Smarandache idempotents in certain types of group rings

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‍ءؤظاعى زاŮ†Ůƒؤى سليَمانى‏

đ?‘şđ?’Žđ?’‚đ?’“đ?’‚đ?’?đ?’…đ?’‚đ?’„đ?’‰đ?’† đ?’Šđ?’…đ?’†đ?’Žđ?’‘đ?’?đ?’•đ?’†đ?’?đ?’•đ?’” đ?’Šđ?’? đ?’„đ?’†đ?’“đ?’•đ?’‚đ?’Šđ?’? đ?’•đ?’šđ?’‘đ?’†đ?’” đ?’?đ?’‡ đ?’ˆđ?’“đ?’?đ?’–đ?’‘ đ?’“đ?’Šđ?’?đ?’ˆđ?’” Parween Ali Hummadi

Shadan Abdulkadr Osman

College of Science Education

College of Science

University of Salahaddin

University of Salahaddin

Erbil – Kurdistan Region- Iraq

Erbil- Kurdistan Region - Iraq

đ??€đ??›đ??Źđ??­đ??Ťđ??šđ??œđ??­: In this paper we study S-idempotents of the group ring ℤ2 G where is a finite cyclic group of order đ?‘›. We give a condition on such that every nonzero idempotent element of the group ring ℤ2 G is Smarandache idempotent and we find Smarandache idempotents of the group ring đ?’ŚG, where đ?’Ś is an algebraically closed field of characteristic 0 and G is a finite cyclic group. Keywords: Idempotent, S-idempotent, group ring, algebraically closed field. idempotent and we find the number of S-idempotent element. In section two we study S-idempotents of the group ring đ?’ŚG where đ?’Ś is an algebraically closed field of characteristic 0 and G is a finite cyclic group, we show that every non trivial idempotent is S-idempotent.

đ??ˆđ??§đ??­đ??Ťđ??¨đ???đ??Žđ??œđ??­đ??˘đ??¨đ??§: Smarandache idempotent element in rings introduced by Vasantha Kandasamy [1]. A Smarandache idempotent (S-idempotent) of the ring â„› is an element 0 ≠đ?‘Ľ ∈ â„› such that 1) đ?‘Ľ 2 = đ?‘Ľ 2) There exists đ?‘Ž ∈ â„›\ {0, 1, đ?‘Ľ} i) đ?‘Ž2 = đ?‘Ľ and ii) đ?‘Ľđ?‘Ž = đ?‘Ž đ?‘Žđ?‘Ľ = đ?‘Ž or đ?‘Žđ?‘Ľ = đ?‘Ľ (đ?‘Ľđ?‘Ž = đ?‘Ľ). She introduced many Smarandache concepts [2]. Vasantha Kandasamy and Moon K. Chetry discuss S-idempotents in some type of group rings [3],. A prime number đ?‘? of the form đ?‘? = 2đ?‘˜ −1 where đ?‘˜ is a prime number called Mersenne prime [4]. In section one of this paper we study S-idempotents of the group ring ℤ2 G where đ??ş is a finite cyclic group of order đ?‘›. If đ?‘› = 2đ?‘?, đ?‘? is a Mersenne prime, we show that every nonzero idempotent element is S-

1. S-idempotents of ℤđ?&#x;? đ??† In this section we study Sidempotents in the group ring ℤ2 G where G is a finite cyclic group of order đ?‘›, specially where n=2p, p is a Mersenne prime (i.e. đ?‘? = 2đ?‘˜ −1 for some prime đ?‘˜). Theorem 1.1. The group ring ℤ2 G where G= g | g m = 1 is a cyclic group of an odd order đ?‘š >1, has at least two non trivial idempotent elements, moreover no non trivial idempotent element is Sidempotent. Proof: Consider the element

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‍ءؤظاعى زاŮ†Ůƒؤى سليَمانى‏

� = g + g 2 + g 3 +‌+ g

đ?‘š −1 2

+g

đ?‘š −1 +1 2

+‌

+ g đ?‘š −1 , of ℤ2 đ??ş. Since the coefficient of each g i , đ?‘– = 1 , ‌ , đ?‘š is in ℤ2 , đ?›ź 2 = g 2 +g 4 + ‌+g đ?‘š −1 +g +g 3 +‌+ g đ?‘š −2 . Hence đ?›ź 2 = đ?›ź, that is đ?›ź is an idempotent element, so (1 + đ?›ź) is also an idempotent element. It remains to show that no idempotent element of ℤ2 đ??ş is an Sidempotent. Suppose đ?›ź = đ?‘Ž1 + đ?‘Ž2 g+ đ?‘Ž3 g 2 +‌+ đ?‘Žđ?‘š −1+1 g 2

đ?‘š −1 2

+

‌+ đ?‘Žđ?‘š g , is a non trivial S-idempotent. Thus đ?›ź is different from 0 and 1, moreover there exists đ?›˝ in ℤ2 đ??ş\{0,1, đ?›ź} such that đ?›˝ 2 = đ?›ź, let đ?›˝ = đ?‘?1 + đ?‘?2 g + đ?‘?3 g 2 + ‌ + đ?‘š −1

