[FIELDS] EXISTENCE OF THE POWER-SERIES 89 
form (36.) which converges for |z| < It follows therefore that 
PS 
34 +1° 
equation (34.) is satisfied by a convergent power-series of the 
form (35.). 
Since vg = 28 w =28(n+b) we see that equation (31.) is also 
satisfied by a convergent power-series. In arriving at this result we 
have assumed that fractional exponents do not appear in the co- 
efficients of equation (31.). If however these coefficients involve the 
1 1 
single irrationality ge the substitution =se gives us an equation in 
(€, v ) in which fractional exponents do not appear and whose ampli- 
tude is evidently still 1. By the preceding reasoning then this 
equation is satisfied by a converging power-series in €, that is the 
equation (31) is satisfied by a power-series involving only integral 
powers of ge. Since yv=m+..... + Up +d, =02° +a’ 20 + ,,.., — 
a&-D ¢®D y, our original equation (1.) is satisfied by a con- 
1 1 
vergent power-series R(ze) in which ze is taken to represent the 
single irrationality which suffices for the representation of the first k 
elements of the above sum, for this irrationality evidently also 
suffices for the representation of all powers of z which appear in the 
coefficients of equation (31.). Since equation (1.) is satisfied by 
1 
v=P(ze) it is also satisfied on putting v=P(eze) where eis any eth 
1 
root of unity. The e series P(eze) constitute a cycle and F(z,v) is 
evidently divisible by each one of the factors of the product 
1 
H{v—P(eze) } and therefore by the product itself and we have 
identically 
1 
40. F(z, v) =U {y—Pleze)}.G(z, v), 
where G(z,v) is a polynomial in v of degree n—e with coefficients 
which are integral power-series in z with integral exponents, the 
coefficient of v"-¢ being 1. 
Applying now to the equation 
G(z,v) =o 
the same process which we have already applied to the equation 
L 
F(z,v) =0 we can split off from G(z,v) a cycle of factors v—P{e ze), 
leaving us an equation of degree 7—e—e in v to which in its turn to 
apply the process. So proceeding it is evident that we can ultimately 
represent the equation (1.) in the form 
41. F(z, v) =(v—P;)(v—P»2)....(v-—Pn) =0 
where the symbols P; represent power-series in 3 involving integral or 
fractional exponents, none of them however being negative, and 
where furthermore these power-series group themselves into a 
number of cycles. 
