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For a nonzero finitely generated indecomposable module $P$ that is a projective module over a right Artinian ring $A$, its top $P/PJ(A)$ is a simple module. The Jacobson radical $J(A)$ is a nilpotent ideal, so the top is nonzero; the quotient is a semisimple module over the semisimple ring $A/J(A)$. If it decomposed, a nontrivial idempotent of its endomorphism ring would lift by projectivity to an endomorphism of $P$. The Hopkins-Levitzki theorem gives finite composition length, and the Fitting lemma makes the lift either invertible or nilpotent. Its induced map on the top would have the same property, impossible for that nontrivial idempotent.

Right multiplication by $e_t=h_t/H_{\lambda }$ on the group algebra is an idempotent with image $A{e}_t$. Every diagonal entry in the permutation basis is the coefficient of the identity in $e_t$, namely $1/H_{\lambda }$. Thus its rank equals its trace, giving $\dim S^{\lambda }=n!/H_{\lambda }$. Triangular standard-tableau ideals and the Robinson–Schensted correspondence identify this dimension with the standard-tableau count.

If idempotents $e_1,\dots ,e_N$ satisfy $e_i{e}_j=0$ for $i>j$, then the left ideals $A{e}_i$ have an internal direct sum. Multiply a relation $\sum z_i=0$ by $e_1$ on the right to obtain $z_1=0$, and continue in order. Pairwise orthogonality in both directions is unnecessary. If the sum has dimension $\dim A$, it is the whole algebra.

The vertex projective module of a path algebra $P(i)=kQ{e}_i$ has at vertex $j$ the span of paths from $i$ to $j$. Each arrow appends itself to such paths. It is a projective module because it is a summand of the free module $kQ$. Its endomorphism ring is $(e_{i}kQ{e}_i{)}^{\mathrm{op}}$, acting by right multiplication. The closed-path corner of a path algebra is a domain, so this ring has no nontrivial idempotents and $P(i)$ is indecomposable. Directed cycles may make these projectives infinite-dimensional.

For a Young tableau $t$, let $r_t$ sum its row permutations and $c_t$ be the signed sum of its column permutations in the group algebra of the symmetric group. The product $h_t=c_t{r}_t$ is a Young symmetrizer. The other order is another conventional realization. It satisfies $h_t^2=H_{\lambda }{h}_t$, where $H_{\lambda }$ is the hook product of a partition, so division by that scalar gives a primitive idempotent in characteristic zero. Acting on a tensor power constructs a Schur module.