SMARANDACHE-GALOIS FIELDS W. B. Vasantha Kandasamy Deartment of Mathematcs Indan Insttute of Technology, Madras Chenna - 600 036, Inda. E-mal: vasantak@md3.vsnl.net.n Abstract: In ths aer we study the noton of Smarandache-Galos felds and homomorhsm and the Smarandache quotent rng. Galos felds are nothng but felds havng only a fnte number of elements. We also roose some nterestng roblems. Keywords: Smarandache rng, Smarandache-Galos feld, Smarandache feld homomorhsm, Smarandache quotent rng Defnton [2]: The Smarandache rng s defned to be a rng A such that a roer subset of A s a feld (wth resect wth the same nduced oeratons). By roer set we understand a set ncluded n A, dfferent from the emty set, from the unt element f any, and from A. Defnton : A fnte rng S (.e. a rng havng fnte number of elements) s sad to be a Smarandache-Galos feld f S contans a roer subset A, A S such that A s a feld under the oeratons of S. Clearly we know every fnte feld s of characterstc and has n elements, 0<n<. Examle : Let 0 = {0,, 2, 3, 4, 5,..., 9} be the rng of ntegers modulo 0. 0 s a Smarandache-Galos feld. For the set A = {0, 5} s a feld for 5 2 = 5 acts as a unt and s somorhc wth 2. Examle 2: Let 8 = {0,, 2,..., 7} be the rng of ntegers modulo 8. 8 s not a Smarandache-Galos feld, for 8 has no roer subset A whch s a feld. Thus we have the followng nterestng theorem. Theorem 2: n s not a Smarandache feld for any rme and for any n. Proof: n s the rng of ntegers modulo n. Clearly n s not a feld for r. s = 0 (mod n ) when r + s = n. Now any q n f not a multle of wll
generate n under the oeratons addton and multlcaton. If q s a multle of (even a ower of ) then t wll create zero dvsors. So n cannot have a roer subset that s a feld. Theorem 3: Let m be the rng of ntegers modulo m. m =... t, t >, where all are dstnct rmes. Then m s a Smarandache-Galos feld. Proof: Let m be the rng of ntegers modulo m. Let m =... t, for every rme under addton and multlcaton wll generate a fnte feld. So m s a Smarandache-Galos feld. Examle 3: Let 6 = {0,, 2,...,5}. Clearly {0, 2, 4} s a feld wth 4 2 = 4 (mod 6) actng as the multlcatve dentty. So {0, 2, 4} s a feld. Smlarly {0, 3} s a feld. Hence 6 s a Smarandache-Galos feld. Examle 4: Let 05 = {0,, 2,...,04}be the rng of ntegers modulo 05. Clearly A = {0, 7, 4, 2, 28,..., 98} s a feld wth 5 elements. So 05 s a Smarandache-Galos feld. Examle 5: Let 24 = {0,, 2,...,23} be the rng of ntegers modulo 24. {0, 8, 6} s a feld wth 6 as unt snce 6 2 = 6 and {0, 8, 6} somorhc wth 3. So 24 s a Smarandache-Galos feld. Note that 24 = 2 3.3 and not of the form descrbed n Theorem 3. Examle 6: 2 = {0,, 2,..., }. A = {0, 4, 8} s a feld wth 4 2 = 4 (mod 2) as unt. So 2 s a Smarandache-Galos feld. Theorem 4: Let m be the rng of ntegers wth m = α 2. Let A = α α2 α {,2, L,( ),0}. Then A s a feld of order 2 wth α. α = α for some α and 2 α acts as a multlcatve unt of A. Proof: Let m and A be as gven n the theorem. Clearly A s addtvely and multlcatvely closed wth 0 as addtve dentty and α as multlcatve dentty. We now ose the followng roblems: α αt Problem : m s the rng of ntegers modulo m. If m =,..., t wth one of α =, t. Does t mly m has a subset havng elements whch forms a feld?
