Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Motivation (you can ignore this part): A problem in Hartshorne (II.5.8c) asks to show that if we have a coherent sheaf $\mathscr{F}$ on a reduced noetherian scheme $X$, and the function

$$\varphi(x)=\dim_{k(x)} \mathscr{F}_x \otimes_{\mathcal{O}_x} k(x)$$

is constant on $X$ (where $k(x)$ is the residue field), then in fact $\mathscr{F}$ is locally free. (This is a homework problem for me.) I think that if $X$ is affine (say $=\operatorname{Spec}A$), then $\mathscr{F}$ will actually be free, not just locally. It will certainly be true that $\mathscr{F}$ will be the sheaf associated with a finitely generated $A$-module $M$, so there will be an exact sequence

$$0\rightarrow K\rightarrow A^m \rightarrow M\rightarrow 0$$

for some $m\in\mathbb{N}$, and because $A$ is noetherian, $K$ will be finitely generated. Thus $M$ is a finitely presented module, say by the matrix

$$\begin{pmatrix} g_{11}&\dots& g_{1r}\\ \vdots &\ddots &\vdots \\ g_{m1}&\dots& g_{mr} \end{pmatrix}$$

Then if $x=\mathfrak{p}\triangleleft A$, $\mathscr{F}_x\otimes_{\mathcal{O}_x}k(x) = M_\mathfrak{p}\otimes_{A_\mathfrak{p}}k(\mathfrak{p}) = M\otimes_A k(\mathfrak{p})$ is precisely the module presented by this matrix except with the entries interpreted as elements of $k(\mathfrak{p})$. The condition that $\varphi(x)$ is constant means that the rank of this matrix (with entries interpreted in $k(\mathfrak{p})$) doesn't depend on $\mathfrak{p}$. (Say it always $=s$.) Then the original matrix (entries interpreted in $A$) has the property that no $\mathfrak{p}\triangleleft A$ contains every $s\times s$ minor; however, every $\mathfrak{p}$ contains every $(s+1)\times (s+1)$ minor. The former condition means that the $s\times s$ minors generate the unit ideal in $A$. Since $A$ is presumed to be a reduced ring, the latter condition means that every $(s+1)\times(s+1)$ minor is zero.

Now $\mathscr{F}$ is free iff $M$ is free. So if it's true that $X$ affine $\Rightarrow$ $\mathscr{F}$ is free (under the conditions of the question), then what I want to show is the plausible-to-me-seeming claim that if a matrix fulfills the conditions just described, then it presents a free module. It seems to me that if this is true it will not depend on the noetherian hypothesis on $A$, since all the relevant ideals and modules are already finitely generated. So -

My question: Let $A$ be a commutative ring with unity. Let $M$ be a finitely presented module over $A$. Say that $M$ is presented by an $m\times r$ matrix with the property that the $s\times s$ minors generate the unit ideal in $A$, but all the $(s+1)\times (s+1)$ minors are zero.

Is it true that in this case $M$ is free, of rank $m-s$?

If so, can you give me a hint toward a proof? (I have been having fun with this problem so far so I'd prefer less than a full solution.)

My work so far: I checked this "by hand" in the simplest nontrivial case, that the matrix presenting $M$ is $2\times 1$ and $s=1$: suppose $a,b\in A$ and $M$ is presented by $\begin{pmatrix} a\\b\end{pmatrix}$. The condition on the $s\times s$ minors means $(a,b)=1$, so $\exists f,g\in A$ with $fa+gb=1$. (The condition on the $(s+1)\times(s+1)$ minors doesn't tell us anything because it's already forced by the shape of the matrix.) Then the $A$-linear map

$$ r\mapsto \begin{pmatrix}gr\\-fr\end{pmatrix}$$

is an isomorphism of $A^1\rightarrow M$. It is injective because $gr=-fr=0$ implies that $r=r(fa+gb)=0+0=0$, and it is surjective because for arbitrary $\begin{pmatrix}x\\y\end{pmatrix}$ in $A^2$, take $r=bx-ay$, and then since $fa=1-gb$ and $gb=1-fa$, we have

$$\begin{pmatrix}x\\y\end{pmatrix} - \begin{pmatrix}gr\\-fr\end{pmatrix} = \begin{pmatrix} x-g(bx-ay)\\ y+f(bx-ay)\end{pmatrix}=\begin{pmatrix}(1-gb)x+gay\\ (1-fa)y+fbx\end{pmatrix}=(fx+gy)\begin{pmatrix}a\\b\end{pmatrix}$$

which represents zero in $M$.

