User martin brandenburg - MathOverflow

Analogy between the exterior power and the power set

2013-04-13T16:59:46Z

The symmetric algebra of an object exists in every cocomplete $\otimes$-category. For the category of sets $\mathrm{Sym}(X)$ is the set of multi-subsets of $X$.

The usual definition of the exterior power works in every cocomplete linear $\otimes$-category in which $2$ is invertible. But what about the non-linear case? Are there also "exterior powers" in $\otimes$-categories which are not linear? Of course the usual definition using alternating maps does not work. But isn't it striking that for the cartesian category of sets there is a quite natural candidate, namely the power set? Here are some analogies (here $P(X)$ denotes the power set of $X$ if $X$ is finite; in general it is the set of all finite subsets of $X$; $P_n(X)$ is the set of all subsets of $X$ with $n$ elements):

$P(X) = \coprod_n P_n(X)$ and $\Lambda(M) = \oplus_n \Lambda^n(M)$
$P(X \sqcup Y) = P(X) \times P(Y)$ and $\Lambda(M \oplus N) = \Lambda(M) \otimes \Lambda(N)$

It follows the "categorified Vandermonde identity":

$P_n(X \sqcup Y) = \coprod_{p+q=n} P_q(X) \times P_q(Y)$ and $\Lambda^n(M \oplus N) = \oplus_{p+q=n} \Lambda^p(M) \otimes \Lambda^q(N)$
$(P(X),\cup)$ is a commutative monoid and $(\Lambda(M),\wedge)$ is a graded-commutative algebra, i.e. commutative monoid object in the tensor category of graded modules equipped with with twisted symmetry
If $M$ is free with (ordered) basis $X$, then $\Lambda(M)$ is free with basis $P(X)$, and $\Lambda^n(M)$ is free with basis $P_n(X)$. In particular, $\dim \Lambda^n(M)=\dim P_n(X)$.
If $T$ is a commutative monoid, then homomorphisms $P(X) \to T$ correspond to maps $f : X \to T$ with $f(x)^2=f(x)$, and if $A$ is a graded-commutative algebra, then homomorphisms $\Lambda(M) \to A$ correspond to homomorphisms of modules $f : M \to A_1$ with $f(x)^2=0$ or rather $f(x)f(y)+f(y)f(x)=0$ in the context of $\otimes$-categories (so these conditions are not the same, but both use $f(x)^2$).

Therefore I would like to ask: Is there a notion of exterior algebra for certain cocomplete $\otimes$-categories, including categories of modules and the category of sets? In the latter case, do we get the power set?

Answer by Martin Brandenburg for Applications of Govorov-Lazard Theorem?

2013-05-05T08:08:22Z

I cannot offer any deep theorem, but I stumbled upon the following basic question: Let $C$ be a AB5 abelian $R$-linear tensor category. There is a notion of flat object in $C$. Since directed colimits and coproducts of monos are mono, one can see that flat objects are closed under directed colimits as well as coproducts. Now, there is a cocontinous tensor functor $\mathsf{Mod}(R) \to C, M \mapsto M \otimes_R 1$. Does it preserve flat objects? In fact, this follows immediately from Lazard's Theorem.

Let's try to prove this directly: If $X \to Y$ is a monomorphism in $C$, we have to prove that $M \otimes_R X \to M \otimes_R Y$ is a monomorphism in $C$. But how can we use flatness when $X,Y$ are no $R$-modules? Besides, even if $X,Y$ are $R$-modules, this is not clear at all, since $\mathsf{Mod}(R) \to C$ doesn't preserve monos (it only preserves colimits, in particular epis). I don't think that there is a direct proof.

Remark that Lazard's Theorem transforms a statement of the form "for all modules ... is exact" into a statement of the form "there exists a direct system of modules such that ... is an isomorphism". The latter is easier to check and invariant under many functors, whereas the first would require the functor to be essentially surjective or something like that in order to lift the test module.

Families of local rings coming from a locally ringed space

2013-04-26T09:54:37Z

Let $X$ be a topological space. For every point $x \in X$ let $R_x$ be a local ring. Under what (necessary / sufficient / necessary and sufficient) conditions is there a sheaf ${\cal O}_X$ such that $(X,\mathcal{O}_X)$ is a locally ringed space with $\mathcal{O}_{X,x} \cong R_x$ for all $x \in X$?

If $X$ is discrete, we need no conditions, and we can take the sheaf $\mathcal{O}_X(U)=\prod_{x \in U} R_x$ . But in general, a necessary condition is that $x \prec y$ gives a ring homomorphism $R_x \to R_y$, and that these are compatible in the sense that $x \mapsto R_x$ extends to a functor from the specialization preorder of $X$ to the category of rings. We will also have $\mathcal{O}_X(U) \subseteq \varprojlim_{x \in U} R_x$ , but probably no equality.

Is Dependent Choice equivalent to the statement that every PID is factorial?

2010-09-24T16:07:22Z

In this question, it was asked if AC is needed in the proof of the well-known fact that every principal ideal domain is factorial. As KConrad and Joel David Hamkins have pointed out, only DC, the axiom of dependent choice, is needed, and in the comment box the question was raised if the statement is actually equivalent to DC. But this was not answered so far.

So I ask it here, is the statement equivalent to DC? It would not surprise me since there are already many algebraic equivalents of AC (existence of vector bases, existence of maximal ideals). Also note that the requirement of ideals being principal is somewhat analoguous to the assumption of entireness of the relation in DC, and the representation as a product of irreducibles is somewhat analoguous to the conclusion of DC.

If the statements turns out to be weaker, I would be also interested in a proof which is the set-theoretical equivalent (perhaps it's DC-fin as in Joel's answer?).

