Timeline for Freeness of tensor product
Current License: CC BY-SA 3.0
11 events
when toggle format | what | by | license | comment | |
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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 |