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I'm studying on the book "Cohen-Macaulay rings" of Bruns-Herzog

(Here's a link and an image of the page in question for those unable to use Google Books.)

At page 17 it talks about minimal free resolution, but it doesn't give a proper definition (or I'm misunderstanding the one it gives), could you give me a definition?

And if $(R,\mathfrak{m},k)$ is a Noetherian local ring, $M$ a finite $R$-module and

$F.:\cdots\rightarrow F_n\rightarrow F_{n-1}\rightarrow\cdots\rightarrow F_1\rightarrow F_0\rightarrow 0$

a finite free resolution of $M$. The it is minimal if and only if $\varphi_i(F_i)\subset\mathfrak{m}F_{i-1}$ for all $i\geq1$. Why?

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Google books links are a poor substitute for writing out (or posting an image) of the statement. Not everyone will get free access to the images (it sometimes depends on the country from which you are connecting), and at least I am forced to click through the book to get to whatever page it is you are mentioning. Making the reader work harder to understand your question makes it more difficult to get useful replies. –  Arturo Magidin Jun 14 '11 at 19:05
    
Is $F_0 = M$ in your complex? If so this is going to mess up your numbering the $Ext$ groups et al. –  jspecter Jun 14 '11 at 20:12
    
@Arturo: The link to the book is only an help if you want to see how it is explained in the book. But to answer my questio you only need what I wrote. –  Jacob Fox Jun 14 '11 at 20:28
    
@Jspecter: no I think that $F_0\neq M$ –  Jacob Fox Jun 14 '11 at 20:29
    
@Jacob: And I'm pointing out that it is far more helpful to write out (or post an image) than to direct people to a link that often does not work. Google group links are highly sensitive to country of location and other issues that are not immediately obvious; not everyone will see the same thing you are seeing through such a link. –  Arturo Magidin Jun 14 '11 at 20:31
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2 Answers

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I can't see that book online, but let me paraphrase the definition from Eisenbud's book on commutative algebra. See Chapter 19 page 473-477 for details.

Let $R$ be a Noetherian local ring with maximal ideal $\mathfrak{m},$ then

Definition: A free resolution of a $R$-module $M$ is a complex

$$\mathcal{F}: ...\rightarrow F_i \rightarrow F_{i-1} \rightarrow ... \rightarrow F_1 \rightarrow F_0$$

with trivial homology such that $\mathbf{\text{coker}}(F_1 \rightarrow F_0) \cong M$ and each $F_i$ is a free $R$-module.

He then defines a minimal resolution as follows:

Definition: A complex

$$\mathcal{F}: ...\rightarrow F_i \rightarrow F_{i-1} \rightarrow ...$$

over $(R,\mathfrak{m})$ is minimal if the induced maps in the complex $\mathcal{F}\otimes R/\mathfrak{m}$ are each identically $0$. (Note that this is equivalent to the condition that $Im(F_i \rightarrow F_{i-1}) \subset \mathfrak{m}F_{i-1}$)

after this he proves the fact that a free resolution $\mathcal{F}$ is minimal if and only if a basis for $F_{i-1}$ maps into a minimal set of generators for $\mathbf{\text{coker}}(F_i \rightarrow F_{i-1}).$

The proof is via a straightforward appeal to Nakayama's Lemma.

From the way your question is worded, I would guess that your text has taken the latter of these equivalent conditions as the definition of a minimal free resolution and proved it to be equivalent to the former. But I can't be sure without seeing the text.

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This answer is a slight reformulation of jspecter's; perhaps it will help.


A minimal free resolution is one in which each free module has the minimal number of generators. (Hence the name.)

Here is how you make it:

Start with $M$, f.g. over $R$. Its minimal number of generators is $\dim M/\mathfrak m M$. So we can take a free module $F_0$ with this number of generators and a surjection $F_0 \to M$ (but we can't do this with any free module of smaller rank).

If this map is an isomorphism, we're done.

Otherwise, note that since $F_0/\mathfrak m F_0 \to M/\mathfrak m M$ is a surjection between $R/\mathfrak m$-vector spaces of the same dimension, it is an isomorphism, and so the kernel of the surjection $F_0 \to M$ is contained in $\mathfrak m F_0$. Now apply the same process that we used to construct $F_0 \to M$ to this kernel, to obtain a map of free modules $F_1 \to F_0$ whose cokernel is $M$, now with both $F_0$ and $F_1$ being free on the minimal possible number of generators.

The same argument as above will show that the kernel of $F_1 \to F_0$ is contained in $\mathfrak m F_1$.

We now continue inductively, and so produce a minimal free resolution of $M$.

As a biproduct of the construction, we find that each map $F_i \to F_{i-1}$ has image lying in $\mathfrak m F_{i-1}$. Equivalently, each map $F_i\to F_{i-1}$ reduces to the zero map modulo $\mathfrak m$.

Now, as jspecter points out, there is a converse to the preceding remark: any free resolution with the property that $F_{i+1} \to F_{i}$ has image lying in $\mathfrak m F_{i}$ for every $i \geq 0$ (or equivalently, with the property that the maps $F_{i+1} \to F_{i}$ reduce to $0$ mod $\mathfrak m$ for $i \geq 0$) is a minimal free resolution, in the sense that the $i$th stage (for $i \geq 0$), the number of generators of $F_i$ is equal to the minimal number of generators of the kernel of the map $F_{i-1} \to F_{i-2}$. (Here we agree that $F_{-1} = M$ and that $F_{-2} = 0$.)

(In jspecter's answer, things are phrased in terms of of cokernels rather than kernels. But we are saying the same thing: since the $F_i$ form a complex, the cokernel of $F_i \to F_{i-1}$ is the same as the kernel of $F_{i-1} \to F_{i-2}$. I have given a formulation in terms of kernels just because I find it slightly more intuitive.)

As jspecter also says, in the book you are reading, the definition of minimal resolution is almost surely the one I give at the beginning of this answer. The book is then using the preceding remark and its converse to give the alternative characterization of minimal resolutions that you asked about.

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:Sorry, I have a question. From your argument the minimal free resolution of $M$ exists for any finitely generated $R$-module $M$ ? –  Arsenaler Oct 16 '12 at 14:33
    
@msnaber: Dear msnaber, Yes. Regards, –  Matt E Oct 16 '12 at 17:17
    
Thanks Emerton! –  Bombyx mori Jan 20 '13 at 23:20
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