PS: Of course, this question takes place in ZF ;).

Answer by Martin Brandenburg for Using the Yoneda embedding to talk about exactness in an additive category

2013-01-23T18:52:20Z

Here is an example: Let $A \to B \to C \to 0$ be a sequence in $\mathcal{C}$. Then it becomes exact in $\widehat{\mathcal{C}^{\mathrm{op}}}^{\mathrm{op}}$ (where $\widehat{\mathcal{C}} := [\mathcal{C}^{\mathrm{op}},\mathsf{Ab}]$) after applying $\mathcal{C} \hookrightarrow \widehat{\mathcal{C}^{\mathrm{op}}}^{\mathrm{op}}, X \mapsto \hom(X,-)$ if and only if $B \to C$ is a cokernel of $A \to B$ (immediate). On the other hand, it becomes exact in $\widehat{\mathcal{C}}$ after applying $\mathcal{C} \hookrightarrow \widehat{\mathcal{C}}, X \mapsto \hom(-,X)$ if and only if it is isomorphic to a sequence of the form $F \oplus E \xrightarrow{p} C \oplus E \xrightarrow{q} C \to 0$, where $E,F,C \in \mathcal{C}$, $q$ is the projection and $p$ is the projection $F \oplus E \to E$ followed by the inclusion $E \to C \oplus E$ (exercise). Thus we get a split exact complex. Therefore, exactness in $\widehat{\mathcal{C}}$ applies the one in $\widehat{\mathcal{C}^{\mathrm{op}}}^{\mathrm{op}}$, but not vice versa.

Similarily, $B \to C \to 0$ becomes exact in $\widehat{\mathcal{C}^{\mathrm{op}}}^{\mathrm{op}}$ if and only if $B \to C$ is an epimorphism, and it becomes exact in $\widehat{\mathcal{C}}$ if and only if $B \to C$ is a split epimorphism.

By duality, $0 \to A \to B \to C$ becomes exact in $\widehat{\mathcal{C}}$ if and only if $A \to B$ is a kernel of $B \to C$, and it becomes exact in $\widehat{\mathcal{C}^{\mathrm{op}}}^{\mathrm{op}}$ if and only if it is a split exact complex.

Combining these results, we see that $0 \to A \to B \to C \to 0$ becomes exact in $\widehat{\mathcal{C}^{\mathrm{op}}}^{\mathrm{op}}$ if and only if it becomes exact in $\widehat{\mathcal{C}}$ if and only if it is a split exact complex.

Even for abelian $\mathcal{C}$, this is not the "correct" notion of exactness. I doubt that there is anything interesting which can be said about sequences of the form $A \to B \to C$.

Edit: For abelian $\mathcal{C}$ the sequence $A \to B \to C$ becomes exact in $\widehat{\mathcal{C}^{\mathrm{op}}}^{\mathrm{op}}$ iff $A \to B \to C$ is exact and the image of $B \to C$ is a direct summand of $C$. See mathunderflow 360234.

Answer by Martin Brandenburg for Separable and algebraic closures?

2010-06-07T06:34:01Z

First of all, there is not the algebraic/separable closure. Choices have to be made. However, if an algebraic closure $k^{\mathrm{alg}}$ of $k$ is fixed, inside it there is a unique separable closure $k^{\mathrm{sep}}$ of $k$, namely the subfield consisting of the separable elements over $k$.

Ignoring the failure of uniqueness, you can regard $k^{\mathrm{alg}}$ as the biggest algebraic extension of $k$, whereas $k^{\mathrm{sep}}$ is the biggest galois extension of $k$. The latter is because $k^{\mathrm{sep}}$ is easily seen to be normal. In particular, you can apply Galois theory and relate the group theory of the absolute Galois group $\mathrm{Gal}(k^{\mathrm{sep}}/k)$ with the field theory of Galois extensions of $k$. The algebraic closure is too big to make Galois theory work.

Obviously $k$ is perfect if and only if $k^{\mathrm{alg}} = k^{\mathrm{sep}}$. Finite fields and fields of characteristic $0$ (in particular number fields) are perfect. But what is the difference in the other cases? Let $p = \mathrm{char}(k) > 0$. Then $k^{\mathrm{alg}} / k^{\mathrm{sep}}$ is purely inseparable, i.e. for every $a \in k^{\mathrm{alg}}$ there is some $n \geq 1$ such that $a^{p^n} \in k^{\mathrm{sep}}$. In other words, this field extension is given by adjoining all $p^n$-th roots. A consequence of this is that the restriction map $\mathrm{Aut}_k(\overline{k}) \to \mathrm{Gal}(k^{\mathrm{sep}}/k)$ is an isomorphism.

Actually one can show that the canonical map $k^{\mathrm{sep}} \otimes_k k^{\mathrm{perf}} \to k^{\mathrm{alg}}$ is an isomorphism, where $k^{\mathrm{perf}}=\cup_{n \geq 0} k^{1/p^n}$ is the perfect hull of $k$.

Answer by Martin Brandenburg for Fixed point theorems

2013-04-10T13:02:10Z

The Arithmetic fixed point theorem (see also MO/30874) states that if $F$ is a formula in number theory with only one free variable $v$, then there is a sentence $A$ such that number theory can prove $A \Leftrightarrow F_v(\underline{[A]})$. An immediate application is Gödel's Theorem.

IBN for algebraic theories

2013-04-07T01:38:00Z

Let us say that a finitary algebraic theory $\tau$ has IBN (invariant basis number) if the free functor $F : \mathsf{Set} \to \mathsf{Mod}(\tau)$ reflects the isomorphism relation: If $S,T$ are sets with $F(S) \cong F(T)$, then $S \cong T$. When $\tau$ is the theory of $R$-modules for some ring $R$, then this is the usual IBN property of $R$ (at least when we restrict to finite sets).

