Defenition: Let $\Lambda$ be an algebra. $\Lambda$ is called basic if $\Lambda \simeq \coprod\limits_{i=1}^nP_i$, where $P_i$ are nonisomorphic indecomposable projective modules.
Problem: Let $\Lambda$ be a basic artin $R$-algebra, let $\Lambda'$ be an R-subalgebra of $\Lambda$.
Prove that $\Lambda'$ is also basic.
My reasoning: It is known that if $\Lambda$ is basic then $\Lambda/r$ is basic (where $r$ is the radical of $\Lambda$). Hence, $r$ is projective and, therefore, $\Lambda$ is hereditary.
Let that $\Lambda' \simeq \coprod\limits_{j=1}^mM_j$, where $M_j$ are indecomposable modules. We shall show that $M_j$ are nonisomorphic projective modules.
If we consider a natural embedding of $\Lambda'$ into $\Lambda$ we will get $M_j$ are projective.
Question: How can we show that $M_j$ are nonisomorphic?
I know that Jacobson radical of $\Lambda$ equals nilradical, maybe it could help?