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(@orlando)
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Joined: 1 year ago
Posts: 12
Topic starter

Hello,

I'm beginner and i have a question. If i understand

(1) the classifying topos for commutative ring is the topos of presheaf on the (opposite) category of finitly presented ring.

(2) If i add an axiom to the ring theory i get a subtopos.

(3) Example : local ring and the Zariski topos.

My question is : if i add the axiom $\forall e, e^2 = e \to (e=1 \text{ or } e=0)$, what is the classifying topos ? I have in mind a to find a topology, for example let R a ring a covering familly is a familly of localisation  $(R \to R_i)$ with  $R_i = R[e_i^{-1}]$, $\sum e_i = 1$, and $e_i e_j = \delta(i,j) e_i$  (the Kronecker symbol).

(@elio-pivet)
New Member
Joined: 1 year ago
Posts: 3

@orlando

I'm a beginner too, but I know Olivia Caramello gave in her book a general method to compute the classifying topos of a quotient theory of a theory of presheaf type such as what you just described. The idea would be to associate a homomorphism of models to the formula you add. The homomorphism of models would then generate a sieve in the category of finitely presented models, which then would generate a topology. I didn't do the computation though, so I don't know if you would obtain the topology you suggest.

(@orlando)
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Joined: 1 year ago
Posts: 12
Topic starter

Hello @elio-pivet,

Thx, i thinck, in Olivia's book, you're are mentioning  page 88 " quotient and subtopos " ?  For the moment, i don't understand the procedure (i take time to understand).

(@jens-hemelaer)
Eminent Member
Joined: 2 years ago
Posts: 20

First, we need to translate the axioms that you suggest to geometric sequents. In axioms for a geometric theory, both the universal quantifier $\forall$ and the implication $\Rightarrow$ are not allowed. But instead we can look at the sequent:
$e^2 = e \vdash (e=0) \vee (e=1)$

This is like you said but we avoided $\forall$ and $\Rightarrow$. Now, how to translate this to a Grothendieck topology?

First step is to see the translation between formulas and objects of your category, in this case (the opposite of) the category of finitely presented rings. I think that the theory of rings is a cartesian theory, and then the syntactic site for the theory has as objects precisely the finitely presented models (in Set) of the theory. In our case, this means that the objects of the syntactic site are the finitely presented rings. The morphisms are morphisms of rings, but in the syntactic site the morphisms go in the opposite direction.

For example, the formula $(e^2=e)$ has one free variable and corresponds to the ring $\mathbb{Z}[x]/(x^2=x)$. I changed the variable name to avoid confusion. Similarly, $e=0$ corresponds to the ring $\mathbb{Z}[x]/(x=0)$ and $e=1$ corresponds to the ring $\mathbb{Z}[x]/(x=1)$. These two last rings are both isomorphic to $\mathbb{Z}$.

The sequent $e^2 = e \vdash (e=0) \vee (e=1)$ then corresponds to the sieve generated by the projections
$\mathbb{Z}[x]/(x^2=x) \to \mathbb{Z}[x]/(x=0)$ and $\mathbb{Z}[x]/(x^2=x) \to \mathbb{Z}[x]/(x=1)$.
The Grothendieck topology that you are looking for is now the smallest Grothendieck topology that contains this sieve as a covering sieve. For this see e.g. the Elephant D3.1.10 or towards the end of page 93 in "Theories, Sites, Toposes".

We can construct new covering sieves as follows. Suppose you have a ring $R$ and an idempotent $e \in R$. We would like to show that $R \to R[e^{-1}]$ and $R \to R[(1-e)^{-1}]$ together generate a covering sieve. This is almost the covering sieve that you first suggested. Note that $R[e^{-1}] = R/(e=1)$ and $R[(1-e)^{-1}] = R/(e=0)$ so quotients and localizations are the same thing here.

Consider the ring morphism $f : \mathbb{Z}[x] \to R$ that sends $x$ to $e$. If we pullback the map $\mathbb{Z}[x] \to \mathbb{Z}[x]/(x=0)$ along $f$, then we get the ring morphism $R \to R/(e=0)$. Similarly, if we pullback the map $\mathbb{Z}[x] \to \mathbb{Z}[x]/(x=1)$ along $f$, then we get the ring morphism $R \to R/(e=1)$. So we see that the sieve generated by $R \to R/(e=0)$ and $R \to R/(e=1)$ is a covering sieve, because it is the pullback of a covering sieve.

