This is Exercise 2.20 in Atiyah-Macdonald.

How can we prove that $\varinjlim (M_i \otimes N) \cong (\varinjlim M_i) \otimes N$ ?

Atiyah-Macdonald give a suggestion, they say that one should obtain a map $g \colon (\varinjlim M_i )\times N \longrightarrow \varinjlim (M_i \otimes N)$, but I don't know how to do this.



2 Answers 2


One way is to use the adjunction property of the tensor product. For any $R$-module $P$, we have the following sequence of canonical $R$-module isomorphisms:

$$\begin{eqnarray}\hom(\varinjlim(M_i \otimes N), P) \cong&& \varprojlim \hom(M_i \otimes N, P) \\ \cong && \varprojlim \hom(N, \hom(M_i, P)) \\ \cong && \hom(N, \varprojlim \hom(M_i, P)) \\ \cong&& \hom(N,\hom(\varinjlim M_i, P)) \\ \cong&& \hom((\varinjlim M_i) \otimes N, P)\end{eqnarray}$$

and thus $$\hom(\varinjlim(M_i \otimes N), -) \cong \hom((\varinjlim M_i) \otimes N, -).$$

By the Yoneda lemma, $$\varinjlim(M_i \otimes N) \cong (\varinjlim M_i) \otimes N.$$

Of course, I made no use of the properties of the tensor product, other than its left-adjointness. The same argument shows that a functor which is left adjoint commutes with direct limits. Dually, a functor which is right adjoint commutes with inverse limits.

The proof which A&M suggest is basically just an expanded version of the same argument - see if you can figure it out!

Addendum: I'm not going to write the detailed proof from first principles (it's long) but I'll expand a bit on the hint in A&M. Here is the hint as it is in my edition, where they write $P=\varinjlim(M_i \otimes N)$ and $M=\varinjlim M_i$:

[...] For each $i\in I$, let $g_i:M_i \times N\to M_i\otimes N$ by the canonical bilinear mapping. Passing to the limit we obtain a mapping $g: M\times N \to P$. Show that $g$ is $A$-bilinear and hence defines a homomorphism $\phi: M\otimes N \to P$. Verify that $\psi\circ \phi$ and $\phi \circ \psi$ are identity mappings.

Here $\phi: P \to M\otimes N$ has been defined earlier as the limit of all the canonical homomorphisms $\mu_i\otimes 1 : M_i\otimes N \to M \otimes N$ where $\mu_i$ is the canonical homomorphism $M_i \to M$.

It's a pretty good hint! The only difficulty is the part in bold. We do not know how to take the limit of the $M_i \times N$ because we are not considering those as $R$-modules in the construction of the tensor product. We are considering $M_i \times N$ as a suitable domain for bilinear maps. Obtaining the map $\psi$ will be done in a few stages:

  1. Obtain bilinear maps $M_i \times N \to P$ by composing $g_i$ with the canonical homomorphism $M_i\otimes N \to P$;
  2. Convert these to homomorphisms $M_i \to \hom(N,P)$ by using the fact that a bilinear map $M_i \times N \to P$ is the same thing as family of homomorphism $N \to P$ parametrized $R$-linearly by the elements of $M_i$;
  3. Show that the homomorphisms $M_i \to \hom(N,P)$ thus obtained commute with the structure maps $\mu_i$ of the direct system of the $M_i$'s, and therefore obtain a homomorphism $M \to \hom(N,P)$ by the universal property of the direct limit.
  4. Convert this homomorphism back to a bilinear map $M \times N \to P$. Use the universal property of the tensor product to factor this map through $M\otimes N \to P$, which is the $\psi$ you need.

Then you must verify that $\psi$ and $\phi$ are inverses to each other.

  • 2
    $\begingroup$ I don't think the Yoneda Lemma is mentioned in chapter 2! $\endgroup$
    – user38268
    Mar 28, 2012 at 22:15
  • 1
    $\begingroup$ It's not, unfortunately. (However we don't even need it in full generality - we only need that $\hom(V,-) \cong \hom(V',-) \Rightarrow V \cong V'$.) $\endgroup$ Mar 28, 2012 at 22:18
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    $\begingroup$ It follows directly from the Yoneda lemma! We can put it in this way: the objects $M$ and $M'$ both represent the same functor (namely the functor $\hom(M,-)\cong \hom(M',-)$). And representing objects are unique up to isomorphism. If $\hom(M,-)\cong \hom(M',-)$, then in particular $\hom(M,M)\cong \hom(M',M)$. The image of the identity $1_M$ is a map $M' \to M$; it is the desired isomorphism. Check out any book on category theory, or the wikipedia page on the Yoneda lemma. In any case I'll write a more fleshed-out proof later. $\endgroup$ Mar 28, 2012 at 22:41
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    $\begingroup$ @Jr. No, I don't think it's essential to the book, but it's nonetheless an important construction in algebra and geometry. It's an example of a colimit, dual to the notion of limit. I expanded a bit on the hint in the book. Good luck! $\endgroup$ Mar 29, 2012 at 2:43
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    $\begingroup$ Maybe this is a bit silly, but It seems to me that you're taking the "wrong" variable to the other side. Writing $L = {-} \otimes N$ and $R = \hom(N,{-})$ you get: $$\begin{align*} \hom(L\varinjlim M_i, P) & = \hom(\varinjlim M_i, R P) \\ & = \varprojlim\hom(M_i,RP) \\ & = \varprojlim\hom(LM_i,P) \\ & = \hom(\varinjlim LM_i, P)\end{align*}$$ requiring one commutation less (and only in the first variable of $\hom$). This doesn't detract from your nice answer! $\endgroup$
    – t.b.
    Aug 7, 2012 at 15:39

