MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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) $?

share|cite|improve this question
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
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
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.)

share|cite|improve this answer
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 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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.