Dimension of quotients of Noetherian rings Suppose that $A$ is a Noetherian ring containing a field $k$ and consider an ideal $I$ with $J =\sqrt{I}$ the radical of $I$. Suppose that $A/J$ is a finite-dimensional vector space over $k$. I am trying to prove that this implies that $A/I$ is also a finite-dimensional vector space over $k$. 


I know that since $A$ is Noetherian there exists $m$ such that $J^m \subset I$ and so I was thinking of considering the filtration $J^m \subset \cdots \subset J^2 \subset J \subset A$ and then saying something like $\dim_k J^i/J^{i+1} = \dim_k A/J$ for all $i$ up to $i = m$. Then I would claim that $\dim_k A/I \le m \cdot \dim_k A/J$ using exact sequences like
$$
0 \rightarrow J/J^2 \rightarrow A/J^2 \rightarrow  A/J \rightarrow 0.
$$
If this works, then $A/I$ is finite-dimensional. My issue is with the statement that $\dim_k J^i/J^{i+1} = \dim_k A/J$ which I think is incorrect. Is this the right approach? It seems like something is missing.
$\dim_k J^i/J^{i+1}&lt;\infty$ for all $i\ge 1$ for the following reason: $J^i/J^{i+1}$ is an $A/J$-module (since $J(J^i/J^{i+1})=0$), $J^i$ is finitely generated (since $A$ is noetherian), therefore $J^i/J^{i+1}$ is a finitely generated $A/J$-module. Furthermore, since $A/J$ is a finitely generated $k$-module we can deduce that $J^i/J^{i+1}$ is a finitely generated $k$-module and thus $\dim_k J^i/J^{i+1}&lt;\infty$.  

Now, step by step, one can prove that $\dim_kA/J^i&lt;\infty$ for all $i\ge 1$.
Hint: You don't need $\dim_k J^i/J^{i + 1}$ to be independent of $i$, you just need it to be finite for each $i$ up to $m$.  Use the Nakayama lemma.