EDIT: All my codes are over the field $\mathbb{F}_2$.
I am holding some old lecture notes that contain a solution to the following problem:
Given an cyclic error correcting code $C$ of length $n$, rate $r$ and distance $d$, construct a cyclic error correcting code $C'$ of length $2n$, rate $r$ and distance $d$.
The solution there only works if $n$ is odd.
I think it's easy to solve it for any value of $n$ in the following way:
Let $C'$ contain exactly all words of the form $(a_0,b_0,\dotsc,a_{n-1},b_{n-1})$ were both $(a_0,\dotsc,a_{n-1})$ and $(b_0,\dotsc,b_{n-1})$ are codewords of $C$. In other words, I "interleave" the code with itself.
The dimension of the above code is twice the dimension of the original code, and so the rate is the same as in the original code. The distance remains the same too because the lowest weight word is achieved by interleaving the $0$ word with a lowest weight nonzero word. Finally, $C'$ is cyclic because a cyclic shift (by $1$ place) "swaps the roles" of $a$ and $b$ and does a cyclic shift on one of them. I think another way to see it is that if $h(x)$ is the generator polynomial of $C$ then $h^2(x)$ is the generator polynomial for $C'$ and it is in particular cyclic.
Is my construction and proof correct?
I am asking because I suspect I am missing something because the lecture notes only solve the problem (stated exactly as I put it) for odd values of $n$, and that's a reason to suspect the solution is not that simple.
To answer your request, yes, it looks like all your reasoning is valid.
Most of it holds for fields with characteristic different from 2, although it's not so clear what the generator polynomial would be.
It's possible the construction in your notes has some extra property for some other use later, or perhaps the author genuinely overlooked this solution.