The Stacks project

22.14 Projective modules over algebras

In this section we discuss projective modules over algebras analogous to Algebra, Section 10.77. This section should probably be moved somewhere else.

Let $R$ be a ring and let $A$ be an $R$-algebra, see Section 22.2 for our conventions. It is clear that $A$ is a projective right $A$-module since $\mathop{\mathrm{Hom}}\nolimits _ A(A, M) = M$ for any right $A$-module $M$ (and thus $\mathop{\mathrm{Hom}}\nolimits _ A(A, -)$ is exact). Conversely, let $P$ be a projective right $A$-module. Then we can choose a surjection $\bigoplus _{i \in I} A \to P$ by choosing a set $\{ p_ i\} _{i \in I}$ of generators of $P$ over $A$. Since $P$ is projective there is a left inverse to the surjection, and we find that $P$ is isomorphic to a direct summand of a free module, exactly as in the commutative case (Algebra, Lemma 10.77.2).

We conclude

  1. the category of $A$-modules has enough projectives,

  2. $A$ is a projective $A$-module,

  3. every $A$-module is a quotient of a direct sum of copies of $A$,

  4. every projective $A$-module is a direct summand of a direct sum of copies of $A$.


Comments (2)

Comment #2156 by shom on

In the Algebras and modules section, in the line " Then we can choose a surjection by choosing a set of generators of over . Since P is projective there is a left inverse to the surjection"

should be and perhaps the generating set should be instead of (to avoid confusion).


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Your email address will not be published.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 09JZ. Beware of the difference between the letter 'O' and the digit '0'.