Let $R$ be a right Noetherian ring (actually it is left Noetherian as well) and $S=R[x]$ the polynomial ring in one (commuting) variable. If $X$ is the set of all monic polynomials then $X$ is a right denominator set in $S$, so we can form the ring $SX^{-1}=R(x)$.
Now, if $f:R[x]\to R(x)$ is the (not necessarily injective) homomorphism associated with the localization, then obviously we can induce two functors:
$f_\ast(-)=-\otimes_SR(x)$, the extension of scalars; and
$f^\ast$, the restriction of scalars.
Now, I would like to show $f_\ast(f^\ast(M))=M$, i.e. $f_\ast$ is a left inverse of $f^\ast$. It seems to me that this would be true IF $R(x)\otimes_SR(x)\cong R(x)$ since then
$f^\ast(M)\otimes_SR(x)\cong (M\otimes_{R(x)}R(x))\otimes_SR(x)\cong M\otimes_{R(x)}(R(x)\otimes_SR(x))\cong M\otimes_{R(x)}R(x).$
Is this right, and in that case, is it true that $R(x)\otimes_SR(x)\cong R(x)$?