Let $T$ be a left $R$ tilting(partial) module, then any direct sum of copies of $T$ is tilting(partial).

Question

Let $T$ be a left $R$ tilting(partial) module, then any direct sum of copies of $T$ is tilting(partial).

I already find the following equivalences for a tilting and a classical tilting module $T$ which I think would made the proof easier; a left $T$ module over $R$ is tilting if and only if $Gen(T)=T^{\perp}$. Also, a left $T$ module over $R$ is partial tilting if and only if $Gen(T) \subset T^{\perp}$ and $T^{\perp}$ is a torsion class. Let me be clear about all definitions; $T^{\perp}$ consists of all left modules $N$ over $R$ such $$Ext_{R}^{1}(T,N)=0$$.

And $Gen(M)$ is the class of all left modules over $R$ that are $M$ generated, that means $N \in Gen(M)$ if there is an epimorhism $f:M^{(X)} \twoheadrightarrow N$ where $M^{(X)}:= M \oplus M \oplus M\oplus...$, $X$ times where $X$ is a set. Also, we say a class of modues is a torsion class if it is closed under quotients, direct sums and extensions. I already proved that for a tilting module $T$ we have $T^{\perp}$ is closed under extensions and $Gen(T)$ is always closed over direct sums(even if $T$) is not tilting by a question I did yesterday.

Answer 1

I think it's probably easier to do it from definitions because that way you don't need to worry about showing $(T^{(\kappa)})^{\perp}$ is closed under arbitrary direct sums. It is enough to show $T^{(\kappa)}$ is tilting for every cardinal $\kappa$ as every tilting module is partial tilting.

Firstly if $T$ has finite projective dimension then so does $T^{(\kappa)}$ because $\text{Ext}^{n}(T^{(\kappa)},N)\cong \prod_{\kappa}\text{Ext}^{n}(T,N)$ for all $N\in\text{Mod-}R$ and $n<\omega$.

Second, for every $n<\omega$ and cardinal $\lambda$ you have $\text{Ext}^{n}((T^{(\kappa)}, (T^{(\kappa)})^{(\lambda)})\simeq \prod_{\kappa}\text{Ext}^{n}(T,(T^{(\kappa)})^{(\lambda)})=0$ as $T$ is tilting.

Finally, as $T$ is tilting there is the exact sequence $0\to R\to T_{0}\to\cdots\to T_{r}\to 0$ with $T_{j}\in\text{Add}(T)$. But by taking enough direct sums as needed you can show $T_{j}\in\text{Add}(T^{(\kappa)})$.

Notice you do not need to assume $1$-tilting with these definitions, but I think $\text{Gen}(T)=T^{\perp}$ holds only for $1$-tilting.

Answer 2

For both the statements, it is enough to note that for a nonempty index set $I$, $\mathrm{Gen}(T)=\mathrm{Gen}(T^{\oplus I})$ and $T^{\perp}=(T^{\oplus I})^{\perp}$. [This will, in fact, show that $T$ and $T^{\oplus I}$ are equivalent tilting modules in the "tilting" case, meaning that their tilting class is the same.]

$\mathrm{Gen}(T)=\mathrm{Gen}(T^{\oplus I})$ is immediate, as both are the class of all epimorphic images of arbitrary direct sums of $T$.

For $T^{\perp}=(T^{\oplus I})^{\perp}$, note that given a module $M$, we have $$\mathrm{Ext}^{1}_R(T^{\oplus I}, M)\simeq (\mathrm{Ext}^{1}_R(T, M))^{\times I}.$$ The significance of this is that $\mathrm{Ext}^{1}_R(T^{\oplus I}, M)=0$ if and only if $\mathrm{Ext}^{1}_R(T, M)=0$. So $M \in (T^{\oplus I})^{\perp}$ iff $M \in T^{\perp}$.