The Stacks project

Lemma 15.96.7. Let $A$ be a ring. Let $M$, $N_1$, $N_2$ be finite projective $A$-modules. Let $s : M \to N_1 \oplus N_2$ be a split injection. There exists a finitely generated ideal $J \subset A$ with the following property: a ring map $A \to B$ factors through $A/J$ if and only if $s \otimes \text{id}_ B$ identifies $M \otimes _ A B$ with a direct sum of submodules of $N_1 \otimes _ A B \oplus N_2 \otimes _ A B$.

Proof. Choose a splitting $\pi : N_1 \oplus N_2 \to M$ of $s$. Denote $q_ i : N_1 \oplus N_2 \to N_1 \oplus N_2$ the projector onto $N_ i$. Set $p_ i = \pi \circ q_ i \circ s$. Observe that $p_1 + p_2 = \text{id}_ M$. We claim $M$ is a direct sum of submodules of $N_1 \oplus N_2$ if and only if $p_1$ and $p_2$ are orthogonal projectors. Thus $J$ is the smallest ideal of $A$ such that $p_1 \circ p_1 - p_1$, $p_2 \circ p_2 - p_2$, $p_1 \circ p_2$, and $p_2 \circ p_1$ are contained in $J \otimes _ A \text{End}_ A(M)$. Some details omitted. $\square$

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Your email address will not be published.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 0F80. Beware of the difference between the letter 'O' and the digit '0'.