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Suppose that $q=2$ and $n=31$.

$C_0=\{0\}$

$C_1=\{1,2,4,8,16\}=C_2=C_4=C_8=C_{16}$

$C_3=\{3,6,12,24,17\}=C_6=C_{12}$

$C_5=\{5,10,20,9,18\}=C_9=C_{10}$

$C_7=\{7,14,28,25,19\}=C_{14}$

$C_{11}=\{11,22,13,26,21\}=C_{13}$

$C_{15}=\{15,30,29,27,23\}$

$ord_{31}(3)=5$, $x^{31}-1$ has roots in $\mathbb{F}_{2^5}$. Let $\alpha$ be the primitive element in $\mathbb{F}_{2^5}$.

Monic polynomial of $C_1$ will be $(x-1)(x-2)(x-4)(x-8)(x-16)=-1024+1984 x-1240 x^2+310 x^3-31 x^4+x^5\equiv30-30x^2+x^5\mod{13}$

But this monic polynomial isn't right, can someone give me a hint or suggestion to get a right a one so that I can construct the generator polynomial? Thanks

Simple
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  • Because $2^5-1=31$ is a prime number, all the elements of $\Bbb{F}_{32}$ that are not in the prime field $\Bbb{F}_2$ are primitive. So calling anything the primitive element is misguided. Well, Dilip Sarwate already addressed the issues :-) – Jyrki Lahtonen Mar 04 '16 at 05:37
  • @JyrkiLahtonen Why $\mathbb{F}{32}$? Aren't we working on $\mathbb{F}{31}$? – Simple Mar 04 '16 at 05:42
  • The alphabet field is $\Bbb{F}q$ and you said $q=2$. The field we need to work is the extension of the alphabet field that contains a root of unity of order $n=31$. The field $\Bbb{F}{32}$ fits the bill. The field $\Bbb{F}_{31}$ only has roots of unity of order $30$ (its primitive element for example). But, more importantly, it is not an extension of $\Bbb{F}_2$. The order of the field needs to be a power of $q$. – Jyrki Lahtonen Mar 04 '16 at 08:07

1 Answers1

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If by generator polynomial of $C_1$ you mean the minimal polynomial of $\alpha$ over $\mathbb F_2$, that is, the binary polynomial of least degree that has $\alpha$ as a root, then the polynomial is $$(x-\alpha)(x-\alpha^2)(x-\alpha^4)(x-\alpha^8)(x-\alpha^{16}).\tag{1}$$ It is also the minimal polynomial of $\alpha^2$ and of $\alpha^4$ and of $\alpha^8$ and of $\alpha^{16}$. The polynomial is irreducible.

In multiplying out the factors, you need use arithmetic in $\mathbb F_{2^5}$ (and not modulo $13$ or $31$ as you have it) but to set up the arithmetic you need to identify which of the $30$ primitive elements of $\mathbb F_{2^5}$ you are choosing to call $\alpha$. The only way to make this identification is to choose one of the six known irreducible binary quintic polynomials and say that $\alpha$ is one of its roots. Of course, once you have chosen, say, $x^5+x^2+1$ as the polynomial, the immediate need of setting up arithmetic to verify that $(1)$ multiplies out to $x^5+x^2+1$ disappears. You will need the arithmetic to figure out the generator polynomials of $C_3$,$C_5$ etc.

So, how do I know that there are 6 quintics and $x^5+x^2+1$ is one of them? I worked it out in this answer.

Community
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Dilip Sarwate
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  • Just want to make sure I understand the answer, I found that $3$ is a primitive element, then the monic polynomial of $C_1$ will be $(x-3)(x-3^2)(x-3^4)(x-3^8)(x-3^{16})$? – Simple Mar 04 '16 at 05:40
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    I have no idea what you mean by $3$. Are you still doing arithmetic mod $31$? See the comments by @Jyrki on the main question. Also, read his incredibly detailed beautiful answer on a different question to understand the difference between $\mathbb F_{31}$ and $\mathbb F_{32}$. Until you stop using $\mathbb F_{31}$ as a substitute for $\mathbb F_{32}$, you will not be able to make any progress on this topic. – Dilip Sarwate Mar 04 '16 at 14:09