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I'm trying to understand the proof of $R[x]\otimes_R M \cong M[x]$. In this page, he said for any $R$-module $P$, and the bilinear maps $h:R[x]\times M\to M[x]$, $f:R[x]\times M\to P$, if there exists a unique $R$-module homomorphism $\phi:M[x]\to P$ s.t. $\phi h=f$, then by the universal property we have $R[x]\otimes_R M \cong M[x]$.

But the universal property states that for any $R$-module $N$, and the bilinear maps $i:R[x]\times M\to R[x]\otimes_R M$, $g:R[x]\times M\to N$, we have a unique bilinear map $\tilde g:R[x]\otimes_R M\to N$ s.t. $\tilde g i=g$.

I don't know why the universal property implies the isomorphism $R[x]\otimes_R M \cong M[x]$ when we find $\phi$.

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  • $\begingroup$ Put in the commutative triangle $R[x]\times M$, $R[x]\otimes_R M$ and $M[x]$. By applying both universal properties we get $a:R[x]\otimes_R M\to M[x]$ and $b:M[x]\to R[x]\otimes_R M$, such that $a\circ b\circ i=i$. Apply the universal property to the triangle with $R[x]\times M$, $R[x]\otimes_R M$ and $R[x]\otimes_R M$, the uniqueness in particular to conclude that $a\circ b$ is the identity. $\endgroup$ – OR. Jun 28 '17 at 12:28
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I think you mixed up the quantifiers a little bit.

The universal property of the bilinear map $i: R[x]\times M\to R[x]\otimes_R M$ is as follows :

For any bilinear map $g:R[x]\times M\to N$, there exists a linear map $\tilde{g}:R[x]\otimes_R M\to N $ such that $g=\tilde{g}\circ i$, and moreover such a map is unique.

Now the argument in the linked thread is : if you manage to construct a bilinear map $h:R[x]\times M\to M[x]$ with the same universal property, i.e. such that for any bilinear map $f:R[x]\times M\to N$ there exists a unique linear map $\hat{f}:M[x]\to N$ such that $f=\hat{f}\circ h$, then $M[x]\cong R[x]\otimes_R M$. The reason for that is that you can apply the property of $h$ to the bilinear map $i:R[x]\times M\to R[x]\otimes_R M$, thus there must be a linear map $\hat{i}:M[x]\to R[x]\otimes_R M$ such that $\hat{i}\circ h=i$; and you can also apply the universal property of $i$ to the bilinear map $h:R[x]\times M\to M[x]$, thus there must be a linear map $\tilde{h}:R[x]\otimes_R M\to M[x]$ such that $h=\tilde{h}\circ i$.

Now combining the two identities, we find that $\tilde{h}\circ \hat{i}$ is a linear map $M[x]\to M[x]$ such that $h=\tilde{h}\circ i=\tilde{h}\circ \hat{i}\circ h$, and thus by uniqueness we have $\tilde{h}\circ \hat{i}=id_{M[x]}$; and by the same argument $\hat{i}\circ \tilde{h}=id_{R[x]\otimes_RM}$. Thus $\hat{i}$ and $\tilde{h}$ are isomorphisms between $M[x]$ and $R[x]\otimes_R M$.

This is really a standard argument about universal properties; in fact it shows not only that there is an isomorphism but that there is only one isomorphism $\psi$ that also preserves the bilinear maps, in the sense that $\psi\circ i=h$.

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