This should be obvious but I'm not seeing it:

The $\mathfrak T$ be a triangulated category with coproducts and with a compact generator $A$ (that is, the functor $\mathfrak T(A,\_)$ preserves coproducts and the localizing subcategory $\langle A\rangle$ of $\mathfrak T$ generated by $A$ is all of $\mathfrak T$.)

For instance, $\mathfrak T$ could be the derived category of a ring or a ring spectrum.

Let $\langle A\rangle_{\aleph_1}\subseteq\mathfrak T$ be the $\aleph_1$-localizing subcategory of $\mathfrak T$ (that is, the smallest triangulated subcategory containing $A$ and being closed under countable coproducts).

Certainly, if $B$ belongs to $\langle A\rangle_{\aleph_1}$, then $\mathfrak T_*(A,B)$ is a countably generated module over $\mathfrak T_*(A,A)$.

Question: Does the converse hold, that is, do we have $$ \langle A\rangle_{\aleph_1}=\{B\in\mathrm{Ob}(\mathfrak T)\mid\text{$\mathfrak T_*(A,B)$ is countably generated over $\mathfrak T_*(A,A)$}\}\;? $$


1 Answer 1


I don't think what you say is true. Let $k$ be a field, $V$ a $k$-vector space of uncountable dimension and $R=k\oplus V$ the $k$-algebra where $V$ is a square-zero ideal. Consider $\mathfrak T=D(R)$ the derived category of $R$, and $A=R$. Take a non-trivial vector $0\neq v\in V$. The complex $$B=\cdots\rightarrow 0\rightarrow R\stackrel{v}\longrightarrow R\rightarrow 0\rightarrow \cdots$$ is in $\langle A\rangle_{\aleph_1}$, in fact it is in $\langle A\rangle_{\aleph_0}$. In this case $\mathfrak T_*(A,A)=R$ concentrated in degree $0$ and $\mathfrak T_*(A,B)=R/(v)\oplus V[1]$, which is not countably generated since $\dim_kV$ is uncountable.

What you claim is true under some transfinite coherence hypothesis, e.g. it is proven in the literature under the hypothesis that $\mathfrak T_*(A,A)$ is countable.

  • $\begingroup$ Thank you very much for your answer. Luckily, the case that $\mathfrak T_*(A,A)$ is the most important for me. Would it be too bold if I asked you for a reference in the literature? I searched through Neeman's book but could not find it. $\endgroup$
    – Rasmus
    Jun 29, 2013 at 16:16
  • $\begingroup$ You're welcome! Sorry, what case is the most important for you? $\endgroup$ Jun 29, 2013 at 16:24
  • $\begingroup$ The case that $\mathfrak T_*(A,A)$ is countable. Sorry... $\endgroup$
    – Rasmus
    Jun 29, 2013 at 16:25
  • $\begingroup$ It is a consequence of (shameless self-promotion) Theorem 8.1 in arxiv.org/abs/1304.3599 for $\alpha=\gamma=\delta=\aleph_0$. $\endgroup$ Jun 29, 2013 at 16:31
  • $\begingroup$ Thank you for the reference. I am left wondering why $\mathfrak T^{\aleph_1}=\langle A\rangle_{\aleph_1}$. According to Neeman's book $\mathfrak T^{\aleph_1}$ is $\aleph_1$-localizing. $\endgroup$
    – Rasmus
    Jun 29, 2013 at 17:33

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