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Here is the screenshot of the proof on $M\otimes_S N$ being a left $R$-module when $M$ is an $(R,S)$-bimodule and $N$ a left $S$-module, taken from Module Theory by Blykh.

proof of theorem

What I am wondering is why, instead of verifying that $M\otimes_S N$ is a left $R$-module under such action directly, the author defined a balanced map $\vartheta_r$ compatible with the action and induced a $\Bbb Z$-endomorphism $f_r$ on $M\otimes_S N$ from it.

Here are some of my explanations:

When we are settled down to verify the left $R$-module structure on $M\otimes_S N$, the axioms \begin{align*} (r+s)\cdot p&=r\cdot p+s\cdot p \\ (rs)\cdot p&=r\cdot(s\cdot p) \\ 1_R\cdot p&=p \end{align*} where $p\in M\otimes_S N$ are rather trivial. Nonetheless, the last axiom for $p_1,p_2\in M\otimes_S N$ $$r\cdot(p_1+p_2)=r\cdot p_1+r\cdot p_2$$ is nontrivial. The definition of the action relies on a specific representation of each $p\in M\otimes_S N$. It would be hard for us to obtain the desired formula once the representation of $p_1+p_2$ is given differently from the representations of $p_1$ and $p_2$. However, for the sake of endomorphism $f_r$ on $M\otimes_S N$, everything is way simpler, as such axiom follows merely by the linearity of $f_r$.

Not sure if my intuitions are correct. Hope to hear from you guys.

Update: I also noticed that the function $f_r$ ensured the action to be well-defined. It is always a big problem when we define functions on elements that rely on specific structures.

Bernard Pan
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    This is the importance of defining maps out of a tensor product via the universal property: $f_r$ is a $\mathbb Z$-morphism (meaning, is additive), so $r(p_1+p_2) = f_r(p_1+p_2) = f_r(p_1)+f_r(p_2) = rp_1+rp_2$. – azif00 Jul 10 '21 at 21:02

2 Answers2

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The reason for defining a map likes this comes even before checking the axioms of a module: You need to have a well-defined skalar multiplication $(r,p)\mapsto rp$ for a ring element $r$ and every element $p$ of the module. Since elements may have more than one representation in tensor products, it is not entirely obvious that the multiplication $(r,\sum_im_i\otimes_Sn_i)\mapsto\sum_i(rm_i)\otimes_Sn_i$ is well-defined.

To prove that a map from a tensor product is well-defined, you almost always show that the corresponding map from the product of the modules is bilinear, and thus induces a homomorphism from the tensor product by the universal property of the latter.

anankElpis
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This is a really common type of proof when talking about the tensor product. The point is that you can get a lot done by utilizing the universal property of the tensor product, i.e. that a balanced map from the cartersian product induces a morphism from the tensor product.

The axioms of a module that you a verifying are the most standard ones, i.e. given in terms of an action of a ring on an abelian group. An equivalent set of axioms can be formulated in terms of a morphism from the ring into the endomorphism group of the underlying abelian group. That is what the author did here.

  • Thanks! I remember the theorem that $M$ is a left $R$-module if and only if there is a ring homomorphism from $R$ to $\operatorname{End}(M)$, but always find no place to use it. Your answer inspires me with a good application of this theorem. – Bernard Pan Jul 10 '21 at 21:55