đ?‘?đ?‘š −1+1 g 2

đ?‘š −1 2

+ ‌ + đ?‘?đ?‘š g đ?‘š −1 ,

where đ?‘?đ?‘– ∈ ℤ2 . But đ?›ź 2 = đ?›ź, which means that đ?‘Ž1 + đ?‘Ž2 g 2 + đ?‘Ž3 g 4 +‌+ đ?‘Žđ?‘š −1+1 g đ?‘š −1 + 2

+‌ + đ?‘Žđ?‘š g đ?‘š −2 = đ?‘?1 + đ?‘?2 g 2 + đ?‘?3 g 4 + ‌ + đ?‘?đ?‘š −1+1 g đ?‘š −1 +‌ + đ?‘?đ?‘š g đ?‘š −2 . 2

It follows that đ?‘Žđ?‘– = đ?‘?đ?‘– for each (1 i ≤ m). Therefore đ?›ź = đ?›˝, which is an obvious contradiction. The group ring ℤ2 đ??ş , where đ??ş is acyclic group of an odd order may contains more than two idempotent elements as it is shown by the following example. Example 1.1. Consider the group ring ℤ2 đ??ş where đ??ş g| g 7 = 1 is a cyclic group of order 7. By Theorem 1.1, g+g 2 + g 3 + g 4 + g 5 + g 6 and 1 + g + g 2 + g 3 + g 4 + g 5 + g 6 are

2

idempotent elements, In addition 2 4 2 2 4 (g + g + g ) = g + g + g and (1 + g + g 2 + g 4 )2 = 1 + g 2 + g 4 + g, so 1 + g + g 2 + g 4 and g + g 2 + g 4 are idempotent elements. Therefore ℤ2 đ??ş has more than two idempotent elements. The proof of the following result is not difficult. Theorem 1.2. If đ?›ź is an S-idempotent of the group ring ℤ2 đ??ş where đ??ş is a cyclic group of order đ?‘›, then (1 + đ?›ź) is an S-idempotent of ℤ2 đ??ş. Theorem 1.3. The group ring ℤ2 đ??ş, where đ??ş= g | g 2n = 1 is a cyclic group of order 2đ?‘›, đ?‘› is an odd prime, has at least two Sidempotents. Proof: Let đ?›ź = g 2 + g 4 + â‹Ż + g n −1 + g n +1 + â‹Ż + g 2n −2 . Thus đ?›ź 2 = g 4 + g 8 + â‹Ż + g 2n −2 + g 2 + g 6 + â‹Ż + g 2n −4 = đ?›ź. Hence đ?›ź is an idempotent element, so (1 + đ?›ź) is also an idempotent element .We will show that đ?›ź is Sidempotent, so let g

n −1 2

đ?›˝ = g + g n+2 + g 3 + g n +4 + â‹Ż + 3n +1 + g 2 + â‹Ż + g n −2 + g 2n −1 .

It is clear that đ?›˝ 2 = đ?›ź. We claim that đ?›źđ?›˝ = đ?›˝. For this purpose we describe the multiplication đ?›źđ?›˝ by the following array say đ?’œ:

