# finite fields

• Sep 3rd 2010, 07:35 AM
ulysses123
finite fields
Let F=(Z₃[x])/<x²+1>
Calculate (x + 1)⁴ in F. Explain why your calculation shows that F* is cyclic

I have calculated (x + 1)⁴and then divided the result by x^2+1 which gives me x+2

I am not sure if i have done this correctly and i dont see how this shows me that the multiplicative group will be cyclic.
• Sep 3rd 2010, 12:40 PM
tonio
Quote:

Originally Posted by ulysses123
Let F=(Z₃[x])/<x²+1>
Calculate (x + 1)⁴ in F. Explain why your calculation shows that F* is cyclic

I have calculated (x + 1)⁴and then divided the result by x^2+1 which gives me x+2

I am not sure if i have done this correctly and i dont see how this shows me that the multiplicative group will be cyclic.

It seems you wrote $\displaystyle F_9:=Z_3[x]/<x^2+1>=$ the field with 9 elements.
Now , I don't know what is $\displaystyle (x+1)'$ or whatever it was you wrote, but perhaps the intention was to

calculate the powers of $\displaystyle (x+1)+<x^2+1> \in F_9$ and, thus, show that this is an

element of order 8, thus showing that $\displaystyle F_9^*$ is a cyclic group...
But then I don't understand why did you divide $\displaystyle x+1\,\,by\,\,x^2+1$...??

Explain better your symbols and what exactly have you done.

Anyway, it's easy to check that $\displaystyle [(x+1)+<x^2+1>]^4=-1$ and thus $\displaystyle x+1$ indeed has order 8 in the multiplicative group of the field.

Tonio