All Questions
Tagged with abelian-groups ac.commutative-algebra
42 questions
22
votes
1
answer
958
views
Alternating forms on abelian groups
Let $G$ and $H$ be abelian groups. By an alternating form, I mean a bilinear function $A\colon G\times G\to H$ such that $A(x,x)=0$ for all $x\in G$.
Question. If $A\colon G\times G\to H$ is an ...
- 196
4
votes
0
answers
212
views
When does a short exact sequence of abelian groups with $B\cong A\oplus C$ split?
$\hspace{20pt}$Duplicate on stackexchange.
This question, in a way, extends this one. The question is what are some sufficient conditions on the abelian group $B$ so that if $B\cong A\oplus C$ and a ...
- 171
20
votes
5
answers
2k
views
Constructively, is the unit of the “free abelian group” monad on sets injective?
Classically, we can explicitly construct the free Abelian group $\newcommand{\Z}{\mathbb{Z}}\Z[X]$ on a set $X$ as the set of finitely-supported functions $X \to \Z$, and so easily see that the unit ...
- 426
4
votes
2
answers
227
views
Maximal subgroups of finite abelian $2$-groups
Suppose $G$ is a finite abelian $2$-group, and $S$ is a subset of $G$, $\langle S\rangle=G$,$S^{-1}=S$,$e\notin S$. How to determine whether there exists a maximal subgroup $M$ of $G$, such that $S$ ...
- 44.4k
3
votes
1
answer
153
views
A question about freeness of a certain class of abelian groups
Lets call an abelian group $G$, to be semi-free (or SF) if every nonzero subgroup of $G$ is isomorphic to $\mathbb{Z}\times H$ for some abelian group $H$.
Is every semi-free group, a free group? If ...
- 63.9k
3
votes
0
answers
70
views
Admissibility of Ulm's invariants
Let $G$ be a reduced abelian $p$-group. We set $G_0=G$. Let $\alpha$ be an ordinal. Inductively, if $\alpha=\beta+1$ is a successor ordinal, we define
$$G_{\alpha}=pG_{\beta}.$$
If $\alpha$ is a limit ...
- 63.9k
8
votes
2
answers
1k
views
A conceptual proof that bounded index subgroups of a bounded torsion abelian group contain bounded index complemented subgroups
Call an abelian group $G = (G,+)$ $m$-torsion for some natural number $m$ if one has $m \cdot x = 0$ for all $x \in G$. A subgroup $H$ of $G$ is said to be complemented if one can write $G = H \oplus ...
- 55.9k
0
votes
0
answers
90
views
Invariants of primary groups
In Kaplansky's book "Infinite Abelian Groups", an abelian group $G$ is called primary if every element has order power of $p$ for some fixed prime number $p$. It is well-known that every ...
- 63.9k
28
votes
2
answers
863
views
$A^2$ is isomorphic to $A^{(\omega)}$, but not $A$
Is there an abelian group $A$ with $A\not\cong A\oplus A\cong A\oplus A\oplus A\oplus\cdots$ (a direct sum of countably many copies of $A$)?
Edited to add: As no answers are forthcoming, does anyone ...
- 35.2k
4
votes
1
answer
242
views
Quadratic refinements of a bilinear form on finite abelian groups
$\DeclareMathOperator\Hom{Hom}$Let $A$ be a finite abelian group and $\text{Sym}(A)$ the (abelian) group of symmetric bilinear forms over $A$ valued in $\mathbb{R}/\mathbb{Z}$.
A quadratic function on ...
- 9,720
37
votes
1
answer
1k
views
If $A$, $B$ are abelian groups such that $\mathrm{Hom}(A, G) \cong \mathrm{Hom}(B, G)$ for all abelian groups $G$, must $A$ and $B$ be isomorphic?
$\DeclareMathOperator\Hom{Hom}$The question is in the title. If the isomorphism $\Hom(A, G) \cong \Hom(B, G)$ is natural in $G$ then this is just the Yoneda Lemma. If $A$ and $B$ are finitely ...
- 12.9k
4
votes
1
answer
277
views
Is there a good notion of kernels of quadratic forms on abelian groups?
Let $G$ be an abelian group and let $q:G \to \mathbb{Q/Z}$ be a quadratic form, i.e. $q(a)=q(-a)$ and $b(x,y)=q(x+y)-q(x)-q(y)$ is a bihomomorphism. On vector spaces, when people speak about the ...
- 2,928
0
votes
0
answers
162
views
Finite-exponent abelian groups
Let $G$ be an abelian group and $G=\bigoplus_{i=1}^t{{\Bbb{Z}}_{p_i}^{n_i}}^{(\Lambda_i)}$ where each $\Lambda_i$ is a set (at least one of $\Lambda_i$ is infinite). Since $G_{\Bbb{Z}}$ is a finite-...
