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Bruns and Herzog in their book Cohen-Macaulay Rings, page 120 write: "The Hilbert-Burch theorem 1.4.17 identifies perfect ideals of grade $2$ as the ideals of maximal minors of certain matrices".

For the convenience of the community I restate the Hilbert-Burch theorem as given by Bruns and Herzog:

Hilbert-Burch Theorem [1.4.17 in B&H]: Let $R$ be a Noetherian ring and $I$ an ideal with a free resolution $F_{\bullet}:0 \rightarrow R^n \stackrel{\phi}{\rightarrow} R^{n+1} \rightarrow I \rightarrow 0$. Then there exists an $R$-regular element $a$ such that $I = a I_n(\phi)$, ($I_n(\phi)$ is the $n$-th Fitting invariant of $\phi$). If $I$ is projective, then $I=(a)$, and if $\operatorname{projdim}I=1$, then $I$ is perfect of grade $2$. Conversely, if $\phi:R^n \rightarrow R^{n+1}$ is an $R$-linear map such that $\operatorname{grade} I_n(\phi) \ge 2$, then $I=I_n(\phi)$ has the free resolution $F_{\bullet}$.

My question: Now let $I$ be a perfect ideal of grade equal to $2$. Since $I$ is perfect we must have that $\operatorname{projdim} R/I = 2$. How does it follow then from the Hilbert-Burch theorem that $I$ is $I_n(\phi)$ for some $\phi$ and some $n$? How do we construct this $\phi$?

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It follows from the first part of the statement when the ring is local. Consider the following minimal free resolution for $R/I$: $$ 0 \to R^m \to R^n \to R \to R/I \to 0. $$ Since rank is additive, we see that $m + 1 = n$. Then the map $\phi: R^m \to R^n$ can be written as $m \times n$ matrix. The Theorem says $I = aI_n(\phi)$ where $a$ is an regular element. If $a$ is not a unit, then height of $I$ is at most $1$. Therefore, $a$ is a unit and $I = I_n(\phi)$.

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  • $\begingroup$ Dear Youngsu, could you please expand the details of your answer? In particular i) how do we see that the rank of $R/I$ is zero? ii) why have you taken $R$ as the first free module in the free resolution, i.e. why did you not take a free resolution $0 \rightarrow R^m \rightarrow R^n \rightarrow R^k \rightarrow R/I$? iii) why if $a$ is not a unit then $I$ has height at most $1$ and why this can not be true? $\endgroup$ – Manos Mar 20 '14 at 17:53
  • $\begingroup$ @Manos: 1) Since the grade of $I$ is $2$, $I$ contains a regular element. Therefore, $Quot(R) \otimes I \cong Quot(R)$. 2) I guess you could, but then how are you going to apply the Theorem? I wanted to make the kernel of the map $R^k \to R/I$ to be $I$. 3) $aI_n(\phi) \subseteq (a)$ and the height of $(a)$ is at most one if $a$ is not a unit. $\endgroup$ – Youngsu Mar 20 '14 at 19:44
  • $\begingroup$ I am good with 1) and 3). Regarding 2) my question is how do you know that if you choose a projective resolution of $R/I$ to start with $R$, then this projective resolution will have length equal to $2$? Because the projective dimension of $R/I$ is equal to $2$ that means that there exists some projective resolution of length $2$, but not any projective resolution needs to have length equal to $2$. Right? $\endgroup$ – Manos Mar 20 '14 at 22:04
  • $\begingroup$ @Manos: I see. I believe what you mentioned is delicate. In fact I should change my answer since I do not know how to construct such a resolution. My answer only works when the ring is local. Then the minimal free resolution can be used. However, I do not think one can patch them in general. I'm sorry for confusion. $\endgroup$ – Youngsu Mar 20 '14 at 23:43
  • $\begingroup$ Your answer has still some interesting parts, so i will upvote it. But maybe you can add that it is valid for local rings. In fact B&H may have the local case in their mind even though they do not say so explicitely. $\endgroup$ – Manos Mar 21 '14 at 0:14

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