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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

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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
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 $M_{\mathfrak{F}_1} \cong M_{\mathfrak{F}_2}$.

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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...

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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

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