I am reading the book of Coutinho: A primer of Algebraic $D$-modules. In past, I usually study commutative algebra, so I am a freshmen with non-commutative (Weyl) algebra? In Chapter 12 of the Coutinho book, he constructs tensor product of two modules as follows:

Construction (page 109 of Coutinho's book)

Let $R, S$ and $T$ be (general) rings. Let $M$ be an $R$-$S$-bimodule and let $N$ be an $S$-$T$-bimodule. We will define the tensor product of $M$ and $N$ over $S$, denoted by $M \otimes_SN$.

First consider the set of $M \times N$ of all pairs $(u, v)$ with $u \in M$ and $v \in N$. Let $\mathcal{A}$ be the free Abelian group whose basis is formed by the elements of $M \times N$. The elements of $\mathcal{A}$ are formal (finite) sums of the form $$\sum_i a_i(u_i,v_i) \quad (\star)$$ with $a_i \in \mathbb{Z}$, $u_i \in M$, $v_i \in N$. Note that this sum of pairs are mere symbols: the sum is not an element of the direct sum $M \oplus N$. In fact, if we assume that different indices correspond to different pairs in $(\star)$, then the sum is zero iff each $a_i = 0$. If $r \in R$ and $t \in T$, put $$r(u, v) = (ru, v)$$ $$(u, v)t = (u, vt).$$ These are well-defined actions that make an $R$-$T$-bimodule of $\mathcal{A}$. ...

My question: Is $\mathcal{A}$ an $R$-module?

Example: I consider $r=0$. If $\mathcal{A}$ is an $R$-module then $0(u, v) = 0$. But in Coutinho's construction $0 (u, v) = (0u, v) = 1.(0, v) \neq 0$.

I also checked the construction of tensor product in the book "Homology" of S. Maclane, it is different from the Coutinho's one.

  • 1
    $\begingroup$ Why are you asking if it is an $R$-module if you've shown thar it isn't? :-) $\endgroup$ – Mariano Suárez-Álvarez Oct 14 '12 at 4:35
  • $\begingroup$ I think $\mathcal{A}$ is not an $R$-module. But Coutinho claimed it is an $R$-module. $\endgroup$ – Pham Hung Quy Oct 14 '12 at 5:37
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    $\begingroup$ To construct the tensor product you need to take the free abelian group on $M \times N$ and divide by the subgroup generated by the elements $(us, v) - (u,sv)$. In this way you obtain an abelian group; the actions of $R$ and $T$ are obtained by functoriality. $\endgroup$ – Angelo Oct 14 '12 at 8:48
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    $\begingroup$ What's important is the resulting tensor product, with its universal property. The constructions are just for proving its existence, they are never used in proofs. $\endgroup$ – Angelo Oct 14 '12 at 12:02
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    $\begingroup$ To Pham Hung Quy: there no reason to delete your question. $\endgroup$ – Angelo Oct 14 '12 at 17:34

I should finish this question: $\mathcal{A}$ is not an $R$-module as Coutinho's claiming. The construction of tensor product should by taking a free abelian group $\mathcal{A}$ and then modulo a certain subgroup (see the comment of Angelo or S. Maclane: Homology, Chapter V). In the case commutative ring $R$, we can take $\mathcal{A}$ is a free $R$-module.


I suggest to also have a look at Bourbaki's Algebra (edition from 1970 or later), Chapter II, Section 3, especially Paragraph 4 (about multimodule structures on tensor products).


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