5
$\begingroup$

Construct a finite field of 16 elements and find a generator for its multiplicative group. Find all generators of multiplicative group.

Very obvious Construction of a field with 16 elements according to me would be ${\mathbb{F_2}[x]}/{f(x)}$ where $f(x)$ is an irreducible polynomial of order 4 in $\mathbb{F_2}[x]$. I took $f(x)=x^4+x+1$.Elements of the resulting field are $\{0,1,x,x^2,x^3,1+x,1+x^2,1+x^3,x+x^2,x+x^3,x^2+x^3,x+x^2+x^3,1+x+x^2,1+x+x^3,1+x+x^3,1+x^2+x^3,1+x+x^2+x^3\}$.

Now, Question is to find a generator of its multiplicative group.

I have calculated by hand that $<x>$ is the whole multiplicative group. That was very obvious by intuition that $x$ generates. Now the question is what other elements generate that group. I have tried for $x^2$, though i did not calculate whole group,with in few steps i found $x\in<x^2>$ so i concluded $x^2$ generates multiplicative group.

In case of $1+x$ , with in 4/5 steps i got multiplicative identity 1. So $1+x$ does not generate this group.

It would become a mess if i write by hand what subgroup would each element generate?

Is there any better way to find out what all elements generate the multiplicative group?

$\endgroup$
  • 4
    $\begingroup$ Hint: The group is cyclic of order $15$, so if you have one generator, you should be able to find all of them. $\endgroup$ – Thomas Andrews Jul 28 '13 at 14:37
  • 1
    $\begingroup$ Also, if $\langle x\rangle$ generates the entire group then $x^4=x+1$ ought to generate the entire group, so I think you might have made an error somewhere. $\endgroup$ – Thomas Andrews Jul 28 '13 at 14:46
  • $\begingroup$ Oh, yes. I guessed it, but some where i messed it up. Thank You. $\endgroup$ – user87543 Jul 28 '13 at 14:49
  • $\begingroup$ You know that the multiplicative group is iso to $C_{15}$, so it suffices to check (by hand!!) that neither $x^3$ nor $x^5$ is equal to $1$. That's a lot quicker than going up to fifteen. You do generate the log table while doing that, so it didn't go to waste, if you did it that way. $\endgroup$ – Jyrki Lahtonen Jul 28 '13 at 20:26
13
$\begingroup$

If $\langle x\rangle$ is the whole multiplicative group, it is isomorphic to $C_{15}$, so your generators will be $x^a$ for all $a$ satisfying $\gcd(a,15)=1$. There should be $\phi(15)=8$ such values.

$\endgroup$
  • $\begingroup$ :) I did not concentrate on case of $C_{15}$.. Thank You so much for this. I do not have reputation to up-vote this :P Sorry for that but thank you. $\endgroup$ – user87543 Jul 28 '13 at 14:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy