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Let $R$ be a commutative ring, $\mathfrak{p}$ a prime ideal of $R$, $M$ a $R$-module, and $N$ a $R_\mathfrak{p}$-module. Why do we have this isomorphism? $M \mathbin{\otimes_R} N \cong M_\mathfrak{p} \mathbin{\otimes_{R_\mathfrak{p}}} N$

I can prove this by bare hands by taking one map to be defined by $m \otimes n \mapsto (m / 1) \otimes n$ and the other map by $(m / s) \otimes n \mapsto m \otimes (n / s)$. A little work is required to show that the latter is well-defined, but when that is done we have two mutually inverse $R$-linear (and $R_\mathfrak{p}$-linear) maps. But what is the conceptual reason for this isomorphism? Expanding the right hand side a bit, we see that we are saying $M \mathbin{\otimes_R} N \cong (M \mathbin{\otimes_R} R_\mathfrak{p}) \mathbin{\otimes_{R_\mathfrak{p}}} N$ and expanding the left hand side, it seems that what we want to prove is $M \mathbin{\otimes_R} (R_\mathfrak{p} \otimes_{R_\mathfrak{p}} N) \cong (M \mathbin{\otimes_R} R_\mathfrak{p}) \mathbin{\otimes_{R_\mathfrak{p}}} N$ but I see no reason why tensor products over different rings should associate like that...

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    I think that this question is a possible duplicate of: [Help understand canonical isomorphism in localization (tensor products)](http://math.stackexchange.com/questions/41750/help-understand-canonical-isomorphism-in-localization-tensor-products)2011-10-24

2 Answers 2

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Let $A,B$ be associative rings with $1$. Let $X$ be a right $A$-module, $Y$ an $(A,B)$-bimodule, and $Z$ a left $B$-module. Then the obvious functorial morphisms $ (X\otimes_AY)\otimes_BZ \rightleftarrows X\otimes_A(Y\otimes_BZ) $ are inverse isomorphisms.

Here are two references:

  • Bourbaki, Algèbre, II.3.8, Proposition 8, p. 64.

  • Cartan-Eilenberg, Homological algebra, II.5, Proposition 5.1, p. 27.

EDIT. Cartan and Eilenberg don't really give a proof. It doesn't seem easy to find an online proof. So, I thought it might be worth writing such a proof here. I looked at Bourbaki's and Atiyah-MacDonald's proofs. The one below is closer to Atiyah-MacDonald, but I think things get more transparent when one zooms less on the objects themselves, and more on the functors they represent.

Let $A$ and $C$ be rings, let $X$ be a right $A$-module, $Y$ an $(A,C)$-bimodule, and $Z$ a left $C$-module. We must show that there is a (unique) $\mathbb Z$-linear morphism $ \left(X\ \underset{A}{\otimes}\ Y\right)\ \underset{C}{\otimes}\ Z\to X\ \underset{A}{\otimes}\ \left(Y\ \underset{C}{\otimes}\ Z\right) $ satisfying $ (x\otimes y)\otimes z\mapsto x\otimes(y\otimes z).\tag1 $ Let $M$ be a $\mathbb Z$-module. Let $B$ be the $\mathbb Z$-module of those $\mathbb Z$-bilinear maps
$ b:\left(X\ \underset{A}{\otimes}\ Y\right)\times Z\to M $ which satisfy identically $b(\tau c,z)=b(\tau,cz)$, and let $T$ be the $\mathbb Z$-module of those $\mathbb Z$-trilinear maps
$ t:X\times Y\times Z\to M $ which satisfy identically $t(xa,y,z)=t(x,ay,z)$ and $t(x,yc,z)=t(x,y,cz)$.

Consider the $\mathbb Z$-linear map from $B$ to $T$ which attaches to $b$ in $B$ the element $t$ of $T$ defined by $t(x,y,z):=b(x\otimes y,z)$.

Given a $t$ in $T$ we'll define an element $b$ in $B$ by a construction inverse to the one in the previous sentence.

Pick a $z$ in $Z$, and form the $\mathbb Z$-bilinear map $ b_z:X\times Y\to M $ given by $b_z(x,y):=t(x,y,z)$. One checks that $b_z$ induces a $\mathbb Z$-linear map $ \ell_z:X\ \underset{A}{\otimes}\ Y\to M, $ and that $b(\tau,z):=\ell_z(\tau)$ fits the bill.

Put $ F(X,Y,Z):=\left(X\ \underset{A}{\otimes}\ Y\right)\ \underset{C}{\otimes}\ Z,\quad G(X,Y,Z):=X\ \underset{A}{\otimes}\ \left(Y\ \underset{C}{\otimes}\ Z\right). $ The above observations provide a functorial isomorphism $ \text{Hom}_{\mathbb Z}(F(X,Y,Z),?)\simeq\text{Hom}_{\mathbb Z}(G(X,Y,Z),?). $ Yoneda's Lemma gives then a functorial isomorphism $F\to G$, and one easily verifies that it satisfies (1).

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    Dear @Zhen Lin: Here are two references: (1) Bourbaki, **Algèbre**, II.3.8, Prop. 8. (2) Cartan-Eilenberg, **Homological algebra**, II.5, [Prop. 5.1](http://www.archive.org/stream/homologicalalgeb033541mbp#page/n48/mode/1up) p. 27.2011-10-24
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As often, a more general result is easier to understand.
So let us forget about localizations and consider a morphism of commutative rings $\phi:A\to B$, an $A$-module $M$ and a $B$-module $N$.
Every $B$-module $T$ can also be considered as an $A$-module ("forgetful functor", "restriction of scalars"), which we will denote by $T_A$.

We then have a canonical isomorphism of $A$-modules $M\otimes_A (N_A )\stackrel {\sim} {\to} ((M\otimes_A B)\otimes_B N)_A $

sending $m\otimes n \mapsto(m\otimes1)\otimes n $
which specializes to what you want.

The geometric picture is that you have a morphism of affine schemes $f=\phi^*:Y=Spec(B)\to X=Spec(A)$, a quasi-coherent sheaf $\mathcal M=\tilde M$ on $X$ and a quasi-coherent sheaf $\mathcal N=\tilde N$ on $Y$.
The above isomorphism of modules translates into the isomorphism of sheaves of $\mathcal O_X$-Modules

$\mathcal M \otimes_{\mathcal O_X} f_* \mathcal N \stackrel {\sim} {\to} f_*(f^*\mathcal M \otimes_{\mathcal O_Y} \mathcal N) $ In algebraic geometry, this is called the projection formula (it also appears in other contexts ).