Timeline for $A$ is isomorphic to $A \oplus \mathbb{Z}^2$, but not to $A \oplus \mathbb{Z}$
Current License: CC BY-SA 3.0
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| Jan 6, 2023 at 8:12 | comment | added | spin | Perhaps worth mentioning that this answer was turned into the paper "Jeremy Rickard, Pathological abelian groups: a friendly example, J. Algebra 558 (2020), 640-645." MR4102131 | |
| Apr 18, 2020 at 9:02 | comment | added | Martin Brandenburg | Sorry for not accepting this answer earlier. My plan was to understand every part of the answer before accepting it, but that might as well never happen in the next years :) | |
| Apr 18, 2020 at 9:01 | vote | accept | Martin Brandenburg | ||
| Apr 7, 2016 at 17:07 | history | edited | Jeremy Rickard | CC BY-SA 3.0 |
added reference
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| Feb 8, 2016 at 11:12 | comment | added | YCor | @JeremyRickard: indeed. I think I could fix this if I could show that for every injective endomorphism $f$ of $A$ we have $f^{-1}(A_0)=A_0$ (we have one inclusion since $f(A_0)\subset A_0$). I can't now; I'll think twice. (I'm assuming here $D$ finitely generated.) | |
| Feb 8, 2016 at 6:37 | comment | added | Jeremy Rickard | @YCor I've only just really thought about your comment on arbitrary dense subgroups $D$ of the reals. Is there not a problem because the group of bounded sequences of elements of $D$ is not an $R'_D$-module? | |
| Jan 31, 2016 at 18:05 | history | edited | Jeremy Rickard | CC BY-SA 3.0 |
added acknowledgements for proofs that inspired some of this
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| Jan 31, 2016 at 16:20 | comment | added | YCor | Oh, you're right. The set of $k$ such that $B\simeq B\oplus\mathbf{Z}^k$ always has the form $n\mathbf{N}$ (and all $n$ can be achieved by immediate variations of your argument). | |
| Jan 31, 2016 at 15:54 | comment | added | Jeremy Rickard | @YCor If $B\cong B\oplus\mathbb{Z}^2\cong B\oplus\mathbb{Z}^3$, then $B\oplus\mathbb{Z}\cong(B\oplus\mathbb{Z}^2)\oplus\mathbb{Z}\cong B$. | |
| Jan 31, 2016 at 15:51 | comment | added | YCor | PS a motivation for my discussion about more general subgroups is the following question: is it possible to find an abelian group $B$ such that $B$ is isomorphic to $B\oplus\mathbf{Z}^k$ for $k=2,3$ but not $k=1$? In the current example $B$ is isomorphic to $B\oplus\mathbf{Z}^k$ iff $k$ is even. | |
| Jan 31, 2016 at 15:49 | comment | added | YCor | I'm convinced. A remark: while the $\varphi_{ij}$ can be defined directly (not using Lemmas 1,2), just by evaluation of $\varphi$ on finitely supported sequences, the proof of Lemma 5 really makes use at least of Lemma 1. | |
| Jan 31, 2016 at 15:44 | history | edited | Jeremy Rickard | CC BY-SA 3.0 |
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| Jan 31, 2016 at 15:38 | history | edited | Jeremy Rickard | CC BY-SA 3.0 |
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| Jan 31, 2016 at 15:34 | comment | added | Jeremy Rickard | I think I've finished now. Comments/corrections/requests for clarification welcome. | |
| Jan 31, 2016 at 15:30 | history | edited | Jeremy Rickard | CC BY-SA 3.0 |
finished writing proof
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| Jan 31, 2016 at 12:42 | comment | added | YCor | Great! Your arguments are actually shorter than mine especially for Lemma 2; I didn't think of using Lemma 1 to prove Lemma 2 (since I proved Lemma 2 before). It seems that until (4) everything works with an arbitrary dense subgroups $D$ of the reals. Then, considering $R_D=\{r\in\mathbf{R}:rD\subset D\}$, and $R'_D$ the commutant of $R_D$ in the group of group automorphisms of $D$, (5) shows that after precomposition with a right shift, we have a $R'_D$-module homomorphism. | |
| Jan 31, 2016 at 11:44 | comment | added | Jeremy Rickard | @YCor It sounds as though we've just chosen different proofs of the standard case to try to generalize. Actually, it wasn't quite null sequences that I needed to use. | |
| Jan 31, 2016 at 11:43 | comment | added | Jeremy Rickard | I've included the start of the proof, and will finish later. I've put the proof of (1) of my "sketch" first, as that seems to be what people are most interested in. | |
| Jan 31, 2016 at 11:40 | history | edited | Jeremy Rickard | CC BY-SA 3.0 |
started including proof
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| Jan 31, 2016 at 3:09 | comment | added | YCor | Oh, I have the argument for the remainder of (1), and is again an adaptation of the usual argument. This second part of (1) (which translates in (2) into the finiteness of rows) is really needed in the "cook up" of (4). I have written details for my own sake, but would rather let Jeremy shot first (but I can send by email to anybody interested). Btw I have no use of null sequences. | |
| Jan 31, 2016 at 1:26 | comment | added | YCor | Btw all details I could check also work if one considers the additive subgroup $D$ generated by $1$ and some irrational $t$. When $t$ is not an quadratic algebraic number, it makes a little difference at the end of the proof, in which case the module argument has to be performed over the base ring $M_2(\mathbf{Z})$. When $t$ is quadratic, one considers the ring $R$ of those real numbers $r$ such that $rD\subset D$ and it works the same way. | |
