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My question is related to this youtube video around the time 31:45. I'm studying Commutative Algebra. The textbook I use is “Introduction to Commutative Algebra” by Atiyah. I want to understand Proposition 3.10 enter image description here and the proof of 3.10 need to understand the following statement: enter image description here

Could anyone give me the whole proof (with details) of this statement? Because I have no idea of how to prove it.

Elliot Yu
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Rose Sun
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  • This should follow from the fact that $N\otimes_B(B\otimes_AM)\cong N\otimes_AM$ for any $B$-module $N$. – Kenta S Sep 01 '22 at 15:03
  • Please do not rely only on images to state your questions. Besides being unsearchable and space consuming, ones that have problems with blurring, distortion and shading artifacts like these are simply hard to read. (but thank you for at least rotating them upright.) There isn't any reason not to type out at least the second image, which is what you're asking about specifically. – rschwieb Sep 01 '22 at 17:15

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If $M,B$ are $A$-modules, you can consider the extension of the scalars and $M_B=B\otimes_A M$ is a $B$-module.

If you take $R$ a $B$-module, you can consider the restriction of the scalars and see $R$ as an $A$-module.

From the commutativity and the associativity of the tensor product, follows $$M_B\otimes_B R=(M\otimes_A B) \otimes_B R=M\otimes_A( B \otimes_B R)\simeq M\otimes_A R.$$ So you can apply equivalently $\_\otimes_A M$ or $\_\otimes_BM_B$ and they are functors with the same properties.

FreeFunctor
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