How is a dimension of an ideal defined in $R$

Question

Suppose I consider the ring $R=\Bbb Z_2[x]/\langle x^9+1\rangle$

$x^9+1$ has a complete factorization into irreducible polynomials like

$x^9+1=(x+1)(x^2+x+1)(x^6+x^3+1)$

Now $I=\langle x+1\rangle /\langle x^9+1\rangle $ is an ideal in $R$

I got a question

What is the dimension of $I$ in $R$?

Can someone please tell me how is a dimension of an ideal defined in $R$ and what will be the answer?

Answer 1

A naïve interpretation of dimension in this case would be the $\mathbb Z_2$ dimension. In your case the original ring $R$ is $9$ dimensional over $\mathbb Z_2$, and $I$ has codimension $1$ in $R$ (since $(x-1)$ has codimension $1$ in $\mathbb Z_2[x]$.) So in that case it would be $8$ dimensional.

But dimension could also refer to the Krull dimension of $R/I$. That seems rather dull though since proper quotients of a polynomial ring over a field are all Artinian, so that the Krull dimension is always $0$.

None of it seems related to the factorization you gave, and without more context it's hard to guess what was intended.