Does an ideal of finite codimension in a finitely generated algebra have always to be finitely generated? I have been reading a book on Lie Algebras ("Álgebras de Lie" by San Martin) and there is this exercise in the chapter on universal enveloping algebras with a claim that I can not prove:

Suppose that $A$ is a finitely generated associative unital algebra 
  and $I$ is a bilateral ideal in $A$. If $\dim A/I$ is finite then
  
  
*
  
*$I=Aa_1A + Aa_2A + \dots + Aa_kA$, for some $a_1$, $\dots$ , $a_k\in I$;
  
*$\dim A/(I^q)$ is finite for all $q\in\mathbb Z_+$.

I have never had any course on noncommutative algebra or so, but it seems to me that (the item 1) is kind of a Noetherian property for noncommutative algebras. That's why I found it kind of cool.
Any insight in the case where $A$ is the enveloping algebra of a finite dimensional Lie algebra $\mathfrak g$ and $I$ is the kernel of a representation $\tilde\rho\colon A\to\mathfrak{gl}(V)$, induced by a finite dimensional representation $\rho\colon\mathfrak g\to\mathfrak{gl}(V)$, is welcome.
I got from the book "Noncommutative Noetherian Rings" by McConnel and Robson that the enveloping algebra of a finite dimensional lie algebra is Noetherian (Section 1.7). So, I get item 1 for this case. But I still don't know how to prove 2. 
Solving this exercise is a hobby to me. So, references to general long readings, long explanations and indirect solutions are very welcome.
 A: Let $y_1,\dots, y_n$ be elements of $A$ which form a basis for $A/I$.  
Take a finite generating set $z_1,\dots, z_r$ for $A$.  We can write $$z_i= (\sum_j d_{i}^j y_j) + z_i'$$ 
with the $d_i^j$ in the ground field and $z_i'$ in $I$.  The set $\{y_i\} \cup \{z_i'\}$ forms a new finite generating set for $A$.  
Define elements of the ground field $c_{ij}^k$ by 
$$ y_iy_j=\sum_k c_{ij}^k y_k \qquad (\textrm{in } A/I)$$.  
Therefore $y_iy_j - \sum_k c_{ij}^k y_k$ is an element of $I$.  
Let $J$ be the ideal generated by these $n^2$ elements, together with the elements $z_i'$.  
Suppose that $J$ is strictly contained in $I$.  So there is some element $t$ in $I$ but not in $J$.  Write $t$ as a polynomial in $y_1,\dots, y_n$ and the elements $z_i'$.  In $A/J$, we can use the relations in $J$ to kill the $z_i'$ elements and simplify the products, so we wind up with a linear combination of $y_1,\dots, y_n$.  But since $y_1,\dots, y_n$ form a basis for $A/I$, their linear combination only lies in $I$ if it is zero.  Thus our element $t$ is zero in $A/J$, i.e., $t\in J$.  So $I=J$.
This establishes 1.
Example: suppose we wanted to simplify $y_1y_2y_3$ in $A/J$.  We know $y_1y_2-\sum c_{12}^k y_k \in J$  This tells us that in $A/J$, $$y_1y_2y_3= \sum_k c_{12}^ky_ky_3.$$  Now, we look at the elements of $J$ which tell us about multiplying $y_3$ times $y_k$ for each $k$.  We conclude that in $A/J$,
$$ y_1y_2y_3= \sum _k \sum_m c_{12}^kc_{3k}^m y_m.$$
Now $A/I^2$ is an extension of $A/I$ (finite-dimensional) and $I/I^2$ (finite-dimensional because $I$ is finitely generated and $A/I$ is finite-dimensional), so finite-dimensional.  Continuing in the same way, $A/I^q$ is finite-dimensional.  
