a question about derived functor tor

Question

let $R$ be a $k$-algebra,$k$ is a commutative ring, ($R$ is flat $k$ module),B is $R$ module,any C is $k$ module then prove:

$R\bigotimes_k {\mathrm {Tor}}_n^k（B,C）\cong {\mathrm {Tor}}_n^R(B,R\bigotimes_kC)$

What I don't understand is when $n$ is $0$,it means $R\bigotimes_k (B\bigotimes_kC)\cong B\bigotimes_R(R\bigotimes_kC)$,for $B\bigotimes_R(R\bigotimes_kC)\cong （B\bigotimes_R R）\bigotimes_kC\cong B\bigotimes_kC$，then $R\bigotimes_k (B\bigotimes_kC)\cong B\bigotimes_kC$ could you tell me why the isomorphism is true.

Answer 1

I also think A.G's idea yesterday,but i find B and $R\bigotimes _k C$ are both $R$ module,and I ask some friends for help. They don't tell me why, but they tell me this is true and use $R$ is flat $k$ module.

Answer 2

Firstly, there is an isomorphism $$R\otimes_{k} {\mathrm {Tor}}_n^k(B, C)\cong {\mathrm {Tor}}_n^{k}(B, R\otimes_kC)$$ To show this isomorphism, we just need the fact that $R$ is a flat $k$-module.

Moreover, we have the following proposition (Flat base change for Tor) $${\mathrm {Tor}}_{n}^{k}(B, R\otimes_{k}C)\cong {\mathrm {Tor}}_{n}^{R}(B\otimes_{k}R, R\otimes_{k}C)$$

Even though $B$ is a $R$-module, we don't have that $B\otimes_{k}R\cong B\otimes_{R}R\cong B$.

If the isomorphism you provide is true, then we must have $${\mathrm {Tor}}_{n}^{R}(B, R\otimes_{k}C)\cong {\mathrm {Tor}}_{n}^{R}(B\otimes_{k}R, R\otimes_{k}C)$$

Now we can let $C=k$, and $n=0$, we have $$B\otimes_{R}R\otimes_{k}k\cong B\otimes_{k}R\otimes_{R}R\otimes_{k}k$$ which implies $$B\cong B\otimes_{k}R$$

As I said above, this is not true in general.