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Let $A$ be a finite algebra over field i.e. $A=\sum_{i=1}^{n}ke_i$, $\{I_\alpha\}$ set of all maximal ideals of $A$.

I know that $\#\{I_\alpha\}<\infty$ and $r(A)$ is nilpotent i.e. $r(A)^n=0$ for some $n\in\mathbb{N}$.

Am I right that:

1th: $$r(A)^n=(\cap_{\alpha}I_{\alpha})^n=(\prod_{\alpha}I_{\alpha})^n=\prod_{\alpha}I_{\alpha}^n=\cap_{\alpha}I_{\alpha}^n$$ because all $I_\alpha$ are termwise coprime and all $I_{\alpha}^{n}$ are termwise coprime too.

2nd: by chinese remainder theorem $$A=A/r(A)^n=A/\cap_{\alpha}I_{\alpha}^n=\prod_{\alpha}A/I_{\alpha}^n$$

3th: is it true that this decomposition isn't unique?

Thanks a lot

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A finite dimensional commutative algebra is a direct product of local rings (your A/I^n), and I think the decomposition is unique in a reasonably strong sense. This sort of thing needs a lot of adjustment in the non-commutative case. Generally speaking, direct sum decomposition of finite-dimensional (finite-length) modules is quite strongly unique by Fitting's theorem or Krull-Remak-Schmidt. –  Jack Schmidt Oct 3 '12 at 17:31
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Sadly the noncommutative version refuses to give us local rings in general :( –  rschwieb Oct 3 '12 at 19:00
    
@JackSchmidt: sorry for stupid question, but is it true that finite dimensional algebra over field is only vector space with some extra structure? –  Aspirin Oct 3 '12 at 19:13
    
Yes, it is a vector space over the field, which also has the structure of a ring. –  Rankeya Oct 3 '12 at 20:32

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