How many cyclic linear codes are in n^ Trivial, but can't find information anywhere on it.
How many cyclic linear codes (other than $\{00...0\}$ and $K^n$) are there if $n = 17$? And if $n = 136$?
so in $n=17$ there are $10$ codes. in $n=136$ it is $(x+17)^8= 80$? 
 A: Find all the idempotents of $x^{17}+1$, then you have the numbers of the linear cyclic codes ($-2$ in this case because you want only the proper ones). They are $6$.
After that, for corollary $4.4.4$: let $n=2^rs$ where s is odd, you find that $(2^r)^z-2 = 6$, you resolve and you find that $x^{17}$ is a product of $3$ irreducible polynomials. For $n=136$ you need to solve $(2^r)^z-2$, where $z=3$ and $r=3$.
A: We have $x^{17}+1=f_1(x)f_2(x)f_3(x)$, where $f_i(x)\in\Bbb{F}_2[x]$, $i=1,2,3,$ are the irreducible factors that you found in your previous question. By CRT we get that
$$
\Bbb{F}_2[x]/(x^{17}+1)\cong\Bbb{F}_2[x]/(f_1(x)) \oplus \Bbb{F}_2[x]/(f_2(x))\oplus \Bbb{F}_2[x]/(f_3(x))
$$
as modules over $\Bbb{F}_2[x]$. The binary cyclic linear codes of length 17 are exactly the submodules of this. Because all the summands here are simple and pairwise non-isomorphic, with each of them you either take it or leave it. That's three binary choices - a total of $8$. You excluded the trivial cases (take all or leave all), so the answer is $8-2=6$.
When the length is $8\cdot17=136$ it is more complicated. We have
$$
(x^{136}+1)=(x^{17}+1)^8=f_1(x)^8f_2(x)^8f_3(x)^8
$$ 
and by CRT (they are still pairwise coprime)
$$
\Bbb{F}_2[x]/(x^{136}+1)\cong\Bbb{F}_2[x]/(f_1(x)^8) \oplus \Bbb{F}_2[x]/(f_2(x)^8)\oplus \Bbb{F}_2[x]/(f_3(x)^8).
$$
It still holds that we want to count the number of submodules of this. It still holds (the three indecomposable idempotents can be used to prove this) that any submodule $M$ is of the form $M=M_1\oplus M_2\oplus M_3$, where $M_i$ is a submodule of $\Bbb{F}_2[x]/(f_i(x)^8)$. But these summands are no longer simple,
so they have several submodules.
As the ring $\Bbb{F}_2[x]$ is a PID, all those possible submodules are cyclic. The generator has to be a factor of $f_i(x)^8$, so it is necessarily of the form
$(f_i(x)^j)/(f_i(x)^8)$ for some $j=0,1,\ldots,8$. 
So this time there are $8+1=9$ choices for the submodule of $\Bbb{F}_2[x]/(f_i(x)^8)$. We need to make three such choices, so that gives us a total of $9^3=729$ binary linear cyclic codes of length 136. Again, you leave out the trivial cases, which leaves us $727$ non-trivial binary cyclic linear codes of length 136.

Clarifying the simple case of $f_1(x)=x+1$ and cyclic codes of length 8. The above theory tells us that as $x^8+1=(x+1)^8$ there are nine binary cyclic linear codes of length 8. The seven non-trivial ones are generated by
$(x+1)^j, j=1,2,\ldots,7$, and have respective dimensions $8-j$.
In the general case we have one summand for each irreducible factor of $x^n+1$.
The number of choices for that summand is one plus the multiplicity of the factor. Here the multiplicities are all one, when $n=17$ (resp. all eight, when $n=136$), so the number choices per summand was $1+1=2$ (resp. $8+1=9$).
