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I have to find a primitive element of $\mathbb{F}_{16}^{\times}$. I defined $\mathbb{F}_{16} = \frac{\mathbb{F}_{2}[a]}{\langle a^4+a+1 \rangle}$.

I tried to prove that $a^3$ and $a^5 \neq 1$.

I got $a^3=a^3 \neq 1$ and $a^5 = a^2 + a \neq 1$.

But, when I try to check if $a^{15} = 1$, I get $a^{15} = a^3 + a^2 \neq 1$. I checked several times and can not get it, it may be something I am missing.

Are my calculations right?

How can I find the primitive element? Which is the primitive element?

EvaMGG
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    You must have $a^{15}=1$, whatever $a$ is a primitive element or not, since $\mathbb{F}_{16}^*$ has order $15$. Moreover, there is no "the primitive element" : since $\varphi(15)=8$, you can find $8$ primitive elements. – TheSilverDoe Feb 22 '21 at 20:26
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    If you'll include your calculation for $a^{15}$ in the question we can help you spot your mistake. – Robert Shore Feb 22 '21 at 20:28
  • Also, there are many choices of primitive elements. You'll find $a$ is one, but also that there are others as well. – ndhanson3 Feb 22 '21 at 20:29
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    $a^{15} = (a^5)^3 = a^3 (a+1)^3 = a^3 (a^3 + a^2 + a + 1) = a^4(a^2 + a + 1) + a^3 = (a^3 + a^2 + a) + (a^2 + a + 1) + a^3 = 1$. – diracdeltafunk Feb 22 '21 at 20:30
  • or $a^{15}=(a^5)^3=(a^2+a)^3=(a^2)^3+(a^2)^2a+a^2a^2+a^3=a^6+a^5+a^4+a^3$ $=a^3+a^2+a^2+a+a+1+a^3=1$ – J. W. Tanner Feb 22 '21 at 20:35
  • Compare your calculation with the table I once produced for referrals like this. If you cannot spot the error, please comment, and we can figure it out. – Jyrki Lahtonen Feb 22 '21 at 21:02

1 Answers1

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in ${GF(2^4)}$ The polynomial ${1+x+x^4}$ has 8 primitive elements.

  • Condition: ${x^i \in GF(2^r)}$ is primitive iff ${gcd(i, 2^{n}-1)=1)}$, Here, ${2^4-1=15}$ ,so we've possible candidates are ${\{1,2,4,7,8,11,13,14\}}$.
  • you can check, an element ${\alpha \in GF(2^r)}$ is primitive iff ${\alpha ^m \ne 1}$ for ${1 \leq m < 2^r -1}$
  • Let's take ${x^4=\alpha}$ then ${\alpha^2=x^8=1+x^2, \alpha^{14}=1+x^3,\alpha^{15}=x+x^4=x+1+x=1}$, so we don't have any exponent m less than 15 which gives 1. So, 4 is a primitive element.
  • Now, lets check ${x^5=\alpha}$, now ${\alpha^2=1+x+x^2, \alpha^3= x^{15}=1}$, here m=3 (<15), hence 5 in not a primitive element.
SSA
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