Similarly, you can show that the sieves you suggest (with multiple idempotents) also must be covering sieves. This then shows that the Grothendieck topology you suggest is right: it is the Grothendieck topology corresponding to the extra axiom $e^2 = e \vdash (e=0) \vee (e=1)$.

(@elio-pivet)
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Joined: 1 year ago
Posts: 3

@orlando Hi, I was mostly thinking about page 281, especially the first half, which expresses a general, almost algorithmic way to compute a classifying topos, I believe it's the same thing Jens Hemelaer suggested above.

(@jens-hemelaer)
Eminent Member
Joined: 2 years ago
Posts: 20

@elio-pivet Yes, I agree, I didn't notice that part. In the notation of page 281, the model $M_{\{e . e^2=e\}}$ should for example correspond to the ring $\mathbb{Z}[x]/(x^2=x)$.

(@orlando)
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Joined: 1 year ago
Posts: 12
Topic starter

Thank you @jens-hemelaer very much for your calculations which seems to confirm my intuition and the references. It's really nice, the automatic procedure. I'm taking a little time to read your message.

@elio-pivet : ok thx again !

(@orlando)
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Joined: 1 year ago
Posts: 12
Topic starter

Hello,

I try to make another  baby example. I take the theory of ring  with the axiom $\vdash \exists a,b, a^2+b^2 = 1$ (it's ok ?)

The conversion give : $Z \to Z[a,b] / (a^2+b^2=1)$ and a cover of a ring R is just $\phi : R \to V$ s.t factor in $R \to R[a,b] / (a^2+b^2=1)$

Now if i understand, the generic ring $\mathbf{A}^1$ must satisfy localy this axiom and that is trivial cause you can extend the base$R \to R[a,b] / (a^2+b^2=1)$ and you have the solution $(a,b) \pmod{(a^2+b^2=1)}$

But i have to check that $\mathbf{A}^1$ is a sheaf and this is not trivial ! I have to sheafify $\mathbf{A}^1$ or i made something wrong ? 🤔

(@jens-hemelaer)
Eminent Member
Joined: 2 years ago
Posts: 20

Hi @orlando,

Nice example. I agree with your notation, and I agree that the sieve generated by $R \to R[a,b]/(a^2+b^2=1)$ is a covering sieve for the corresponding Grothendieck topology. I don't understand the condition on the generic ring, do you have a reference?

In this example, the axiom $\vdash \exists a,b, a^2+b^2=1$ is a theorem of the theory of rings (you can take $a=1$ and $b=0$). So that means that the corresponding Grothendieck topology is trivial... To see this directly, you have to prove that the covering sieve generated by $R \to R[a,b]/(a^2+b^2=1)$ is the maximal sieve (meaning it contains the identity morphism $R \to R$). To prove this, you can take the composition of the map $R \to R[a,b]/(a^2+b^2=1)$ with the map $R[a,b]/(a^2+b^2=1) \to R$ that sends $a$ to $1$ and $b$ to $0$.

If you are interested in another example, then you can look instead of the theory of rings at the theory of $\mathbb{R}$-algebras. The classifying topos is the topos of presheaves on the opposite of the category of finitely presented $\mathbb{R}$-algebras. To the theory of $\mathbb{R}$-algebras, you can then add the axiom $\vdash \exists a, a^2=-1$.

(@orlando)
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Joined: 1 year ago
Posts: 12
Topic starter

I try latex with the symbol dollar. (that ok but the preview doesn't work for me).

hum i'm stupid ...  thx @jens-hemelaer.  I Don't want my axiom  a theorem (i d'ont see the point $(1,0)$ my bad)! And in fact, the a covering sieve for a ring $R$  contain   $id : R \to R$ because $R \to R[a,b] /(a^2+b^2-1) \to R$ the second given by the point $(1,0)$ ... so that trivial in this case ! So $\mathbf{A}^1$ is a sheaf cause all is a sheaf ! Good !

i can do the same thing with $a^2+b^2 = 3$ (it's not a theorem). or you're example.

