User mattia talpo - MathOverflow

Coherence of the monoid algebra of a non-finitely generated monoid

2011-11-30T21:39:25Z

Let $P$ be an integral, sharp, finitely generated commutative monoid (say even torsion-free and saturated if you like), and consider the "rational cone" $P_\mathbb{Q}\subseteq P^{gp}\otimes_\mathbb{Z} \mathbb{Q}$ generated by $P$ (i.e. the submonoid of elements of the form $\frac{m}{n} p$ where $m, n \in \mathbb{N}$ and $p\in P$).

Consider the monoid algebra $R=\mathbb{Z}[P_\mathbb{Q}]$. I'm interest in the answer to the

Question:

is $R$ coherent?

By coherent here I mean that every finitely generated ideal $I\subseteq R$ is finitely presented, i.e. if $I=\langle r_1,\dots, r_k \rangle$, then the kernel of the obvious map $R^k\to R$ with image $I$ is finitely generated.

Notice that in general (and I think almost always) $R$ will not be noetherian. For example if $P=\mathbb{N}$, then $R=\mathbb{Z}[\mathbb{Q}_+]=\mathbb{Z}[t^q | q>0]$ with the obvious relations $t^qt^{q'}=t^{q+q'}$ for $q,q' \in \mathbb{Q}_+$, and this is clearly not noetherian.

To elaborate a bit, consider $\frac{1}{n} P \subseteq P_\mathbb{Q}$ and the rings $A_n=\mathbb{Z}\left[\frac{1}{n} P\right]$, with in particular $A_1=\mathbb{Z}[P]$. There are obvious maps $f_{n,m}: A_n\to A_m$ when $n | m$, induced by the inclusions $\frac{1}{n} P\subseteq \frac{1}{m} P$.

$\{A_n,f_{n,m}\}$ form a directed system of rings, and the inductive limit is $R$.

One case in which I know that $R$ is coherent is when the transition maps $f_{n,m}$ are flat (this happens for example for $P=\mathbb{N}^r$).

Notice also that $R$ depends only on $P_\mathbb{Q}$, so for example if there is another submonoid $P'\subseteq P_\mathbb{Q}$ with the same properties as $P$ and such that $P'_\mathbb{Q}=P_\mathbb{Q}$ , and the transition maps for $P'$ are flat, then $R$ is coherent.
This is what happens for the monoid $P=\langle (2,0), (1,1), (0,2) \rangle_\mathbb{N} \subseteq \mathbb{Z}^2$ (for which the transition maps are not flat), and in this case $P'=\langle (2,0),(0,2)\rangle\subseteq P$ does the trick.

An example in which something like this should not happen is $P=\mathbb{N}^4/(e_1+e_2=e_3+e_4)$, and so $A_1=\mathbb{Z}[x,y,z,w]/(xy-zw)$.

In general, a sufficient condition seems to be the following: for every $n$ and every ideal $I\subseteq A_n$ there exists an $m$ such that $n|m$ and $$ Tor_1^{A_m}(I_m,R)=\varinjlim Tor_1^{A_m}(I_m,A_k)=0 $$ where $I_m=IA_m\subseteq A_m$ is the extension of $I$.

Question:

is this true/plausible?

Thanks in advance for any comment!

Motivation:

All of this comes from root stacks of logarithmic schemes. If the answer to the first question is affirmative, I would be happy because then an "infinity root stack" will have coherent structure sheaf, coherent sheaves on it will be exactly those which are locally of finite presentation, and those have good "approximation" properties w.r.t. inverse limits.
[the case of the question is the "universal" one, in some sense]

Answer by Mattia Talpo for Deformation of Line bundles over dual numbers

2011-11-08T17:41:02Z

One important point is that $X$ and $X'$ have the same topological space, since $i:X'\rightarrow X$ is a nilpotent immersion. In particular $M$ is already a sheaf on $|X|=|X'|$ (of $\mathcal{O}_X$-modules). Since $\mathcal{O}_{X'}=\mathcal{O}_X\oplus \mathcal{O}_X[\varepsilon]$ , to give $M$ the structure of a sheaf of $\mathcal{O}_{X'}$ -modules you only need to define the action of $\varepsilon$, as Mike Skirvin points out in his comment.

So you define multiplication by $\varepsilon$ by the composition $M\stackrel{g}{\to} L\stackrel{f}{\to} M$ of the two maps given by the extension (notice that this is $\mathcal{O}_X$-linear, and its square is zero). This makes $M$ a sheaf of $\mathcal{O}_{X'}$ -modules, and moreover using the local criterion of flatness you can check that it is flat over the base $D$.

