3
$\begingroup$

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

  • 0
    Yes, it is a vector space over the field, which also has the structure of a ring.2012-10-03

0 Answers 0