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CONTEXT

The exercises read:

Exercise 1.22: Let $R=K[x]$. Use the structure theorem for finitely generated modules over a principal ideal domain to show that every finitely generated $R$-module has a finite free resolution.

Exercise 1.23: Let $R=K[x]/(x^n)$. Compute a free resolution of the $R$-module $R/(x^m)$, for any $m\leq n$. Show that the only $R$-modules with finite free resolutions are the free modules.

Let me start with Exercise 1.22. Let $M$ be an $R$-module. Using the structure theorem for finitely generated modules over principal ideal domains, we write

\begin{equation*} M \simeq R^m \oplus R/(r_1) \oplus \ldots R/(r_n). \end{equation*}

By taking generators of $M$, it is clear that there is a surjective map from $R^{n+m}$ to $M$. Let $K$ denote the kernel of this map. Then we find the following finite free resolution of M

\begin{equation*} \mathcal{F}\colon 0 \rightarrow K \rightarrow R^{n+m} \rightarrow M \rightarrow 0, \end{equation*}

concluding the exercise.

My problems begin with Exercise 1.23. Here I try to mimc the approach above, and consider the finite free resolution

\begin{equation} \label{eqn:1} \mathcal{F}\colon 0 \rightarrow (x^m) \rightarrow R\rightarrow R/(x^m) \rightarrow 0. \end{equation}

The idea here being, again, to consider a surjective map from a free $R$-module to our module, and then take the kernel to construct an exact sequence. In page 45 of Eisenbud, he suggests doing something similar for the general case, in fact, he says:

It is easy to see that every module has a free resolution and, if $R$ is graded, that every graded module has a graded free resolution. To construct one, begin by taking a set of generators for $M$ and map a free module onto $M$ by sending the free generators of the free module to the given generators of $M$. Let $M_1$ be the kernel of this map, and repeat the procedure, now starting with $M_1$.

My answer to Exercise 1.23 is wrong, and the solution can be seen here and here. However, I fail to see what is wrong with my approach, and I think this is because I have trouble understanding free modules and free resolutions properly. So here are my questions.

QUESTIONS

Why is $R/(x^m)$ not a free $R$-module? The set $\{1\}$ is clearly a generating set, why is it not a basis?

How does my free resolution for Exercise 1.23 fail? I would argue that it is exact and both $(x^m)$ and $R$ are free $R$-modules, so, what am I missing here?

Why does my approach work for Exercise 1.22 but not for Exercise 1.23?

Why does my approach fail in general compared to the one explained by Eisenbud?

Josu P. Z.
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    Regarding the first question: ${1}$ is not a basis, because it is not $R$-linear independent. Take $x^m\in R$ (which is nonzero, since $m<n$). Then $0 \cdot 1 = 0 = x^m \cdot 1$ in $R/(x^m)$, so there are two different linear combinations of your basis giving the same element... – Jonas Linssen Aug 17 '23 at 10:05
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    Some basic things to think about can be the following. Assume $k$ is, say, the real numbers. Then $k[x]/(x^m)$ is a finite-dimensional $k$-vector space. Can this be the case for a general free $k[x]$-module? Now, if $R$ is any ring and $I$ is a left ideal, then $R/I$ is a left $R$-module generated by the class of $1$, as you say. Does this imply that every quotient $R/I$ is a free $R$-module? What happens if we take, say, $R=\mathbb Z$ and $I= 2\mathbb Z$? – MathsIsFun Aug 17 '23 at 10:19

1 Answers1

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As said in the comments, $R/(x^m)$ is not a free $R$-module despite having the generating set $\{1\}$. This is because $0\cdot 1$ and $x^m\cdot 1$ are two different $R$-linear combinations that give the same element, so $\{1\}$ is not an $R$-linearly independent set, hence not an $R$-basis of $R/(x^m)$.

The reason why the exact sequence

\begin{equation*} \mathcal{F}\colon 0 \rightarrow (x^m) \rightarrow R \rightarrow R/(x^m) \rightarrow 0 \end{equation*}

fails to be a free resolution of $R$ is very similar. Indeed, the problem here is that $(x^m)$ is not a free $R$-module, because of the same argument above: $0\cdot x^m$ and $x^{n-m}\cdot x^m$ are two different $R$-linear combinations that give the same element, so $\{x^m\}$ is not a $R$-linearly independent set, hence not an $R$-basis of $(x^m)$.

The approach for Exercise $1.22$ works because in this case we are working with modules over principal ideal domains. Here we have the nice theorem that submodules of free modules are free, which guarantees that $K$ is a free module. However, in Exercise $1.23$ we work over Noetherian rings, and here it is not true that submodules of free modules are free, so the same approach fails to give a free resolution.

In the explanation of Eisenbud, I failed to interpret that in general $M_1$ may not be free or even finitely generated. One way to fix this problem is to replace $M_1$ with some module $N_1$ that is free and that contains $M_1$, and then look for appropriate homomorphisms that make the sequence exact.

This can be done in this particular example. Instead of considering the inclusion map from $(x^m)$ to $R$, one can consider the map from $R$ to $R$ given by multiplication by $x^m$. This has kernel $(x^{n-m})$, and to continue the sequence we consider the map from $R$ to $R$ given by multiplication by $x^{n-m}$. Continuing this way, we obtain the free resolution

\begin{equation*} \mathcal{F}\colon \ldots \xrightarrow{\cdot x^{n-m}} R \xrightarrow{\cdot x^m} R \xrightarrow{\cdot x^{n-m}} R \xrightarrow{\cdot x^m} R \rightarrow R/(x^m) \rightarrow 0. \end{equation*}

Josu P. Z.
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