2
$\begingroup$

I was asked to prove that $\varinjlim{M_i}\otimes\varinjlim{N_i}\cong \varinjlim{(M_i\otimes N_i)}$, where $M_i $ and $N_i $ are $R$-modules. By left adjointness of the tensor product, it is easy to see that $$\varinjlim{M_i}\otimes\varinjlim{N_i}\cong \varinjlim ({M_i}\otimes\varinjlim{N_i})\cong \varinjlim \varinjlim (M_i\otimes N_j)$$

I put different indices $i,j$ to make clearer that they refer to two differents inductive limits on a filtered poset $I$. But how can i get from that an expression with just "one" direct limit? Moreover, I was wondering if it were true just in this case (so with tensor product) or it was something more categorical and being true in general when one takes more limits on the same family. Thank you in advice!

$\endgroup$
1
  • $\begingroup$ You could argue by universal properties. For instance, you can show that $\varinjlim{(M_i\otimes N_i)}$ fulfills the universal property that defines $\varinjlim{M_i}\otimes\varinjlim{N_i}$, or the other way around. Should be a straing-forward (albeit maybe a bit long) diagram chase exercise. $\endgroup$ – Arthur Apr 26 '17 at 18:19
1
$\begingroup$

I presume your direct limits are over $\Bbb N$. You have in essence a family of modules $A_{i,j}$ indexed over $\Bbb N\times\Bbb N$ with maps $\phi_{i,j,k,l}:A_{i,j}\to A_{k,l}$ when $i\le k$ and $j\le l$. From directedness the limit consists of objects $(i,j,a)$ with $a\in A_{i,j}$ with $(i,j,a)=(i',j',a')$ when there are $k$, $l$ such that $\phi_{i,j,k,l}(a)=\phi_{i',j',k,l}(a')$. Now each $(i,j,a)$ can be represented as a $(k,k,b)$ where $k=\max(i,j)$. Also if $(i,j,a)=(i',j',a')$ then $a$ and $a'$ will become equal in some $A_{l,l}$. So the direct limit over $\Bbb N\times\Bbb N$ is the same as the direct limit of the diagonal elements $A_{i,i}$.

To summarise, the diagonal is cofinal in $\Bbb N\times \Bbb N$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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