‍ءؤظاعى زاŮ†Ůƒؤى سليَمانى‏

g3

g5

‌

g n −2

gn

g n +2

‌

g 2n −3

g 2n −1

g n +4

g n +6

‌

g 2n −1

g 2n +1

g 2n +3

‌

g n −2

gn

g5

g7

‌

gn

g n +2

g n +4

‌

g 2n −1

g

g n +6

g n +8

‌

g

g3

g5

‌

gn

g n +2

â‹Ž

â‹Ž

â‹ą

â‹Ž

â‹Ž

â‹Ž

â‹ą

â‹Ž

â‹Ž

3n +5 2

‌

g

n +7 2

‌

g

g

3n +1 2

5n −9 2

g

5n −5 2

g

5n −1 2

‌

g

3n −7 2

g

3n −3 2

g

3n +1 2

‌

5n −5 2

g

5n −1 2

g

5n +3 2

3n −3 2

g

3n +1 2

g

5n +3 +1 2

n +3 2

g

3n +5 2

g

3n +9 2

‌

g

n +7 2

g

n +11 2

‌

g

â‹Ž

g đ?’œ = g

g

g

â‹Ž

â‹Ž

â‹ą

g n −2

gn

‌

g 2n −1

g

gn

7n −11 2

g

7n −7 2

g

5n −9 2

g

5n −5 2

‌

g

7n −7 2

g

7n −3 2

‌

g

5n −5 2

g

5n −1 2

â‹Ž

â‹Ž

â‹ą

â‹Ž

â‹Ž

g 2n −7

g 2n −5

g 2n −3

‌

g n −8

g n −6

‌

g n −6

g n −4

g n −2

‌

g 2n −7

g 2n −5

g n +2

‌

g 2n −5

g 2n −3

g 2n −1

‌

g n −6

g n −4

g

g3

‌

g n −4

g n −2

gn

‌

g 2n −5

g 2n −3

g

g3

‌

g p −4

g p −2 gp ‌ g 2p −5 g 2p −3 the fourth rows in this array, according to That is đ?’œ = [đ?‘Žđ?‘–đ?‘— ] đ?‘› −1 Ă—(đ?‘› −1) , where đ?‘Žđ?‘–đ?‘— is the same argument it remains only the summand of đ?›źđ?›˝ which is equal to the p +2 p +4 product of the đ?‘–th summand of đ?›˝ with g +g . Proceeding in this manner we will get the (đ?‘? − 3)th and the (đ?‘? −1)th the đ?‘—th summand of đ?›ź. This means đ?‘› −1 đ?‘› −1 rows, and adding their terms it remains đ?›źđ?›˝ = đ?‘–=1 đ?‘— =1 đ?‘Žđ?‘–đ?‘— . If we take the first only g 2p −3 + g 2p −1 . Thus we get and the third rows of this array we will đ?›źđ?›˝ = g + g n+2 + g 3 + g n +4 + â‹Ż + see that g i occurs twice for each đ?‘– except n −1 3n +1 (đ?‘– = 1, 3). By adding the terms of this two g 2 + g 2 + â‹Ż + g n −2 +g 2n −1 = đ?›˝. Hence đ?›ź is S-idempotent. By Theorem 1.2, rows it remains only g + g 3 (observing (1+ đ?›ź) is also S-idempotent. This complete that the coefficient of each g i , i=1, 2, ‌,m the proof. is in ℤ2 ). Again by adding the second and

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‍ءؤظاعى زاŮ†Ůƒؤى سليَمانى‏ 2

Lemma 1.4. In ℤ2 đ??ş, where đ??ş = g | g 2p = 1 , đ?‘? is a Mersenne prime (i.e. đ?‘? = 2đ?‘˜ −1 for some đ?‘˜+1 prime đ?‘˜) g 2đ?‘™ = g 2 đ?‘™ and the elements 2 3 đ?‘˜âˆ’1 đ?‘˜ of đ?’Ž = { g 2đ?‘™ , g 2 đ?‘™ , g 2 đ?‘™ , ‌ , g 2 đ?‘™ , g 2 đ?‘™ } are distinct for each odd number l less than đ?‘?. Proof: Since 2đ?‘˜ +1 đ?‘™ − 2đ?‘™ = 2đ?‘™ 2đ?‘˜ − 1 = 2đ?‘™đ?‘?, 2đ?‘˜ +1 đ?‘™ ≥ 2đ?‘™(mod2đ?‘?), which implies that đ?‘˜+1 t g 2đ?‘™ = g 2 đ?‘™ . Now suppose that g 2đ?‘™ = g 2 đ?‘™ (for some 1< đ?‘Ą ≤ đ?‘˜). This means 2t đ?‘™ ≥ 2đ?‘™ (mod2đ?‘?), hence(2đ?‘˜ − 1)|đ?‘™( 2đ?‘Ąâˆ’1 − 1)yields either ( 2đ?‘˜ − 1)| đ?‘™ or (2đ?‘˜ − 1)|(2đ?‘Ąâˆ’1 − 1). But (2đ?‘˜ − 1)|đ?‘™ contradicts the hypothesis that đ?‘™ < đ?‘?, and if 2đ?‘˜ − 1 | (2đ?‘Ąâˆ’1 − 1), hence đ?‘˜ < đ?‘Ą − 1, contradiction with 1< đ?‘Ą ≤ k.

�i = g 2� + g 2 � + ‌ + g 2 where l is an odd number.

Let ℤ2 đ??ş be a group ring where G = g | g 2p = 1 is a cyclic group of order 2đ?‘?, đ?‘? is a Mersenne prime. Then every 2 element of the form đ?›ź = g 2đ?‘™ + g 2 đ?‘™ + â‹Ż + đ?‘˜ g 2 đ?‘™ , is an S-idempotent (l is an odd number). 2 đ?‘˜ Proof: Let đ?›ź = g 2đ?‘™ + g 2 đ?‘™ + â‹Ż + g 2 đ?‘™ . By Lemma1.4, all elements in 2đ?‘™ 22 đ?‘™ 2đ?‘˜ đ?‘™ đ?’Ž = {g , g , ‌ , g } are distinct, moreover đ?‘˜+1 g 2đ?‘™ = g 2 đ?‘™ . Hence đ?›ź 2 = đ?›ź. Now, let đ?›˝ = g đ?‘™ + g t 2 + g t 3 + â‹Ż + g t đ?‘˜ and đ?‘Ľđ?‘– , đ?‘– ≼ 2 be the smallest positive integer such that đ?‘Ľđ?‘– < 2đ?‘?. Thus đ?‘Ľđ?‘– ≥ 2i đ?‘™ (mod2đ?‘?), this means đ?‘Ľđ?‘– = 2i đ?‘™ −2đ?‘?r, for some r ∈ ℤ+. Define đ?‘Ąđ?‘– by 1 1 đ?‘Ľ if đ?‘Ľ is odd 2 ≤ đ?‘– ≤ đ?‘˜ đ?‘– 2 2 đ?‘– đ?‘Ąđ?‘– = 1 1 đ?‘Ľ + đ?‘? if 2 đ?‘Ľđ?‘– is even 2 ≤ đ?‘– ≤ đ?‘˜ . 2 đ?‘–

