is any indecomposable projective-injective A-module a direct summand of tilting module
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$\begingroup$ You are probable asking if such an indecomposable is a direct summand of every tilting module (for it surely is a direct summand of some tilting module: namely, the free tilting module $A$) if this were true, then a selfinjective algebra would have no nontrivial (basic) tilting modules. Is that true? $\endgroup$– Mariano Suárez-ÁlvarezCommented Jan 17, 2012 at 22:27
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1$\begingroup$ Indeed, that is true (for silly reasons: a tilting module has projective dimension at most 1, so it is in fact projective because the algebra is selfinjective) There are in general lots of tilting complexes, which is where I was getting at—but that's irrelevant here! $\endgroup$– Mariano Suárez-ÁlvarezCommented Jan 18, 2012 at 1:51
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2 Answers
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Yes, any indecomposable projective-injective $A$-module $P$ is (isomorphic to) a direct summand of every tilting $A$-module $T$. Just check that $T \oplus P$ is also a tilting module, and then use the fact that any two tilting $A$-modules have the same number of isomorphism classes of idecomposable summands.
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Alternatively, if $P$ is your module and $T$ is a tilting module, you know (from the definition of being tilting) that there is a short exact sequence $0\to P\to T_0\to T_1\to 0$ with $T_0$ and $T_1$ in $\operatorname{add} T$, because $P$ is an indecomposiable projective. But then, since it is also an indecomposable injective, the short exact sequence splits and $P$ is a direct summand of $T_0$. Now what you want follows from Krull-Schmidt's theorem.
answered Jan 17, 2012 at 23:55