# Question about finitely generated projective modules

Is it true that every finitely generated projective R-module is a direct sum of the direct summands of R? I am not assuming anything nice about R.

Edit: To be more clear, let's say you are given a ring, and want to find all finitely generated projective R-modules, up to isomorphism. Is it enough to find the direct summands of R and generate all the others from these by taking direct sums?

-
It is certainly true that every projective module is a direct summand of a free $R$-module. –  Zhen Lin Feb 5 '13 at 22:41
@Zhen Lin Right. I guess what I'm wondering is can you get direct summands of RxR, say, that didn't come from direct sums of those from R? –  Joe Feb 5 '13 at 22:43

Of course No. When $R$ has only trivial idempotents (for example when $R$ is an integral domain), $R$ has no nontrivial direct summands. So take any integral domain with nontrivial Picard group, for example $\mathbb{Z}[\sqrt{-5}]$.

-

The answer is "Yes" if the Krull-Schmidt theorem (KST) holds for $R$, i.e. if every f.g. $R$-module is the unique direct sum of indecomposable submodules (unique up to isomorphism and order). Examples for such rings are Artinian rings, complete local rings or group rings of finite groups over a field.

For, write $R=I_1 \oplus \cdots \oplus I_n$ with each $I_i$ indecomposable. Note that $I_i$ is projective (since its a direct summand of a free module).

Let $P$ be f.g. projective. There is an f.g. $R$-module $Q$ and $m < \infty$ with

$$P \oplus Q = R^m=I_1^m \oplus \cdots \oplus I_n^m.\tag{1}$$

We can also apply KST to $P,Q$, i.e. we have $P=P_1 \oplus \cdots \oplus P_p$ and $Q=Q_1 \oplus \cdots \oplus Q_q$ with $P_i,Q_j$ indecomposable and obtain

$$R^m = P_1 \oplus \cdots \oplus P_p \oplus Q_1 \oplus \cdots \oplus Q_q.\tag{2}$$

Hence $(1),(2)$ are two decompositions of $R^m$ into indecomposeables. By uniqueness, each $P_i$ is a direct sum of (some of) the modules $I_1,...,I_n$ and consequently, the same holds for $P=P_1 \oplus \cdots \oplus P_p$.

-