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How to prove, that tensor products commute with direct limits, if the main ring is not the same?

For every $i$ we have modules $L_i$ and $M_i$ over a ring $A_i$, and for every $i \geq j$ homomorphisms $f^i_j: L_i \rightarrow L_j$, $g^i_j: M_i \rightarrow M_j$, $u^i_j: A_i \rightarrow A_j$, such that $f^i_j (al) = u^i_j(a)f^i_j(l)$, $g^i_j(am) = u^i_j (a) g^i_j (m)$ for every $a \in A_i,\, l\in L_i, \, m \in M_i$. To prove, that $\varinjlim (L_i \otimes_{A_i} M_i) = (\varinjlim L_i)\otimes_{\varinjlim A_i} (\varinjlim M_i)$.

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I assume that you mean directed colimits. The correct notation is not $\lim$, but rather $\varinjlim$ or $\mathrm{colim}$. – Martin Brandenburg Sep 9 '13 at 15:35
up vote 4 down vote accepted

Let $L=\lim_i L_i$, $M=\lim_i M_i$ and $A=\lim_i A_i$.

For every $i$, we have canonical maps $$L_i\otimes_{A_i} M_i\to L\otimes_{A_i} M \twoheadrightarrow L\otimes_A M.$$ They pass to the inductive limit $$ f: \lim_i (L_i\otimes_{A_i} M_i) \to L\otimes_A M.$$ For all $i$, we have an $A_i$-bilinear map by composing $$ L_i\times M_i \to L_i\otimes_{A_i} M_i\to \lim_i (L_i\otimes_{A_i} M_i),$$ hence an $A_i$-bilinear map $$ L\times M\to \lim_i (L_i\otimes_{A_i} M_i)$$ which is $A$-bilinear. Hence we get an $A$-linear map $$ f: L\otimes_A M\to \lim_i (L_i\otimes_{A_i} M_i).$$ We check directly that $f$ and $g$ are inverse to each other.

Edit Add proof of the above claim.

Proof: To check that $g\circ f=\mathrm{Id}$, it is enough to check the equality holds for vectors of the form $x\otimes y$ with $x\in L, y\in M$ because they generate $L\otimes_A M$. For all $i$, denote by $r_i : L_i\to L$, $s_i: M_i\to M$ the canonical maps. Then there exist $i, x_i\in L_i$ and $y_i\in M_i$ such that $x=r_i(x_i)$ and $y=s_i(y_i)$. By construction, $f(x\otimes y)$ is the image of $x_i\otimes y_i\in L_i\otimes_{A_i} M_i$ in $\lim_i (L_i\otimes_{A_i} M_i)$. On the other hand, again by construction, $g$ takes this image to $r_i(x_i)\otimes s_i(y_i)=x\otimes y$ in $L\otimes M$. So $g\circ f=\mathrm{Id}$.

The equality $f\circ g=\mathrm{Id}$ is proved similarly.

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In what sense $L_i \otimes_A M_i$? $L_i$ and $M_i$ are modules over $A_i$, and there is a homomorphism $A_i \rightarrow A$. So there is no a restriction of scalars from $A_i$ to $A$. – user46336 Oct 28 '12 at 18:24
@user46336: sorry, I exchanged some indexes. Will edit. – user18119 Oct 28 '12 at 19:52
Why are they inverse? The main problem is to show, that $f: lim(L_i \otimes_{A_i} M_i) \rightarrow L \otimes_A M$ is a monomorphism. – user46336 Oct 29 '12 at 6:01
@user46336: I added the details. The point is to find the inverse rather to check the map is injective (in general hard to do directly with tensor products). – user18119 Oct 29 '12 at 10:19

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