Let $\mathbb{F}$ be a field of characteristic $0$, and $p$ a prime number.
Let $f \in \mathbb{F}[x]/(x^p - 1)$ be given by $f(x) = \sum_{i=0}^{p-1}a_ix^i$, with $a_i \in \mathbb{Z}$.
Let $q$ be a prime number; we need to find a field, $K$, of characteristic $q$, a polynomial $h(x) = \sum_{i=0}^{p-1}b_ix^i \in K[x]/(x^p - 1)$ s.t:
$|\{b_i :b_i \neq 0\}| := wt(h) \leq wt(f)$
$ \dim_K((h)) \leq \dim_{\mathbb{F}}((f))$, where $(f)$ denotes the ideal generated by $f$ in $\mathbb{F}[x]/(x^p - 1)$.
I thought of using:
$ \dim_{\mathbb{F}}((f)) = p - \deg(\gcd_{\mathbb{F}}(f, x^p -1))$
and, though it may not be relevant,
if $L/K$ is a field extension,$f,g \in K[x]$, then $\gcd_K(f,g) = \gcd_L(f,g)$
Here's how I began; we define $K = \mathbb{Z}/ q\mathbb{Z}$ a finite field of characteristic $q$, and $h = f \pmod q$.
Then showing the first inequality is simple; if $a_i \neq_q 0$, specifically $a_i \neq 0$ and so $wt(h) \leq wt(f)$.
I was thinking that the quoted results can assist with the second, but I haven't managed to see it yet. Any ideas?
Here's an attempt:
We take $K = \mathbb{F}_q$. Define $h = f$ mod $q$.
As above, if $h = \sum_{i = 0}^{p-1}b_ix^i$ and $b_i \neq_q 0$, then specifically $a_i \neq 0$ and so $wt(h) \leq wt(f)$.
For the dimension inequality, we use the first stated result;
Also we use:
if $g_1 | f$ in $\mathbb{F}$ then $g_1g_2 = f$, then $g_1g_2 = h$ mod $q$. Meaning $g_1|h$ in $\mathbb{F}_q$.
Suppose that $g = \sum_{i=0}^n c_ix^i = gcd_{\mathbb{F}}(f, x^p - 1)$, then $g$, mod $q$, divides $h$ and $g$, mod $q$, divides $x^p - 1$ in $\mathbb{F}_q$. Then $deg(g$ mod $q) \leq gcd_{\mathbb{F}_q}(h, x^p - 1)$.
To end, let $(\sum_{i=0}^n c_ix^i)(\sum_{i=0}^m a_jx^j) = x^p -1$ with $n + m = p$. If $c_n =_q 0$ then $deg(x^p - 1 $ mod $q) < p $ in contradiction.
So we have $gcd_{\mathbb{F}}(f, x^p - 1) \leq gcd_{\mathbb{F}_q}(f, x^p - 1)$ and this proves the claim.