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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) $?

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No. The relations among the generators will typically satisfy further relations, which necessitates taking longer resolutions. –  Jack Huizenga Dec 11 '12 at 2:33
    
I see. Then may I know under what conditions on $ I $ will such a free resolution exist? –  Leonard Dec 11 '12 at 2:40
    
If $Tor_{i}^R(I,R/{\mathfrak{m}}) = 0$ for $i > 1$. –  Sasha Dec 11 '12 at 3:05
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Look up Hilbert's syzygy theorem on Wikipedia (it has an apostrophe, spoiling its linkability, alas). –  Allen Knutson Dec 11 '12 at 3:14
    
@Sasha: Your comment is intriguing. Could you kindly direct me to some references where this identity is proven? –  Leonard Dec 11 '12 at 7:03

2 Answers 2

up vote 1 down vote accepted

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.)

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BTW, another way to see that there is no length 2 resolution is to note that the resolution given (which is a truncation of the Koszul resolution of $R/\mathfrak{m}$) is a "minimal free resolution," since the maps manifestly vanish mod $\mathfrak{m}$. –  Daniel Litt Dec 11 '12 at 6:38

The resolution that you are asking for exists for a special class of ideals, namely the perfect codimension 2 ideals. Here perfect means that a finite resolution exists, and codimension 2 means roughly (in any reasonable situation of classical algebraic geometry, at least) that $\dim R/I = \dim R - 2$.

The Hilbert--Burch theorem http://en.wikipedia.org/wiki/Hilbert%E2%80%93Burch_theorem classifies all such ideals: the generators you speak of are the maximal minors of an $n \times (n+1)$ matrix. Conversely, any such ideal is perfect (and its resolution has the form you mentioned) if and only if it has codimension 2.

Remark: this is one of a few special cases where you can classify the structure of an ideal just by how its resolution looks. Some others of note come from Koszul complexes (ideals generated by a regular sequence) and the Buchsbaum--Eisenbud complex (codimension 3 Gorenstein ideals). Anyway I think this is a really neat subject and questions like the one you're asking (in the comment) are a nice gateway into the subject.

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