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Are all submodules of free modules free? I would like a reference to a proof or counterexample please.

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closed as off-topic by Ricardo Andrade, YCor, Karl Schwede, Chris Godsil, Andreas Blass Feb 22 '14 at 14:03

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I won't downvote since someone already answered, but I don't think is appropriate for MO. – Steve D Mar 3 '10 at 4:26
Another example you might be interested in is take the ring R = ZxZ. Certainly R as a module over itself is free; take Z as a submodule. It's clearly not free, since it's not the direct sum of copies of R, yet it is projective. – Jason Polak Mar 3 '10 at 16:00
+1, since the answers made lemonade (e.g. KConrad's reference for Pete Clark, in the comments). – Scott Morrison Mar 3 '10 at 18:39
@TravisBemrose Trouble is that the $R$-module map $R \to \mathbb{Z} \times \{0\}$ sending the identity $1 \in R$ (i.e. $(1, 1)$) to $(1, 0)$ is not an isomorphism, since $(0, 1)$ gets sent to $0$. – Todd Trimble Nov 13 '15 at 19:11
@TravisBemrose: principal ideals generated by zero divisors are not free.... – Jason Polak Nov 16 '15 at 13:41
up vote 30 down vote accepted

Вот общий пример: неглавной идеал в кольце $A$. Кольцо $A$ -- свободный $A$-модуль. Идеал в кольце -- подмодуль, а он тоже свободный $A$-модуль только в случае, что он главной идеал: ненулевые элементы $a$ и $b$ в кольце удовлетворяют нетривиальному $A$-линейному соотношению $c_1a + c_2b = 0$, где $c_1 = b$ и $c_2 = -a$, поэтому если существует базис, то мощность является одним.

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To paraphrase: "An ideal in a ring is a submodule of a free module over that ring. Any two elements $a$ and $b$ of the ideal satisfy the relation $(b)a+(-a)b=0$, so if the ideal has a basis (i.e. is a free module), it must be principal." – Anton Geraschenko Mar 3 '10 at 6:25
To continue the paraphrase: "Therefore any nonprincipal ideal, as a module over the ring, is a counterexample." Principal ideals can be counterexamples too if the ring isn't a domain: if a is a zero divisor (and not 0 itself) then the principal ideal (a) is not a free module. – KConrad Mar 3 '10 at 6:37
Дорогой колега, откуда вы так хорошо пишете по-русски? Во всяком случае, поздравляю! Кстати, отвечать по-русски - прекрасная идея: я только что проголосовал за вас – Georges Elencwajg Mar 3 '10 at 9:22
यह तो मुझे भी आता है| – Chandan Singh Dalawat Mar 4 '10 at 10:41
@JonBeardsley: Tout à fait, mais c'est un bel hommage à la diversité de nos langues. – ACL Feb 22 '14 at 9:32

No, for a general ring, yes for PIDs. Take as a counter example the ring of integers mod 4 as a module over it self.

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The first sentence is correct, but the second is not: the integers mod 4 is not a counterexample. Any ring A is a free A-module since {1} is a basis: it spans and has no nontrivial A-linear relations. – KConrad Mar 3 '10 at 6:25
He probably meant "take 2ℤ/4 as a module over ℤ/4." – Anton Geraschenko Mar 3 '10 at 6:28
I thought he meant, "take Z/4 over itself as an example of a free module that has a non-free submodule." – Jonas Meyer Mar 3 '10 at 6:35
I'm going out on a limb to admit my ignorance here, but: what is a standard reference that shows that a submodule of a free module over a PID is free? (Note that I did not say finitely generated.) – Pete L. Clark Mar 3 '10 at 17:04
A reference is Rotman's Advanced Modern Algebra, pp. 650--651. The proof uses the axiom of choice in the form of the well-ordering principle. – KConrad Mar 3 '10 at 17:45

For any (unitary commutative) ring $A$, the following are equivalent: (i) submodules of free $A$-modules are free, (ii) any ideal is free as $A$-module, (iii) the ring $A$ is a principal ideal domain. The proof is already clear form the above discussion. Therefore, if counterexamples exist, conterexamples must exists as ideals of the ring. As for the counterexamples already given here, (1) the ideal $(X,Y)$ of the polynomial ring $k[X,Y]$ is not even flat, as noted by Robin Chapman, (2) the ideal $2\mathbb{Z}/4\mathbb{Z}$ in the ring $\mathbb{Z}/4\mathbb{Z}$ is not flat, since it is generated by a non-zero nilpotent element, (3) the ideal $\mathbb{Z}\times 0$ in the ring $\mathbb{Z}\times \mathbb{Z}$ is projective but not free, similar result holds for the ideal $\mathbb{Z}/2\mathbb{Z}$ or $\mathbb{Z}/3\mathbb{Z}$ in the ring $\mathbb{Z}/6\mathbb{Z}$. I am trying to find an ideal $I$ of a ring $A$ such that $I$ is flat but not projective as $A$-module. Obviously, such a conterexample esists only if the ring $A$ is non-Noetherian. Maybe non-Noetherian prufer domains can work. Could someone give me such an example, i.e., to find an ideal $I$ of a ring $A$ such that $I$ is flat but not projective as $A$-module?

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@Konstantin Ardakov: You are right. Thanks a lot for the detailed explaination! – Joy-Joy Jan 30 '11 at 15:39
You're welcome. – Konstantin Ardakov May 26 '12 at 19:26
See Martin's answer there (a reference to Lam's book on modules and rings):… – ACL Mar 9 at 8:48

To answer ding8191's question: let $A$ be any valuation ring of height $1$ which is not a discrete valuation ring. That is to say, let $F$ be a field and suppose that there is a function $v : F \to \mathbb{R} \cup \{ \infty \}$ such that $v(xy) = v(x) + v(y), v(1) = 0, v(0) = \infty$ and $v(x + y) \geq \min $ $\{ v(x), v(y) \}$ (thus $v$ is a valuation), such that $v(F)$ is not a discrete subgroup of $\mathbb{R}$, and let $A = \{ x \in F :v(x)\geq 0\}$ (the valuation ring of $F$).

Then the maximal ideal $\mathfrak{m}$ in this ring is a direct limit of principal ideals and is therefore flat. If $\mathfrak{m}$ was projective, then it would have to be free by Kaplansky's Theorem (all projectives over any local ring are free), but since $\mathfrak{m}$ is contained in $F$ it has to have rank at most $1$, hence $\mathfrak{m}$ is principal. But this would force the valuation on $A$ to be discrete.

You can find this material in Chapter VI of Bourbaki's "Commutative Algebra" (Hermann, 1972). In particular, Lemma $1$ of $\S 3.6$ says that if $A$ is a valuation ring (in a more general sense than I defined above), then every torsion-free $A$-module is flat, and Proposition $9$ of the same subsection implies that $A$ is a discrete valuation ring whenever $A$ is a valuation ring of height $1$ with principal maximal ideal.

To give a concrete example, consider the field $F$ of Puiseux series and let $A$ be its valuation ring.

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Another example: let $R=k[x,y]$ where $k$ is a field. Then $R$ is a free module over itself, and the ideal $I$ of $R$ generated by $x$ and $y$ is not only not free over $R$, it is not even flat over $R$.

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For a ring R, if e is an idempotent element diferent from zero and one, then Re is projective module which isn't free.

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There exist examples of rings $R$ with a proper direct summand isomorphic to $R$ itself (as a, say, right $R$-module). The absence of this pathology is called direct finiteness of the ring and it does not come for free. Of course, this does not happen on commutative rings! – Simone Virili Sep 28 '12 at 15:18

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