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Can we express an arbitrary matrix ring $M_n(R[x])$ over the polynomial ring $R[x]$, where $R$ is a commutative ring with unity, as a direct product of other matrix rings (nontrivial)? I am finding similar to the fundamental theorem of finitely generated abelian groups. Do I need to modify some conditions about $n$ and $R$ in order to use some concepts in module theory?

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Ring factorizations of $M_n(S)$ for a ring $S$ correspond to central idempotents in $M_n(S)$, which are in turn necessarily contained in the diagonal copy of $S$. (These determine the ring factorizations of $S$ into pieces. (See this for the role of central idempotents, and this for the center of a matrix ring.)

So it depends on what the central idempotents of $R[x]$ are. Under some conditions the central idempotents must even be contained in $R$, for example, when $R$ is reduced (has no nilpotent elements.)

Without further specification of context, it is hard to say anything more.

Looks like even more is true than I remembered: as long as $R$ is commutative, the idempotents of $R[x]$ are contained in $R$, according to mathoverflow (Thanks Geoffry Trang for the comment)

So knowing the factorizations of $R$ will already tell you everything about factorizations of $M_n(R[x])$.

rschwieb
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    According to MathOverflow, idempotents in $R[x]$ must be constant polynomials (contained in $R$) for any commutative ring $R$. – Geoffrey Trang Apr 08 '24 at 14:45
  • @GeoffreyTrang I thought that might be the case, but didn't know the proof offhand. The one i was thinking relied on a grading of $R[x]$ and it was obvious if $R$ is reduced the idempotents are constant. Thank you! – rschwieb Apr 08 '24 at 14:59
  • What if $R=\mathbb{Z}_n$ where $n=p_1^{n_1}p_2^{n_2}p_3^{n_3}...p_k^{n_k}$? –  Apr 09 '24 at 00:13
  • @JordanG What about it? The Chinese Remainder Theorem tells you everything you'd want to know about its idempotents – rschwieb Apr 09 '24 at 13:44
  • I mean, what is a possible factorization of the matrix ring over $\mathbb{Z}n[x]$? Is it the product of matrix rings over $\mathbb{Z}{p_i^{n_i}}[x]$, $i=1,2,...,k$? –  Apr 09 '24 at 14:04
  • @JordanG The foregoing discussion says the decompositions of $\mathbb Z_n[x]$ are exactly those decomposing $Z_n$, so yes. At least, I assume you mean $\mathbb Z_n$ as the quotient of $\mathbb Z$ by $(n)$. – rschwieb Apr 09 '24 at 14:15
  • Thank you! I also appreciate if you drop some topics/resources in algebra about the first paragraph of your answer, especially about the relationship of the factorization of a matrix ring and the factorization of its base ring. –  Apr 09 '24 at 14:54
  • @JordanG I put in two links. – rschwieb Apr 09 '24 at 15:59