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I am having a problem with exercise 8.9 in A Term of Commutative Algebra by Altman and Kleiman.

The exercise goes as follows:

Let $R$ be a ring, $R′$ an $R$-algebra, $M$, $N$ two $R′$-modules. Show there is a canonical $R$-linear map $\tau : M\otimes_R N \to M \otimes_{R′} N$.

Let $K \subset M \otimes_R N$ denote the $R$-submodule generated by all the differences $(x′ m) \otimes n − m \otimes (x′ n)$ for $x′ \in R′$ and $m \in M$ and $n \in N$. Show $K$ is equal to $\ker(\tau)$, and $\tau$ is surjective. Show $\tau$ is an isomorphism if $R′$ is a quotient of $R$.

They have given the following solution ((8.3), (8.6) and (8.8) are pictured below):

The canonical map $\beta′ : M \times N \to M \otimes_{R′} N$ is $R′$-bilinear, so $R$-bilinear. Hence, by (8.3), it factors: $\beta′ = \tau\beta$ where $\beta : M \times N \to M \otimes_R N$ is the canonical map and $\tau$ is the desired map.

Set $Q := (M \otimes_R N )/K$. Then $\tau$ factors through a map $\tau′ : Q \to M \otimes_{R′} N$ since each generator $(x′ m) \otimes n − m \otimes (x′ n)$ of $K$ maps to $0$ in $M \otimes_{R′} N$.

By (8.8), there is an $R′$-structure on $M\otimes_R N$ with $y ′ (m \otimes n) = m \otimes (y ′ n)$, and so by (8.6)(1), another one with $y ′(m \otimes n) = (y ′ m) \otimes n$. Clearly, $K$ is a submodule for each structure, so $Q$ is too. But on $Q$ the two structures coincide. Further, the canonical map $M \times N \to Q$ is $R′$-bilinear. Hence the latter factors through $M \otimes_{R′} N$, furnishing an inverse to $\tau ′$. So $\tau ′ : Q \xrightarrow{\sim} M \otimes_{R′} N$. Hence $\ker(\tau)$ is equal to $K$, and $\tau$ is surjective.

Finally, suppose $R′$ is a quotient of $R$. Then every $x′ \in R′$ is the residue of some $x \in R$. So each $(x′ m) \otimes n − m \otimes (x′n)$ is equal to $0$ in $M \otimes_R N$ as $x′ m = xm$ and $x′ n = xn$. Hence $\ker(\tau)$ vanishes. Thus $\tau$ is an isomorphism.

(8.3)

(8.6)

(8.8)

Now to my questions: What do they mean by saying that $Q$ is a submodule? It is a quotient, so how can it be a submodule? And what is the meaning of the statement "on $Q$ the two structures coincide"?

Thank you for any inputs.

Student G
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1 Answers1

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I don't believe $Q$ is a "sub-module", but maybe it could be seen as one. On the other hand, it is easy to see why on $Q$ the two structures coincide. They're telling you how $M\otimes_{R}N$ has two structures as a $R'$-module; however, in $Q$ we have $$ x'((m\otimes n) + K) = (x'm) \otimes n + K = m \otimes (x'n) + K = x'((m\otimes n) + K) $$ where the second equality holds because $(x'm) \otimes n - m \otimes (x'n)\in K$. I hope this helps.