Symmetric polynomial centers have primitive split content idempotents

The nilpotent ideal of the content summand. For $1\le k\le d$ let $a_k$ be the value of ${\sigma }_k$ on any word of content $\alpha$ (independent of the order of the word). On the summand $p_{\mathrm{w}\mathrm{o}\mathrm{r}\mathrm{d}}A$ inside $eA$ write $x_r={\mathrm{w}\mathrm{o}\mathrm{r}\mathrm{d}}_r+N_r$ with commuting nilpotents $N_r$ by [F1]. Choose $M$ with $N_r^M=0$ for all $r$; then every product of more than $d(M-1)$ monomials of positive degree in the $N_r$ vanishes, because among more than $d(M-1)$ factors some $N_r$ occurs at least $M$ times (finite pigeonhole). Hence every polynomial in the $N_r$ with zero constant term is nilpotent, and in particular $N_k:=e{\sigma }_k(x)-a_{k}e$ is nilpotent on every $p_{\mathrm{w}\mathrm{o}\mathrm{r}\mathrm{d}}A$; since there are finitely many words, a common exponent makes $N_k$ nilpotent in $eA$. The $N_k$ commute, and the ideal $J$ they generate in the commutative ring $eZ(A)$ is nilpotent by the same degree count.

Elementary-symmetric generation without division. For a homogeneous symmetric polynomial, order its finitely many monomials lexicographically with $x_1>\cdots >x_d$. Its leading exponent vector $(b_1,\dots ,b_d)$ has $b_1\ge \cdots \ge b_d$: otherwise swapping an inverted pair produces a larger monomial with the same coefficient, contrary to maximality. The product ${\sigma }_1^{b_1-b_2}{\sigma }_2^{b_2-b_3}\cdots {\sigma }_d^{b_d}$ has leading monomial $x_1^{b_1}\cdots x_d^{b_d}$ with coefficient $1$. Subtract its multiple by the leading coefficient. The remaining polynomial is symmetric and has a smaller leading monomial of the same total degree; the finite set of degree-$m$ monomials makes repetition terminate. Apply this to each homogeneous component. Every symmetric polynomial is therefore a polynomial in the ${\sigma }_k$, over any field and without division by factorials.

The center of the content summand is a local ring. By step 1.2 every symmetric polynomial is a polynomial in ${\sigma }_1,\dots ,{\sigma }_d$, so the symmetric-center hypothesis gives $eZ(A)=Fe+J\mathrm{.}$ Since $e\mathrm{\ne }0$ and $J$ is nilpotent, $Fe\cap J=0$: if $ce\in J$ with $c\mathrm{\ne }0$, nilpotence would give $0=(ce{)}^M=c^{M}e$, contradicting $e\mathrm{\ne }0$. Therefore $eZ(A)\mathrm{/}J\cong F$, a field, and an idempotent $b$ has residue $0$ or $1$ in this field. In the first case $b\in J$ is nilpotent, and $b=b^M=0$; in the second case the same argument applied to the idempotent $e-b\in J$ gives $b=e$. Thus every idempotent of $eZ(A)$ is $0$ or $e$.

The nonzero content idempotents are exactly the primitive central idempotents. Step 2.1 shows that each $e_{\alpha }\mathrm{\ne }0$ is a primitive idempotent of $Z(A)$, hence primitive central in $A$. Conversely, let $b$ be a primitive central idempotent of $A$. The $e_{\alpha }$ are central, pairwise orthogonal and sum to $1$, so $b={\sum }_{\alpha }b{e}_{\alpha }$; if two terms $b{e}_{\alpha }$ were nonzero, they would be orthogonal nonzero idempotents summing to the primitive idempotent $b$, a contradiction, so exactly one term equals $b$. This nonzero $b{e}_{\alpha }$ is an idempotent of $e_{\alpha }Z(A)$, hence equals $e_{\alpha }$ by step 2.1, proving $b=e_{\alpha }$. Hence the nonzero $e_{\alpha }$ are precisely the primitive central idempotents, and the blocks of $A$ are exactly the nonzero content summands. No algebraic closure, splitting-simple-module theorem, power-sum generation, division by $d!$, or level-one restriction is used.