Problem 2: If m s as n Problem, can m contan any other subset other than the one mentoned n there to be a feld? Further we roose the followng roblem. Problem 3: Let m be the rng of ntegers modulo m that s a Smarandache- Galos feld. Let A m be a subfeld of m. Then rove A /m and A s a rme and not a ower of rme. A natural queston now would be: Can we have Smarandache-Galos felds of order n where s a rme? When we say order of the Smarandache-Galos feld we mean only the number of elements n the Smarandache Galos feld. That s lke n Examle 3 the order of the Smarandache-Galos feld s 6. The answer to ths queston s yes. Examle 7: Let [x] be the olynomal rng n the varable x over the feld ( a rme). Let (x) = o + x +... + n x n be a reducble olynomal of degree n over. Let I be the deal generated by (x) that s I = (x). Now [x] = R I = (x) s a rng. Clearly R has a roer subset A of order whch s a feld. So ther exsts Smarandache-Galos feld of order n for any rme and any ostve nteger n. Examle 8: Let 3 [x] be the olynomal rng wth coeffcents from the feld 3. Consder x 4 + x 2 + 3 [x] s reducble. Let I be the deal generated by x 4 + x 2 [x] +. Clearly R = 3 = {I, I +, I +2, I + x, I + 2x, I + x +, I + x + 2, I I + 2x +, I + 2x + 2, I + x 2, I + x 3,..., I + 2x + 2 + 2x 2 + 2x 3 } havng 8 elements. Now { I, I +, I +2 } R s a feld. So R s a Smarandache-Galos feld of order 3 4. Theorem 5: A fnte rng s a Smarandache rng f and only f t s a Smarandache-Galos feld. Proof: Let R be a fnte rng that s a Smarandache rng then, by the very defnton, R has a roer subset whch s a feld. Thus R s a Smarandache- Galos feld. Conversely, f R s a Smarandache-Galos feld then R has a roer subset whch s a feld. Hence R s a Smarandache rng. Ths theorem s somewhat analogous to the classcal theorem "Every fnte ntegral doman s a feld" for "Every fnte Smarandache rng s a Smarandache-Galos feld".
Defnton 6: Let R and S be two Smarandache-Galos felds. φ, a ma from R to S, s a Smarandache-Galos feld homomorhsm f φ s a rng homomorhsm from R to S. Defnton 7: Let R and S be Smarandache Galos felds. We say φ from R to S s a Smarandache-Galos feld somorhsm f φ s a rng somorhsm from R to S. Defnton 9: Let m be a Smarandache feld. A m be a subfeld of m. Let r A such that r 0, r 2 = r (mod m} acts as the multlcatve dentty of A. Defne m {0,,2, =..., r-}. We call m the Smarandache quotent rng and the oeraton on modulo r. m = {0,,..., r-} s usual addton and multlcaton Theorem 9: Let m be a Smarandache-Galos feld. A m be a subfeld of m. m the Smarandache quotent rng need not n general be a Smarandache rng or equvalently a Smarandache-Galos feld. Proof: By an examle. Take 24 = {0,, 2,..., 23} be the rng of ntegers modulo 24. Let A = {0, 8, 6}; 6 2 = 6 (mod 24) acts as multlcatve dentty for A. 24 = {0,, 2,..., 5}. Clearly 24 s not a Smarandache rng or a Smarandache-Galos feld. Thus, motvated by ths we roose the followng: Problem 4: Fnd condtons on m for m to have ts Smarandache quotent rng to be a Smarandache rng or Smarandache-Galos feld. Examle 0: 2 = {0,,..., } s the rng of ntegers modulo 2. A = {0, 4, 8} s a feld wth 4 2 = 4 (mod 2) as multlcatve dentty. 2 = {0,, 2, {0,4,8} 3} (mod 4) s not a Smarandache-Galos feld or a Smarandache rng. Examle : 2 = {0,, 2,..., 20} s the rng of ntegers modulo 2. A = {0, 7, 4} s a subfeld. 2 {0,2,...,6}mod 7 = s not a Smarandache-Galos feld. Let B = {0,3,6,9,2,5,8} 2. Clearly B s a feld wth 5 2 = 5 (mod 2)
as a multlcatve unt. Now, 2 = {0,,2,...,4} s a Smarandache- {0,3,6,9,2,5,8} Galos feld. Thus we have the followng nterestng: Problem 5: Let m be the Smarandache rng. Let A be a subset whch s a feld. When does an A exst such that m s a Smarandache-Galos feld? A References: [] I. N. Hersten, Tocs n Algebra, New York, Blasdell, 964. [2] Padlla, Raul. Smarandache Algebrac Structures, Bulletn of Pure and Aled Scences, Delh, Vol. 7 E., No., 9-2, 998; htt://www.gallu.unm.edu/~smarandache/alg-s-txt.txt.