If $\tau \to \sigma$ is a homomorphism and $\sigma$ satisfies IBN, then also $\tau$ satisfies IBN. Besides the classical example of vector spaces, this gives lots of examples for IBN theories (abelian groups, modules over commutative rings $\neq 0$, groups and Lie algebras (using abelianization), monoids, semigroups, quasigroups, loops, magmas, commutative variants of them, etc.). One can show IBN for (commutative) $R$-algebras, where $R \neq 0$ is a commutative ring. Benjamin Steinberg has remarked in the comments that $\tau$ has IBN when there is a $\tau$-module with finite cardinality $>1$. This gives lots of further examples.

A directed colimit of IBN theories $\tau = \mathrm{colim}_i \tau_i$ is IBN for finite sets and therefore IBN by E: If $F_\tau(S) \to F_\tau(T)$ is a homomorphism, it is given by a map $S \to F(T)$, which factors through some $S \to F_{\tau_i}(T)$. Similarily the other way round. That $F_\tau(S) \to F_\tau(T) \to F_\tau(S)$ is the identity, already holds for the factorizations if $i$ is big enough. Therefore it suffices to consider finitely presented theories.

Now I have several questions:

A. Has the IBN property for algebraic theories in general been studied in the literature?

B. What are further interesting examples of IBN or $\neg$ IBN (beyond module categories)?

C. What about the infinitary theory of compact Hausdorff spaces? If $X,Y$ are sets such that their Stone-Čech compactifications $\beta(X),\beta(Y)$ are homeomorphic, does it follow $X \cong Y$? (answered by Benjamin Steinberg: Yes)

D. Do nontrivial commutative algebraic theories satisfy IBN? In other words, is the rank of a free module over a nontrivial generalized ring à la Durov well-defined? This should be crucial for the theory of generalized schemes, right?

E. Is there some algebraic theory which satisfies IBN for finite sets, but not IBN for arbitrary sets? (answered by Joseph Van Name: No).

F. When $|S| < \kappa$, then $F(S)$ is $\kappa$-presentable. Is there some $\tau$ which satisfies IBN, but $F(S)$ is $\kappa$-presentable for some $\kappa < |S|$?

Answer by Martin Brandenburg for Are the projection morphisms from a product of varieties necessarily open?

2013-04-07T01:54:12Z

Yes. More generally, flat morphisms locally of finite presentation are universally open (EGA IV₂, Théorème 2.4.6).

Profinite groups as étale fundamental groups

2012-12-01T09:23:06Z

Does every profinite group arise as the étale fundamental group of a connected scheme?

Equivalently, does every Galois category arise as the category of finite étale covers of a connected scheme?

Not every profinite group is an absolute galois group of a field (the only finite ones have order $1$ or $2$ by Artin-Schreier). Therefore we cannot restrict to spectra of fields. Perhaps one first has to check if every finite group arises as a fundamental group of a scheme. I don't even know enough examples to answer this question for cyclic groups. At least order $3$ is possible (see here, Remark 2).

If the answer turns out to be no, then I would like to know which profinite groups arise as fundamental groups.

Answer by Martin Brandenburg for Stone–Čech Compactification of $\mathbb{Z}$ with Fürstenberg Topology

2013-03-27T15:33:58Z

Edit. The comments and the other answers reveal that my proof has some gap. But I won't delete it, instead I've rewritten it as an attempt to prove $\beta \mathbb{Z} = \hat{\mathbb{Z}}$. I hope that the failure of this naive proof motivates to read the more sophisticated answers.

When $\mathbb{Z}$ is equipped with the Fürstenberg topology, do we have $\beta \mathbb{Z} = \hat{\mathbb{Z}}$?

The Fürstenberg topology is the subspace topology induced by the profinite completion $\hat{\mathbb{Z}} = \lim_{n>0} \mathbb{Z}/n$. The embedding $\mathbb{Z} \to \hat{\mathbb{Z}}$ is dense, hence for every compact Hausdorff space $X$ we get an injective map $\hom(\hat{\mathbb{Z}},X) \to \hom(\mathbb{Z},X)$. The question is whether it is surjective, because this would mean that $ \hat{\mathbb{Z}}$ satisfies the defining universal property of $\beta \mathbb{Z}$.

Let $f : \mathbb{Z} \to X$ be a continuous map. This means that for every $a \in \mathbb{Z}$, every open subset $U \subseteq X$ containing $f(a)$ already contains $f(a+n \mathbb{Z})$ for some $n>0$. Let $a=(a_1,a_2,\dotsc) \in \hat{\mathbb{Z}}$, i.e. $a_n \equiv a_m \bmod n$ for $n|m$. Since $X$ is compact, the net $(f(a_n))_{n>0}$ (using divisibility for the indices) has a convergent subnet, say $(f(a_{n(i)}))_{i \in I} \longrightarrow x$ .

Actually any two subnets have the same limit: Assume that $(f(a_{m(j)}))_{j \in J} \longrightarrow y$ . Choose open neighborhoods $U,V$ of $x,y$, it is enough to prove $U \cap V \neq \emptyset$ since $X$ is Hausdorff. For large $i$ we have that $f(a_{n(i)}) \in U$, and for large $j$ we have $f(a_{m(j)}) \in V$. Choose $b>0$ with $f(a_{n(i)} + b \mathbb{Z}) \subseteq U$ and $f(a_{m(j)} + b \mathbb{Z}) \subseteq V$. We may assume $n(i),m(j)|b$. For $p=n(i) m(j)$ we have $a_p \equiv a_{n(i)} \bmod n(i)$, hence $a_p \equiv a_{n(i)} \bmod b$. Similarily we get $a_p \equiv a_{m(j)} \bmod b$. Hence $f(a_p) \in U \cap V$.