Now since $|X|=|X'|$ and $\mathcal{O}_X=\mathcal{O}_{X'}/(\varepsilon\cdot\mathcal{O}_{X'})$ , you have $i^*(M)=M\otimes_{\mathcal{O}_X'}\mathcal{O}_X=M/(\varepsilon \cdot M)=M/f(L)=L$, since by definition of the action of $\varepsilon$, $\varepsilon \cdot M=f(L)$.

Answer by Mattia Talpo for Does Hom(X,Spec R) = Hom(R, O(X)) hold for algebraic stacks?

2011-10-31T14:56:02Z

I think this is true.

Take a smooth atlas $U\to X$, and notice that sections of $\mathcal{O}_X$ (which you can see as morphisms of quasi-coherent sheaves $\mathcal{O}_X \to \mathcal{O}_X$) correspond to sections of $\mathcal{O}_U$, such that the two restrictions to $U_1:=U\times_X U$ by means of the two projections to $U$ coincide.

This is also true of morphisms out of $X$: a morphism $X\to T$ where $T$ is a scheme corresponds to a morphism $U\to T$ such that the two compositions $U_1\to U\to T$ coincide (basically because $Hom(-,Spec R)$ is a sheaf, as you said).

This reduces the question to the fact that morphisms $U\to Spec R$ such that the two compositions $U_1\to Spec R$ coincide correspond to morphisms $R\to \mathcal{O}_U(U)$ such that the two compositions $R\to \mathcal{O}_{U_1}(U_1)$ coincide.

Answer by Mattia Talpo for universal categorical quotient

2011-10-27T09:05:42Z

I assume that by "universal categorical quotient" you mean a morphism $f\colon X\to Y$ that is a categorical quotient, and stays so after any base change $Y'\to Y$.

Then the implication (1)$\Rightarrow$(2) is obvious, and for the other, take any $G$-invariant morphism $g\colon X'\to Z$, and an affine covering $U_i$ of $Y$. Call $f'\colon X'\to Y'$ the base change of $f$.

For any $i$ the map $f'^{-1}(U_i)\to U_i$ will be a categorical quotient by (2) since it is an affine base-change of $f$, and so the restriction $g|_{f'^{-1}(U_i)} \colon f'^{-1} (U_i) \to Z$, being $G$-invariant, will factor through $U_i$. Since this factorization is canonical, those with different $i$'s will be compatible in the intersections (cover it with affines), and in the end you get a (unique) factorization $X'\to Y'\to Z$ for $g$, showing that $f'$ is a categorical quotient.

Answer by Mattia Talpo for Twisted forms and $\check{H}^1$

2011-04-21T15:14:09Z

If your $Y$ is an object over $X$ of a stack in the étale topology, then you can use the cocycles as descent data to get a twisted form (because by definition of a stack all descent data are effective).

Some examples other than the ones you mentioned are quasi-coherent sheaves and affine morphisms of schemes (i.e. quasi-coherent sheaves of algebras).

This is far from a necessary condition on $Y$, I don't know if there is a sharper characterization.

Answer by Mattia Talpo for Immerse an affine schemes into $A^n_S$

2011-02-15T00:10:27Z

If by "immersion" you mean closed immersion (and I think you do), then the answer is no, unless the morphism $X\to S$ is affine: a closed immersion is affine, and $\mathbb{A}^n_S\to S$ is also affine, so if your immersion exists, then the composite $f:X\to S$ is affine.

On the other hand if $f$ is affine, then $f_*\mathcal{O}_X$ has a surjective morphism from some $\mathcal{O}_S[x_1,\dots,x_n]$ (as sheaves of $\mathcal{O}_S$-algebras), and the induced map between the relative spectra gives your closed immersion.

(notice that the fact that $X$ is an affine scheme is irrelevant)

Answer by Mattia Talpo for algebraic local charts

2011-01-27T19:31:44Z

What you're looking for is the étale topology (I think).

The fact is that Zariski opens are way too big to grasp stuff which should be more local than "Zariski local", for example any two smooth points on two $n$-dimensional varieties over $\mathbb{C}$ have isomorphic completed local rings (power series ring in $n$ indeterminates), but you will never find two isomorphic zariski open neighborhoods, unless the varieties are birational!

The way to go more locally is to consider étale morphisms to a scheme (which should morally be locally invertible, but they aren't in the algebraic context) to be "open subsets" of that scheme. Of course this doesn't make sense unless you modify the notion of topology, and in fact this is what led to the notion of Grothendieck topologies.

