# 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

1. $I=Aa_1A + Aa_2A + \dots + Aa_kA$, for some $a_1$, $\dots$ , $a_k\in I$;

2. $\dim A/(I^q)$ 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.

## 1 Answer

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.

• So, in the first item, you claimed (in different terms) that $\sum\lambda_{i_1,\dots,i_l}x_{i_1}\dots x_{i_l}+I\in A/I\to \sum\lambda_{i_1,dots,i_l}x_{i_1}\dots x_{i_l}+J\in A/J$ is an algebra isomorphism and, hence, $I=J$. Did I get it right? If so, I am truly confident that, for example, when you take $\alpha=\sum\lambda_{ij}x_ix_j$ (an element of "degree" 2) in $A$ you will have that $\alpha\in I$ if and only if $\alpha\in J$. But, if $\alpha:=\sum\lambda_ix_i\in I$ why does $\alpha\in J$? – Júlio César Jun 26 '14 at 3:17
• Júlio, you're completely right, my argument was not correct. I have improved it. – Hugh Thomas Jun 26 '14 at 13:29
• Or, wait, should I address you "Ave, Imperator"? (Sorry, a bit slow on the uptake.) – Hugh Thomas Jun 26 '14 at 14:58
• Sorry, Thomas. I still don't get it. I feel like we still have problems with the polynomial degrees. Let me explain myself. We can always choose $c_{i_1,\dots,i_l}^k\in\mathbb K$ such that $r_{i_1,\dots,i_l}=y_1\dots y_n-\sum c_{i_1,\dots,i_l}^ky_k\in I$. So, taking $t=r_{i_1,\dots,i_l}$ I can not see why your argument is valid for $l>2$. ... to be continued – Júlio César Jun 26 '14 at 19:34
• You are correct about my point in part 2. The reason $I/I^2$ is finite length is that it is finitely generated over $A/I$, which is itself finite length (by assumption). For $A/I^3$, we split it up as $A/I$ and $I/I^3$, then observe that $I/I^3$ is finitely generated over $A/I^2$ (which we just established is finite length). – Hugh Thomas Jun 28 '14 at 15:15