Is the Tensor Product of Flat Modules Flat?

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SUMMARY

The tensor product of finitely many flat modules over a commutative ring is indeed flat. This conclusion is established through the properties of flat modules and their behavior under tensor products. The discussion emphasizes the importance of understanding the foundational concepts of module theory and commutative algebra to grasp this result fully. No solutions were provided by other participants, highlighting the complexity of the topic.

PREREQUISITES
  • Understanding of flat modules in module theory
  • Familiarity with tensor products in algebra
  • Knowledge of commutative rings and their properties
  • Basic concepts of module homomorphisms
NEXT STEPS
  • Study the properties of flat modules in detail
  • Explore the construction and applications of tensor products
  • Investigate the role of commutative rings in module theory
  • Learn about module homomorphisms and their significance in algebra
USEFUL FOR

Mathematicians, algebraists, and students studying module theory and commutative algebra will benefit from this discussion, particularly those interested in the properties of flat modules and tensor products.

Euge
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Here is this week's POTW:

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Prove that the tensor product of finitely many flat modules over a commuative ring is flat.
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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!
 
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No one answered this week's problem. You can read my solution below.
By induction, it suffices to consider the tensor product of two flat modules, $M$ and $N$, over commutative ring $R$. Given a short exact sequence $0 \to X' \to X \to X'' \to 0$ of $R$-modules, tensoring with $N$ yields a short exact sequence $0 \to N \otimes_R X' \to N \otimes_R X \to N \otimes_R X'' \to 0$. Tensoring the latter sequence with $M$ results in the short exact sequence $0 \to M \otimes_R (N \otimes_R X') \to M \otimes_R (N \otimes_R X) \to M \otimes_R (N \otimes_R X'') \to 0$; by naturality of the associativity isomorphism, the sequence $0 \to (M \otimes_R N) \otimes_R X' \to (M \otimes_R N) \otimes_R X \to (M \otimes_R N) \otimes_R X'' \to 0$ is exact, as desired.
 

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