Timeline for $A$ is isomorphic to $A \oplus \mathbb{Z}^2$, but not to $A \oplus \mathbb{Z}$
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8 events
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| Jan 3, 2016 at 19:05 | comment | added | Martin Brandenburg | Well, the inverse is given by $\mathbb{Z}^{\mathbb{N}} \oplus \mathbb{Z}^{\oplus \mathbb{N}} \stackrel{\cong}{\longrightarrow} A$, $(a,b) \mapsto c$ with $c(2k)=a(k)$ and $c(2k+1)=a(k)+b(k)$. So $A \cong A \oplus \mathbb{Z}$ holds (and $A$ is not free, by the way). | |
| Jan 3, 2016 at 17:30 | comment | added | Gerhard Paseman | I don't think so, but I am low on coffee, so I do not trust my own judgment on this. Gerhard "Ask Me After A Cup" Paseman, 2016.01.03 | |
| Jan 3, 2016 at 17:24 | comment | added | Martin Brandenburg | But this group is isomorphic to $\mathbb{Z}^{\mathbb{N}} \oplus \mathbb{Z}^{\oplus \mathbb{N}}$ via $a \mapsto ((a(2k))_{k \geq 0},(a(2k+1)-a(2k))_{k \geq 0})$, right? So it won't work. | |
| Jan 3, 2016 at 17:23 | comment | added | Gerhard Paseman | That, or variations. I haven't gone through the proof to see how it works, but it seems to me the order properties of $\omega$ are part of what pulls it through. Thus pick some poset P as in my comment, and an appropriate subalgebra of a Pth power. Gerhard "Hand-waving At A High Level" Paseman, 2016.01.03 | |
| Jan 3, 2016 at 15:01 | comment | added | Martin Brandenburg | So the suggestion is the abelian group $A=\{a \in \mathbb{Z}^{\mathbb{N}} : \exists n \forall k \geq n ~ ( a(2k)=a(2k+1))\}$, right? My first question: Is $A$ free? | |
| Jan 3, 2016 at 8:09 | comment | added | Eric Wofsey | I suspect that some version of this would give an example of a ring $A$ such that $A\cong A\times\mathbb{Z}^2$ but $A\not\cong A\times\mathbb{Z}$, but it seems like it would take a lot more work to prove that $A\not\cong A\times\mathbb{Z}$ as a group (if it is even true). Note that in the Boolean case, $A$ is isomorphic to $A\oplus B$ as an $\mathbb{F}_2$-module, though not as an $\mathbb{F}_2$-algebra. | |
| Jan 3, 2016 at 3:30 | comment | added | Gerhard Paseman | This suggests to me a (likely unoriginal) notion of directed power. Let P be a poset (well-founded, say), and for an algebra B consider this algebra raised to the power of P. Picking the right poset may allow similar constructions where the power is isomorphic to some subpowers (using fragments of P) and not others. I leave it for others to consider. Gerhard "May Have More Energy Tomorrow" Paseman, 2016.01.02 | |
| Jan 3, 2016 at 3:21 | history | answered | Gerhard Paseman | CC BY-SA 3.0 |