Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 138 / 5 / b

We prove $(i)\Rightarrow (iii)$, the main step in Ext-connected components determine blocks. Suppose a block of a finite-dimensional algebra $B=Re$ had more than one component of the graph of nonsplit extensions between its simple modules. Partition its simple types into one component and the union of the others. There are no cross-extensions, so part (a) decomposes the left regular $B$-module as $B=U_1\oplus U_2$. Both summands are nonzero: every simple $B$-module is a quotient of the regular module and therefore occurs among its composition factors.

The maximality in (a) makes these summands canonical. Every right multiplication is a left $B$-module endomorphism, so it preserves them. Thus they are two-sided ideals, and their projection $p:B\to U_1$ commutes with both left and right multiplication. Putting $f=p(e)$ givesSince both summands are nonzero, $f\ne 0,e$. This contradicts the primitivity of the central block idempotent $e$. There is therefore just one Ext component within each block, proving that simples in the same block satisfy $\boxed{S\sim T}$.

We prove $(ii)\Rightarrow (i)$. The central block idempotents split every finite-dimensional module into its block components. A nonzero indecomposable representation, in particular an indecomposable projective module, has only one such component; otherwise these give a nontrivial direct-sum decomposition.

Every Jordan–Hölder factor of a module in a block belongs to that block, since its central idempotent acts as identity on submodules and quotients. Thus any two simple factors of the same projective indecomposable are in the same block. Applying this along the proposed chain yields $\boxed{S,T\text{ belong to the same block}}$.

We prove $(iii)\Rightarrow (ii)$. Consider one nonsplit short exact sequencewith $U,W$ simple. Lift the canonical surjection $P(W)\to W$ from the projective cover through $q$, obtaining $f:P(W)\to V$. Its image maps onto $W$. The intersection $\mathrm{im}f\cap U$ is either zero or $U$, since $U$ is simple. If it were zero, $q$ would identify $\mathrm{im}f$ with $W$, splitting the extension. Hence $U\subset \mathrm{im}f$, and surjectivity onto $W$ then implies $\mathrm{im}f=V$.

Therefore both $U$ and $W$ occur as Jordan–Hölder factors of the indecomposable projective $P(W)$. Replacing every edge in an Ext chain by this observation proves (ii). Together with the preceding two implications, this provesThe repeated word “projective” in the printed version of (ii) has no mathematical effect.