Combining the last two lemmas we deduce that in the group ring ℤ2 đ??ş, where đ??ş is a cyclic group generated by g of order 2đ?‘?, đ?‘? is a Mersenne prime (i. e. đ?‘? = 2đ?‘˜ −1 for some prime đ?‘˜), if

If

4

1 2

đ?‘Ľđ?‘– is odd, then (g t i )2 = g 2 i

= g 2 � . Hence � 2 = �. If 2i �

1 2

i−1 đ?‘™âˆ’ pr

2

đ?‘Ľđ?‘– is even,

then (g t i )2 = g , and đ?›˝ 2 = đ?›ź for each (2 ≤ đ?‘– ≤ đ?‘˜ ). We will show that đ?›źđ?›˝ = đ?›˝. For this purpose as before we describe the multiplication đ?›źđ?›˝ in the following array say đ?’œ:

đ?‘š = đ?‘˜ , then đ?›ź = g + g + â‹Ż + g + g p +1 + â‹Ż + g 2p −2 , can be partitioned to sum of đ?‘š elements say đ?›ź1 , đ?›ź2 , ‌ , đ?›źm each đ?›źi (1 ≤ đ?‘– ≤ đ?‘š) is of the form 2

đ?‘˜

+ g2 đ?‘™ ,

Theorem 1.6.

Lemma 1.5. If đ?‘?= 2đ?‘˜ −1 is a Mersenne prime, then đ?‘˜ | (2đ?‘˜ − 2). Proof: Since đ?‘˜ is prime, according to Fermat’s Little Theorem, đ?‘˜ | 2đ?‘˜ − 2 .

2đ?‘˜ −2

đ?‘˜ −1 đ?‘™

p −1

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‍ءؤظاعى زاŮ†Ůƒؤى سليَمانى‏

g t 2 +8đ?‘™ â‹Ż

g t 2 +2

đ?‘˜âˆ’2 đ?‘™

g t 2 +2

đ?‘˜âˆ’1 đ?‘™

g t 2 +2

đ?‘˜đ?‘™

g t 3 +2đ?‘™

g t 3 +4đ?‘™

g t 3 +8đ?‘™ â‹Ż

g t 3 +2

đ?‘˜âˆ’2 đ?‘™

g t 3 +2

đ?‘˜âˆ’1 đ?‘™

g t 3 +2

đ?‘˜đ?‘™

đ?’œ = g t 4 +2đ?‘™

g t 4 +4đ?‘™

g t 4 +8đ?‘™ â‹Ż

g t 4 +2

đ?‘˜âˆ’2 đ?‘™

g t 4 +2

đ?‘˜âˆ’1 đ?‘™

g t 4 +2

đ?‘˜đ?‘™

â‹ą

â‹Ž

â‹Ž

g t đ?‘˜ −1 +2đ?‘™ g t đ?‘˜ −1 +4đ?‘™ g t đ?‘˜ −1 +8đ?‘™ â‹Ż g t đ?‘˜ −1 +2 g t đ?‘˜ +2đ?‘™

g t đ?‘˜ +4đ?‘™

g t đ?‘˜ +8đ?‘™ â‹Ż

g t đ?‘˜ +2

đ?‘˜âˆ’2 đ?‘™

đ?‘˜âˆ’2 đ?‘™

where đ?‘Žđ?‘–đ?‘— is the summand of đ?›źđ?›˝ which is equal to the product of the đ?‘–th summand of đ?›˝ with đ?‘—th summand of đ?›ź. This means đ?›źđ?›˝ = đ?‘˜đ?‘–=1 đ?‘˜đ?‘—=1 đ?‘Žđ?‘–đ?‘— .We complete the proof by the following three steps. Step 1: Considering the first and the đ?‘˜th column in this array we claim that đ?‘Ž1đ?‘— = đ?‘Ž đ?‘— +1 đ?‘˜ ...(1), for each 1 ≤ j ≤ k − 1 , equivalently k (2 j +1)đ?‘™ g = g t j+1 +2 đ?‘™ .

