Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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...

share|improve this question
    
associativity of the tensor product is dealt with in the early part of Cartan and Eilenberg; I don't have my copy here so I can't give a precise reference. –  mt_ Oct 24 '11 at 20:36
1  
@Zhen Lin: tensor product is definitely associative over different rings. this is more clearly necessary over non-commutative rings, where the modules are all bi-modules anyways. –  Jack Schmidt Oct 24 '11 at 20:41
    
I think that this question is a possible duplicate of: Help understand canonical isomorphism in localization (tensor products) –  Amitesh Datta Oct 24 '11 at 22:48

2 Answers 2

up vote 6 down vote accepted

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).

share|improve this answer
    
Thanks for explicitly stating the claim. Can you also provide a reference? –  Zhen Lin Oct 24 '11 at 21:22
1  
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 p. 27. –  Pierre-Yves Gaillard Oct 24 '11 at 21:38

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 ).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.