Az Eszterházy Károly Tanárképző Főiskola Tudományos Közleményei. 2004. Sectio Mathematicae. (Acta Academiae Paedagogicae Agriensis : Nova series ; Tom. 31)
KOSTRA, J. and VAVROS, M., On transformation matrices connected to normal bases in rings
On transformation matrices connected to normal bases in rings 17 Let (at, Q2 5 • • •) a n) be a normal basis of semiorder R. Let (ßl,ß 2,...,ß n) = (ai,0 2,.--,Ön) • A be a normal basis of submodule S C R- Then ßi = ( nb - 1 )o 2 + (nb - 1 )o 3 + • • • + (nb - 1 )a ß 2 = (nb - l)a 1 + (nb - 1 )a 3 + h {nb - I )o ßn = (nb - 1)o 1 + (nb - l)a 2 + • • • + (nb - l)a„_i . From the above it follows that for all i,j ßißj = (nb - l) 2 • (&!<*! + b 2a 2 + • • • + b na n), where bi E 7L for all i. By the expression of A1 we have for any i,j ßißj = Clßl + • • • + C nß n (n - 1) If b = 1 (mod n — 1), then coefficients Cj £ S, and S is a subsemiorder of the semiorder R. Clearly the same holds for A • U, where U is a unirnodular circulant. matrix of degree n. Remark 2. Matrix A = circ3((), 5, 5) from Examples 1, 2 was obtained from matrix A' = circ 3(-3,2,2) and 2 ^ 1 (mod 2). Remark 3. If in the above Theorem 1 a + (n— 1)6 = — 1, then if b = -1 (mod n— 1 ) matrix A transforms a normal basis of any semiorder R to a normal basis of subsemiorder S C R. The previous Theorem 1 gives the way to find a circulant matrix A of arbitrary degree for which there exist semiorders R\,R 2 such that A transforms a normal basis of Ri to a normal basis of submodule Si C Ri and 6'j is a semiorder and S 2 is not a ring and so S 2 is not a semiorder. Example 3. Let £11 be an 11-th primitive root of units and let (^i, s2, £3,64, £5), where £1 = C11 + Cii\ £2 = Ci 1 + Cid = Ci 1 + C111 £4 = Ci 1 + CID £5 = C11 + Cm be a, normal integral basis of the field A = Q +(Ci 1) ove r Q- The field K — Q(Cii + Cii ) is the maximal real subfield of Q(Cn)« Let A' = circ 5(a, b, 6, 6, b) = circ 5(—7,2, 2,2,2), a + 46 = 1, b £ 1 (mod 4). Let A = circ 5(0, 56 - 1, 56- I, 56- 1, 56 - 1) = circ 5(0, 9,9,9,9),
