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 $\phi$.

  • $\begingroup$ 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

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 $\psi\circ i=h$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.