Hence $\tilde{f}(a) := $(the limit of some subnet of $f(a_n)$) gives a well-defined map $\hat{\mathbb{Z}} \to X$. Clearly it agrees with $f$ on constant sequences. But now the problem seems to be that $\tilde{f}$ is not continuous ...

Answer by Martin Brandenburg for Self-containing structures

2013-03-24T03:22:54Z

What about fractals? Or more general, objects with some kind of self similarity? They often arise as fixed points of a self-map. In category theory these are known as initial algebras or terminal coalgebras. For example, $\mathbb{N}=1+\mathbb{N}$, $[0,1] = [0,1] {\cup}_{0 \sim 1} [0,1]$, and the set of binary trees $T$ satisfies $T=1+T^2$. Another example which comes to my mind: There are abelian groups $A$ with $A^3 \cong A$ and $A^2 \not\cong A$. I have to admit that this does not quite fit to your question, since you want that "$X \in X$", but in the above examples we have "$X \subsetneq X$". So let me add something else:

The category of small categories, functors and morphism of functors $\mathsf{Cat}$ is a category (which is not small).
The class of ordinal numbers $\mathrm{On}$ is with $\in$ the well-order of all isomorphism classes of small well-orders. Of course it is not small.
Quite similar and trivial, but the von Neumann universe is just the set (or class if you don't work with universes) of all small sets.
If $X$ is a set, the set of topologies on $X$ carries a topology. A subbase is given by those topologies containing some fixed subset of $X$.
Uniform convergence spaces seem to be filters of filters(?).
This paper discusses the graph of all graphs on $n$ vertices on page 8.
The Rado graph contains a copy of every finite or countably infinite graph.

nonhausdorff dimension

2010-01-08T21:16:17Z

if $X$ is a topological space, a first step in making $X$ hausdorff is taking the quotient $H(X)=X/\sim$, where $\sim$ is the equivalence relation generated by: if $x,y$ cannot be seperated by disjoint open sets, then $x \sim y$. observe that $X$ is hausdorff, when $X \to H(X)$ is an isomorphism, and that for every hausdorff space $K$ the map $Hom(H(X),K) \to Hom(X,K)$ induced by the projection $X \to H(X)$ is bijective.

by a fairly general categorical argument, we can construct from this the free functor from topological spaces to hausdorff spaces (i.e. it's left adjoint to the forgetful functor): for ordinal numbers $\alpha$, define the functor $H^\alpha$ (together with natural transformations $H^{\alpha} \to H^{\beta}, \alpha < \beta$) by $H^0 = id, H^{\alpha+1} = H \circ H^\alpha$ and $H^\alpha = colim_{\delta < \alpha} H^\delta$. for every topological space $X$ there is an ordinal number $\alpha$ such that $H^\alpha(X) = H^{\alpha+1}(X)$, then $H^\alpha(X)$ is the free hausdorff space associated to $X$. define $h(X)$, the "nonhausdorff dimension" to be the smallest such ordinal number $\alpha$. every ordinal number arises as a nonhausdorff dimension(!).

I've came up with this with a friend and we don't know of any literature about it. perhaps someone of you has already seen it elsewhere? there are some further questions: every $H^\alpha(X)$ is a quotient of $X$, but how can we describe the equivalence relation explicitely? what is the intuition for a space $X$ to have nonhausdorff dimension $\alpha$? are there known classes of topological spaces whose nonhausdorff dimension can be bounded? and of course: is there some use for the nonhausdorff dimension? ;-)

Answer by Martin Brandenburg for + functor (used to construct sheafification)'s property

2013-03-22T01:21:02Z

Exterior powers in tensor categories

2011-07-18T10:02:39Z

Let $\mathcal{C}$ be a cocomplete $R$-linear tensor category. Many notions of commutative algebra may be internalized to $\mathcal{C}$. For example an algebra is an object $A$ in $\mathcal{C}$ together with morphisms $e : 1 \to A$ (unit) and $m: A \otimes A \to A$ (multiplication) satisfying the usual laws. The $n$-th symmetric power $\text{Sym}^n(X)$ of an object $X$ is the quotient of $X^{\otimes n}$ by $x_1 \otimes ... \otimes x_n = x_{\sigma(1)} \otimes ... \otimes x_{\sigma(n)}$, so formally it is defined as a coequalizer of the $n!$ symmetries $X^{\otimes n} \to X^{\otimes n}$. Then $\text{Sym}(X) = \bigoplus_{n\geq 0} \text{Sym}^n(X)$ is an algebra object and $\text{Sym}$ is in fact left adjoint to the forgetful functor $\mathsf{CAlg}(\mathcal{C}) \to \mathsf{CAlg}(\mathcal{C})$.

But now what about the exterior power $\Lambda^n(X)$? It is clear how to define $X^{\otimes n}$ modulo $x_1 \otimes ... \otimes x_n = \text{sgn}(\sigma) \cdot x_{\sigma(1)} \otimes ... \otimes x_{\sigma(n)}$ in this context, which one might call the anti-symmetric power $\mathrm{ASym}^n(X)$. The correct definition of the exterior power also has to mod out $... \otimes a \otimes ... \otimes a \otimes ... = 0$. But I have no idea how to internalize this to $\mathcal{C}$, even for $n=2$. The reason is that there is no morphism $X \to X \otimes X$ which acts like $a \mapsto a \otimes a$. Another idea would be to define $\Lambda(X)$ as a graded-commutative algebra object with the usual universal property, classifying morphisms $f$ on $X$ which satisfy something like $f(x)^2=0$, but again it is unclear how to formulate this in $\mathcal{C}$.

