Let $R$ denote the ring $\mathbb{F}_l[x]/(x^p -1)$, where $p$ is some prime such that $r := \mathrm{ord}_p(l) < \epsilon p$ for $0 < \epsilon < 1$.
It is known that if $s = \frac{p-1}{r}$, an integer, then we have that $R \simeq \mathbb{F}_l \oplus (\mathbb{F}_{l^r})^s$.
I want to show there exists an ideal $I_p$ in $R$ s.t. $\frac{\epsilon p}{2} \leq \dim(I_p) < \epsilon p$.
Here's what I tried:
Each ideal of the form $I_k := \{(0,..,0,y,0,..,0) | y\in \mathbb{F}_{l^r}\}$ is of dimension $r$ (where $1 < k$ denotes the kth coordinate).
Since $\frac{p-1}{r}(r) + r = p-1+r \geq p > \epsilon p$ there exist $n_0 \in \mathbb{N}$ minimal s.t $n_0 \leq \frac{p-1}{r}$ with $\epsilon p \leq n_0r + r$.
Then we have $\epsilon p \leq n_0r + r \leq 2n_0r$ and so $\frac{\epsilon p}{2} \leq n_0r.$
Now it can't be that $\epsilon p \leq n_0r$ since then we'd have:
$(n_0 - 1)r + r = n_0r \geq \epsilon p$ in contradiction to $n_0$ being minimal with this property.
At last we have that $I_p := \oplus_{k=1}^{n_0}I_k$ is an ideal of dimension $n_0r$. (We're summing $n_0$ ideals whose degree is $r$, forgive the weird notation.)
Is this correct?