# Rings over which every module is free

We know that modules over skewfields are free. Is the converse true? In other words, is it true that a nontrivial ring over which every module is free is a skewfield?

If the ring A is commutative, then writing that for any proper ideal I of A, the A-module A/I is free yields the result. What about the general case?

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Interesting question! I think you mean "not free" in the second line. – David Zureick-Brown Oct 14 '09 at 22:20
@David: no, he means free. The point is that A/I is free by assumption, but that implies that I=0 or I=A. So A must be a field. – Anton Geraschenko Oct 14 '09 at 22:23
@Anton: Right. My thought was that A/I is not a free A module for a non-trivial ideal I, which is essentially the same. – David Zureick-Brown Oct 14 '09 at 22:45

The non-commutative case isn't very different. Assume for reasonableness that your ring is Noetherian (I refuse to think about non-Noetherian rings on principle). Your argument gives that all two-sided ideals just be the whole algebra or trivial. So, your algebra is simple.

Another consequence is that every exact sequence splits. Thus, if you have a pair of left ideals $I \subset J$, you can choose left ideals $K_1,K_2$ such that $J=I \oplus K_1$ and $R=J \oplus K_2$. Thus, if you had a infinite descending chain of ideals, you would have a complementary ascending chain of ideals, which is impossible, since your ring is Noetherian. Thus, your ring is Artinian.

By Artin-Wedderburn, your algebra is a matrix algebra over a skew field. But every matrix ring except 1 x 1 has a non-free projective module. So A is a skew-field.

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To apply the Artin-Wedderburn theorem, you also need the ring to be artinian. How to prove this? – Benoit Jubin Oct 15 '09 at 19:15
Ok, now I fixed it. – Ben Webster Oct 15 '09 at 20:23
I object to your assertion that you refuse to think about non-Noetherian rings, especially for this problem. As pointed out in ohdarkdevil's response below, if every module is free, then every module is projective. This is equivalent to the ring being semisimple. So it is automatically Noetherian. – Mark Hovey Nov 23 '09 at 23:26

Yes it is true that a ring (with unit) over which ever module is free is a division ring.

An easy way to show this is to say that if every module is free, then in particular, every module is projective and hence the ring is noetherian. Then you proceed to show that it is impossible for the ring to have a non-trivial ideal, using the fact that noetherian rings have invariant basis numbers.

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What's a basis number? – David Zureick-Brown Oct 14 '09 at 22:42
Invariant basis number means that the rank of a free module is well-defined, i.e. any two bases have the same size. – Eric Wofsey Oct 14 '09 at 23:23

Caution: the zero ring is not a division ring, but all of its modules are free! But other arguments already given here show that any nonzero ring over which every right module is free is a division ring.

I'll pipe in with one more approach, a personal favorite. Assume that every right R-module is free and that R ≠ 0. Let S be a simple right R-module (at least one exists because R is nonzero: choose any maximal right ideal M of R and let S = R/M). Then S is free, so choose a basis for S. Pick any element x of the basis. Because S is simple and x ∈ S∖{0} we have S = xR, and because x was a basis element we have xR ≅ R as right R-modules. In particular, RR ≅ SR is simple, from which it follows that R is a division ring.

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"and because x was a basis element we have xR ≅ R as right R-modules" — why is it true? What is the isomorphism? – o2genum Dec 26 '13 at 19:13
@o2genum: A basis for a right $R$-module is a linearly independent spanning set. In particular, linear independence implies that any element $x$ of a basis has right annihilator equal to zero. Thus the map $R \to xR$ sending $r \mapsto xr$ is an isomorphism. – Manny Reyes Dec 26 '13 at 23:07

Here's a more concrete version of the Ohdarkdevil's argument. Since every module is free, every short exact sequence splits. Suppose R is not a division ring, so it has some nontrivial left ideal I. Then R splits as I \oplus R/I. Now I is nonzero and free, so it contains a direct summand J isomorphic to R. We can thus write R itself as a direct sum R \oplus M, for M=I/J \oplus R/I. But now iterating this gives that R has a direct summand that is an infinite direct sum of copies of M. This is impossible since any direct summand of R, being a quotient of R, must be finitely generated.

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Iterating this construction gives that for any n, M^n is a direct summand of R, but how does this imply that R has an infinite direct sum of copies of M as a direct summand? – Benoit Jubin Oct 15 '09 at 19:07
They're not just arbitrary M^n direct summands, they're nested. The M^n is naturally the first n coordinates of the M^{n+1}, and then the union of all of them will be a direct sum of infinitely many Ms. – Eric Wofsey Oct 15 '09 at 19:40