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*}

Lemma 10.84.3. Let $P$ be a projective module over a local ring $R$. Then any element of $P$ is contained in a free direct summand of $P$.

Proof. Since $P$ is projective it is a direct summand of some free $R$-module $F$, say $F = P \oplus Q$. Let $x \in P$ be the element that we wish to show is contained in a free direct summand of $P$. Let $B$ be a basis of $F$ such that the number of basis elements needed in the expression of $x$ is minimal, say $x = \sum _{i=1}^ n a_ i e_ i$ for some $e_ i \in B$ and $a_ i \in R$. Then no $a_ j$ can be expressed as a linear combination of the other $a_ i$; for if $a_ j = \sum _{i \neq j} a_ i b_ i$ for some $b_ i \in R$, then replacing $e_ i$ by $e_ i + b_ ie_ j$ for $i \neq j$ and leaving unchanged the other elements of $B$, we get a new basis for $F$ in terms of which $x$ has a shorter expression.

Let $e_ i = y_ i + z_ i, y_ i \in P, z_ i \in Q$ be the decomposition of $e_ i$ into its $P$- and $Q$-components. Write $y_ i = \sum _{j=1}^{n} b_{ij} e_ j + t_ i$, where $t_ i$ is a linear combination of elements in $B$ other than $e_1, \ldots , e_ n$. To finish the proof it suffices to show that the matrix $(b_{ij})$ is invertible. For then the map $F \to F$ sending $e_ i \mapsto y_ i$ for $i=1, \ldots , n$ and fixing $B \setminus \{ e_1, \ldots , e_ n\} $ is an isomorphism, so that $y_1, \ldots , y_ n$ together with $B \setminus \{ e_1, \ldots , e_ n\} $ form a basis for $F$. Then the submodule $N$ spanned by $y_1, \ldots , y_ n$ is a free submodule of $P$; $N$ is a direct summand of $P$ since $N \subset P$ and both $N$ and $P$ are direct summands of $F$; and $x \in N$ since $x \in P$ implies $x = \sum _{i=1}^ n a_ i e_ i = \sum _{i=1}^ n a_ i y_ i$.

Now we prove that $(b_{ij})$ is invertible. Plugging $y_ i = \sum _{j=1}^{n} b_{ij} e_ j + t_ i$ into $\sum _{i=1}^ n a_ i e_ i = \sum _{i=1}^ n a_ i y_ i$ and equating the coefficients of $e_ j$ gives $a_ j = \sum _{i=1}^ n a_ i b_{ij}$. But as noted above, our choice of $B$ guarantees that no $a_ j$ can be written as a linear combination of the other $a_ i$. Thus $b_{ij}$ is a non-unit for $i \neq j$, and $1-b_{ii}$ is a non-unit—so in particular $b_{ii}$ is a unit—for all $i$. But a matrix over a local ring having units along the diagonal and non-units elsewhere is invertible, as its determinant is a unit. $\square$


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 0592. Beware of the difference between the letter 'O' and the digit '0'.