Im trying to prove whether the $Functor$: $R[X]\otimes_R\square: Mod_R \to Mod_R$ is an $Exact \space Functor$, i dont know anything about the ring $R$. ( If i understand right, this is equivalent to show that the module $R[X]$ (over $R$) is $flat$.)
If this $Functor$ is indeed exact, My attempt was to show that this functor preserves injective maps, and this is the point where I need some help:
Let's say I have the injective map $i:M \to N$ between two left $R-Modules$.
So, the induced map would be $i*:R[X]\otimes_R M \to R[X]\otimes_R N$. To show this map is injective, let $r[x]\otimes_Rm \in ker(i*)$. How can I use $i$ injectiveness to show that the element $r[x]\otimes_Rm$ is zero ?
Thanks a lot!
If you know about Tor groups, Tor commutes with direct sums and $R[x]$ is free as an $R$ module, so $Tor(R[x],-)=0$ always, and hence it's flat.