Help understand canonical isomorphism in localization (tensor products)

Yes, the map you construct is an isomorphism. It might be easiest to verify this by first using the canonical isomorphism $A_p\otimes_A M \cong M_p$, so that one then has the following simple chain of canonical isomorphisms: $$ M_P \otimes_{A_P} N_P \cong (A_P\otimes_A M)\otimes_{A_P} (A_P\otimes_A N) \qquad$$ $$\cong (A_P\otimes_A M) \otimes_A N \cong A_P\otimes_A (M\otimes_A N) \cong (M\otimes_A N)_P.$$


@user6495 Since I cannot leave comments, I will clarify the question you asked (I hope Matt E does not mind). The point is that the tensor product is commutative and associative. Therefore, we can write $(A_p\otimes_A M)\otimes_{A_p} (A_p\otimes_A N)\cong (M\otimes_A (A_p\otimes_{A_p} A_p))\otimes_A N\cong (M\otimes_A A_p)\otimes_A N\cong (A_p\otimes_A M)\otimes_A N$. The associativity of the tensor product used here is in the general sense of bimodules; $A_p$ is an $(A,A_p)$-bimodule.

Tags:

Localization

Commutative Algebra

Tensor Products

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