If $A=k[x_1,\dots,x_n]/I$ is a finitely generated $k$-algebra with $A\cong k^m$ as modules, what can we say about $I$?

Solution 1:

Question: "However finding necessary conditions is more difficult. The part that I understand less is how should I be in order to make A a free k-module, while how to make A a finitely generated module is more clear. Is a too general setting to say something valid, or there is actually some necessary condition?"

Answer: To parametrize cofinite ideals $I \subseteq A:=k[x_1,..,x_n]$ is a much studied problem in algebra: An ideal $I \subseteq A$ is cofinite iff $dim_k(A/I)<\infty$, meaning $A/I \cong k^n$ as $k$-module for some integer $n$.

Comment: "However finding necessary conditions is more difficult."

Using the Chinese remainder lemma you may prove that $I$ is cofinite iff there is a finite set of maximal ideals $\mathfrak{m}_1,..,\mathfrak{m}_l$, integers $n_1,..,n_l \geq 1$ and ideals

$$\mathfrak{m}_j^{n_j} \subseteq I_j \subseteq A$$

with $I=\prod_j I_j$. The ideals $I_j$ are called "$\mathfrak{m}_j$-squeezed ideals". If $k$ is algebraically closed it follows $\mathfrak{m}_i=(x_1-a_1,..,x_n-a_n)$ with $a_i \in k$.

Note: If $A$ is a regular $k$-algebra of finite type, it follows for any maximal ideal $\mathfrak{m}$ that $\mathfrak{m}^n/\mathfrak{m}^{n+1} \cong Sym^n_k(\mathfrak{m})$, hence we have "good control" on powers of maximal ideals.

Cofinite ideals appear in the study of the "Hilbert scheme of points". The Hilbert scheme of points (of length $n$) $Hilb^n(X)$ is a parameter space parametrizing length $n$ subschemes $Z_n \subseteq X$ of a fixed scheme $X$. These parameter spaces appear in various situations (the n!-conjecture etc).

Applications of the Chinese remainder Theorem to the study of the Hilbert scheme of points and $(\mathfrak{m},l)$-squeezed ideals.

Comment: "The part that I understand less is how should I be in order to make A a free k-module, while how to make A a finitely generated module is more clear. Is a too general setting to say something valid, or there is actually some necessary condition?"

Answer: Since $k$ is a field it follows (if you accept Zorns lemma) that any $k$-algebra $B$ has a basis. Hence any such $B$ is a free $k$-module.