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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

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    $\begingroup$ Yes for finitely generated algebras. More generally if $A$ is an arbitrary noetherian ring, it has a finite number of minimal prime ideals. This implies that the number of idempotents is finite. This is standard material; I guess it's in most commutative algebra textbooks. $\endgroup$
    – YCor
    Apr 23 '17 at 23:29
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    $\begingroup$ Of course there's no decomposition for arbitrary commutative algebras: maybe the simplest example is the (Boolean) algebra of continuous functions from the Cantor set to $\mathbf{Z}/2\mathbf{Z}$. $\endgroup$
    – YCor
    Apr 23 '17 at 23:31
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    $\begingroup$ For general commutative rings the substitute is to view the ring as the global sections of a sheaf of indecomposable rings on its Pierce spectrum $\endgroup$ Apr 24 '17 at 1:05
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    $\begingroup$ anyway after reading twice, the question seems to be definitely about f.g. algebras. $\endgroup$
    – YCor
    Apr 24 '17 at 7:49
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    $\begingroup$ thanks for the commentaries, of course, the question is about f.g algebras. Thanks for the nice example with continuous functions. $\endgroup$ Apr 24 '17 at 21:28
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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.

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