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