The category I'm working in is $\operatorname{Mod}A$ for some unitary ring $A$.

I'm looking for a reference on when certain subcategories of $\operatorname{Mod}A$ give rise to torsion pairs. Let $\mathcal{T}$ be a strict full subcategory of $\operatorname{Mod}A$ which is closed under quotients, extensions and coproducts. Then when is the pair $(\mathcal{T}, \mathcal{T}^{\perp_{0}})$ a torsion pair in $\operatorname{Mod}A$?

Edit: $\mathcal{T}^{\perp_{0}} = \{ X \in \operatorname{Mod}A \: | \: \operatorname{Hom}_{A}(\mathcal{T}, X) = 0 \}$

Obviously, $\operatorname{Hom}(X, Y) = 0$ for all $X \in \mathcal{T}$ and $Y \in \mathcal{T}^{\perp_{0}}$.

Also, by definition, $\operatorname{Hom}(\mathcal{T}, Y) = 0$ if and only if $Y \in \mathcal{T}^{\perp_{0}}$.

The other way is less clear to me, namely $\operatorname{Hom}(X, \mathcal{T}^{\perp_{0}}) = 0$ if and only if $X \in \mathcal{T}$. The 'if' part is obvious, but the 'only-if' part I can't show.

Any help on seeing why this is true? Alternatively, why it's not true and which further conditions I need to make it true? Or just a reference to any of these.

  • $\begingroup$ What does $\perp_0$ mean? Perpendicular with respect to Hom but not necessarily with respect to higher Ext? $\endgroup$ – Jeremy Rickard May 24 '19 at 13:23
  • $\begingroup$ @JeremyRickard Pardon me, forgot to specify that. Yes, hom-orthogonal. $\endgroup$ – Auclair May 24 '19 at 20:04

Take $Y$ to be the sum of submodules of $X$ that are in $\mathcal{T}$. By the coproduct and quotient hypothesis, this is in $\mathcal{T}$.

Now if $Z\in\mathcal{T}$ and $Z\to X/Y$, then the image is a submodule of $X/Y$ that is in $\mathcal{T}$ (as a quotient of $Z$), therefore so is its pullback to $X$ (it is an extension with $Y$), therefore the image is included in $Y$; therefore $Z\to X/Y$ is $0$.

Therefore $X/Y\in\mathcal{T}^{\bot_0}$; therefore if $\hom(X,\mathcal{T}^{\bot_0})=0$, $X/Y=0$ so $X=Y\in\mathcal{T}$.


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