I have no reference for my remark. The generic model for ring theory is the ring $\mathbf{A}^1 : \text{f.p.Ring} \to \text{Ring}$ $R \mapsto R$ (the presheaf). If a take a new axiom (a new theory $\mathbb{T}$) i get  a  topology $J$  on $\text{f.p.ring}^\text{op}$.  And $\text{Sh}(\text{f.p.ring}^\text{op}, J)$ is " the "classify topos for $\mathbb{T}$. So i have a generic (universal) model, so a model in $\text{Sh}(\text{f.p.ring}^\text{op}, J)$ so a sheaf that satisfy $\mathbb{T}$.

I'm not sure but :

1/ If the topology $J$ is subcanonical this model is $\mathbf{A}^1$ .

2/ now if the topology is not canonical, what is this generic model ? I guest it's sheafifycation of $\mathbf{A}^1$ for $J$ ? (perhaps a computation in a example is hard ?).

(@jens-hemelaer)
Eminent Member
Joined: 2 years ago
Posts: 20

Thanks, I understand now what you mean with the remark. And yes, the generic model will be the sheafification of the original generic model. The reason is that geometric morphisms to the subtopos are precisely the morphisms to the original topos that factorize through the subtopos. So in the other direction, the functor that sends the generic model to a concrete model factorizes through the sheafification.

But as you say, sheafification is not necessary here if the topology is subcanonical. For example, on finitely presented rings the flat topology or any coarser topology is subcanonical.

In your new example, to show that the topology is subcanonical, it would be enough to show that the morphism $\mathbb{Z}\to\mathbb{Z}[x,y]/(x^2+y^2=3)$ is flat. I can't see at the moment whether this is the case or not.

Regarding your second question, I'm not sure but I suspect that at least in some cases explicitly computing the sheafification of the generic model is difficult.

(@orlando)
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Joined: 1 year ago
Posts: 12
Topic starter

And yes, the generic model will be the sheafification of the original generic model. The reason is that geometric morphisms to the subtopos are precisely the morphisms to the original topos that factorize through the subtopos. So in the other direction, the functor that sends the generic model to a concrete model factorizes through the sheafification.

Perfect, Diaconescu's equivalence ! I don't finish to read the proof for the moment !

Yes ! I'm also thinking about  Flatness ! It's a difficult notion for me but https://arxiv.org/pdf/1611.02942.pdf   end of the page 466 (it's in French) : $c(f)$ is the ideal generated by the coefficient of $f = x^2+y^2 -3$ so it's the ideal generated by $1$ and the theorem apply, but we have to check injectivity of $\mathbb{Z} \to \mathbb{Z}[x,y] / (x^2+y^2 =3)$.

I try to add the axiom $2 = 0$ for the fun ! It's a very fun game  ! I think the generic model is the ring $R \mapsto R / 2R$ ... not sure for the moment !

(@jens-hemelaer)
Eminent Member
Joined: 2 years ago
Posts: 20

Hi! To show injectivity: if $\mathbb{Z} \to \mathbb{Z}[x,y](x^2+y^2=3)$ is not injective, then $(x^2+y^2-3)$ would contain a nonzero integer $n$. But this would mean that $n = (x^2+y^2-3)g$ for some $g \in \mathbb{Z}[x,y]$, which is impossible.

I like the reference! Alternatively, you can show that $\mathbb{Z}[x,y](x^2+y^2=3)$ is a free $\mathbb{Z}$-module, by taking as a basis the monomials $x^i y^j$ with $i \in \{0,1\}$ and $j \geq 0$ (and free modules are flat). The idea behind it seems to be the same as in the reference.

I'd be interested to hear what you find out with other axioms. Yes, very fun game!

(@orlando)
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Joined: 1 year ago
Posts: 12
Topic starter

I'll explain my project to you. (my english is bad !)

1. For the moment learn a little theory by building small examples. The idea is to observe concrete objects to see things more abstract (and this theory may seem a little abstract).

