$T^{-1}(N \otimes_A M) \cong T^{-1}N \otimes_A M$ Let $A \to B$ be a homomorphism of rings, let $S$ (resp. $T$ ) be a multiplicative subset in $A$ (resp. $B$) such that the image of $S$ in $B$ is contained in $T$ , and let $N$ be a $B$-mod.
How to show $T^{-1}(N \otimes_A M) \cong T^{-1}N \otimes_A M$ ?
By the universal property of tensor of module, I have the homomorphism such that $T^{-1}N \otimes_A M \ni (n/t) \otimes m \mapsto (n\otimes m)/t \in T^{-1}(N \otimes_A M)$.
However, in the case of the converse direction, I have no idea to construct a homomorphism except for struggling to prove the well-definedness by taking the elements.
Is there clear way ?
 A: We can construct mutually inverse isomorphisms of $B$-modules as follows:
The map
$$
  N × M \to (T^{-1} N) ⊗_A M \,,
  \quad
  (n, m) \mapsto (n / 1) ⊗ m
$$
is $A$-linear in the second argument and $B$-linear in the first, and therefore induces a $B$-linear map
$$
  N ⊗_A M \to (T^{-1} N) ⊗_A M \,,
  \quad
  n ⊗ m \mapsto (n / 1) ⊗ m \,.
$$
Every element of $T$ acts invertibly $T^{-1} N$ and therefore also invertibly on $(T^{-1} N) ⊗_A M$.
We get therefore an induced homomorphism of $T^{-1} B$-modules
$$
  φ
  \colon
  T^{-1}(N ⊗_A M) \to (T^{-1} N) ⊗_A M \,,
  \quad
  (n ⊗ m)/t \mapsto (n / t) ⊗ m \,.
$$
Let us now construct the inverse of $φ$, which we shall call $ψ$.
For every element $m$ of $M$, we have the map
$$
  N \to N ⊗_A M \to T^{-1} (N ⊗_A M) \,,
  \quad
  n \mapsto n ⊗ m \mapsto (n ⊗ m) / 1 \,.
$$
This map is $B$-linear, and every element of $T$ acts invertibly on $T^{-1}(N ⊗_A M)$.
We get therefore an induced homomorphism of $T^{-1} B$-modules
$$
  ψ_m
  \colon
  T^{-1} N \to T^{-1} (N ⊗_A M) \,,
  \quad
  n / t \mapsto (n ⊗ m) / t \,.
$$
By combining these maps $ψ_m$, we get a map
$$
  (T^{-1} N) × M \to T^{-1} (N ⊗_A M) \,,
  \quad
  ((n / t) , m) \mapsto (n ⊗ m) / t \,.
$$
This map is $T^{-1} B$-linear in the first coordinate (since these are the maps $ψ_m$), and $A$-linear in the second coordinate.
We get therefore an induced homomorphism of $T^{-1} B$-modules
$$
  ψ
  \colon
  (T^{-1} N) ⊗_A M \to T^{-1} (N ⊗_A M) \,,
  \quad
  (n / t) ⊗ m \mapsto (n ⊗ m) / t \,.
$$
The two homomorphisms $φ$ and $ψ$ are mutually inverse.
The homomorphism $φ$ is therefore an isomorphism, with inverse given by $ψ$.

We could also show the isomorphism $T^{-1} (N ⊗_A M) ≅ (T^{-1} N) ⊗_A M$ by using

*

*that localizing at $T$ is isomorphic to extension of scalars from $B$ to $T^{-1} B$, which is turn is isomorphic to $T^{-1} B ⊗_B (-)$; and


*the associativity of the tensor product.
We get from these ingredients the chain of isomorphisms of $T^{-1} B$-modules.
$$
  T^{-1} (N ⊗_A M)
  ≅
  (T^{-1} B) ⊗_B (N ⊗_A M)
  ≅
  ((T^{-1} B) ⊗_B N) ⊗_A M
  ≅
  (T^{-1} N) ⊗_A M \,.
$$
These isomorphisms can explicitly be described as
$$
  (n ⊗ m)/t
  \enspace\mapsto\enspace
  (1 / t) ⊗ (n ⊗ m)
  \enspace\mapsto\enspace
  ((1 / t) ⊗ n) ⊗ m
  \enspace\mapsto\enspace
  (n / t) ⊗ m \,.
$$
The overall isomorphism from $T^{-1} (M ⊗_A N)$ to $(T^{-1} N) ⊗_A M$ is therefore the same as in the previous approach (namely $φ$).
