tailieunhanh - Finite rings and Wilson’s theorem

In this paper we consider the product of all elements in the group of units in a finite ring and we generalize Wilson’s theorem to finite rings. As an application, we study some generalizations of Wilson’s theorem on residually finite Dedekind domains. And we also give some examples for such rings. | Turkish Journal of Mathematics Research Article Turk J Math (2013) 37: 571 – 576 ¨ ITAK ˙ c TUB doi: Finite rings and Wilson’s theorem 1 Yasuyuki HIRANO,1,∗ Manabu MATSUOKA2 Naruto University of Education, 748 Nakashima Takashima Naruto Narutocity Tokushima 772-8502, Japan 2 Kuwanakita-Highschool, 2527 Shimofukayabe Kuwana Mie 511-0808, Japan Received: • Accepted: • Published Online: • Printed: Abstract: In this paper we consider the product of all elements in the group of units in a finite ring and we generalize Wilson’s theorem to finite rings. As an application, we study some generalizations of Wilson’s theorem on residually finite Dedekind domains. And we also give some examples for such rings. Moreover we study some generalizations of Wilson’s theorem on rings of matrices over a finite commutative ring. Key words: Finite rings, Wilson’s theorem, residually finite Dedekind domains 1. Introduction Wilson’s theorem asserts that (p − 1)! ≡ −1 (mod p) for any prime p. C. F. Gauss generalized this theorem as follows: the product of the positive integers × · · · × . By Proposition 1, R∗! = 1 if and only if only one is of even order. Since R is a finite commutative ring, R is a direct sum of local rings. So assume that R∗! = 1 and that R = R1 ⊕ · · · ⊕ Rm , where each Ri is a local ring. Then only one of the Ri ’s, say R1 , is of even order. By Lemma 1, all R2 , . . . , Rm are finite fields of characteristic 2. Since R1 is a finite commutative local ring, characteristic of R1 is pk for some prime p and some positive integer k . If p is an odd prime, then R1 is of type (A). Next assume that p = 2 . Then R1 /J(R1 ) ∼ = GF (2s ) for some positive integer s. Since R∗ ! = 1 , J(R1 ) = 0 . Suppose k = 1 . If J(R1 )h = 0 and J(R1 )h+1 = 0 . Then J(R1 )h = J(R1 )h /J(R1 )h+1 is a vector space over GF (2s )(∼ = R1 /J(R1 )), each |J(R1 )h | is a power of 2s . More .