If this is not possible at all, which additional structure on $\mathcal{C}$ do we need in order to define exterior powers within them? Is this some categorified $\lambda$-ring structure? This structure should be there in the case of usual module categories (over rings or even ringed spaces). Of course there is no problem when $2 \in R^*$, because then the exterior power equals the anti-symmetric power. The question was also discussed in a blog post.

Here is a more specific (and a bit stronger) formulation: Is there some $R[\Sigma_n]$-module $T$, such that for every $R$-module $X$, we have that $T \otimes_{R[\Sigma_n]} X^{\otimes n} \cong \Lambda^n X := X^{\otimes n}/(... \otimes x ... \otimes x ...)$?

Concerning the "hidden extra structure" in the case of modules: Let the base ring be $\mathbb{Z}$, or more generally a ring $R$ in which $r^2 - r \in 2R$ for all $r \in R$; this includes boolean rings such as $\mathbb{F}_2$ and also $\mathbb{Z}/n$. If $X$ is an $R$-module, then there is a well-defined(!) homomorphism $X^{\otimes~ n-1} \to \text{ASym}^n(X), x_1 \otimes ... \otimes x_n \mapsto x_1 \wedge x_1 \wedge ... \wedge x_n$, and its cokernel is $\Lambda^n(X)$.

Principal maximal ideals in Z[x]/(F)

2013-03-18T10:58:55Z

Is there some irreducible $F \in \mathbb{Z}[x]$ such that $\mathbb{Z}[x]/(F)$ has no principal maximal ideal? Equivalently, is it possible that the $1$-dimensional integral domain $\mathbb{Z}[x]/(F)$ has no prime element?

The maximal ideals have the form $(p,f)$, where $p \in \mathbb{Z}$ is a prime and $f \in \mathbb{Z}[x]$ is a monic polynomial such that $f \bmod p$ is an irreducible factor of $F \bmod p$. In particular, $F$ should be reducible modulo every prime number. See here for a classification of biquadratic polynomials with this property. But this does not suffice, as $F=x^4+1$ shows. This is irreducible, reducible modulo every prime number, but $(2,x+1)=(x+1)$ in $\mathbb{Z}[x]/(x^4+1)$. A better candidate seems to be $F=x^4-10x^2+1$, I have checked $(p,f) \neq (f)$ for some primes $p$.

I suspect that the question is connected with the class group of the curve $V(F) \subseteq \mathbb{A}^1_{\mathbb{Z}}$?

Background: The question is equivalent to the question if $\mathbb{Z}[x]$ wins in the game of integral domains, which is a simplification of the game of rings by Will Sawin.

What about schemes built up out of graded rings?

2013-03-14T01:00:22Z

Toen-Vaquié construct a category of schemes relative to some complete cocomplete closed symmetric monoidal category $C$. Affine schemes correspond by definition 1:1 to commutative monoid objects in $C$. For $C=\mathsf{Ab}$ we get the usual category of schemes, where affine schemes correspond to commutative rings. For $C=\mathsf{Set}$ one gets one of the various definitions of schemes over $\mathbb{F}_1$. One of the drawbacks of this quite general theory is that schemes are defined via their functors on commutative monoid objects, without any geometric incarnation.

Question. What happens when $C$ is the category of graded abelian groups (equipped either with the usual symmetry, or with the twisted symmetry)? Here affine schemes correspond to (graded) commutative rings. Is there any connection with the usual Proj construction? Is there any more geometric interpretation of these schemes? For example one might hope for a fully faithful functor into the category of locally ringed spaces. Is this category of schemes something new at all?

Baer's criterion for projective modules

2013-03-12T10:20:17Z

Let $R$ be a commutative ring. If necessary, assume that $R$ has any convenient properties you like.

Is there some $R$-module $Q$ such that an $R$-module $P$ is projective if and only if $\hom_R(P,Q) \to \hom_R(P,Q')$ is surjective for all quotients $Q'$ of $Q$?
If yes, can $Q$ be chosen as a cogenerator?
What happens when we restrict to finitely generated $P$?

Neither $Q=R$ nor $Q=\hom_{\mathbb{Z}}(R,\mathbb{Q}/\mathbb{Z})$ work. This question is a follow-up of math.SE/325495.

K-Theory as a special $\lambda$-ring

2011-09-15T16:03:58Z

I wonder if there is a nice and short proof that the $K$-theory of a topological space is a special $\lambda$-ring without invoking the splitting principle and alike. Is it possible to show directly that $\lambda^k(V \otimes W)$ and $P_k(\lambda^1(V),...,\lambda^k(V),\lambda^1(W),...,\lambda^k(W))$ are stabily equivalent, without making unnatural choices? The same question for $\lambda^i(\lambda^k(V))$ and $P_{ik}(\lambda^1(V),...,\lambda^{ik}(V))$. It would be nice if there is some natural isomorphism on the level of vector spaces, which then may be glued to an isomorphism between vector bundles, perhaps with some extra summands on both sides which vanish in K-theory. Note that an affirmative answer would, in particular, answer this question by Darij Grinberg.

I hope that my question is clear enough although it is not precise. On the other hand, there is a precise generalization: Let $A$ be a topological ring, perhaps a Banach algebra. Is the Grothendieck ring of topological $A$-module bundles over a fixed $X$ a special $\lambda$-ring?

