In the following, all algebras are finite dimensional and unital and $\mathbb{K}$ is an algebraically flosed field.
I'm trying to understand some things about the concept of direct sum of algebras. Given $\mathbb{K}$-algebras $A_1$ and $A_2$ with identities $e_1$ and $e_2$ respectively, we define the direct sum $A_1\oplus A_2$ to be the direct sum of the corresponding vector spaces with multiplication given by $$(a_1+a_2)(b_1+b_2)=a_1b_1+a_2b_2.$$
If $M$ is an $A_1\oplus A_2$-module, then $(e_1,e_2)M$ gives me a decomposition $e_1 M\oplus e_2 M$, where I can see $e_iM$ as an $A_i$-module. Now I want to understand the projective indecomposable $A_1\oplus A_2$-modules. First, I claim that $P$ is an indecomposable $A_1\oplus A_2$ module, if and only if $P=P_1\oplus 0$ or $P=0\oplus P_2$ for some indecomposable $A_i$-module $P_i$. The if part is pretty obvious. For the only if part, suppose $P=Q_1\oplus Q_2$ and then I could write $$P=(Q_1\oplus 0) \oplus (0\oplus Q_2)$$ and $P$ would be decomposable. Moreover, $P_i$ above is indecomposable since otherwise if e.g. $P_1=M\oplus N$ I would have $$P=(M\oplus N) \oplus 0 = (M\oplus 0) \oplus (N\oplus 0).$$ (I realize that I am a bit handwavy here with the $\oplus$ sign and I hope I'm not doing something very wrong.)
Now, an indecomposable $P$ is projective if and only if $P_i$ above is projective too. I think this is pretty clear and can be seen by taking the corresponding natural projection (for the only if part) and injection (for the if part). Please correct me if I'm wrong.
So, if these are indeed the indecomposable projective $A_1\oplus A_2$-modules, I want to find the homomorphisms between them. If I have two indecomposable projective $A_1\oplus A_2$-modules of the form $P\oplus 0$ and $Q\oplus 0$ then the homomorphisms from $P\oplus 0$ to $Q\oplus 0$ are the same as the homomorphisms from $P$ to $Q$. But what about a homomorphism $f$ from $P\oplus 0$ to $0\oplus Q$? I would guess that there is only the zero-homomorphism, since you can just compose $f$ with the projection on both each summand and get $0$. But I'm not really sure that I have not forgotten anything. Is my reasoning correct?
Sorry for the long and boring question, I just want to be sure that I have understood things correctly.
