St. Petersburg Mathematical Journal, 19. cilt,495-765. sayfalarAmerican Mathematical Society, 2008 |
Kitabın içinden
82 sonuçtan 1-3 arası sonuçlar
Sayfa 695
... implies that b1t1 + c1tз = 0 ( mod p ) . Since t1tą – t2t3 = 1 , the second relation implies that ( b2 − b1 ) / 2 = tз ( b1t2 + c1t1 ) ( mod p3 ) . - Thus , if ( b2b1 ) / 2 is divisible by p , then we have the congruences - = biti citз ...
... implies that b1t1 + c1tз = 0 ( mod p ) . Since t1tą – t2t3 = 1 , the second relation implies that ( b2 − b1 ) / 2 = tз ( b1t2 + c1t1 ) ( mod p3 ) . - Thus , if ( b2b1 ) / 2 is divisible by p , then we have the congruences - = biti citз ...
Sayfa 779
... implies that ; to , whence df 0. Let fy denote the form in Эп ato i f of minimal degree such that dfN 0. Observe that either of the conditions sand s ' implies ( 1 ) s aƒN + Ət ƏƒN + 1 Ət in O / ( so , to ) N + 1 ; in this ring we put N ...
... implies that ; to , whence df 0. Let fy denote the form in Эп ato i f of minimal degree such that dfN 0. Observe that either of the conditions sand s ' implies ( 1 ) s aƒN + Ət ƏƒN + 1 Ət in O / ( so , to ) N + 1 ; in this ring we put N ...
Sayfa 919
... implies that k € = ax + b for some a Є Rˇ , b Є R , σ Aut ( R ) , and for all x € R. Since 0f * = 0 and 1f * = k , we conclude that b = 0 and ak . Thus , xfk = kx ° , x E R. By the choice of k , this implies that σ leaves fixed each ...
... implies that k € = ax + b for some a Є Rˇ , b Є R , σ Aut ( R ) , and for all x € R. Since 0f * = 0 and 1f * = k , we conclude that b = 0 and ak . Thus , xfk = kx ° , x E R. By the choice of k , this implies that σ leaves fixed each ...
İçindekiler
A Brudnyi and D Kinzebulatov Uniform subalgebras of L on the unit | 495 |
N A Vavilov Can one see the signs of structure constants? | 519 |
Waldemar Hołubowski A new measure of growth for groups and algebras | 545 |
Telif Hakkı | |
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