I started to look at the next simplest case I could think of: I took $A=k[a,b,c,d]/(a+b+c+d-1,ad-bc)$ for $k$ some field, and was thinking about the module presented by the matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}$. I would like to show this matrix is isomorphic to $A^1$. I still haven't thought about minors bigger than $1\times 1$ generating the unit ideal, so I realized in the interest of time I should ask for help.

A last thought: Suddenly while writing this I realized that the result I want is strongly reminiscent of the Quillen-Suslin theorem, as described in Michael Artin's Algebra. This makes me suspicious that the result I want won't be true in full generality. Any thoughts on the relation of my question to the Quillen-Suslin theorem would be appreciated as well.

share|improve this question
    
+1 for the lucidity of your question and for presenting your pleasantly detailed and explicit "work so far". –  Georges Elencwajg Mar 31 '13 at 10:27

2 Answers 2

up vote 3 down vote accepted

No, it is not true that locally free implies free on an affine scheme. Here are some details.

Consider a noetherian ring $A$ and the corresponding affine scheme $X=\text {Spec(A)}$.
A finitely generated module $M$ over $A$ corresponds to a coherent sheaf $\tilde M=\mathcal F$ on $X$.
We have the equivalences $$M \;\text {is projective}\iff \mathcal F \;\text {is locally free}\quad (I)$$ and $$M \;\text {is free}\iff \mathcal F \;\text {is free} \quad (II)$$

However the conditions $(II)$ are much stronger than $(I)$:

If the rank of $M$ (or equivalently of $\mathcal F$) is one, for example, the corresponding projective modules constitute the Picard group $\text {Pic}(A)$, denoted $\text {Pic}(X)$ in the geometric setting.
So every non-zero element of $\text {Pic}(A)$ yields a non-free locally free coherent sheaf of rank one.
But do there exist rings with $\text {Pic} (A)\neq 0$ ?
Sure: I could give you examples but I prefer to impress you by stating Claborn's unbelievable (but true!) result:

Given an arbitrary abelian group $G$, there exists a Dedekind domain $A$ with $\text {Pic}(A)=G.$

Edit on your "last thought"
Indeed Quillen-Suslin is very relevant: they proved that on $\mathbb A^n_k$ ($k$ a field) , corresponding to $A=k[T_1,\ldots, t_n]$, $(I)$ and $(II)$ are equivalent: every locally free sheaf on affine space is free .
This is analogous to the result that every topological vector bundle is trivial on $\mathbb R^n$, but much more difficult.
(And as a parenthetical remark, it is a crying shame that Suslin, one of the greatest living algebraists/algebraic geometers never got a Fields medal.)

share|improve this answer

Even for $A$ noetherian is not true that $M$ is free.

Proposition 1.4.10 from Bruns and Herzog, Cohen-Macaulay Rings, says the following:

Let $A$ be a noetherian ring and $M$ a finitely generated $A$-module with a finite free presentation $F_1\stackrel{\varphi}\to F_0\to M\to 0$. TFAE:

(i) $I_s(\varphi)=R$ and $I_{s+1}(\varphi)=0$;

(ii) $M$ is projective and $\operatorname{rank}M=\operatorname{rank}F_0-s$. (Here $I_t(\varphi)$ denotes the ideal generated by the $t\times t$ minors of the matrix $\varphi$.)

This shows that in your case $M$ is projective of finite rank $m-s$.

share|improve this answer

Your Answer

 
discard

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.