Lemma 15.65.18. Given ring maps $R \to A \to B$ with $A \to B$ faithfully flat and $K \in D(A)$ the tor amplitude of $K$ over $R$ is the same as the tor amplitude of $K \otimes _ A^\mathbf {L} B$ over $R$.

Proof. This is true because for an $R$-module $M$ we have $H^ i(K \otimes _ R^\mathbf {L} M) \otimes _ A B = H^ i((K \otimes _ A^\mathbf {L} B) \otimes _ R^\mathbf {L} M)$ for all $i$. Namely, represent $K$ by a complex $K^\bullet$ of $A$-modules and choose a free resolution $F^\bullet \to M$. Then we have the equality

$\text{Tot}(K^\bullet \otimes _ A B \otimes _ R F^\bullet ) = \text{Tot}(K^\bullet \otimes _ R F^\bullet ) \otimes _ A B$

The cohomology groups of the left hand side are $H^ i((K \otimes _ A^\mathbf {L} B) \otimes _ R^\mathbf {L} M)$ and on the right hand side we obtain $H^ i(K \otimes _ R^\mathbf {L} M) \otimes _ A B$. $\square$

There are also:

• 2 comment(s) on Section 15.65: Tor dimension

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).