Lemma 46.8.2. Let $A$ be a ring.
A module is pure projective if and only if it is a direct summand of a direct sum of finitely presented $A$-modules.
For any module $M$ there exists a universally exact sequence $0 \to N \to P \to M \to 0$ with $P$ pure projective.
Proof. First note that a finitely presented $A$-module is pure projective by Algebra, Theorem 10.82.3. Hence a direct summand of a direct sum of finitely presented $A$-modules is indeed pure projective. Let $M$ be any $A$-module. Write $M = \mathop{\mathrm{colim}}\nolimits _{i \in I} P_ i$ as a filtered colimit of finitely presented $A$-modules. Consider the sequence
For any finitely presented $A$-module $P$ the map $\mathop{\mathrm{Hom}}\nolimits _ A(P, \bigoplus P_ i) \to \mathop{\mathrm{Hom}}\nolimits _ A(P, M)$ is surjective, as any map $P \to M$ factors through some $P_ i$. Hence by Algebra, Theorem 10.82.3 this sequence is universally exact. This proves (2). If now $M$ is pure projective, then the sequence is split and we see that $M$ is a direct summand of $\bigoplus P_ i$. $\square$
Your email address will not be published.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)