Here's an idea for a proof that doesn't involve Yoneda:

To show $(\varinjlim M_n) \otimes N \cong \varinjlim(M_n \otimes N)$ it is enough to verify that $(\varinjlim M_n) \otimes N $ satisfies the universal property of $\varinjlim(M_n \otimes N)$. This means that if $(M_n, f_{nm})$ is the direct system with limit $(\varinjlim M_n, f_n)$ and $(Y, g_n)$ is any $R$-module such that $g_n = g_m \circ (f_{nm} \otimes id_N)$ for all $n \geq m$ then there exists a unique homomorphism $\alpha : (\varinjlim M_n) \otimes N \to Y$ such that the following diagram commutes (diagonal arrows omitted):

$$\begin{matrix} (\varinjlim M_n) \otimes N & \xrightarrow{\alpha} & Y \\ \left\uparrow{f_n \otimes id_N}\vphantom{\int}\right. & & \left\uparrow{g_m}\vphantom{\int}\right.\\ M_n \otimes N& \xrightarrow{f_{nm} \otimes id_N} & M_m \otimes N \end{matrix}$$

Define $\alpha$ as follows: for $m \otimes n \in (\varinjlim M_n) \otimes N $ there exist $k$ and $m_k \in M_k$ such that $f_k(m_k) = m$. Set $\alpha (m \otimes n) = g_k (m_k \otimes n)$. We need to show that $\alpha$ is well-defined, an $R$-module homomorphism, makes the diagram commute and is unique.

(i) $\alpha$ is well-defined: Let $k,j$ be such that $f_k \otimes id_N (m_k \otimes n) = m \otimes n = f_j \otimes id_N (m_j \otimes n)$.

We want to show that $g_k (m_k \otimes n) = g_j (m_j \otimes n)$. Since we have $f_k (m_k) \otimes n = m \otimes n = f_j (m_j ) \otimes n $ we have $f_k (m_k) = f_j (m_j)$. This means that $m_k$ and $m_j$ map to the same thing in $\varinjlim M_n$ hence by the properties of $\varinjlim M_n$ there exists an $l \geq j,k$ such that $f_{kl}(m_k) = f_{jl}(m_j)$. For the morphisms of $Y$ we have that $g_i = g_j \circ (f_{ij} \otimes id_N)$ where $i \leq j$ hence (as pointed out by t.b. in the comments): $$\begin{align} g_k (m_k \otimes n) &= g_l \circ (f_{kl} \otimes id_N) (m_k \otimes n) \\ &= g_l (f_{kl} (m_k) \otimes n) \\ &= g_l (f_{jl} (m_j) \otimes n) \\ &= g_l \circ (f_{jl} \otimes id_N) (m_j \otimes n) \\ &= g_j (m_j \otimes n) \end{align}$$

(ii) $\alpha$ is an $R$-module homomorphism, that is, $\alpha (m \otimes n + m^\prime \otimes n^\prime) = \alpha (m \otimes n) + \alpha (m^\prime \otimes n^\prime)$:

For $m \otimes n, m^\prime \otimes n^\prime \in (\varinjlim M_n) \otimes N$ there exist $j,k$ such that $m \otimes n = f_{k}(m_k) \otimes n$ and $f_{j} (m_j) \otimes n^\prime$ so that $\alpha (m \otimes n) = g_k (m_k \otimes n)$ and $\alpha (m^\prime \otimes n^\prime) = g_j(m_j \otimes n^\prime) $. Then for all $l \geq k,j$: $g_j(m_j \otimes n^\prime) = g_l (f_{jl}(m_j) \otimes n^\prime)$, similarly for $k$ and hence $$ \alpha (m \otimes n) + \alpha (m^\prime \otimes n^\prime) = g_k(m_k \otimes n) + g_j(m_j \otimes n^\prime) = g_l (f_{kl}(m_k) \otimes n) + g_l (f_{jl}(m_j) \otimes n^\prime) = g_l(f_{kl}(m_k) \otimes n + f_{jl}(m_j) \otimes n^\prime)$$

