The Stacks project

Lemma 15.6.8. With $A, A', B, B', I$ as in Situation 15.6.1.

  1. Let $(N, M', \varphi )$ be an object of $\text{Mod}_ B \times _{\text{Mod}_ A} \text{Mod}_{A'}$. If $M'$ is flat over $A'$ and $N$ is flat over $B$, then $N' = N \times _{\varphi , M} M'$ is flat over $B'$.

  2. If $L'$ is a flat $B'$-module, then $L' = (L \otimes _{B'} B) \times _{(L \otimes _{B'} A)} (L \otimes _{B'} A')$.

  3. The category of flat $B'$-modules is equivalent to the full subcategory of $\text{Mod}_ B \times _{\text{Mod}_ A} \text{Mod}_{A'}$ consisting of triples $(N, M', \varphi )$ with $N$ flat over $B$ and $M'$ flat over $A'$.

Proof. In the proof we will use Lemma 15.6.4 without further mention.

Proof of (1). Set $J = \mathop{\mathrm{Ker}}(B' \to B)$. This is an ideal of $B'$ mapping isomorphically to $I = \mathop{\mathrm{Ker}}(A' \to A)$. Let $\mathfrak b' \subset B'$ be an ideal. We have to show that $\mathfrak b' \otimes _{B'} N' \to N'$ is injective, see Algebra, Lemma 10.38.5. We know that

\[ \mathfrak b'/(\mathfrak b' \cap J) \otimes _{B'} N' = \mathfrak b'/(\mathfrak b' \cap J) \otimes _ B N \to N \]

is injective as $N$ is flat over $B$. As $\mathfrak b' \cap J \to \mathfrak b' \to \mathfrak b'/(\mathfrak b' \cap J) \to 0$ is exact, we conclude that it suffices to show that $(\mathfrak b' \cap J) \otimes _{B'} N' \to N'$ is injective. Thus we may assume that $\mathfrak b' \subset J$. Next, since $J \to I$ is an isomorphism we have

\[ J \otimes _{B'} N' = I \otimes _{A'} A' \otimes _{B'} N' = I \otimes _{A'} M' \]

which maps injectively into $M'$ as $M'$ is a flat $A'$-module. Hence $J \otimes _{B'} N' \to N'$ is injective and we conclude that $\text{Tor}_1^{B'}(B'/J, N') = 0$, see Algebra, Remark 10.74.9. Thus we may apply Algebra, Lemma 10.98.8 to $N'$ over $B'$ and the ideal $J$. Going back to our ideal $\mathfrak b' \subset J$, let $\mathfrak b' \subset \mathfrak b'' \subset J$ be the smallest ideal whose image in $I$ is an $A'$-submodule of $I$. In other words, we have $\mathfrak b'' = A' \mathfrak b'$ if we view $J = I$ as $A'$-module. Then $\mathfrak b''/\mathfrak b'$ is killed by $J$ and we get a short exact sequence

\[ 0 \to \mathfrak b' \otimes _{B'} N' \to \mathfrak b'' \otimes _{B'} N' \to \mathfrak b''/\mathfrak b' \otimes _{B'} N' \to 0 \]

by the vanishing of $\text{Tor}_1^{B'}(\mathfrak b''/\mathfrak b', N')$ we get from the application of the lemma. Thus we may replace $\mathfrak b'$ by $\mathfrak b''$. In particular we may assume $\mathfrak b'$ is an $A'$-module and maps to an ideal of $A'$. Then

\[ \mathfrak b' \otimes _{B'} N' = \mathfrak b' \otimes _{A'} A' \otimes _{B'} N' = \mathfrak b' \otimes _{A'} M' \]

This tensor product maps injectively into $M'$ by our assumption that $M'$ is flat over $A'$. We conclude that $\mathfrak b' \otimes _{B'} N' \to N' \to M'$ is injective and hence the first map is injective as desired.

Proof of (2). This follows by tensoring the short exact sequence $0 \to B' \to B \oplus A' \to A \to 0$ with $L'$ over $B'$.

Proof of (3). Immediate consequence of (1) and (2). $\square$


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 0D2I. Beware of the difference between the letter 'O' and the digit '0'.