- 12.9k
0
votes
0
answers
72
views
countable direct sum of cyclic abelian $p^{2}$ groups
Let $G={{\Bbb{Z}}_{p^{2}}}^{(\aleph)}$ (countable direct sum of copies of ${\Bbb{Z}}_{p^2}$). It is clear that every subgroup of $G$ is a homomorphic image of $G$. Now this is my question:
Is it true ...
- 56.9k
6
votes
2
answers
388
views
Abelian groups such that $A \cong \mathrm{End}(A)$ and "complete rings"
Motivation: for any ring $R$ there is the natural monomorphism $\mathrm{in} \colon R \to \mathrm{End}(R_{add}): r \mapsto (x \mapsto rx)$, where $R_{add}$ is an additive abelian group ( rings are ...
- 3,041
6
votes
4
answers
2k
views
Is a torsion-free abelian group finitely generated, if all of its localizations at primes $p$ are finitely generated over $\mathbb{Z}_p$?
Background: When proving that the group of $k$-isogenies $\mathrm{Hom}_k(A,B)$ between two abelian varieties is finitely generated, one first shows that the Tate map $$\mathbb{Z}_\ell\otimes_{\mathbb{...
- 63.9k
3
votes
1
answer
474
views
Structures of subgroups of a finite abelian p-group
$\newcommand\la{\langle}\newcommand\ra{\rangle}$Let $G=\mathbb{Z}/p^{i_1}\times\cdots\times\mathbb{Z}/p^{i_r}$ with $i_1\leq\ldots\leq i_r$ be a finite abelian $p$-group. Then there can be many ...
- 37.4k
6
votes
0
answers
346
views
Uncountable Mittag-Leffler condition?
Let $(X_\alpha)_{\alpha <\kappa}$ be an inverse system of abelian groups.
If $\kappa = \omega$ (or by extension if $\kappa$ is of countable cofinality), then the Mittag-Leffler condition is a ...
- 64k
11
votes
1
answer
498
views
Example of an uncountable sequence of abelian groups with nonvanishing $\varprojlim^2$?
$\DeclareMathOperator{\op}{\mathrm{op}}\DeclareMathOperator{\Ab}{\mathsf{Ab}}\DeclareMathOperator{\Vect}{\mathsf{Vect}}$Question 1: What is an example of a sequence $(X_\alpha)_{\alpha<\kappa}$ of ...
- 3,411
5
votes
0
answers
190
views
Can an infinite abelian $p$-group be tall and thin?
Does there exist an abelian $p$-group $A$ with countable Ulm invariants and uncountable height?
Here by height, I mean the minimal ordinal $\rho$ such that $p^\rho A$ is divisible [1]. For an ordinal ...
- 64k
6
votes
1
answer
454
views
Which abelian groups are $\varprojlim^1$ groups?
Question 1: Let $\mathcal A$ be an abelian group. Does there exist an inverse system $(A^n)_{n \in \mathbb N} = (\cdots \to A^n \to A^{n-1} \to \cdots \to A^0)$ such that $\varprojlim^1 A^\bullet \...
- 4,600
3
votes
0
answers
98
views
Hales' generalization of the stacked bases theorem (seeking a proof)
In his paper Analogues of the stacked bases theorem, published in the proceedings of a 1976 conference, A.W. Hales claimed some interesting generalizations of the stacked bases theorem for abelian ...
- 2,992
7
votes
1
answer
268
views
Additive group of local rings
Is there a theory or characterization for those finite $p$-groups that can be considered as the additive group of a finite local commutative ring with identity?
- 15.9k
1
vote
1
answer
289
views
Inverse limit of $p^n$-torsion abelian groups
Let $p$ be a prime and let $\{A_n\}_{n > 0}$ be an inverse limit of abelian groups such that $A_n$ is $p^n$-torsion with $A_n/p^{n - 1} \cong A_{n - 1}$ (these isomorphisms are part of the data). ...
- 34
0
votes
1
answer
309
views
exact short sequence of divisible groups splits? [closed]
Let $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ be a short exact sequences of divisible abelian groups. Does then the sequence splits?
- 11.3k
4
votes
2
answers
507
views
Co-finite type abelian groups
Suppose $B$ is an abelian group such that for every integer $n\ge 1$, the $n$-torsion subgroup $B[n]$ is finite.
Let $B_{\rm tor} = \varinjlim_{n\ge 1} B[n]$ be the torsion subgroup of $B$.
Is it ...
- 63.9k
2
votes
0
answers
176
views
Trivial Tate modules
Let $A$ be an abelian group, and $p$ a prime.
I'll call $$T_p(A) := \text{Hom}_{\mathbf{Z}}(\mathbf{Q}_{p}/\mathbf{Z}_{p}, A).$$
If $A$ is finite, then $T_p(A)$ is trivial, but the converse is not ...
