Proving (1+a2)(1+a3)...(1+an)=0 for Any Finite Field F

[mansi]

let F= { 0=a1, a2,a3,a4...an} be a finite field. show that
(1+a2)(1+a3)...(1+an) = 0.

when n is odd, it's simple since 1 belongs to F. then odd number of elements are left( they're distinct from 0 and 1). at least one of them, say x,must have itself as its inverse. x^2 = 1 implies x=-1 ( since x != 1)...so the result is true.

for n even and the case where none of them are inverses of themselves, can somebody suggest a solution?

 

[matt grime]

One of the a_r must be 1, which is distinct from 0, so one of the terms is 2, which is zero in char 2, and wlog a_2 = -1 in any other field.

It's a field. If elements multiply together to give 0 one of them must be zero, ie, 1+a_2=0 after relabelling. a_2= -1, the inverse of 1 which isn't zero and hence must be in the list you gave.

 

[mansi]

one of the terms is 2, which is zero in char 2, and wlog a_2 = -1 in any other field.

didn't get this...please elaborate

 

[matt grime]

You get that the product is zero if and only if one of the factors is zero, since it is a field?

So it suffices to show that one of the 1+a's is zero.

But this is true since 1 has an additive inverse, -1 (which equals 1 in a field of char 2)

 

[mansi]

thanks...that makes it clear...
but here's another question does this mean i could have done this problem without considering separate cases?

 

[matt grime]

Yes, absolutely. In any field there is a non-zero element, x, such that 1+x=0. The result is also true in any finite ring with unit, any finite domain too.