# Is a finitely generated projective module a direct summand of a *finitely generated* free module?

Let $R$ be a (not necessarily commutative) ring and $P$ a finitely generated projective $R$-module. Then there is an $R$-module $N$ such that $P \oplus N$ is free.

Can $N$ always be chosen such that $P \oplus N$ is free and finitely generated?

Equivalently: Is there always a finitely generated $N$ such that $P \oplus N$ is free?

If the answer is "no": What can be said about the rings $R$ such that this property is true?

Sure. Projective modules $P$ have the property (and actually this is an equivalent characterization) that every epimorphism $F \to P$ splits. Now choose a finite generating system of $P$, this lets you choose $F$ finitely generated free. Of course every direct summand of $F$ is a quotient of $F$ and therefore also finitely generated.

• Thanks for your answer. This was easier than I thought! I hope you don't mind me re-posting your solution with a few added details. Feel free to comment. – azimut Aug 6 '13 at 13:38
• Thanks also for your comment about $F = \operatorname{im}\varphi \oplus\ker\psi$ in general abelian categories (why did you delete it?). Sometimes I have the feeling that I should learn a bit more coregory theory... – azimut Aug 6 '13 at 17:49

To completely understand the +1 answer of Martin Brandenburg, I had to add a few details for myself. I decided to document the result in this answer:

Let a finite generating system of $P$ be given by $v_1,\ldots,v_n$. Set $F = R^n$ and $\varphi : F \to P$, $(x_1,\ldots x_n) \mapsto \sum_{i = 1}^n x_i v_i$. Since $P$ is projective, the epimorphism $\varphi$ splits, meaning that there is a monomorphism $\psi : P \to F$ such that $$\varphi\circ\psi = \operatorname{id}_P.$$

Now we check that $$F = \operatorname{im}\psi \oplus \ker\varphi.$$

1. To show that $F = \operatorname{im}\psi + \ker\varphi$, let $v\in F$. Define $x = \psi(\varphi(v))$ and $y = v - x$. Obviously, $x\in\operatorname{im}(\psi)$ and because of $$\varphi(y) = \varphi(v-x) = \varphi(v) - (\underbrace{\varphi\circ\psi}_{=\operatorname{id}_P}\circ\varphi)(v) = \varphi(v) - \varphi(v) = \mathbf{0},$$ $y \in \ker(\varphi)$. So $v = x + y \in \operatorname{im} \psi + \ker\varphi$.
2. To show that the sum is direct, let $v\in\operatorname{im}\psi \cap \ker\varphi$. So there is a $w\in P$ with $v = \psi(w)$, and $\mathbf{0} = \varphi(v) = (\varphi\circ\psi)(w) = w$. Hence $v = \psi(\mathbf{0}) = \mathbf{0}$ and $\operatorname{im}\psi \cap \ker\varphi = \{\mathbf{0}\}$.

Application of the homomorphism theorem to the monomorphism $\psi$ yields $P\cong\operatorname{im}\psi$, so $$P \oplus \ker\varphi \cong F = R^n$$ is finitely generated and free.