for some r ∈ ℤ+. If 1

1 2

đ?‘Ľđ?‘— +1 is odd,

1

1 2

đ?‘˜âˆ’1 đ?‘™

đ?‘˜ −1 đ?‘™

g t đ?‘˜ −1 +2 g t đ?‘˜ +2

đ?‘˜đ?‘™

đ?‘˜đ?‘™

for some đ?‘&#x; , đ?‘ ∈ ℤ+. Thus

then 2 đ?‘Ľđ?‘— +1 = 2j đ?‘™ − đ?‘?đ?‘&#x; is odd (this hold only if r is odd), hence đ?‘Ąđ?‘— +1 = 2j đ?‘™ − đ?‘?đ?‘&#x;. So, đ?œ” = 2j đ?‘™ − đ?‘?đ?‘&#x; + 2k đ?‘™ − 2j đ?‘™ – đ?‘™ ≥ 0 (mod 2đ?‘?). Therefore (2j + 1)đ?‘™ ≥ t j+1 + 2k đ?‘™ ( mod 2p ). This yields (1). If

â‹Ž

g t đ?‘˜âˆ’1 +2 g t đ?‘˜ +2

= [đ?‘Žđ?‘–đ?‘— ]đ?‘˜Ă—đ?‘˜ ,

therefore by adding the terms of the first row and the đ?‘˜th column it remains k only đ?‘Ž1đ?‘˜ = g đ?‘™ 2 +1 . Step 2: Consider the subarray â„Ź = đ?‘?đ?‘–đ?‘— đ?‘˜ −1Ă—đ?‘˜âˆ’1 of đ?’œ = đ?‘Žđ?‘–đ?‘— đ?‘˜Ă—đ?‘˜ , where đ?‘?đ?‘–đ?‘— = đ?‘Ž đ?‘–+1 đ?‘— for each 1 ≤ đ?‘–, đ?‘— ≤ đ?‘˜ − 1 , by neglecting the first row and the đ?‘˜th column, we will show that đ?‘?đ?‘–đ?‘— = đ?‘?đ?‘—đ?‘– ‌(2), for all 1 ≤ i, j ≤ k − 1 such that (i ≠j), j i equivalently g t (i+1)+2 đ?‘™ = g t (j+1)+2 đ?‘™ . Let đ?œ” = đ?‘Ąđ?‘–+1 + 2j đ?‘™ − đ?‘Ąđ?‘— +1 − 2i đ?‘™. Now, đ?‘Ľđ?‘–+1 = 2i+1 đ?‘™ − 2đ?‘?đ?‘&#x; and đ?‘Ľđ?‘— +1 = 2j+1 đ?‘™ − 2ps,

Let đ?œ” = đ?‘Ąđ?‘— +1 + 2k đ?‘™ − 2j + 1 đ?‘™. Now, đ?‘Ľđ?‘— +1 ≥ 2j+1 đ?‘™ (mod 2đ?‘?), thus đ?‘Ľđ?‘— +1 = 2j+1 đ?‘™ − 2đ?‘?đ?‘&#x;,

g đ?‘™(2

2 k +1

g t 2 +4đ?‘™

â‹Ž

g đ?‘™(2

gđ?‘™

g t 2 +2đ?‘™

â‹Ž

â‹Ż

k −1 +1)

g 5đ?‘™

â‹Ž

g 9đ?‘™

k −2 +1)

g 3đ?‘™

đ?‘?đ?‘&#x; and 1

1 2

1 2

đ?‘Ľđ?‘— +1 = 2j đ?‘™ − đ?‘?đ?‘ . If

đ?‘Ľđ?‘–+1 = 2i đ?‘™ − 1 2

đ?‘Ľđ?‘–+1 and

đ?‘Ľđ?‘— +1 are even, hence 2i đ?‘™ − đ?‘?đ?‘&#x; and 2j đ?‘™ − đ?‘?đ?‘ are even ( this hold only if đ?‘&#x; and đ?‘ are even), it follows đ?‘Ąđ?‘–+1 = 2i đ?‘™ − đ?‘?đ?‘&#x; + đ?‘? and đ?‘Ąđ?‘— +1 = 2j đ?‘™ – đ?‘?đ?‘ + đ?‘?. So, đ?œ” = đ?‘ − đ?‘&#x; đ?‘? ≥ 0 (mod 2đ?‘?). Hence đ?‘Ąđ?‘–+1 + 2j đ?‘™ ≥ đ?‘Ąđ?‘— +1 + 2i đ?‘™ mod 2p). This 1 1 yields (2). If 2 đ?‘Ľđ?‘–+1 and 2 đ?‘Ľđ?‘— +1 are odd, it is clearly đ?œ” = đ?‘ − đ?‘&#x; đ?‘? ≥ 0 (mod 2đ?‘?). Hence đ?‘Ąđ?‘–+1 + 2j đ?‘™ ≥ đ?‘Ąđ?‘— +1 + 2i đ?‘™ mod 2đ?‘?). 2

đ?‘Ľđ?‘— +1

is even, then 2 đ?‘Ľđ?‘— +1 = 2j đ?‘™ − đ?‘?đ?‘&#x; is even (this hold only if r is even), hence t j+1 = 2j đ?‘™ − đ?‘?đ?‘&#x; + đ?‘?. So, đ?œ” = 1 − r đ?‘? + đ?‘™đ?‘? ≥ 0 (mod 2đ?‘?). Hence (2j + 1)đ?‘™ ≥ đ?‘Ąđ?‘— +1 + 2k đ?‘™ (mod 2p). This also yields (1). This implies that đ?‘Ž1đ?‘— + đ?‘Ž đ?‘— +1 đ?‘˜ = 0 mod 2p ),

