Let $p,l \in \mathbb{N}$ be prime, and assume that $ord_p(l) = p-1$; that is $l$ is a primitive pth root of unity.
It is known that in this case the ring $R := \mathbb{F}_l[x]/(x^p-1)$ is with
$R \simeq \mathbb{F}_l[x]/(x-1) \oplus \mathbb{F}_l[x]/(\frac{x^p-1}{x-1})$.
Hence this ring $R$ has only two non-trivial ideals.
I'd like to know what their dimensions are.
My thinking is this -
the ideal $I_1 = \{(x,0): x \in \mathbb{F}_l[x]/(x-1) \}$ of $\mathbb{F}_l[x]/(x-1) \oplus \mathbb{F}_l[x]/(\frac{x^p-1}{x-1})$, is of dimension $p-1$ because the dimension of the quotient ring $[\mathbb{F}_l[x]/(x-1) \oplus \mathbb{F}_l[x]/(\frac{x^p-1}{x-1})] / I_1$ is $dim_{\mathbb{F}_l}(R) - dim_{\mathbb{F}_l}(\mathbb{F}_l[x]/(x-1)) = p -1$.
Similarly $dim(I_2) = p - (p-1) = 1$.
Is this a correct argument, can it be made more rigorous and is this what is meant by the dimension of these ideals?
I don't know the full context, but normally I would expect "dimension" refers to the dimension of the ideal itself (as a vector space), not the dimension of the quotient by the ideal. So your dimensions are backwards: $I_1$ has dimension $1$, since it is a $1$-dimensional vector space over $\mathbb{F}_l$, and $I_2$ has dimension $p-1$ similarly.