"Étalification" of a scheme

2012-06-24T10:26:54Z

Let $X$ be a scheme. Does the forgetful functor

$$\mathrm{EtSch}/X \to \mathrm{Sch}/X$$

have a right adjoint $Z \mapsto \tilde{Z}$? One might call $\tilde{Z}$ the étalification of $Z$. So this is an étale $X$-scheme together with an $X$-morphism $\tilde{Z} \to Z$, which induces for every étale $X$-scheme $Y$ a bijection $\hom_X(Y,\tilde{Z}) \cong \hom_X(Y,Z)$.

If $X$ is the spectrum of a field $k$, the answer is yes, using 1) the equivalence of sites between étale $k$-schemes and continuous $G$-sets, where $G$ is the absolute Galois group of $k$, 2) the fact that every sheaf on $G$-sets is representable. Explicitly: If $Z$ is a $k$-scheme, let us denote by $Z_{\mathrm{sep}}$ the subset of all points $z \in Z$ such that $k(z)/k$ is finite and separable. Then $$\tilde{Z} = \coprod\limits_{z \in Z_{\mathrm{sep}}} \mathrm{Spec}(k(z)).$$ For example, $\widetilde{\mathbb{A}^1}=\coprod_{\alpha \in k} \mathrm{Spec}(k)$ and $\widetilde{\mathbb{A}^2}$ is the coproduct of spectra of the form $(k[x]/(p)[T])/(q)$, where $p \in k[x]$ is irreducible and $q$ is some irreducible separable polynomial over $k[x]/(p)$.

Perhaps this construction is well-known, therefore I've put the reference request tag.

Open idempotents in modules over a local ring

2013-03-06T11:51:13Z

Let $R$ be a local ring. By an open idempotent I mean an $R$-module $F$ equipped with a homomorphism $e : F \to R$ such that $e \otimes F = F \otimes e$ is an isomorphism $F \otimes F \cong F$ (this notion in a monoidal category is due to Drinfeld, Boyarchenko). For example, $0 \to R$ and $R \to R$ are open idempotents.

Are these the only ones? It is not hard to see this under the assumption that $F$ is finitely generated, using Nakayama. But I don't know what happens in the general case.

Answer by Martin Brandenburg for Local ring of product of varieties

2013-03-06T10:16:04Z

Let $k$ be a field. The fiber product of two $k$-schemes $X,Y$ (or even locally ringed spaces, see here) has as points triples $(x,y,\mathfrak{p})$, where $\mathfrak{p}$ is a prime ideal in $\kappa(x) \otimes_k \kappa(y)$, or equivalenty a prime ideal $\mathfrak{q}$ in $\mathcal{O}_{X,x} \otimes_k \mathcal{O}_{Y,y}$ which restricts to the maximal ideals in both factors. The stalk of the structure sheaf at $(x,y,\mathfrak{p})$ is $(\mathcal{O}_{X,x} \otimes_k \mathcal{O}_{Y,y})_\mathfrak{q}$ . If $x,y$ are $k$-rational points, we have $\mathfrak{p}=0$ and therefore $\mathfrak{q}$ is the kernel of the natural map $\mathcal{O}_{X,x} \otimes_k \mathcal{O}_{Y,y} \to k$, $f \otimes g \mapsto f(x) \otimes g(y)$. So the stalk of $X \times_k Y$ at a rational point $(x,y)$ is a localization of the tensor product of the stalks, where those elements $\sum_i f_i \otimes g_i$ with $\sum_i f_i(x) \cdot g_i(y) \neq 0$ are inverted.

Answer by Martin Brandenburg for Is every projective $\mathbf{Z}[x]$-module free?

2013-03-03T03:37:32Z

When $R$ is a PID, then every finitely generated projective $R[x]$-module is free. As Steven already said, this is due to Seshadri. Here is the reference:

Seshadri, C.S., Triviality of vector bundles over the affine space $K^2$, Proc. Nat. Acad. Sci. USA 44 (1958), 456-458.

Colimits of quasi-coherent sheaves on a ringed space

2013-02-28T05:01:32Z

Recall from the stacks project that a sheaf of modules $F$ on a ringed space $X$ is called quasi-coherent if there is an open covering $\{U_i\}$ such that each $F|_{U_i}$ has a presentation, i.e. is the cokernel of a homomorphism between free $\mathcal{O}_{U_i}$-modules. The global section functor $\mathsf{Qcoh}(X) \to \mathsf{Mod}(\Gamma(X,\mathcal{O}_X))$ has a left adjoint $M \mapsto \widetilde{M}$, the sheaf of modules associated to $M$. If $X$ has a fundamental system of quasi-compact neighborhoods, every quasi-coherent sheaf is locally associated to a module, and the converse is always true.

My first question is: Is there any detailed treatment of the categorical properties of $\mathsf{Qcoh}(X)$ which goes beyond the basics treated in the stacks project? More specifically, I wonder if $\mathsf{Qcoh}(X)$ is a cocomplete symmetric monoidal category, when the underlying space of $X$ has nice topological properties. This is well-known to be true when $X$ is a scheme, but what about the case that $X$ is, say, a smooth manifold equipped with the sheaf of smooth functions?

The only thing to check is that $\mathsf{Qcoh}(X)$ is stable under colimits computed in $\mathsf{Mod}(X)$. When $X$ has a fundamental system of quasi-compact neighborhoods, it is not hard to reduce this to directed colimits.

Question. Assuming that the underlying space of $X$ is nice enough, is $\mathsf{Qcoh}(X)$ closed under directed colimits?

There is a remark in the stacks project which says that this fails for general $X$ (without proof, can someone give an example?), but I am not really interested in weird spaces. Let me explain why the naive proof attempt doesn't work: Say we are given a sequence $F_0 \to F_1 \to \dotsc$ of quasi-coherent sheaves on $X$. Given $x \in X$, for every $n \in \mathbb{N}$ there is an open neighborhood $U_n$ of $x$ such that $F_n|_{U_n}$ has a presentation. The problem is that $\cap_n U_n$ is not open.