5

‍ءؤظاعى زاŮ†Ůƒؤى سليَمانى‏

This also establishes (2). If 1

1 2

idempotents in ℤ2 đ??ş, where đ??ş is a cyclic group of order 2đ?‘?, đ?‘? a Mersenne prime, then đ?›ź1 + đ?›ź2 is S-idempotent. Proof: Let đ?›ź1 , đ?›ź2 be two distinct basic S-idempotents in ℤ2 đ??ş, so there exist đ?›˝1 and đ?›˝2 such that đ?›˝1 2 = đ?›ź1 , đ?›ź1 đ?›˝1 = đ?›˝1 , đ?›˝2 2 = đ?›ź2 and đ?›ź2 đ?›˝2 = đ?›˝2 . Now, đ?›˝1 + đ?›˝2 2 = đ?›˝1 2 + đ?›˝2 2 = đ?›ź1 + đ?›ź2 , and (đ?›ź1 + đ?›ź2 ) đ?›˝1 + đ?›˝2 = đ?›ź1 đ?›˝1 + đ?›ź1 đ?›˝2 + đ?›ź2 đ?›˝1 + đ?›ź2 đ?›˝2 = đ?›˝1 + đ?›˝2 + đ?›ź1 đ?›˝2 + đ?›ź2 đ?›˝1 . We show that đ?›ź1 đ?›˝2 + đ?›ź2 đ?›˝1 = 0. By describing the multiplications đ?›ź1 đ?›˝2 and đ?›ź2 đ?›˝1 by the two arrays đ?’œ and respectively and using similar argument of Theorem 1.6, we get đ?’œ + â„Ź = 0 that is đ?›ź1 đ?›˝2 + đ?›ź2 đ?›˝1 = 0 . Therefore đ?›ź1 + đ?›ź2 is an S-idempotent.

đ?‘Ľđ?‘–+1 is odd

and 2 đ?‘Ľđ?‘— +1 is even, it is also clear that đ?œ”= đ?‘ − đ?‘&#x; − 1 đ?‘? ≥ 0 (mod 2p). Thus đ?‘Ąđ?‘–+1 + 2j đ?‘™ ≥ đ?‘Ąđ?‘— +1 + 2i đ?‘™ mod 2p).This also 1

1

yields (2). If 2 đ?‘Ľđ?‘–+1 is even and 2 đ?‘Ľđ?‘— +1 is odd, thus by using similar argument we get đ?‘Ąđ?‘–+1 + 2j đ?‘™ ≥ t j+1 + 2i đ?‘™ (mod 2đ?‘?). This also yields (2). For all cases we get đ?‘?đ?‘–đ?‘— + đ?‘?đ?‘—đ?‘– = 0 1 ≤ đ?‘–, đ?‘— ≤ đ?‘˜ − 1 . Step 3: From Step 1 and Step 2 we get that đ?›źđ?›˝ = đ?‘Ž1đ?‘˜ + đ?‘˜âˆ’1 and it is not đ?‘–=1 đ?‘?đ?‘–đ?‘– difficult to show that đ?›źđ?›˝ = đ?›˝ which means that đ?›ź is an S-idempotent. We call an S-idempotent of ℤ2 đ??ş 2 đ?‘˜ of the form đ?›ź = g 2đ?‘™ + g 2 đ?‘™ + â‹Ż + g 2 đ?‘™ , where đ?‘™ is an odd number a basic Sidempotent.

Theorem 1.8. If đ?›ź1 , đ?›ź2 , ‌ , đ?›źn are đ?‘› basic S-idempotents in ℤ2 đ??ş where đ??ş is a cyclic group of order 2đ?‘?, đ?‘? is a Mersenne prime, then đ?›ź1 + đ?›ź2 + â‹Ż + đ?›źn is S-idempotent. Proof: Follows from Theorem 1.7.

[

Example 1.2. Consider the group ring ℤ2 đ??ş where đ??ş = g | g 62 = 1 is a cyclic group of order 62 (i.e. đ?‘? =31 and đ?‘˜ = 5). By Theorem 1.7, if đ?‘™ = 1, then đ?›ź = g 2 + g 4 + g 8 + g 16 + g 32 and đ?›˝ = g + g 33 + g 35 + g 39 + g 47 . It is clear that đ?›˝ 2 = đ?›ź . Let us describe the multiplication đ?›źđ?›˝ by the following array say đ?’œ: g3

g5

g9

g 17

g 33

g 35

g 37

g 41

g 49

g3

đ?’œ = g 37

g 39

g 43

g 51

g5 .