A good introduction to this stuff is Milne's book "étale cohomology", first chapters (I hope this is what you were looking for).

Answer by Mattia Talpo for Quasi-coherent of finite presentation: Redundant?

2010-09-21T16:22:29Z

Well, a finitely presented sheaf of $\mathcal{O}_X$-modules is a cokernel of a map between quasi-coherent $\mathcal{O}_X$-modules, so it is itself quasi-coherent. So yes, the "quasi-coherent" is redundant in your sentence.

Answer by Mattia Talpo for Intuition for the satellite of a functor

2010-06-28T20:27:52Z

I don't know if this can be considered as "intuition", anyway

another way to think about derived functors is the following: given an abelian category $\mathcal{A}$, you can define its derived category $\mathcal{D}(\mathcal{A})$ (or its variants of bounded complexes in one or both directions). You get $\mathcal{D}(\mathcal{A})$ by "localizing" the homotopy category of complexes $\mathcal{K}(\mathcal{A})$, where the objects are complexes in $\mathcal{A}$ and maps are maps of complexes up to homotopy, at the system of quasi-isomorphisms.

There is a natural localization functor $\pi_A:\mathcal{K}(\mathcal{A})\to \mathcal{D}(\mathcal{A})$. If $F:\mathcal{A}\to \mathcal{B}$ is your additive functor between abelian categories, and $\mathcal{K}(F):\mathcal{K}(\mathcal{A})\to \mathcal{K}(\mathcal{B})$ is the induced functor, it is natural to ask for an "extension" of $\mathcal{K}(F)$ to the derived category $\mathcal{D}(\mathcal{A})$, with values in $\mathcal{D}(\mathcal{B})$. In other words this would be a functor $RF:\mathcal{D}(\mathcal{A})\to \mathcal{D}(\mathcal{B})$, such that $\pi_B \circ \mathcal{K}(F) = RF \circ \pi_A$.

This is not possible to find in general, and the problem is that $\mathcal{K}(F)$ may not send quasi-isomorphisms into quasi-isomorphisms. The best you can ask for is a functor $RF:\mathcal{D}(\mathcal{A})\to \mathcal{D}(\mathcal{B})$, with a natural transformation $\eta:\pi_B \circ \mathcal{K}(F) \to RF\circ \pi_A$ having a universal property among such functors and natural transformations (this is a particular case of a Kan extension).

Such an $RF$ (unique up to isomorphism) is called the (total) right derived functor of $F$. One way to think about it is as "the" functor between the derived categories which approximates $\mathcal{K}(F)$ in the best possible way. You can recover the single derived functors $R^iF$ by taking the cohomology objects of $RF$. In most cases the derived functor is constructed by using resolutions, by injective (or projective, if you're defining left derived functors) objects, or more generally by suitable subcategories of $\mathcal{A}$.

In your case, if you also assume that $F$ is right exact, then the satellite functors coincide with the (left) derived ones, and so I guess that it follows that they can be calculated by the formulas you wrote.

The idea of the "best approximation" of $\mathcal{K}(F)$ on the derived category seems very natural to me, and a satisfactory answer to the question "why derived functors". If you were asking specifically about satellites, then I don't know.

Answer by Mattia Talpo for How do I describe the GL_n torsor attached to a smooth morphism of relative dimension n?

2010-06-09T08:31:51Z

I don't know about jets and you already got an answer regarding the bundle of automorphisms, anyway if you want a $GL_{n,Y}$-torsor over $Y$ that gives you back your original vector bundle $Z=V(\mathcal{E})=Spec_Y Sym(\mathcal{E})\to Y$ when you apply the associate bundle construction with $\mathcal{O}_Y^n$, you should take the bundle of local frames of $Z$, that is $P=\underline{Isom}_Y(\mathbb{A}^n_Y,Z)\to Y$, where $\underline{Isom}$ is the scheme representing the sheaf of isomorphisms. This is a $GL_{n,Y}$-torsor over $Y$ by the action of $GL_{n,Y}$ on $\mathbb{A}^n_Y$, and if you want a sheaf $\mathcal{F}$ of $\mathcal{O}_Y$-algebras such that $P=Spec_Y(\mathcal{F})$, it seems reasonable (but i didn't really check) that you can take $\mathcal{F}=\underline{Isom}_{\mathcal{O}_Y-\text{alg}}(Sym(\mathcal{E}),\mathcal{O}_Y^n[x_1,..,x_n])$.