CommunityBot
- 1
6
votes
1
answer
389
views
Transcendence degree of the fraction field of $k[G]$ for torsion free abelian group $G$
Let $k$ be a field of characteristic $p$ and $G$ be a torsion free abelian group . Then the group ring $k[G]$ is an integral domain , let $k(G)$ denote its field of fractions . Then can we say ...
- 28.8k
6
votes
2
answers
249
views
Two abelian groups, each being direct factor of the other
Let $M$ and $N$ be two abelian groups. Suppose that $M$ is a direct factor of $N$ (i.e. there are homomorphisms $i:M\rightarrow N$ and $p:N\rightarrow M$ such that $p\circ i=id_M$) and $N$ is also a ...
- 63.9k
14
votes
1
answer
696
views
$\mathbb{Z}$-module structure of the subring generated by an algebraic number
Let $a$ and $b$ be algebraic numbers which are not necessarily algebraic integers. Is there some invariant that allows us to determine whether $\mathbb Z[a]$ and $\mathbb Z[b]$ are isomorphic as $\...
- 149k
8
votes
2
answers
501
views
Exact sequence of $n$th powers of abelian groups
Let $A,B,C$ be finitely generated abelian groups. Assume that there is an exact sequence $$0 \to C \to A^n \to B^n \to 0,$$where $A^n = A \oplus \dotsc \oplus A$ as usual. It is not assumed that $A^n \...
- 149k
3
votes
0
answers
46
views
What is an example of an integral domain with a module that is 1-separable but not separable?
Let R be an integral domain. All modules under discussion are torsion free unital left R-modules.
An R-module is completely decomposable if it is the direct sum of rank 1 submodules.
An R-...
- 29
7
votes
1
answer
373
views
On describing a sort of "well-behaved" subgroups of a free abelian group
I found this question when I tried to figure out what kind of subgroups of a free abelian group behave just as well as in the finitely generated case.
Let $M$ be a free abelian group and $N$ a ...
- 35.2k
29
votes
0
answers
877
views
The field of fractions of the rational group algebra of a torsion free abelian group
Let $G$ be a torsion free abelian group (infinitely generated to get anything interesting). The group algebra $\mathbb{Q}[G]$ is an integral domain. Let $\mathbb{Q}(G)$ be its field of fractions.
...
CommunityBot
- 1
4
votes
2
answers
715
views
The center of a(n endomorphism) ring is a PID
Let $A$ be a torsion-free Abelian group, and let $\mbox{End}(A)$ be its endomorphism ring. Denote by $R\doteq\mathbf{Z}(\mbox{End}(A))$ the center of $\mbox{End}(A)$. What would be necessary and/or ...
- 111
2
votes
1
answer
205
views
Generalized height of elements in abelian groups
In the book Infinite abelian groups Vol. I by L. Fuchs, on page 154, the notion of the generalized $p$-height of an element in an abelian group is defined, as follows:
Let $A$ be an abelian group ...
- 35.2k
5
votes
0
answers
171
views
Result of Larsen and Lunts on rationality of power series with coefficients in a free abelian group
Let $G$ be a free abelian group (not necessarily finitely-generated) and $F$ be the fraction field of the group ring of $G$. Let $\Theta$ be the set of power series in $F[[t]]$ such that each nonzero ...
CommunityBot
- 1
9
votes
1
answer
995
views
constructive Serre classes
A Serre class (of abelian groups) is a class of abelian groups closed under subgroups, quotients, and extensions. For instance, finitely generated groups and finite groups are both Serre classes.
...
- 66.8k
2
votes
1
answer
194
views
Name and references for a "twisted" addition in a ring
This question may seem a little unmotivated, but it actually isn't something that just occurred to me out of nowhere. It stems from a discussion I had the other day with a friend and is directly ...
- 2,213
13
votes
1
answer
1k
views
Classification of symtrivial modules over a PID
Let us call a module $M$ over a commutative ring $R$ symtrivial if the symmetry $M \otimes M \to M \otimes M, a \otimes b \mapsto b \otimes a$ equals the identity (the same notion applies to arbitrary ...
CommunityBot
- 1
4
votes
4
answers
630
views
A question about the additive group of a finitely generated integral domain
Let $R$ be an integral domain of characteristic 0 finitely generated as a ring over $\mathbb{Z}$. Can the quotient group $(R,+)/(\mathbb{Z},+)$ contain a divisible element? By a "divisible element" I ...
CommunityBot
- 1
9
votes
3
answers
3k
views
Why are divisible abelian groups important?
I just quote wikipedia:
"Divisible groups are important in understanding the structure of abelian groups, especially because they are the injective abelian groups."
I am asking for detail ...
- 57.6k