SEMI-SIMPLE ALGEBRAS 
§ 66] 
107 
Since each s r = o, we have jyj = o. Hence 7j — o, 
since F has no modulus. Thus /(co) = o>” = o. Since 
every root of this characteristic equation of xy is zero, 
the corollary in § 64 shows that xy is nilpotent for every 
y, whence x is zero or properly nilpotent. 
But if F has a modulus the prune n, y n need not be 
zero, although yj=o(j <n). Take 7«*=— 1. Then 
/(co) = co”— i = (w — 1)” (mod n), 
so that all the roots are 1 and s r =o (mod n) for every r. 
To show that not merely our proof, but also the 
theorem itself, may fail for a modular field, take n = 2 in 
what precedes and consider the algebra (1, e) where e 2 = 0, 
over the field of classes of residues of integers modulo 2. 
The first matrix of x = £-\-r]e has the diagonal elements 
£. Hence the trace of every xis 2^=0 (mod 2). The 
elements 1 and 1 +e are not nilpotent, although the traces 
of their products by every y were seen to be zero. 
66. To make an important application of the pre 
ceding theorem, consider 
Relations (9) evidently imply 
(24) tax~ ) ^x+y tx J T^y • 
Hence if the right trace of UiUj is r#, 
n 
(25)