Question

Abstract algebra.

Answer 1

can you define a field?

Answer 2

a field is commutative division ring

Answer 3

ive got no clear thought on this. my class work never got past groups. so i lack any practical experience in the matter.

Answer 4

HI!!

Answer 5

i have a suggestion
try it with a simple polynomial of degree say 2 and see why it works

Answer 6

if you want a bigger hint i can provide one, but i think it will pretty much give it away

Answer 7

already considered that, too much trouble :)

Answer 8

half a degree is all im willing to process at the moment

Answer 9

Is it okay if i divide f(a)by a^n?

Answer 10

first off the notation is a bit odd, because they use \(a_i\) for the coefficients, and \(a\) for the zero

Answer 11

so lets use \(b\) and you see that \[a_0+a_1b+a_2b^2+...+a_nb^n=0\]
the see what you get when you multiply this by \(b^{-n}\) which i guess is another way of saying "divide by \(b^n\)"

Answer 12

half a degree is better than none! it is what i have after all...

Answer 13

thanks:)

Answer 14

\[\color\magenta\heartsuit\]

Answer 15

consider \(g(x)=x^{-n}f(x^{-1})\) hence \(g(a)=0\implies a^{-n}f(a^{-1})=0\) and using the fact \(a\ne0\) we conclude \(f(a^{-1})=0\). so now just work backwards

Answer 16

err \(g(x)=x^n f(x^{-1})\) and so \(a^n f(a^{-1})=0\)