It remains to prove the theorem for algebras A 
having a modulus. By §38, A— N is semi-simple and 
has a modulus. 
First, let A— N be simple. By §55, A=MXB, 
where M is a simple matric algebra and B is an algebra 
having a modulus, but no further idempotent element. 
By § 77, B = D+Nj, where D is a division algebra and 
N z is zero or the maximal nilpotent invariant sub 
algebra of B. By §56, N = MXN Z . By §52, MxD 
is simple and is not a zero algebra of order 1. Hence 
A=Mx{D-\-N x ) is the sum of the simple algebra 
M XD and N. 
Second, let A— N be semi-simple, but not simple. 
By § 57, A=N'-\-S, where N'^N and S is the direct 
sum of algebras A t , . . . . , A h where each At is of 
the type MXB just discussed and hence is the sum of a 
simple algebra K { and where Ni is zero or the maximal 
nilpotent invariant sub-algebra of A,- if it exists. More 
over, N = N'+'ZN{. Hence A—K+N, where K = HKi 
is a direct sum of simple algebras, no one a zero algebra 
of order 1, and hence is semi-simple and not a zero algebra 
of order 1 (§ 40). 
79. Complex algebras. Any algebra over the field 
C of all complex numbers a-\-bi is called complex. 
A complex division algebra D is of order 1 and is 
generated by its modulus. For, if/(co) =0 is the equation 
of lowest degree satisfied by an element x of D, /(co) is 
not a product of polynomials /i(co) and / 2 (w) each of 
degree ^ 1, since fi{x)f 2 (x)=o implies that one of fjpc) 
and f 2 {x) is zero in the division algebra D. But if /(«) 
is of degree >1, it is a product of two or more linear 
PRINCIPAL THEOREM ON ALGEBRAS [chap, viii