14

If $M$ is a module over a commutative ring $R$ with $1$, does $M\oplus M$ free, imply $M$ is free? I thought this should be true but I can't remember why, and I haven't managed to come up with a counterexample.

I apologize if this has already been answered elsewhere.

rschwieb
  • 153,510
JessicaB
  • 2,067
  • Even though it doesn't directly relate, this is a good read regardless: http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/stablyfree.pdf – Alex Youcis Oct 02 '13 at 14:31

3 Answers3

8

This is not true, in general. The modules $M$ which occur as direct summands of a free module are precisely the projective modules, which are not in general free.

The simplest counter-example is probably the ideal generated by $2$ and $1+\sqrt{-5}$ in the ring $\mathbf Z[\sqrt{-5}]$, which is projective over $\mathbf{Z}[\sqrt{-5}]$ but not free.

If you would like a geometric example where a similar situation fails, consider the tangent bundle of the sphere. It is not free, for instance because it has no nonvanishing section (hairy ball theorem). But its sum with the normal bundle of the sphere (sitting in $\mathbf R^3$) is free, because it is just the restriction to the sphere of the tangent bundle of $\mathbf R^3$ (which is free). (Of course, these are not the same as modules over a ring, but the phenomenon is the same.)

Edit: Ah, I just noticed you had written $M\oplus M$, rather than $M\oplus N$. It turns out that the example I provided above in $\mathbf Z[\sqrt{-5}]$ works for that as well (because the Grothendieck group of a Dedekind domain (or more precisely of its category of finite projective modules) is isomorphic to its ideal class group (summed with a copy of $\mathbf Z$). I believe that this is what Vahid below was trying to say: a module whose square is free, in this case, is simply a module whose class in $K_0$ is $2$-torsion).

Bruno Joyal
  • 54,711
  • A module $M$ is called projective if there is a module $N$ such that $M\oplus N$ is a free module. –  Oct 02 '13 at 14:10
  • 2
    Could you please explain why the example you provided " above in $\mathbf Z[\sqrt{-5}]$ works for that as well" ? – Georges Elencwajg Oct 03 '13 at 18:05
  • 2
    Ah, I have found an explanation now: $I\oplus I\simeq R\oplus (I\cdot I)$ [ Cf. our friend Pete Clark's Notes, Lemma 20.17] and $I\cdot I=I^2\simeq R$ because the class number of your ring $R=\mathbf Z[\sqrt{-5}]$is $2$. That formula $I\oplus J=R\oplus I\cdot J$ for fractional ideals $I,J$ in a Dedekind ring $R$ seems to be well worth remembering. – Georges Elencwajg Oct 03 '13 at 18:59
  • @GeorgesElencwajg Thank you for the fleshed out explanation. I believe that this may come from the fact that the Grothendieck group of a Dedekind domain (or more precisely of its category of finite projective modules) is isomorphic to its ideal class group (summed with a copy of $\mathbf Z$). – Bruno Joyal Oct 03 '13 at 19:53
  • 1
    All this was explained also at MathOverflow by Mariano Suarez-Alvarez (using the same Dedekind ring example, with class number $2$), but unfortunately the post was deleted! – user43208 Oct 03 '13 at 20:09
  • @user43208 It really is the simplest example! – Bruno Joyal Oct 03 '13 at 20:10
  • You don't have to know that the class number is $2$. Just calculate $I^2=(2)$. See also my answer. – Martin Brandenburg Oct 03 '13 at 21:11
4

This would mean that there is no element of order 2 in a $K$-group, which is clearly not correct. To find an example, you can try to find a ring $R$ such that $K_0(R)$ has an element of order 2. As an example related to my research interest, I can say $K_0(C(\mathbb{RP}^2))= \mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}$, see Karoubi 1978, IV.6.47.

  • @HSN: The zeroth $K$-group of the 2-dimensional real projective space is $\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}$ not $M\oplus M$. –  Oct 02 '13 at 14:07
  • 2
    I guess I'm just missing the point. What exactly are $M$ and $M\oplus M$ here? – HSN Oct 02 '13 at 14:09
  • See the definition of $K_0(R)$ for a ring $R$. –  Oct 03 '13 at 21:53
2

For invertible ideals $I,J$ of a Dedekind ring $R$, it is well-known that $I \oplus J \cong R \oplus I \cdot J$ (see for example May's Notes on Dedekind rings, Prop 6.4). For the ideal $I = (2,1+\sqrt{-5})$ of $R=\mathbb{Z}[\sqrt{-5}]$ we have $I^2=(4,2+2 \sqrt{-5},-4+2 \sqrt{-5})=(2) \cong R$. Hence $I \oplus I \cong R^2$ is free, but $I$ is not free, since it is invertible and not principal.

Martin Brandenburg
  • 163,620