An isomorphism between Tor over different rings.

Question

I have an exercise which is:

Let $k$ be a commutative ring, and $R$ is a $k$-algebra which is flat as a $k$-module. Prove that if $B$ is an $R$-module (and hence a $k$-module), then $$R \otimes_k \mathrm{Tor}_n^k(B,C) \simeq \mathrm{Tor}_n^R(B,R\otimes_k C)$$ for all $k$-modules $C$ and all $n \geq 0$.

I attempted the following. I took a resolution $\mathbf{P}_{\bullet}$ of $C$ as a $k$-module and tensorized with $B$, yielding $$ \cdots \to B \otimes_k P_n \to \cdots \to B \otimes_k P_1 \to B \otimes_k P_0 \to 0.$$ The homology of this becomes $\mathrm{Tor}^k_n(B,C)$. Then I tried following two roads:

1. Considering the sequence $$0 \to Im(d_{n+1}\otimes Id) \to \ker(d_n \otimes Id) \to \mathrm{Tor}_n^k(B,C) \to 0,$$ which is exact, and passing $R \otimes_k \square$. Since, $R$ is $k$-flat, this will yield an exact sequence. However I wasn't able to continue in this road.
2. Tensorizing $\mathbf{P}_{\bullet}$ with $R$ as a $k$-module, obtaining $$ \cdots \to R \otimes_k P_n \to \cdots \to R \otimes_k P_1 \to R \otimes_k P_0 \to 0,$$ which are all $R$-modules, and is a resolution of $R \otimes_k C$. However, I don't know that this resolution is projective over $R$ (actually, I think this is false, since tensorizing with $B \otimes_R\square $ would make $R$ vanish). Therefore, this road got blocked as well.

I would appreciate some help.

Answer 1

Take an exact sequence of $k$-modules $0\to X\to Y\to C\to 0$, where $Y$ is projective.

Then also $R\otimes_k Y$ is projective as $R$-module, because $Y$ is a direct summand of a free $k$-module and this implies $R\otimes_k Y$ is a direct summand of a free $R$-module. Now we have the long exact sequence $$\DeclareMathOperator{\Tor}{Tor} \dots\to\Tor_{n+1}^k(B,Y)\to\Tor_{n+1}^k(B,C)\to \Tor_n^k(B,X)\to\Tor_n^k(B,Y)\to\Tor_n^k(B,C)\to\\ \dots\to\Tor_2^k(B,Y)\to\Tor_2^k(B,C)\to\Tor_1^k(B,X)\to \Tor_1^k(B,Y)\to\Tor_1^k(B,C)\to \\ B\otimes_k X\to B\otimes_k Y\to B\otimes_k C\to 0 $$ For $n>0$, we essentially have $$ \Tor_{n+1}^k(B,C)\cong \Tor_n^k(B,X) $$ and, obviously, $$ R\otimes_k\Tor_{n+1}^k(B,C)\cong R\otimes_k\Tor_n^k(B,X) $$ (note that here we don't need flatness of $R$).

Similarly, we get, for $n>0$, $$ \Tor_{n+1}^R(B,R\otimes_k C)\cong \Tor_n^R(B,R\otimes_k X) $$ (here we use the fact that $R\otimes_k Y$ is projective).

Notice also that we have a commutative diagram $$\require{AMScd} \begin{CD} R\otimes_k (B\otimes_k X) @>>> R\otimes_k (B\otimes_k Y) @>>> R\otimes_k (B\otimes_k C) @>>> 0 \\ @VVV @VVV @VVV \\ B\otimes_k (R\otimes_k X) @>>> B\otimes_k (R\otimes_k Y) @>>> B\otimes_k (R\otimes_k C) @>>> 0 \end{CD} $$ where the vertical arrows are isomorphisms. Here we're using flatness of $R$, which is also used to build the homomorphism between the two long exact sequences.

This implies $R\otimes_k\Tor_1^k(B,C)\cong\Tor_1^k(B,R\otimes_k C)$. This starts off the standard dimension shifting argument.

Answer 2

(Disregard my previous answer, I misread the question.) I think your second idea is correct.

I don't know that this resolution is projective over $R$.

Base change $R\otimes_k -$ always takes projective $k$-modules to projective $R$-modules, so this is true. ($P$ is projective over $k$ if and only if $P\oplus Q \cong k \left<X\right>$ is a free $k$-module for some $Q$; after taking base change, you see that $(R\otimes_k P) \oplus (R\otimes_k Q) \cong R \left<X\right>$ is a free $R$-module on the same set of generators.)

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I think you had in mind something like the following argument. If $P_\bullet \twoheadrightarrow C$ is a resolution of $C$ by projective $k$-modules, then $$\tag{*} \operatorname{Tor}^k_n (B,C) \cong H_n (B\otimes_k P_\bullet).$$ Now $R\otimes_k P_\bullet \twoheadrightarrow R\otimes_k C$ is a resolution of $R\otimes_k C$ by projective $R$-modules (it is a resolution since $R\otimes_k -$ is exact by our assumption, and $R\otimes_k P_n$ are projective, since base change preserves projectives). We have $$\tag{**} \operatorname{Tor}_n^R (B, R\otimes_k C) \cong H_n (B\otimes_R R\otimes_k P_\bullet).$$ It remains to put together (*) and (**): $$R\otimes_k \operatorname{Tor}^k_n (B,C) \cong R\otimes_k H_n (B\otimes_k P_\bullet) \cong H_n (R\otimes_k B\otimes_k P_\bullet) \cong H_n (B\otimes_R R\otimes_k P_\bullet) \cong \operatorname{Tor}_n^R (B, R\otimes_k C)$$ ($R\otimes_k -$ is exact, hence commutes with taking homology).

Answer 3

As stated in the comments, this problem comes from Rotman's Introduction to Homological Algebra, it is Exercise 6.16. I do believe there is a (rather large) typo in it. I believe it should read:

Let $k$ be a commutative ring, and $R$ is a $k$-algebra which is flat as a $k$-module. Prove that if $B$ is an $R$-module (and hence a $k$-module), then $$\mathrm{Tor}^k_n(B,C)≃\mathrm{Tor}^R_n(B,R⊗_kC),$$ for all $k$-modules $C$ and all $n≥0.$