and only if
.
Proof: Assume
. If
, then
. Since the order of the multiplicative group of
is p − 1. If
, then the irreducible polynomial of θ over
is
. Hence
and so
.
Conversely, let
. If
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
. Hence
. If
then
hence
, so
.
Corollary 3. If p ≠ 2 and θ is a primitive kth root of unity, then
if and only if
.
Remark. We observe that since membership of
in
depends only on p and k, we have that either
or
.
Lemma 4. Let θ be a primitive kth root of unity in
, the algebraic closure of the rationales Q. Let
, 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 (,) denotes the highest common factor. Let S be the valuation ring of
con-
taining the ring
, and let M be the maximal ideal of S. Then
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 consult. In particular, it is worth recalling that each valuation ring is integrally closed in its quotient field K, and so, if
,
, then
(see [7] - [9] ). Moreover, each valuation ring is a local ring which means that for each
,
as well. Expression obtained for the real and imaginary parts of the roots of unity over complex number is 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
,
, we obtain a special case of the quadratic reciprocity law, namely:
or![]()
Corollary 6. Assume (2, q) = 1. Then
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
, 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
which is again a particular case of the quadratic reciprocity Law.
Corollary 8. Let (q, 34) = 1. Then
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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
, then
.
Indeed set
where θ is a primitive 17th root of unity in
. Since
we see
and
. On the other hand one checks easily that
, hence
. We climb that also
is a square in
. To show this consider
and
. Then
. Moreover we have
. Thus
. Since
we see that both
. Hence
too. Since
or
we see that
or
. Since
we see that both element
and
belong to
. Combining corollary 8 with the considerations above, we obtain.
Corollary 9. Suppose that (q, 34) = 1, Then
and
both belong to
if
.
Remark ( [10] [11] ). 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 c1 and c4 in
we deduce that
, similarly
. From this follows that
and
.
Theorem 10. Suppose (34, q) = 1, Then
if and only if
.
Proof. If
then also
and
. Indeed ![]()
if either
and
or
with r even. In the first case
and therefore
, 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
or
we see that
and
. Hence
. So
.
We want to prove that
then
. It is enough to exclude possibilities
. Suppose that
, Then
. Thus
iff
that this is contradiction.
Corollary 11. Assume (34, q) = 1. If
, then
if and only if
and
or
and
.
Therefore the inclusion
depends only on q(mod 136). we now focus attention on
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
.
Definition. Let
, 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
iff one of the following holds :
(i)
where [,] denotes the least common multiple.
(ii) k has the form 8m + 4 and ![]()
(iii) k has the form 8m + 4 and ![]()
Proof. Assume
. Then
and set
so that θ = c + is. For case (i), Let
. Then
and by corollary 3:
. Therefore
and so
. Hence
and thus
.
Case (ii), Let
and
. Then
too, and thus
. Therefore
. On the other hand
Since
and so
, with
, implies that
.
Case (iii),
,
and is belong to
. In this case
and so
. Now
whence
. But
and so
Therefore
Hence
and so
. Therefore
. So
and thus k is even and
. There- fore
,
and
. It is easily seen that these three condition are
equivalent to k = 8m + 4 and
for some m.
Corollary 13. For any k, either
or
.
Proof. As
depends only on q and k and not particular primitive root chosen, finally, we determine how many distinct values of
and
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.
NOTES
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*Corresponding author.