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Timeline for Freeness of tensor product

Current License: CC BY-SA 3.0

11 events
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Dec 8, 2016 at 10:22 vote accept M. Livesey
Dec 7, 2016 at 12:58 answer added Jeremy Rickard timeline score: 8
Dec 6, 2016 at 12:11 comment added M. Livesey @EhudMeir It is also not necessarily projective. Again, Jeremy's example works as a counterexample. If $\mathbb{Z}G$ were projective as a $Z(\mathbb{Z}G)$-module then $\mathbb{F}_2G$ would be projective as a $Z(\mathbb{F}_2G)$-module. However, since $G$ is a $2$-group $Z(\mathbb{F}_2G)$ is a local ring and so any finite dimensional projective $Z(\mathbb{F}_2G)$-module is a direct sum of copies of $Z(\mathbb{F}_2G)$. Now see Jeremy's comment for a contradiction.
Dec 6, 2016 at 11:49 comment added kneidell @JeremyRickard Thanks for the example, I was unaware of that
Dec 6, 2016 at 11:48 comment added Ehud Meir can it still be a projective $Z(\mathbb{Z}G)$-module? that would be enough.
Dec 6, 2016 at 11:44 comment added Jeremy Rickard @kneidell No. For example, if $G$ is dihedral of order $8$ then $Z(\mathbb{Z}G)$ has $\mathbb{Z}$-rank $5$, so $\mathbb{Z}G$ can't be a free $Z(\mathbb{Z}G)$-module.
Dec 6, 2016 at 11:33 comment added kneidell Isn't $\mathbb{Z}G$ free as a $Z(\mathbb{Z}G)$-module? I would guess that if we take $\mathbf{T}\subseteq G$ to be a transversal set of $G/Z(G)$ it would be obvious that $\mathbf{T}$ spans $\mathbb{Z}(G)$ over $Z(\mathbb{Z}G)$. So your claim should follow from the general fact that a tensor product of free modules over a commutative ring is again free..
Dec 6, 2016 at 10:45 comment added M. Livesey Yes. It seems obvious that there is no torsion but I can't prove it.
Dec 6, 2016 at 10:43 comment added Geoff Robinson So the only issue is whether there is torsion?
Dec 6, 2016 at 10:37 history edited YCor
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Dec 6, 2016 at 9:35 history asked M. Livesey CC BY-SA 3.0