Projective and injective modules are compatible with block decompositions of algebras in an obvious way. Projective and injective modules are compatible with block decompositions of algebras in an obvious way.
$A$ is a $k$-algebra and $b$ is an idempotent in $Z(A)$.
(i)Let $P$ be a projective(resp. injective) $A$-module. Then $bP$ is a projective(resp. injective) $Ab$-module.
(ii) Let $Q$ be a projective (resp. injective) $Ab$-module. Then $Q$ remains projective(resp. injective) as an $A$-module.
I really cannot see how this is obvious. And I'm only able to come up with a proof for the projective case in (i). For (ii), is $Q$ even an $A$-module? Any hints would be appreciated!
 A: Think of $A$ as $A_1\times A_2$ where $A_1=bA$ and $A_2=(1-b)A$. Then every $A$-module
$M$ has the form $M=M_1\times M_2$ where $M_i$ are $A_i$-modules (if you insist, $M_1=bM$ and $M_2=(1-b)M$).
This way you can identify the $A_1$-module $N$ with the $A$-module $N\times\{0\}$.
When is $M=M_1\times M_2$, projective/injective? I claim it's injective
iff each $M_i$ is injective over $A_i$. So $M$ is injective if given an
injection $i:N\to P$, then the induced map $i^*:\text{Hom}_A(P,M)\to\text{Hom}_A(N,M)$
is surjective.
But$$\text{Hom}_A(N,M)=\text{Hom}_{A_1\times A_2}(N_1\times N_2,M_1\times M_2)=\text{Hom}_{A_1}(N_1,M_1)\times\text{Hom}_{A_2}(N_2,M_2)$$
etc. Then $i^*$ is onto iff both $\text{Hom}_{A_i}(P_i,M_i)\to\text{Hom}_{A_i}(N_i,M_i)$
are onto. Now for $i$ to be an injection it's necessary  and sufficient that the induced
maps $i_1:N_1\to P_1$ and $i_2:N_2\to P_2$. It follows that $M$ is injective
iff each $M_i$ is injective over $A_i$.
Identifyin the $A_1$-module $N$ with the $A$-module $N\times\{0\}$, since $\{0\}$
is injective over $A_2$ then $N$ is injective over $A$ iff it is injective over $A_1$.
