The Stacks project

Lemma 15.60.6. Let $R$ be a ring. Let $A$, $B$ be $R$-algebras. The following are equivalent

  1. $A$ and $B$ are Tor independent over $R$,

  2. for every pair of primes $\mathfrak p \subset A$ and $\mathfrak q \subset B$ lying over the same prime $\mathfrak r \subset R$ the rings $A_\mathfrak p$ and $B_\mathfrak q$ are Tor independent over $R_\mathfrak r$, and

  3. For every prime $\mathfrak s$ of $A \otimes _ R B$ the module

    \[ \text{Tor}_ i^ R(A, B)_\mathfrak s = \text{Tor}_ i^{R_\mathfrak r}(A_\mathfrak p, B_\mathfrak q)_\mathfrak s \]

    (where $\mathfrak p = A \cap \mathfrak s$, $\mathfrak q = B \cap \mathfrak s$ and $\mathfrak r = R \cap \mathfrak s$) is zero.

Proof. Let $\mathfrak s$ be a prime of $A \otimes _ R B$ as in (3). The equality

\[ \text{Tor}_ i^ R(A, B)_\mathfrak s = \text{Tor}_ i^{R_\mathfrak r}(A_\mathfrak p, B_\mathfrak q)_\mathfrak s \]

where $\mathfrak p = A \cap \mathfrak s$, $\mathfrak q = B \cap \mathfrak s$ and $\mathfrak r = R \cap \mathfrak s$ follows from Lemma 15.60.3. Hence (2) implies (3). Since we can test the vanishing of modules by localizing at primes (Algebra, Lemma 10.23.1) we conclude that (3) implies (1). For (1) $\Rightarrow $ (2) we use that

\[ \text{Tor}_ i^{R_\mathfrak r}(A_\mathfrak p, B_\mathfrak q) = \text{Tor}_ i^ R(A, B) \otimes _{(A \otimes _ R B)} (A_\mathfrak p \otimes _{R_{\mathfrak r}} B_\mathfrak q) \]

again by Lemma 15.60.3. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 08HX. Beware of the difference between the letter 'O' and the digit '0'.