Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

If two torsion theories on a ring localize the ring to the same extension ring, I can find no reason that their "meet" in the lattice of torsion theories must also localize to the same ring. I cannot find anything in Golan's encyclopedia that addresses questions like this.

Does anyone have a counter-example?

Here is a weaker question, not directly related to torsion theories.

Is there an example of the following:

A ring homomorphism R $\to$ S , S-modules P and Q , R-monomorphisms M $\to$ P and M $\to$ Q such that the image of each is an essential R-submodule, but such that the image of M in P $\times$ Q has no essential extension within the product that is an S-submodule

share|improve this question
    
When you say "the same" extension ring, do you mean that the localizations are isomorphic (perhaps in a way that preserves the homomorphism from the original ring into the localization)? –  Manny Reyes Dec 4 '12 at 16:15
    
I had in mind simply isomorphic. But I have no reason to think the property of the meet holds even if the isomorphism is compatible with the structural localization homomorphisms –  Carl Weisman Dec 4 '12 at 16:47
    
Can you remind me what the lattice structure on the torsion theories is? Is it the same as the inclusion ordering on the associated Gabriel filters? I don't have Golan's book handy. –  Manny Reyes Dec 7 '12 at 15:09
    
I believe the lattice structure is indeed the inclusion ordering on Gabriel filters, and the greatest lower bound corresponds to the intersection of the filters. –  Carl Weisman Dec 8 '12 at 13:46

2 Answers 2

up vote 1 down vote accepted

I also assume that by torsion theory we mean hereditary torsion theory. I use Bo Stenström's book Rings of Quotients (especially chapter IX §2 later on). And today I work with right modules.

If by "the same" you really mean nothing more than that they are isomorphic as rings, the following should be an easy counterexample.

Take $k$ a field, let $R = k \times k$ and $e:= (1,0)$. The Gabriel topologies $\mathfrak{F}_1 := \lbrace R, (e) \rbrace$ and $\mathfrak{F}_2 := \lbrace R, (1-e) \rbrace$ with corresponding torsion theories $t_1(M) = (1-e)M$ and $t_2(M) = eM$ have meet $\mathfrak{F}_0 = \lbrace R \rbrace$ with torsion theory $t_0 = 0$.

The localisations are given by $pr_1 : R \rightarrow k = R_{\mathfrak{F}_1} = eR$ and $pr_2 : R \rightarrow k = R_{\mathfrak{F}_{2}} = (1-e)R$,
but the localisation for $\mathfrak{F}_0$ is given by $id: R = R_{\mathfrak{F}_0}$.

Remark that the $R_{\mathfrak{F}_i}$ for $i =1,2$ really are just isomorphic as rings. There is no isomorphism compatible with the localisations, in fact they are not isomorphic as $R$-modules.


If, on the other hand, we have an isomorphism $j: R_{\mathfrak{F}_1} \xrightarrow{\sim} R_{\mathfrak{F}_2}$ such that for the localisations $\psi_i: R \rightarrow R_{\mathfrak{F}_i}$ we have $j \circ \psi_1 = \psi_2$, then I think the meet will be isomorphic, i.e. we then have $R_{\mathfrak{F}_1 \wedge \mathfrak{F}_2} \simeq R_{\mathfrak{F}_i}$. Here is a sketch of a proof.

First note that because $t_i(R) = ker (\psi_i)$, we have $t_1(R) = t_2(R) =: t(R)$. Set $\bar R := R/t(R)$ and fix an injective hull $E(\bar R)$ of the right $R$-module $\bar R$. It is known that

$R_{\mathfrak{F}_i} \cong \lbrace x \in E(\bar R) | (\bar R : x) \in \mathfrak{F}_i\rbrace $

where $(\bar R:x) := \lbrace r \in R: xr \in \bar R \rbrace$. Using this identification, the maps $\psi_i$ become induced by $R \twoheadrightarrow \bar R \hookrightarrow E(\bar R)$, and the map $j$ becomes $R$-linear and fixes $\bar R$. This implies $(\bar R : x) = (\bar R : j(x))$ for all $x \in R_{\mathfrak{F}_1}$. On the other hand for these $x$ we have $(\bar R : x) \in \mathfrak{F}_1$ and $(\bar R : j(x)) \in \mathfrak{F}_2$, so by symmetry it turns out that a) the localisations are really equal as $R$-submodules of $E(\bar R)$, b) for $x \in E(\bar R)$, we have $(\bar R : x) \in \mathfrak{F}_1 \Leftrightarrow (\bar R : x) \in \mathfrak{F}_2 \Leftrightarrow (\bar R : x) \in \mathfrak{F}_1 \wedge \mathfrak{F}_2$, and thus
$R_{\mathfrak{F}_1} = R_{\mathfrak{F}_2} = R_{\mathfrak{F}_1 \wedge \mathfrak{F}_2}$ as $R$-modules (and rings).
By the way, the same argument should now go through for any $R$-module $M$ with localisation-compatible isomorphism $M_{\mathfrak{F}_1} \cong M_{\mathfrak{F}_2}$.

share|improve this answer

As you mention Golan, I guess that all your torsion theories are hereditary. Let $\tau_1$ and $\tau_2$ be t.t. on Mod$(R)$ and $\phi_1:R\to R_1$, $\phi_2:R\to R_2$ the two loc. of $R$. The fact that there exists an isomorphism $\phi:R_1\to R_2$ s.t. $\phi\phi_1=\phi_2$, means that $-\otimes_RR_1$ is naturally eq. to $-\otimes_RR_2$. If $\tau_1$ and $\tau_2$ are perfect then these functors coincide with the localization functors. Thus, in such case, $M\in \mathcal T_{\tau_1}$ (the torsion class of $\tau_1$) iff $M\otimes_RR_1=0$ iff $M\otimes_RR_2=0$ iff $M\in \mathcal T_{\tau_2}$. So $ \mathcal T_{\tau_1}=\mathcal T_{\tau_2}$, that is, $\tau_1=\tau_2$.

If your torsion theories are not perfect I do not remember if $\ker(-\otimes_RR_1)=\mathcal T_1$ holds true, if so you should be able to proceed as above...

share|improve this answer
    
An equivalent definition of a perfect torsion theory is that localization is equivalent to base change. –  Carl Weisman Dec 15 '12 at 13:52
    
yes, I was not saying that the localization is equivalent to $-\otimes_RR_1$, I was just saying that maybe, when these two functors are different, maybe they have the same kernel... do you have some easy counterexample? –  Simone Virili Dec 15 '12 at 14:16
    
Let $R& = $K[X,Y]$ for a field K. Let $F$ be the Gabriel filter whose only member is the unit ideal. Let $G$ be the one comprising all ideals not contained in a proper principal ideal. Both localize $R$ to itself. But, while $F$ localizes everything to itself, $G$ localizes $K$, as an $R$-module, to $0$. –  Carl Weisman Dec 17 '12 at 13:41
    
Let R = K[X,Y] for a field K. Let F be the Gabriel filter whose only member is the unit ideal. Let G be the one comprising all ideals not contained in a proper principal ideal. Both localize R to itself. But, while F localizes everything to itself, G localizes K, as an R-module, to 0. –  Carl Weisman Dec 17 '12 at 13:45

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.