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



torsion theories localizing the base ring to the same ring
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.




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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 $M\_{\mathfrak{F}\_1} \cong M\_{\mathfrak{F}\_2}$.