2. Construct the tangent bundle of a scheme using topos.

I explain 2. (just the idea, not technical, perhaps it's wrong).

a. consider the ring $\mathbf{B} : \text{f.p.ring} \to \text{f.p.ring}$ given by $R \mapsto R[\varepsilon]$ with $\varepsilon ^2 = 0$. So $\mathbf{B}$ is a ring in the topos $\left[ \text{f.p.ring}, \text{Set} \right]$ and we have a geometric morphism :  $(f^\star,f_\star) :\left[ \text{f.p.ring}, \text{Set} \right] \to \left[ \text{f.p.ring}, \text{Set} \right]$. For me $f^\star$ is given in this case by the formula : $\mathfrak{X} \mapsto \left[ R \mapsto \mathfrak{X}(R[\varepsilon])\right]$ (i write a proof, i have to verify details).

Let's try with the functor $\mathbb{S}^1 : R \mapsto \{ (x,y) \in R^2 \mid x^2+y^2=1 \}$, so the corresponding functor is $R \mapsto \{ (x,y) \in (R[\varepsilon])^2 \mid x^2+y^2=1 \}$. Write $x = a+ \varepsilon h_a$ and $y = b + \varepsilon h_b$, putting in the equation give $R \mapsto \{ (a,b,h_a,h_b) \in R^4 \mid a^2+b^2=1 \text{ and } 2ah_a+2bh_b = 0\}$

This the fiber bundle of $\mathbb{S}^1$. I denote lthe left adoint $f^\star$ by $\mathcal{T}$ !

b. But $\mathbf{B}$ is also a local ring in Zariski topos $\text{Zar}$ so that give a geometric morphism $(g^\star,g_\star) : \text{Zar} \to \text{Zar}$. Here i'm not clear but we can show that $f^\star(X)$ is a Zariski sheaf if $X$ is a Zariski sheaf. and i think that prove that $g^\star$ is just the restriction of $f^\star$. So $\mathcal{T} : \text{Zar} \to \text{Zar}$ respect  colimit.

c. Now what is a scheme : it's just a Zariski sheaf $X$ satisfy conditions (i can explain). There is $\mathfrak{U}_i \subset X$ so that $\mathfrak{U}$ is open and affine and $\bigcup \mathfrak{U}_i = X$ (in a sense of Zariski). But $\mathcal{T}$ respect colimit so if $X$ is a scheme, $\mathcal{T}X$ is also a scheme and the gluing data are the same but for $\mathcal{T}X$ we have to interpret the gluing data not in the localy ringed topos $(\text{Zar},\mathbf{A}^1)$ but in $(\text{Zar},\mathbf{B})$.

d. Conclusion : THe fiber bundle of a scheme $X$ is $X$ interpreted in a other localy ringed Topos ! (for me it's nice and really concrete, i can calculate in a exemple (it's boring but i play this game with $\mathbb{P}^1$).

Do you think that coherent ?

Another point ! I just consider a ring objet in $\left[ \text{f.p.Ring}, \text{Ens} \right]$ but the have plenty of ring in this topos for example $(R,a,b) \mapsto R / (a^2+b^2-1)$. Where $(R,a,b)$ stand for a ring $R$ and $(a,b) \in R^2$ so a $\mathbf{Z}[X,Y]$-algebra.

I'm ok with the freedom of $\mathbf{Z}[X,Y] / (X^2+Y^2=3)$ and for injectivity.The injectivity of $R \mapsto R[X,Y] / (X^2+Y^2=3)$  say that$\mathbf{A}^1$ is a separated presheaf (for our topology) and the flatness so that all descent data glue. The lemma is http://pub.math.leidenuniv.nl/~edixhovensj/teaching/2015-2016/TAG/pol2.pdf page $6$.

For the axiom, today i will try to add the axiom $0 = 2 \text{ or$2$is invertible}$  but the case of $2=0$ is interesting because we get $\mathbb{F}_2$-algebra theory

(@jens-hemelaer)
Eminent Member
Joined: 2 years ago
Posts: 20

Hi @orlando,

The constructions you are describing about the tangent bundle make a lot of sense, and it looks interesting to replace the ring object $\mathbf{B}$ with another ring object.

Do you have an idea already about the last axiom that you mention?

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