Let $k$ be a field and $A$ a finitely generated algebra over $k$. I know that if $A$ is finite dimensional as a vector space over $k$, then $A$ can be decomposed as a product of indecomposable $k$-algebras: $$A=A\_1\times\ldots\times A\_n$$ My question is if there is a similar decomposition of an arbitrary $k$-algebra. 


I'm not sure how to approach this problem because of the existence of descending chains, for example in $k[x]$ there is a chain: $$\langle x\rangle\supseteq\langle x^2\rangle\supseteq\langle x^3\rangle\supseteq\ldots$$ and the existence of idempotents with arbitrary degrees. I also tried to show the existence of an $N\in\mathbb{N}$ such that every idempotent in $k[x\_1\ldots,x\_n]/\langle f\_1,\ldots,f\_m\rangle$ can be seen as a sum of idempotents in $k[x\_1\ldots,x\_n]/\langle f\_1,\ldots,f\_m,x\_1^N,\ldots,x\_n^N\rangle$. But I have not had luck, so any help would be appreciated.


Thanks, Luis



Decomposition of finitely generated algebras
As I cannot mark a comment as an answer, I'll follow the ideas of YCor and Benjamin Steinberg to answer this question.


Let $A$ an arbitrary noetherian ring, then the boolean algebra of idempotents of $A$, say $Idem(A)$, (I choose this way to use the suggestion of Benjamin) is noetherian. Moreover, this boolean algebra is also artinian, because any descending chain defines an ascending chain (taking complements). Now is easy to see that $Idem(A)$ is an atomic boolean algebra and the set of atoms is finite. For the last assertion, if we suppose that $\{e\_n|n\in\mathbb{N}\}$ is a set of atoms then we have the following ascending chain $$e\_1\leq e\_1\vee e\_2\vee e\_1\vee e\_2\vee e\_3\leq\ldots$$ which is a contradiction. So $Idem(A)$ is finite as YCor said. 


