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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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