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Let $R$ be a commutative ring with identity, and $M$ an $R$-module. Then $$R\otimes_R M \simeq M$$ This can be easily shown directly by construction of the isomorphism, namely $r\otimes m \mapsto rm$ for $r \in R$ and $m \in M$. Indeed, let $rm=0$. Then $r \otimes m = r\left(1\otimes m\right) = 1 \otimes rm = 1 \otimes 0 = 0$. Hence the homomorphism is injective (I skip the proof that the map in question is indeed a homomorphism; this is straight forward). On the other hand, for any $m\in M$, $1\otimes m$ gets mapped to $m$, thus showing surjectivity.

My question is, does anyone know a way to prove the same result without explicit construction of the isomorphism, i.e. via the universal property?

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    $\begingroup$ Basically, bilinear maps from $R\times M$ are the same as linear maps from $M$. $\endgroup$ – Angina Seng Jun 10 '18 at 15:33
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Consider the map $\phi: R\times M\to Z$ as a $R$-bilinear map. Also consider the natural $R$-bilinear map $\gamma: R\times M\to M$ given by $(r,m)\mapsto rm$. Note that by $R$-bilinearity $\phi(r,m)=\phi(1,rm)$.

Using these two, we construct $\psi: M\to Z$ as follows: $\psi(m)=\phi(1,m)$. Now it is easy to show that $\phi=\psi\circ \gamma$. Hence, $M$ satisfies the universal property of tensor product, so we must have $M\simeq R\otimes M$.

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  • $\begingroup$ This might be a dumb question, but what about uniqueness of the map $\Psi$? You showed existence of such a map fitting into a commutative diagram. However for the universal property to be satisfied the map must also be unique. Did I overlook this part in your arguments? $\endgroup$ – Schief Aug 11 '18 at 14:41
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    $\begingroup$ If $f$ is any map satisfying $\phi=f\circ \gamma$ then $\phi(1,m)=f(\gamma(1,m))=f(m)$. Meaning, there is exactly one such map, namely $\psi$. $\endgroup$ – Hamed Aug 11 '18 at 14:48

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