Free modules are projective, and projective modules are direct summands of free modules.

Are there examples of projective modules that are not free?

(I know this is not possible for modules of fields.)

Free modules are torsion-free. But is the inverse true?

Are there examples of torsion-free modules that are not free?

Thank you~

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    $\begingroup$ You obtain many geometric (but maybe not very concrete/explicit) examples by considering vector bundles: Let $E \to X$ be a vector bundle over a compact manifold or an affine scheme. Then its module of global sections (over the ring of global functions on $X$) is always projective, and free if and only if the vector bundle is trivial. See Serre–Swan theorem. $\endgroup$ – Ingo Blechschmidt Oct 28 '14 at 15:11

Let $R=\mathbb{Z}/6\mathbb{Z}$. Obviously, $R$ is a free module over itself. Because $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/3\mathbb{Z}\cong R$, we have that $\mathbb{Z}/2\mathbb{Z}$, considered as an $R$-module, is projective, but it cannot be free - any non-trivial direct sum of $R$'s would have at least 6 elements.

The Baer-Specker group is an example of a torsion-free $\mathbb{Z}$-module which is not free.

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    $\begingroup$ Thanks very much. I see the fact that the Baer-Specker group is not free, but I don't understand why. Why isn't $\{1,0,0, \cdots \},\{0,1,0, \cdots \},\{0,0,1, \cdots \}, \cdots$ a basis of this group? Does a basis of a free module need to be of finite cardinality? $\endgroup$ – ShinyaSakai Mar 15 '11 at 14:07
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    $\begingroup$ $\mathbb{Q}$ is also torsion-free but not free over $\mathbb{Z}$ (I think) $\endgroup$ – yoyo Mar 15 '11 at 15:09
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    $\begingroup$ @shinyasakai: an element needs to be written as a finite combination of basis vectors. the basis doesnt have to be finite, but any given expression has to be a finite sum (this isnt analysis, its algebra) $\endgroup$ – yoyo Mar 15 '11 at 15:11
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    $\begingroup$ @yoyo - I know $\mathbb{Q}$ is a much easier example; but I was concerned that ShinyaSakai's questions may be homework, and I wanted to give an example that answered their question but would not be an acceptable homework answer. $\endgroup$ – Zev Chonoles Mar 15 '11 at 16:10

The issue of when projective modules are free is discussed in $\S 3.5.4$ of my commutative algebra notes. In particular one gets very easy (but not very satisfying) examples by looking at disconnected rings: e.g. $\mathbb{C} \times \{0\}$ is quite clearly projective but not free over $\mathbb{C} \times \mathbb{C}$.

It is more interesting to ask for examples over domains. One huge class of examples comes from invertible fractional ideals which are not principal. In particular a Dedekind domain possesses such ideals iff it is not a PID. This is worked out much later in my notes in the section on Dedekind domains (currently $\S 22$, but this is subject to change). However, in $\S 3.5.4$ I take some time to exhibit from scratch a specific projective, nonfree module over $R=\mathbb{Z}[\sqrt{-5}]$.

Note that free implies torsionfree holds for modules over a domain $R$. (In fact the usual definition of "torsionfree module" is only useful over domains, so far as I know. I seem to recall that Lam's book Lectures on modules and rings gives a more sophisticated definition for modules over a general ring...) To be more precise, over any commutative ring $R$, free $\implies$ projective $\implies$ flat, and over a domain flat $\implies$ torsionfree. So any example of a projective nonfree module over a domain is also an example of a torsionfree nonfree module.

Over a general domain there is a vast gap between torsion free modules and flat modules. For instance, a prime ideal $\mathfrak{p}$ in a Noetherian domain is always a torsionfree module, but it is flat only if it has height at most one. However, a torsionfree module over a PID is flat -- see e.g. $\S 3.11$ of my notes -- but need not be free. For instance, let $R$ be any PID which is not a field and let $K$ be its field of fractions. Then $K$ is a nontrivial torsionfree $R$-module which is divisible, hence not free.