Answer by Martin Brandenburg for Awfully sophisticated proof for simple facts

2013-01-25T13:46:50Z

$5/2 = 2 \frac{1}{2}$ since both are the groupoid cardinality of the following action:

Thinking about this, it is actually quite enlightening. For more information, see the wonderful paper From Finite Sets to Feynman Diagrams by John Baez and James Dolan.

Answer by Martin Brandenburg for Structure theorem for etale maps

2013-02-19T01:40:21Z

The degree is $1$ when $A,B$ are finitely generated algebras over an algebraically closed field, yes, but then they are not local (unless they are artinian), and the degree can be arbitrary if $A,B$ are just localizations of finitely generated algebras. Already the "trivial" direction of the classification of étale morphisms shows that your argument is not correct, every degree is possible.

Unitalization internal to monoidal categories

2013-02-16T22:51:39Z

Let $C$ be a monoidal category, not assumed to be symmetric. Assume that the underlying category of $C$ is nice enough, for example cocomplete, perhaps even presentable. A semigroup object in $C$ is a pair $(X,\mu)$ consisting of an object $X \in C$ and a morphism $\mu : X \otimes X \to X$ satisfying the associativity law $\mu \circ (X \otimes \mu) = \mu \circ (\mu \otimes X)$. Does the forgetful functor from monoid objects in $C$ to semigroup objects in $C$ have a left adjoint? In other words, is there an unitalization internal to $C$?

The cases $C=\mathsf{Set}$ and $C=\mathrm{Mod}(R)$ are well-known. More generally, the answer is yes when $\otimes$ preserves coproducts in each variable. Then the unitalization of $(X,\mu)$ is $(1 \oplus X,\mu',\eta)$ with the obvious morphism $\mu' : (1 \oplus X) \otimes (1 \oplus X) = 1 \oplus X \oplus X \oplus X \otimes X \to 1 \oplus X$ and $\eta : 1 \to 1 \oplus X$.

Actually I am interested in the case that $C=(\mathrm{End}(D),\circ,\mathrm{id})$ for a (nice) category $D$, thus I would like to know if every semi-monad can be made into a monad. Here $\otimes$ preserves colimits in the left variable, but not in the right variable. Actually $D$ is even a presentable symmetric monoidal category and $\mathrm{End}(D)$ refers to enriched endofunctors, i.e. I am interested in strong (semi) monads.

Answer by Martin Brandenburg for Using schemes to prove things about rings

2013-02-15T17:00:27Z

Some examples:

A. A noetherian commutative ring has only finitely many minimal prime ideals. This is just a corollary of the easy observation that a noetherian space has only finitely many irreducible components.

B. The tensor product of two reduced (integral) $k$-algebras, where $k$ is an algebraically closed field, is again reduced (integral). After reducing to the finite type case, the argument of the proof is essentially geometric.

C. Diophantine equations, for example Fermat's Last Theorem (classify ring homomorphisms $\mathbb{Z}[x,y,z]/(x^n+y^n-z^n) \to \mathbb{Z}$), are (approximately) solved with the machinery of elliptic curves. The equation $x^2+y^3=z^7$ even needs algebraic stacks (see here)!

D. The classification of boolean rings. Or more generally rings whose elements satisfy a polynomial equation. As compared to the modern proof ( $\underline{\mathbb{F}_2} \to \mathcal{O}_{\mathrm{Spec}(R)}$ is an isomorphism at stalks, hence globally), Stone's original one is quite clumsy. But of course, the historical importance of Stone's work cannot be overestimated. He can be seen as one of the innovators of the ideas of scheme theory.

E. The classification of integral domains generated by a single element. This comes down to the classification of prime ideals of $\mathbb{Z}[X]$, which is best done by looking at the fibers of $\mathrm{Spec}(\mathbb{Z}[X]) \to \mathrm{Spec}(\mathbb{Z})$. As in the examples above, this can also be done purely algebraically, but then it gets clumsy.

F. Define an $R$-module $M$ to be locally free of finite rank if there are elements ${f_i}$ of $R$ generating the unit ideal such that $M_{f_i}$ is free of finite rank over $R_{f_i}$. There is a purely algebraic proof that $M$ is locally free of finite rank if and only if $M$ is flat and of finite presentation, if and only if $M$ is finitely generated projective. However, at least for me as a beginner, it was hard to really grasp what is going on in that proof. But when you view $M_{f_i}$ as the restriction of (the quasi-coherent sheaf associated to) $M$ to (the open subscheme defined by) $R_{f_i}$, every step is clear as crystal. More generally, there are many theorems about modules over commutative rings which are best formulated, understood and proven more generally for quasi-coherent sheaves on a scheme. For another example, see David Lehavi's comment to Emerton's slick proof of the structure theorem for finitely generated modules over a PID.

G. There are non-isomorphic commutative rings $R,S$ such that $R[x]$ and $S[x]$ are isomorphic. The first example was found by Mel Hochster with geometric ideas (see here).

H. Affine algebraic geometry is full of problems, which can be formulated in terms of ring and module theory, but are attacked with algebraic-geometric methods. For a survey, see here. Perhaps I should stop here, because the list will never end ...

