Lemma 10.118.4. Let $R \to S$ be a ring map. Let $0 \to M_1 \to M_2 \to M_3 \to 0$ be a short exact sequence of $S$-modules. Then
\[ U(R \to S, M_1) \cap U(R \to S, M_3) \subset U(R \to S, M_2). \]
Proof. Let $u \in U(R \to S, M_1) \cap U(R \to S, M_3)$. Choose $f_1, f_3 \in R$ such that $u \in D(f_1)$, $u \in D(f_3)$ and such that (10.118.3.1) holds for $f_1$ and $M_1$ and for $f_3$ and $M_3$. Then set $f = f_1f_3$. Then $u \in D(f)$ and (10.118.3.1) holds for $f$ and both $M_1$ and $M_3$. An extension of free modules is free, and an extension of finitely presented modules is finitely presented (Lemma 10.5.3). Hence we see that (10.118.3.1) holds for $f$ and $M_2$. Thus $u \in U(R \to S, M_2)$ and we win. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)