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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.

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Thanks.

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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 limitwe 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:

- Obtain bilinear maps $M_i \times N \to P$ by composing $g_i$ with the canonical homomorphism $M_i\otimes N \to P$;
- 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 $M_i$;
- 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.
- 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.

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.

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