g 41

g 43

g 47

g 55

g9

g 49

g 51

g 55

g

g 17

By combining all previous results concerning the group ring ℤ2 đ??ş, where đ??ş is a cyclic group of order 2đ?‘?, đ?‘? is a Mersenne prime we get the following result Theorem 1.9. Consider the group ring ℤ2 đ??ş where đ??ş is a cyclic group of order 2đ?‘?, đ?‘? is a Mersenne prime. Then 1)Every non trivial idempotent is Sidempotent . 2) The number of non trivial S-idempotents đ?‘?−1 is 2(2m − 1), where đ?‘š = đ?‘˜ . Proof: 1) Follows from Theorems 1.6, 1.7, 1.8 and Theorem 1.2. 2) From Theorems 1.6, 1.7, and 1.8, by using the concepts of probability theory we conclude that the number of Sidempotent in ℤ2 G is

Hence applying Theorem 1.6, we get �� = g + g 33 + g 35 + g 39 + g 47 = �. Theorem 1.7. If �1 and �2 are two basic S-

6

‍ءؤظاعى زاŮ†Ůƒؤى سليَمانى‏ m 1

đ?œ†=2

m 2

+

m m

+ â‹Ż+

2 2m − 1 , where đ?‘š =

đ?‘?−1 đ?‘˜

−1 idempotent elements, let đ?›ź = ni=0 ri g i ∈ đ?’Śđ??ş be an idempotent element. −1 Put đ?›˝ = ni=0 (−ri )g i ∈ đ?’Śđ??ş. Hence

=

.

đ?›˝2 =

−1 = (−1) ni=0 ri g i 2 −1 = ni=0 ri g i = đ?›ź n −1 i n −1 i Now, đ?›źđ?›˝ = i=0 ri g i=0 (−ri )g n −1 i 2 −1 = −1 = ni=0 (−ri )g i = đ?›˝. i=0 ri g

2. S-idempotents in the group ring of a finite cyclic group over a field of characteristic zero In this section,we study the group ring đ?’ŚG where đ?’Ś is an algebraically closed field of characteristic 0 and đ??ş is a finite cyclic group of order đ?‘›. We get that every nontrivial idempotent element in this group ring đ?’ŚG is an S-idempotent element.

0

đ?’“đ?&#x;?

0

đ?’“đ?&#x;?

0

1 3 1 3

1 3

1 3

−1 + 3 i 6

1 3

−1 + 3 i 6

đ?›ź1 = 3 − 1− 3 i

1 3

g−

1 3

2

g ,

�2 = 3 +

2

Example 2.1. Let đ??ş be a cyclic group of order 3, and đ?’Ś is an algebraically closed field of −1 characteristic 0, and let đ?›ź = ni=0 ri g i ∈ đ?’Śđ??ş. If đ?›ź is an idempotent element, then by [5], the values of đ?‘&#x;0 , đ?‘&#x;1 and đ?‘&#x;2 are followings 1 3

1 3

1

−1 + 3 i 6

1 3 1 3

−1 + 3 i 6

−1 + 3 i 6

0

−1 + 3 i 6

1 3

−1 + 3 i 6

−1 + 3 i 6

0

2

1+ 3 i

6 1+ 3 i

1− 3 i

đ?›˝22 =

g

+

+ 6 g and đ?›ź3 = 3 + 6 g+ 6 g . For each ( 1 ≤ đ?‘– ≤ 3 ), đ?›źđ?‘– has three Co-idempotents we denote them by đ?›˝đ?‘–đ?‘— −2 1 1 1 ≤ đ?‘— ≤ 3 . They are đ?›˝11 = 3 + 3 g+ 3 g 2 , đ?›˝12 = đ?›˝21 =

2

3i

3 −2 3

g−

−

3i

3 1− 3 i 6

g2 , g+

đ?›˝13 = −1+ 3 i 6

− 3i 3

2

Recall that � called Smarandache Co-idempotent of � [1]. The following example shows that the Smarandache coidempotent need not be unique in general.

Consider the S-idempotents, 2

(−r)i g i

Therefore every nontrivial idempotent in đ?’Śđ??ş is an S-idempotent.

Theorem 2.1. Let đ?’Ś be algebraically closed field of characteristic 0 and đ??ş is a finite cyclic group of order đ?‘›. Then every nontrivial idempotent element in đ?’ŚG is an Sidempotent. Proof: By [5], đ?’Śđ??ş has 2n − 2 nontrivial đ?’“đ?&#x;Ž

n −1 i=0

+

3i

3

2

−3+ 3 i

1+ 3 i 6

đ?›˝32 =

6

−3− 3 i

g 2 , đ?›˝31 =

−3− 3 i

−3− 3 i

g+

6

g+

6 −2 3

−

3+ 3 i 6

g 2 , đ?›˝23 = 1− 3 i 6

g−

g 2 , đ?›˝33 =

3− 3 i 6 1+ 3 i 6 3+ 3 i 6

g g2 , g+

g2 , respectively. We see that �1 �1� = �1� , �2 �2� = �2� and �3 �3� = �3� , �1� 2 = �1 , �2� 2 = �2 and �3� 2 = �3 , for each (1 ≤ � ≤ 3). 6