I. There are various algebraic constructions and invariants for rings which are best understood in geometric terms, such as the Krull dimension. The associated graded ring $\bigoplus_n I^n / I^{n+1}$ of an ideal $I \subseteq R$ roughly contains the infinitesimal information of $\mathrm{Spec}(R)$ at the closed subscheme $V(I)$. Modules of differentials provide another infinitesimal invariant. For function fields over perfect fields we have the geometric genus (of the corresponding proper normal curve). As always, such invariants are useful for example when one wants to prove that two rings are not isomorphic, replacing painful direct computations (see for example math.SE / 128918, 151714, 296737), but also to provide parameters for a possible classification.

J. Even projective schemes are useful in general ring theory, in particular in the context of homogeneous polynomials, see for example Will Sawin's answer in MO/110250, François Brunault's answer in MO/98043 and Qiaochu Yuan's answer at MO/14076.

K. (recreational) In the game on noetherian rings, a move consists of replacing a ring $R$ by $R/(a)$ for some $0 \neq a \in R$. You win when your oponent gives you the trivial ring. A complete analysis of this game is still out of reach, but the first attempts by Will Sawin and Kevin Buzzard illustrate the usage of algebraic geometry. Actually it is a game on (affine) schemes, where each move replaces $X$ by a closed subscheme $X' \subseteq X$ cut out by a single nontrivial equation.

L. Let $k$ be a field and $A \to C \leftarrow B$ homomorphisms of finitely generated $k$-algebras. Is the fiber product $A \times_C B$ again finitely generated? At first sight this seems to be elementary and should be well-known for decades, but it seems to be an open problem(?). Jakob Scholbach has proven it when $A \to C$ and $B \to C$ are regular (i.e. the ideal is generated by codim many elements), using quite a bit of projective algebraic geometry.

Answer by Martin Brandenburg for Generators of the kernel of multiplication of a ring

2013-02-15T21:32:28Z

If $M,N$ are $S$-modules, then $M \otimes_S N$ is the quotient of $M \otimes_R N$ by the $R$-submodule generated by $ms \otimes n - m \otimes sn$ with $m \in M, n \in N, s \in S$; simply because both sides satisfy the same universal property. For $M=N=S$ the result follows. Actually the same holds when $\mathrm{Mod}(R)$ is replaced by an arbitrary cocomplete symmetric monoidal category, and $S$ by a commutative algebra object.

Answer by Martin Brandenburg for Leray spectral sequence of the inclusion of an open subvariety

2013-02-15T18:56:44Z

For a closed immersion $i : Z \hookrightarrow X$ of schemes the spectral sequence is degenerate and becomes the elementary isomorphism $H^p(X,i_* F) \cong H^p(Z,F)$.

For a morphism $j : U \hookrightarrow X$ of schemes the five term exact sequence associated to the spectral sequence becomes

$0 \to H^1(X,j_* F) \to H^1(U,F) \to \Gamma(X,R^1 j_* F) \to H^2(X,j_* F) \to H^2(U,F).$

In general this cannot be simplified, even if $j$ is an open immersion. But this is useful, for example for the computation of the etale cohomology of $\mu_n$ and $\mathbb{G}_m$ on a curve $X$, where $j$ is the inclusion of the generic point (see Chapter 10 in Tamme's book on etale cohomology).

Comment by Martin Brandenburg

2013-05-04T14:15:15Z

I have found new results about the game of rings. Should I add this to the paper, or write a second part?

Comment by Martin Brandenburg

2013-05-03T15:12:46Z

This is a very concise categorical description!

Comment by Martin Brandenburg

2013-04-27T09:09:00Z

The tangent space of a locally ringed space $X$ at a point $x$ is $(\mathfrak{m}_x/\mathfrak{m}_x^2)^*$, where the dual is taken over the residue field $\kappa(x)$. But probably this doesn't answer your question, since you want to use curves into $X$? If $X$ is a scheme over $k$, the replacement for the "infinitesimal interval" is the spectrum of the ring of dual numbers $k[t]/t^2$.

Comment by Martin Brandenburg

2013-04-26T21:13:17Z

Or better do it for yourself.

Comment by Martin Brandenburg

2013-04-26T19:24:04Z

Your condition states that $\mathcal{O}_{X,x} \to R_x$ is surjective. Injectivity comes for free in the case of preorders, but I think in general we just have to add it as a condition. So it seems to me that $\varinjlim_{x \in U \text{ open}} \varprojlim_{y \in U} R_y \cong R_x$ is a necessary and sufficient condition. Do you agree?

Comment by Martin Brandenburg

2013-04-26T19:18:08Z

Thank you. It's obviously a sheaf (if $x \prec y$ in $\cup_i U_i$, then $x \in U_i$ for some $i$, but then also $y \in U_i$, etc.)

Comment by Martin Brandenburg

2013-04-26T18:29:30Z

"Prove that ..." sounds as if this is a homework. What is the context of this question?

Comment by Martin Brandenburg

2013-04-26T13:56:48Z

The background is a little bit longer, I will send you a mail if you are interested.

Comment by Martin Brandenburg

2013-04-26T13:55:13Z

And probably there will be speculative predictions.

Comment by Martin Brandenburg

2013-04-23T12:58:47Z

Of course not. Please read the FAQ mathoverflow.net/faq

Comment by Martin Brandenburg

2013-04-23T10:51:43Z

For the DVR case, induct on the order of $f$ and use the additivity of the length on short exact sequences.

Comment by Martin Brandenburg

2013-04-22T14:16:38Z

@Emerton: I don't know, but maybe "when can we say" is a little but unspecific.

Comment by Martin Brandenburg

2013-04-22T14:00:18Z

You probably mean that the automorphism groups are isomorphic?

Comment by Martin Brandenburg

2013-04-22T13:28:36Z

math.stackexchange.com/questions/361351

Comment by Martin Brandenburg

2013-04-22T12:58:23Z

$K_n$ commutes with products.