On the other hand, for $m \otimes n + m^\prime \otimes n^\prime$ there exists $i$ such that $m \otimes n + m^\prime \otimes n^\prime = f_i(m_i) \otimes n+ f_i(m_i^\prime) \otimes n^\prime$ and hence $\alpha (m \otimes n + m^\prime \otimes n^\prime) = g_i (m_i \otimes n + m_i^\prime \otimes n^\prime) = g_i (m_i \otimes n) + g_i (m_i^\prime \otimes n^\prime) = \alpha (m \otimes n) + \alpha (m^\prime \otimes n^\prime)$.

Alternative proof:

It's easier to show it like this: exploit the UP of $\varinjlim(M_n \otimes N)$ to get a unique $R$-module homomorphism $\alpha$ and then show that $\alpha$ is an isomorphism. To this end, consider the following diagram:

$$\begin{matrix} \varinjlim (M_n \otimes N) & \xrightarrow{\alpha} & (\varinjlim M_n ) \otimes N \\ \left\uparrow{f_n \otimes id_N}\vphantom{\int}\right. & & \left\uparrow{f_m \otimes id_N}\vphantom{\int}\right.\\ M_n \otimes N& \xrightarrow{f_{nm} \otimes id_N} & M_m \otimes N \end{matrix}$$

Since the diagram commutes it immediately follows that $\alpha$ is the map $m \otimes n \mapsto m \otimes n$ which is an isomorphism.

  • 1
    $\begingroup$ What do you mean by the "diagonal arrows" you omitted? $f_m \otimes {\rm id}_N$? I'm a bit confused by your argument: are you assuming that your directed system is linear? Otherwise I don't understand why there is $f_{kj}$ in the first place. Btw: All your $\varprojlim$'s should be $\varinjlim$'s. $\endgroup$
    – t.b.
    Aug 7, 2012 at 15:55
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    $\begingroup$ No, that's not at all what I'm saying. Your argument works fine if your index system is, say, the natural numbers (which are linearly ordered) and the maps in your system (the bonding maps) are all injective. In general you only have a directed system as an indexing system and no assumptions, so $M_k$ and $M_j$ need not be related at all, but there's an $M_l$ they map into. If your bonding maps are injective then this is enough, otherwise you need to go further up in the directed system to identify $m_k$ and $m_j$ (since they map to the same $m$ they must eventually be identified). $\endgroup$
    – t.b.
    Aug 7, 2012 at 16:25
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    $\begingroup$ Yes, well-definedness is the point of contention. You start with $m_j,m_k$ where $j,k$ are arbitrary. Now by directedness there's $l$ such that $j,k \leq l$. But you don't know that $m' = f_{jl}m_j \in M_l$ and $m'' = f_{kl}m_k \in M_l$ are equal, only that their image in the colimit is the same. So: when are two $m',m'' \in M_l$ identified? If and only if there is ... $\endgroup$
    – t.b.
    Aug 7, 2012 at 17:04
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    $\begingroup$ Never mind my previous comments. $\DeclareMathOperator{id}{id}$ Take $m \in \varinjlim M_i$, pick $m_j \in M_j$ such that $m = f_j(m_j)$ and put $\alpha(m\otimes n) = g_j(m_j \otimes n)$, similarly with index $k$. Now you know that $f_{jl}(m_j) = f_{kl}(m_k)$ for large enough $l \geq j,k$, so $$ \begin{align*} g_j(m_j \otimes n) & = g_l(f_{jl} \otimes \id)(m_j \otimes n) \\ & = g_l(f_{jl}(m_j) \otimes n) \\ & = g_l(f_{kl}(m_k) \otimes n) \\ & = g_l(f_{kl} \otimes \id)(m_k \otimes n) \\ & = g_k(m_k \otimes n) \end{align*}$$ as you wanted. Note: $g_j = g_l \circ (f_{jl} \otimes \id)$! $\endgroup$
    – t.b.
    Aug 8, 2012 at 10:49
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    $\begingroup$ I agree that you get the map $\alpha$ this way (the left-hand upwards arrow is labeled in a rather suspicious way, btw.) but I don't see how it immediately follows that $\alpha$ is an isomorphism, there's at least something to be said... Untangling this brings you back to what you did. Further glitch in the first part: Since you want to verify that $(\varinjlim M_i)\otimes N$ has the universal property of $\varinjlim (M_i \otimes N)$ your maps $g_n$ in the first argument are maps $g_n\colon M_n \otimes N \to Y$, so they have to satisfy $g_m = g_n \circ (f_{nm} \otimes {\rm id})$. $\endgroup$
    – t.b.
    Aug 8, 2012 at 11:32

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