g,

g2 ,

7

‍ءؤظاعى زاŮ†Ůƒؤى سليَمانى‏

đ??ş be a finite group of order đ?‘›. If some prime divisor đ?‘? of đ?‘› is a unit in and 1) đ?‘?3 = đ?‘?−1 or 2) đ?‘? = đ?‘?−1 or 3) đ?‘? = 2. Then the group ring â„›G has S-idempotent. Proof: 1) Since đ?‘? is a prime dividing đ?‘›, and đ?‘? is a unit in then by [7] đ?›ź = đ?‘?−1 đ?‘Ľ ∈đ??ť đ?‘Ľ is a nontrivial idempotent where is a subgroup of đ??ş of order đ?‘?. Let đ?›˝ = p x ∈H x. Then đ?›źđ?›˝ = đ?‘?−1 đ?‘? đ?‘Ľ ∈đ??ť đ?‘Ľ đ?‘Ľ ∈đ??ť đ?‘Ľ = đ?‘? đ?‘Ľ ∈đ??ť đ?‘Ľ = đ?›˝, and đ?›˝ 2 = đ?‘?2 đ?‘Ľ ∈đ??ť đ?‘Ľ 2 = đ?‘?3 đ?‘Ľâˆˆđ??ť đ?‘Ľ = đ?‘?−1 đ?‘Ľâˆˆđ??ť đ?‘Ľ = đ?›ź. Hence đ?›ź is a S-idempotent. 2) we have đ?›ź = đ?‘?−1 đ?‘Ľâˆˆđ??ť đ?‘Ľ is a nontrivial idempotent. Let đ?›˝ = đ?‘Ľâˆˆđ??ť đ?‘Ľ. Then

Theorem 2.2. Let đ?’Ś b an algebraically closed field of characteristic 0 and đ??ş = ℤm Ă— ℤn . Then every nontrivial idempotent element in đ?’ŚG is an S-idempotent. Proof: If đ?‘š, đ?‘› are relatively prime, then the proof is given in Theorem 2.1, since ℤm Ă— ℤn ≅ ℤmn is cyclic. If đ?‘š and đ?‘› are not relatively prime, for each k ,j ∈ đ??ş let (đ?‘˜ , đ?‘—) = g kn +j 0 ≤ đ?‘˜ ≤ đ?‘šâˆ’1 , 0≤đ?‘—≤đ?‘›âˆ’1, and let −1 đ?›ź = mn rđ?‘– g đ?‘– ∈ đ?’Śđ??ş be an idempotent i=0 −1 element [6].Takeđ?›˝ = mn i=0 (−rđ?‘– ) g đ?‘– ∈ đ?’Śđ??ş, then it is clear that

đ?›˝ 2 = đ?›ź and đ?›źđ?›˝ = đ?›˝. Therefore every idempotent element in đ?’Śđ??ş is an S-idempotent.

đ?›źđ?›˝ = đ?‘?−1 and

Finally we concern the group ring â„›G where is an integral domain and đ??ş is a finite group of order đ?‘›. We give a condition under which â„›G contains Sidempotents.

đ?‘Ľ đ?›˝ =

đ?‘Ľ ∈đ??ť 2

đ?‘Ľ = đ?‘Ľ ∈đ??ť đ?‘Ľ = đ?›˝, 2 = đ?‘? đ?‘Ľâˆˆđ??ť đ?‘Ľ = đ?‘Ľâˆˆđ??ť đ?‘Ľ

đ?‘Ľ ∈đ??ť

đ?‘?−1đ?‘Ľâˆˆđ??ťđ?‘Ľ=đ?›ź.

Therefore đ?›ź is a S-idempotent. 3) Since đ?‘? = 2 divides đ?‘›, then đ??ş = 2đ?‘˜ and đ?›ź = 2−1 (1 + g k ). Let đ?›˝ = 1 + g k − đ?›ź. Then it is clear that đ?›˝ 2 = đ?›ź and đ?›źđ?›˝ = đ?›˝. So đ?›ź is an S-idempotent.

Theorem 2.3. Let be an integral domain, and let

References [1] W. B. Vasantha Kandasamy: Smarandache Rings, American Research Press, 2002. [2] W. B. Vasantha Kandasamy: Smarandache special definite algebraic structures, American Research Press, 2009. [3] W. B. Vasantha Kandasamy and Moon K. chetry: Smarandache Idempotents in finite ring and in Group Rings , Scientia Magna. 2005, 2(1), 179- 187. [4] K. H. Rosen: Elementary Number Theory and Its Applications, Addison- Welsey Welsey Longman, 2000. [5] W. S. Park: The Units and Idempotents in the Group Ring of a Finite Cyclic Group , Comm. Korean Math. Soc. 1997, 4 (12), 855- 864. [6] W. S. Park: The Units and Idempotents in the Group Ring �(ℤm × ℤn ), Comm. Korean Math. Soc. 2000, 4 (15), 597- 603. [7] D. B. Coleman: Shorter Notes: Idempotents in Group Rings, Proceeding of the American Math. Soc. 1966, 4 (17), 962. 8

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