Proposition 15

Suppose that the ring A is semisimple.

a) For every $\lambda \in \mathscr{S},e_{\lambda}$ is the unique element of the center Z of A satisfying the relations $(e_{\lambda})_{S_{\lambda}}= 1_{S_{\lambda}}$ and $(e_{\lambda})_{S_\mu}= 0$ for $\mu\not=\lambda$.

b) The indecomposable idempotents in the ring Z are the $e_{\lambda}$, and the minimal ideals of Z are the $Ze_{\lambda}$ for $\lambda \in \mathscr{S}$.

c) Let M be an A-module and $(M_{\lambda})_{\lambda\in\mathscr{S}}$ the family of its isotypical components. The family of projectors associated with the decomposition of M into the direct sum of the $M_{\lambda}$ (VIII, p. 65) is $((e_{\lambda})_M)_{\lambda\in\mathscr{S}}$, and we have $M_{\lambda}=\mathfrak{a}_{\lambda}M$ for every $\lambda \in \mathscr{S}$.

Let $\lambda$ and $\mu$ be distinct in $\mathscr{S}$. We have $e_{\lambda}\in \mathfrak{a}_{\lambda}$, and $\mathfrak{a}_{\lambda}$ is contained in the annihilator $\mathfrak{b}_\mu$ of the A-module $S_\mu$ (VIII, p. 142, Proposition 9). We therefore have $(e_{\lambda})_{S_\mu}= 0$. The relation $(e_{\lambda})_{S_{\lambda}}= 1_{S_{\lambda}}$ follows because we have $1 =\sum_{\nu\in\mathscr{S}}e_{\nu}$. Assertion a) is then a consequence of Proposition 8 of VIII, p. 141.

Let $\lambda$ be in $\mathscr{S}$. The two-sided ideal $\mathfrak{a}_{\lambda}$ of A consists of the elements $x$ such that $x=xe_{\lambda}$; we therefore have $Z\cap \mathfrak{a}_{\lambda}= Ze_{\lambda}$, and therefore b) by Proposition 10, b).

Let us prove c). Let $x$ be an element of M. We have $e_{\lambda}\in \mathfrak{a}_{\lambda}$ for every $\lambda \in \mathscr{S}$. Since the mapping $a\mapsto ax$ from $A_s$ to M is A-linear, we have $\mathfrak{a}_{\lambda}x\subset M_{\lambda}$ (VIII, p. 66, Proposition 5) and, in particular, $e_{\lambda}x\in M_{\lambda}$. We have $1 =\sum_{\lambda\in\mathscr{S}}e_{\lambda}$, hence $x=\sum_{\lambda\in\mathscr{S}}e_{\lambda}x$. Consequently, $e_{\lambda}x$ is the component of $x$ in $M_{\lambda}$.