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Let $k$ be a field. Let $A=\frac{k[X_1,\dots,X_r]}{I}$ for some proper ideal $I$ that, in principle, can be non-principal. Since $k[X_1,\dots,X_r]$ is Noetherian, we can apply the prime decomposition theorem and find primary ideals $q_1 , \dots, q_m$, such that $I = q_1 \cap \dots \cap q_m$.

Now I read here on page 3 that there is another possibility for the prime decomposition: $I=m_1^{\alpha_1}\cdots m_s^{\alpha_s} $, where the $m_i$ are maximal ideals. Is it another version of the prime decomposition theorem? If not, I think in this case, the product is the same thing as the intersection. First, note that $m_1^{\alpha_1}\cdots m_s^{\alpha_s} \subseteq m_1^{\alpha_1}\cap\dots\cap m_s^{\alpha_s}$ always holds. But since the $m_i$ are maximal, the other inclusion should hold; although I don't see why. But most importantly, I don't see why the primary components are maximal ideals in the first place?

Thank you!

Conjecture
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    The second version looks very false. Where did you read it? For a counterexample, consider $A = k[X,Y]$ and $I = (Y - X^2)$. Then $I$ itself is not maximal, and since It is reduced an irreducible then it can't be written as a product of powers of maximal ideals, either. I guess the second statement might hold for ideals defining $0$-dimensional algebraic sets. – Viktor Vaughn Dec 24 '24 at 16:37
  • @ViktorVaughn Thank you for the counterexample. I added where I read the statement in the post. – Conjecture Dec 24 '24 at 18:03
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    Okay, thanks for linking the source. You left out some very important context. In that article, the author is considering a specific setting: $E$ is an $n$-dimensional vector space and $A \subseteq \operatorname{End}(E)$ is a subalgebra of the endomorphism algebra of $E$. Since $\operatorname{End}(E)$ has dimension $n^2$, then $A$ is also finite-dimensional as a $k$-vector space, so $A \cong k[X_1, \ldots, X_r]/I$ is $0$-dimensional as an algebraic set. Here are some links that may be helpful: 1, 2 . – Viktor Vaughn Dec 24 '24 at 18:58
  • Thank you for the helpful links. There is still something I don't understand, how do we get from the decomposition of the radical of $I$ to the decomposition of $I$? I am not sure whether $I$ is radical in this case? – Conjecture Dec 25 '24 at 14:38

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