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Is it true that in the category $R$-Mod of $R$-modules (and $R$-module homomorphisms), the diagram $$ M \longrightarrow M \otimes_{R} N \longleftarrow N,$$ where the arrows are the maps $m \to m \otimes 1$, $n \to n \otimes 1$, is a coproduct diagram?

I know that for commutative rings instead of $R$-modules this is true, but I don't why this is different. A detailed explanation for a noob in category theory would be nice. (not hw, I would just like to get this straight, thank you very much).

Stefan Hamcke
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TJIF
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  • I don't have a proof for you, but Wikipedia claims that this is not the co-product unless you are in the category of commutative $R$-modules, and that the co-product is a free product. http://en.wikipedia.org/wiki/Tensor_product_of_algebras – Unwisdom Feb 28 '14 at 01:52
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    If $M$ is a module, why is there an element $1 \in M$? – John Palmieri Feb 28 '14 at 01:54
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    Unwisdom: what the heck is a commutative $R$-module? – darij grinberg Feb 28 '14 at 02:06
  • Commutative $R$-algebra. Sorry. – Unwisdom Feb 28 '14 at 02:08
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    Independently of this, no -- this is not a coproduct diagram, not even in the category of "unital" (or "coaugmented") $R$-modules which are $R$-modules with a distinguished element $1$. I think a coproduct diagram would be obtained by taking the direct sum of $M$ and $N$ and factoring out the submodule spanned by the difference between the $1$ of $M$ and the one of $N$ (and declaring the projection of said $1$'s to be the new $1$). (If I am not a mistaken, a product diagram would be obtained by taking the direct sum of $M$ and $N$ and declaring its element $\left(1,1\right)$ to be its $1$.) – darij grinberg Feb 28 '14 at 02:10
  • @Darij: Why posting answers as comments? – Martin Brandenburg Feb 28 '14 at 03:46

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There is no way of defining the inclusion maps (what is $1 \in M$?). When $M,N$ are free of ranks $n,m$, we have that also $M \otimes_R N$ is free of rank $n \cdot m$. But the direct sum i.e. coproduct $M \oplus N$ is free of rank $n+m$. So they are not isomorphic when $n,m>0$. In fact, there is a big difference between them.

There is the following connection between tensor products and direct sums (see math.SE/291644): For every $m \in M$ let $i_m : N \to \oplus_{m \in M} N$ denote the inclusion. Similarly, for every $n \in N$ let $j_n : M \to \oplus_{n \in N} M$ denote the inclusion. Then $$M \otimes_R N = \bigoplus_{m \in M} N \oplus \bigoplus_{n \in N} M / \langle i_m(n) - j_n(m) \rangle_{m \in M, n \in N}.$$

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Martin Brandenburg
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