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?


2 Answers 2


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.


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.

  • $\begingroup$ 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. $\endgroup$
    – azimut
    Aug 6, 2013 at 13:38
  • $\begingroup$ 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... $\endgroup$
    – azimut
    Aug 6, 2013 at 17:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.