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  • $\begingroup$ For completeness: the definition Lam gives is that a right module $M_R$ over $R$, $M$ is torsion-free if and only if whenever the left annihilator of an $a\in R$ is trivial, then right multiplication by $a$ on $M$ is injective. That is, $\forall a\in R (\mathrm{ann}_{\ell}(a) = 0 \Rightarrow ma=0\rightarrow m=0)$. $\endgroup$ – Arturo Magidin Mar 15 '11 at 18:35
  • $\begingroup$ Quite at the beginning in your notes on commutative algebra an exercise (1.1.c) asks to check that the product is invertible if and only if both factors are so. How can I check this? $\endgroup$ – C-Star-W-Star Jan 15 '17 at 1:58
  • $\begingroup$ @FreeziiS: The key to this problem is to read the CONVENTION ON EXERCISES that comes before it: that is, the ring is assumed to be commutative. The exercise you refer to is not true without this. $\endgroup$ – Pete L. Clark Jan 15 '17 at 16:31
  • $\begingroup$ @PeteL.Clark: Indeed, I know, the key is a commutative ring which is crucial since for noncommutative rings the assertion becomes wrong. A counterexample for the noncommutative case is provided by the right and left unilateral shift. But, I'm wondering, how to prove this in the commutative case??? $\endgroup$ – C-Star-W-Star Jan 15 '17 at 20:26
  • $\begingroup$ If $xy \in R^{\times}$ then there is $z$ such that $xyz = 1$. Thus $x(yz) = 1$ and $y(xz) = 1$, so $x,y \in R^{\times}$. $\endgroup$ – Pete L. Clark Jan 15 '17 at 20:31

If $k$ is a field and $A = \operatorname{Mat}_n(k) = \operatorname{End}_k(V)$ is the ring of $n\times n$ matrices with coefficients in $k$, then the (simple, left) $A$-module $V = k^n$ of "column vectors" is a projective $A$-module, but it is not free as an $A$-module as soon as $n>1$ (this non-freeness is easy to see by comparing the dimension of $V$ and of $A$ as $k$-vector spaces).

To see the fact that $V$ is projective $A$-module, fix a basis $b_1,\dots,b_n$ for $V$ as a vector space, and for each $j=1,\dots,n$, let $e_j \in A$ be the "idempotent" given by projection on the 1 dimensional linear subspace of $V$ spanned by $b_j$. Form the left ideal $I_j = Ae_j$. The mapping $$X \mapsto Xb_j$$ defines an $A$-module isomorphism $I_j \to V$, and it is straightforward to check that $$A = \bigoplus_{j=1}^n I_j,$$ so indeed $V \simeq I_1$ is a direct summand of a free $A$-module, hence projective.

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  • $\begingroup$ Thanks, +1. This is true in a more general context: if $n>1$ and $R$ is either a commutative unital ring or a division ring, then $R^n$ is a projective but not free $M_n(R)$-module. $$~$$ Proof: As $M_n(R)$-modules, $R^n\!\oplus\!\bigoplus_{i=1}^{n-1}R^n\cong\bigoplus_{i=1}^nR^n\cong M_n(R)$, so $R^n$ is projective. To prove $R^n$ is not free, assume $f\!:R^n\!\rightarrow\bigoplus_{i\in I}M_n(R)\!=\!M_n(R)^{(I)}$ is an isomorphism. Take $R'\!:=\!\{rI;r\!\in\!R\}\!\leq\!M_n(R)$ where $I$ is the identity matrix. $\endgroup$ – Leo Sep 30 '11 at 11:41
  • $\begingroup$ Then $R^n$ and $M_n(R)^{(I)}$ are also $R'$-modules and $f$ is a $R'$-linear isomorphism. However as $R'$-modules, $R^n\!\cong\!(R')^n$ and $M_n(R)^{(I)}\!\cong\!((R')^{n^2})^{(I)}\!\cong\!(R')^{(n^2\cdot I)}$ are free with different rank, a contradiction. $\blacksquare$ $\endgroup$ – Leo Sep 30 '11 at 11:43

An ideal $I$ of an integral domain $R$ is a free $R$-module iff it is generated by one element.

In a Dedekind domain every non-zero ideal is invertible, thus projective. However Dedekind domains are usually not principal ideal domains - algebraic number theory yields many examples.

This way one gets torsion-free, finitely generated, projective modules that are not free.

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Given any two rings $R_\pm$.
Take the ring $R:=R_+\oplus R_-$.

Consider the free $R$-module $F:=R$.
Then it is projective as it is free.

Consider the summand $R$-module $P:=R_+\oplus 0$.
Then it is projective as a summand of a projective.

Now for any finite rings $\# R_\pm<\infty$:
By counting argument the constructed module is not free!

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