I would warmly appreciate it if someone could tell me whether the following question has an affirmative answer. I am new to the field of commutative algebra, so I am simply trying to fill in some (huge) gaps. Thanks!


Let $ (R,{\frak{m}}) $ be a Noetherian local (commutative unital) ring. Let $ I $ be an ideal of $ R $ with minimal generating set $ \lbrace x\_{1},\ldots,x\_{n} \rbrace $, and let $ \beta: R^{n} \rightarrow I $ be the surjective $ R $-linear map defined by $ \beta(r\_{1},\ldots,r\_{n}) = r\_{1} x\_{1} + \cdots + r\_{n} x\_{n} $. Viewing $ I $ as an $ R $-module, does there exist a free resolution of $ I $ of the form
$$
0 \longrightarrow R^{n-1} \stackrel{\alpha}{\longrightarrow} R^{n} \stackrel{\beta}{\longrightarrow} I \longrightarrow 0,
$$
where the map $ \alpha $ is left-multiplication by some matrix $ M \in {\text{M}\_{n \times (n-1)}}(R) $?



A Question About Free Resolutions
No. Consider $\mathfrak{m}:=(x,y,z)\subset k[x,y,z]\_{(x,y,z)}=:R$. Then the kernel of the map $$R^3\to \mathfrak{m}$$ defined by the minimal generating set $x,y,z$ is minimally generated by $$k\_1:=(y, -x, 0), k\_2:=(z, 0, -x), k\_3:=(0, z, -y).$$ But $$zk\_1-yk\_2-xk\_3=0$$
so the submodule of $R^3$ that these generate is not free. Rather, we have a free resolution $$0\to R\to R^3\to R^3\to \mathfrak{m}\to 0.$$
where the middle map is defined by the matrix $(k\_1 ~k\_2 ~k\_3)$ and the first map sends a generator of $R$ to


$$\begin{pmatrix}
z \\
-y\\
-x
\end{pmatrix}.$$


(I should mention--one can know this example will work by "pure thought" using local cohomology; essentially there is a natural way of identifying the local cohomology of $(R, \mathfrak{m})$, with the coherent cohomology of $\mathbb{P}^2$. But this does not always vanish in degree $2$, so there cannot be a length $2$ resolution...this argument also shows that there's no way of, say, choosing different generators to get a shorter resolution.)


