Thread: Show that every element of the 9-element field is a root of f?

Consider the polynomial f(x) = x^9 + 2x. Show that every element of the 9-element field is a root of f.

Where would I begin with this, any help please?

which 9 element field?

Originally Posted by jakncoke

which 9 element field? Also i was under the impression that all finite fields were of prime order.

no, there exists a unique finite field (up to isomorphism) for every prime POWER. 9 is a prime power (3 squared).

note that in a field F of 9 elements we must have 3 = 1+1+1 = 0 (the characteristic of a finite field F must be prime, and the (additive) subgroup generated by 1 must have order a divisor of 9 (by lagrange), the only prime divisor of 9 is 3).

note further that since F -{0} is a group of order 8, we have x⁸ = 1, for all non-zero x.

hence x⁹ + 2x = x(x⁸ + 2). if x = 0, then this is obviously 0.

if x ≠ 0, then x⁸ = 1, so x(x⁸ + 2) = x(1 + 2) = x(3) = x(0) = 0.

Originally Posted by Deveno

no, there exists a unique finite field (up to isomorphism) for every prime POWER. 9 is a prime power (3 squared).

note that in a field F of 9 elements we must have 3 = 1+1+1 = 0 (the characteristic of a finite field F must be prime, and the (additive) subgroup generated by 1 must have order a divisor of 9 (by lagrange), the only prime divisor of 9 is 3).

note further that since F -{0} is a group of order 8, we have x⁸ = 1, for all non-zero x.

hence x⁹ + 2x = x(x⁸ + 2). if x = 0, then this is obviously 0.

if x ≠ 0, then x⁸ = 1, so x(x⁸ + 2) = x(1 + 2) = x(3) = x(0) = 0.

man, your on cocaine or something man, crazy fast and accurate. Can't keep up man.