Answer by Mattia Talpo for Can all induced maps be described categorically.?. (or at least as generally as possible)

2010-05-18T08:32:16Z

Let me add that the answer to your question "when does a map between spaces induce a map in homology/cohomology/homotopy?" is "always" (as long as you stick to continuous maps..).

In fact (say for homology) if you have a continuous $f:X\to Y$, the induced homomorphism of chain complexes $C_\bullet(f):C_\bullet(X)\to C_\bullet(Y)$ sends automatically cycles to cycles and boundaries to boundaries (simply because it is compatible with the differentials of the two complexes, in the sense that $f\circ d_X=d_Y\circ f$).

The fact that it "passes" to homology is now an algebraic fact, namely the fact that if you have four abelian groups $A,B,C,D$ and three homomorphisms $f:A\to B, g:A\to C, h:B\to D$ with $g$ and $h$ surjective (so $C$ is a quotient of $A$ and $D$ is a quotient of $B$), then you can find $f':C\to D$ such that $f'\circ g=h\circ f$ if and only if $f(\ker(g))\subseteq \ker(h)$. (you might want to draw a diagram here :D)

Take $A=Z_n(X), B=Z_n(Y), C=H_n(X)=Z_n(X)/B_n(X), D=H_n(Y)=Z_n(Y)/B_n(Y)$, where $Z_\bullet=$cycles and $B_\bullet=$ boundaries, as usual, and you get your induced map in homology.

Comment by Mattia Talpo

2013-02-06T22:53:12Z

It seems to me that the question was what happens when $Z$ (rather than $X$) is an Artin stack..

Comment by Mattia Talpo

2013-02-02T14:21:44Z

This should be relevant: mathoverflow.net/questions/70520/…

Comment by Mattia Talpo

2012-10-24T16:12:25Z

there's a reason why it's not "geometric geometry" ;)

Comment by Mattia Talpo

2012-01-15T14:15:59Z

Oh ok, I was thinking of the support as a subscheme, not as just a closed subset. Now it makes sense :)

Comment by Mattia Talpo

2012-01-15T12:49:57Z

I'm obviously missing something: why can't you just take the filtration $0\subseteq F$, since if $F$ is supported on $S$ then it is of the form $e_*G$? (of course this doesn't make sense if you want the graded pieces to be locally free, but without this requirement I don't see what's the deal with the filtration..)

Comment by Mattia Talpo

2012-01-09T15:14:49Z

@Martin: ok thanks, I know about the construction you're mentioning, I was just wondering about the symbol. Is there some reason why it should be analogue to an integral?

Comment by Mattia Talpo

2012-01-09T14:04:24Z

Silly off-topic question: why the integral? :S

Comment by Mattia Talpo

2011-12-01T00:46:06Z

"integral" means that the map $P\to P^{gp}$ is injective, and "sharp" that the only invertible element of $P$ is the identity. While we're at it, by "torsion-free" I mean that $P^{gp}$ is torsion-free.

Comment by Mattia Talpo

2011-11-10T10:08:32Z

You're welcome! ;)

Comment by Mattia Talpo

2011-11-09T13:00:18Z

Once you know that it is locally free, it must be of rank one, since its pullback to $X$ is..

Comment by Mattia Talpo

2011-11-08T20:27:28Z

I think so, if say $X$ is loc noetherian. Take an open affine $U=Spec A \subseteq X$ (so the corresponding open of $X'$ is the spectrum of $A'=A[\varepsilon]$) and apply the local criterion of flatness for the nilpotent ideal $(\varepsilon)\subseteq A[\varepsilon]$ to the restriction of $M$ to $U$, call $N$ the corresponding $A'$-module. Namely, since $N/(\varepsilon N)$ is flat over $A$ (loc free even) and $Tor_1^{A'}(N,A)=0$ (check that $\cdot \varepsilon: A\to A'$ stays injective after tensoring with $N$), then $N$ is flat over $A'$, so it is locally free. (maybe there's an easier way..)

Comment by Mattia Talpo

2011-10-31T14:38:37Z

I think that the "category" of morphisms $X\to \mathbb{A}^1$ is literally a set, meaning that it's a category with only identity arrows: natural transformations don't do much, since $\mathbb{A}^1$ is a category fibered in sets..

Comment by Mattia Talpo

2011-10-27T09:34:04Z

Seems you need at least 50 reputation to leave comments.. don't worry anyway! ;)

Comment by Mattia Talpo

2011-10-23T11:53:32Z

algebraic geometry?

Comment by Mattia Talpo

2011-10-07T15:01:23Z

ehm, I have to agree with a-fortiori and Leo Alonso, his example works perfectly fine too.. thanks anyway!