The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

22.11 Projective modules over algebras

In this section we discuss projective modules over algebras and over graded algebras. Thus it is the analogue of Algebra, Section 10.76 in the setting of this chapter.

Algebras and modules. 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 10.76.2).

Graded algebras and modules. Let $R$ be a ring. Let $A$ be a graded algebra over $R$. Let $\text{Mod}_ A$ denote the category of graded right $A$-modules. For an integer $k$ let $A[k]$ denote the shift of $A$. For an graded right $A$-module we have

\[ \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}_ A}(A[k], M) = M^{-k} \]

As the functor $M \mapsto M^{-k}$ is exact on $\text{Mod}_ A$ we conclude that $A[k]$ is a projective object of $\text{Mod}_ A$. Conversely, suppose that $P$ is a projective object of $\text{Mod}_ A$. By choosing a set of homogeneous generators of $P$ as an $A$-module, we can find a surjection

\[ \bigoplus \nolimits _{i \in I} A[k_ i] \longrightarrow P \]

Thus we conclude that a projective object of $\text{Mod}_ A$ is a direct summand of a direct sum of the shifts $A[k]$.

If $(A, \text{d})$ is a differential graded algebra and $P$ is an object of $\text{Mod}_{(A, \text{d})}$ then we say $P$ is projective as a graded $A$-module or sometimes $P$ is graded projective to mean that $P$ is a projective object of the abelian category $\text{Mod}_ A$ of graded $A$-modules.

Lemma 22.11.1. Let $(A, \text{d})$ be a differential graded algebra. Let $M \to P$ be a surjective homomorphism of differential graded $A$-modules. If $P$ is projective as a graded $A$-module, then $M \to P$ is an admissible epimorphism.

Proof. This is immediate from the definitions. $\square$

Lemma 22.11.2. Let $(A, d)$ be a differential graded algebra. Then we have

\[ \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}_{(A, \text{d})}}(A[k], M) = \mathop{\mathrm{Ker}}(\text{d} : M^{-k} \to M^{-k + 1}) \]

and

\[ \mathop{\mathrm{Hom}}\nolimits _{K(\text{Mod}_{(A, \text{d})})}(A[k], M) = H^{-k}(M) \]

for any differential graded $A$-module $M$.

Proof. This is clear from the discussion above. $\square$


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'.