| Jan 31, 2016 at 1:20 | comment | added | YCor | I have checked a number of details, including what's after (1), and one half of (1), namely that, denoting $A_0$ the group of functions with finite support, one has $\mathrm{Hom}(A/A_0,\mathbf{Z})=0$. As in the standard case it consists in proving that in this group every element is sum of a 2-divisible element and a 3-divisible elements, but the argument goes different as an arbitrary Bezout relation is not enough. The remaining part I can't prove now is that given $f:A\to\mathbf{Z}$, there exists $n$ such that $f$ is zero on $A_0(>n)$, where $A_0(>n)=\{x\in A_0: x_n=0, \forall k\le n\}$. | |
| Jan 30, 2016 at 22:52 | comment | added | Martin Brandenburg | Yes, it would be nice to see details on (1). Generally speaking, I suggest that the answer may be rewritten so it only contains the actual proof and not its development (this can be found in the history of the post). | |
| Jan 30, 2016 at 16:43 | comment | added | Pace Nielsen | @JeremyRickard I've spent a non-trivial amount of time trying to prove your example works, so I look forward to your proof, especially part (1). | |
| Jan 30, 2016 at 14:47 | comment | added | Jeremy Rickard | @YCor Tending to zero in the real topology. | |
| Jan 30, 2016 at 14:45 | comment | added | YCor | By the group $N$ of null sequences, you mean those eventually zero, or those tending to zero (in the real topology)? | |
| Jan 30, 2016 at 14:16 | comment | added | Jeremy Rickard | @Adam No, I mean bounded as a real number. | |
| Jan 30, 2016 at 13:39 | comment | added | Adam Przeździecki | Do you mean that $a_n+b_n\sqrt{2}$ is bounded in $\mathbb{Z}[\sqrt{2}]$ if $a_n^2-2b_n^2$ is bounded in $\mathbb{Z}$? | |
| Jan 30, 2016 at 11:49 | history | edited | Jeremy Rickard | CC BY-SA 3.0 |
Added sketch of proof.
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| Jan 8, 2016 at 0:31 | comment | added | YCor | Since some part of the discussion became obsolete after you proved that $A$ is not free, I did some minimal cleaning to ease the reading to future readers (esp. moving the freeness of bounded integral sequences to the end), please feel free to revert to the previous version if you don't like it. | |
| Jan 8, 2016 at 0:29 | history | edited | YCor | CC BY-SA 3.0 |
Did some moves to ease the reading after the edit
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| Jan 7, 2016 at 20:20 | history | edited | Jeremy Rickard | CC BY-SA 3.0 |
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| Jan 7, 2016 at 12:27 | comment | added | YCor | @JeremyRickard: oh yes sorry I egocentrically meant "infinite/geometric/combinatorial group theorists" when I said "group theoretists". I think finite group theorists sociologically live more at the algebraic side, so I expect they rather use the same convention as you. | |
| Jan 7, 2016 at 11:11 | comment | added | Jeremy Rickard | @YCor Yes, by "extension of $\mathbb{Z}^\omega$" I mean $\mathbb{Z}^\omega$ is the quotient. I'm pretty sure that's the convention used by most of the group theorists I hang out with, but maybe it varies between subfields of group theory. | |
| Jan 6, 2016 at 23:18 | comment | added | YCor | Extension of $X$ by $Y$ has 2 usual meanings, the group theorists's convention being that $X$ is the kernel and the algebraic geometer's convention being that $Y$ is the kernel. In your claim that "$A$ is extension of $\mathbb{Z}^\omega$ by the group of bounded integral sequences" if I'm correct you use the second convention. Here you can define the quotient map to be $(a_n+b_n\sqrt{2})_{n\ge 0}\mapsto (b_n)_{n\ge 0}$ (the image is the set of all integral sequences, the kernel is the set of bounded integral sequences). | |
| Jan 6, 2016 at 23:04 | history | edited | Jeremy Rickard | CC BY-SA 3.0 |
Added some remarks.
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| Jan 5, 2016 at 8:27 | comment | added | Martin Brandenburg | @HenrikRüping: Probably no since $\sqrt{2}$ is irrational. At least, I don't see any such isomorphism. | |
| Jan 4, 2016 at 0:52 | comment | added | HenrikRüping | I had a look at the similar example. I would like to look at the subgroup of all integer sequences such that $a_{2n}-a_{2n-1}$ is bounded. This subgroup is isomorphic to the direct sum of all integer sequences and bounded integer sequences. (Look at the even terms and at the differences $(a_{2n+1}-a_{2n})_n$). So this group is not a counterexample. Now I am wondering whether there is a similar isomorphism when we use the condition $a_{2n}-\sqrt{2}a_{2n-1}$, which is essentially this group. | |
| Jan 2, 2016 at 17:04 | comment | added | Martin Brandenburg | @Jeremy: Very interesting idea! | |
| Jan 1, 2016 at 23:24 | comment | added | Pace Nielsen | The subgroup $B$ of finitary sequences is not a direct summand, because $A/B$ has nonzero elements which are infinitely $2$-divisible, but $A$ has no such elements. | |
| Jan 1, 2016 at 22:38 | comment | added | Fedor Petrov | Is subgroup of finitary sequences a direct summand? | |
| Jan 1, 2016 at 20:49 | history | edited | Jeremy Rickard | CC BY-SA 3.0 |
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| Jan 1, 2016 at 18:54 | comment | added | Todd Trimble | The freeness of the group of bounded sequences of integers is a hard theorem due to G. Nöbeling, Verallgemeinerung eines Satzes von Herrn Specker, Invent. Math. 6 (1968), 41-55. I learned this through Google books: books.google.com/… | |
| Jan 1, 2016 at 17:51 | comment | added | Eric Wofsey | How do you prove that the group of bounded sequences of integers is free abelian? | |
| Jan 1, 2016 at 17:07 | history | answered | Jeremy Rickard | CC BY-SA 3.0 |