# Finitely generated projective module

Would anyone can help me how to show that a finitely generated projective module over a local ring and PID are free?

What I know about a finitely generated projective module $M$ over a PID $R$ is isomorphic to $R^k\oplus R/(a_1)\oplus\dots\oplus R/(a_n)$, and for the local ring case I don't know how to start.

-

For the case $R$ is a local ring it's a corollary of Nakayama's lemma.

As the notation in the above link, suppose $M$ is a finite generated projective module over $R$, then, first pick a minimal number of generators, i.e., $M=Rm_1+\cdots +Rm_k$, and $k$ is the minimal number with this property, so we get a decomposition

$$R^k=M\oplus N,$$ then, we are left to prove $N=0$.

First, applying $R/I\otimes-$, where $I$ is the unique maximal ideal in $R$, then we get $$(R/I)^k=M/IM\oplus N/IN,$$ and note that $M/IM$, $N/IN$ are vector spaces over the field $R/I$, so by comparing the dimension, we get $N/IN=0$, i.e., $N=IN$, then, we use the Nakayama's lemma, the Statement 1 in the above link, we get $r\in 1+I$, such that $rN=0$, but $r\not \in I$ and $R$ is local implies $r$ is a unit, so $N=0$.

Remarks. 1) To get the choice of $k$, we can first assume $k=\dim_{R/I}(M/IM)$, then use the Statement 4 in the above link to lift the basis of $M/IM$ to get a minimal set of generators of $M$.

2) A deep theorem of Kaplansky says that any projective modules (not necessarily finitely generated) over a local ring is free.

-
the link seems not working –  WLOG Jun 15 at 19:20

For a PID you are nearly done. Just observe that a projective module has no element of finte order and thus all torsion parts have to vanish.

For local rings this is not trivial and I don't have the time right know to give a full prove. The idea however is to write $$R^n=P\oplus Q$$ for a given projective module $P$ and set $p:R^n\to R^n$ to be the composition

$$p:R^n\xrightarrow{\cong} P\oplus Q \xrightarrow{id\oplus 0}P\oplus Q\xrightarrow{\cong} R^n$$

One can then show that $P\cong im(p)$ and that there exists a basis of $R^n$ such that the matrix representation of $p$ has an identity block and zeros everywhere else. This finishes the proof.

-