The Stacks project

10.85 Projective modules over a local ring

In this section we prove a very cute result: a projective module $M$ over a local ring is free (Theorem 10.85.4 below). Note that with the additional assumption that $M$ is finite, this result is Lemma 10.78.5. In general we have:

Lemma 10.85.1. Let $R$ be a ring. Then every projective $R$-module is free if and only if every countably generated projective $R$-module is free.

Proof. Follows immediately from Theorem 10.84.5. $\square$

Here is a criterion for a countably generated module to be free.

Lemma 10.85.2. Let $M$ be a countably generated $R$-module with the following property: if $M = N \oplus N'$ with $N'$ a finite free $R$-module, then any element of $N$ is contained in a free direct summand of $N$. Then $M$ is free.

Proof. Let $x_1, x_2, \ldots $ be a countable set of generators for $M$. We inductively construct finite free direct summands $F_1, F_2, \ldots $ of $M$ such that for all $n$ we have that $F_1 \oplus \ldots \oplus F_ n$ is a direct summand of $M$ which contains $x_1, \ldots , x_ n$. Namely, given $F_1, \ldots , F_ n$ with the desired properties, write

\[ M = F_1 \oplus \ldots \oplus F_ n \oplus N \]

and let $x \in N$ be the image of $x_{n + 1}$. Then we can find a free direct summand $F_{n + 1} \subset N$ containing $x$ by the assumption in the statement of the lemma. Of course we can replace $F_{n + 1}$ by a finite free direct summand of $F_{n + 1}$ and the induction step is complete. Then $M = \bigoplus _{i = 1}^{\infty } F_ i$ is free. $\square$

Lemma 10.85.3. 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. $\square$

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