Mixing
On the dual group of the group of units of commutative ring
$begingroup$
All rings below are commutative with unity.
For a ring $R$, let $U(R)$ denote its group of units, which is in particular abelian.
Now, let $R$ be a ring. Consider the dual group of $U(R)$ namely $hat {U(R)} := Hom_{mathbb Z} (U(R),mathbb Z)$.
Under what conditions on $ U(R)$ , can we say that there exists a ring $S$ such that $U(S) cong hat {U(R)}$ ?
In other words : Given an abelian group $G$, which is the group of units of some ring, under what conditions on $G$, can we say that $U(S)cong Hom_{mathbb Z} (G,mathbb Z)$, for some ring $S$ ?
My thoughts : Given any torsion free abelian group $G$, I can show that the group of units of the Group ring $(mathbb Z/(2) )[G]$ is isomorphic to $G$ . Since the dual group of a torsion free abelian group is again torsion free, so I'm done in the case of torsion free abelian groups.
I don't know what happens if the group has a torsion.
Are there any other general sufficient or necessary conditions ?
ring-theory commutative-algebra modules homological-algebra abelian-groups
$endgroup$
add a comment |
$begingroup$
All rings below are commutative with unity.
For a ring $R$, let $U(R)$ denote its group of units, which is in particular abelian.
Now, let $R$ be a ring. Consider the dual group of $U(R)$ namely $hat {U(R)} := Hom_{mathbb Z} (U(R),mathbb Z)$.
Under what conditions on $ U(R)$ , can we say that there exists a ring $S$ such that $U(S) cong hat {U(R)}$ ?
In other words : Given an abelian group $G$, which is the group of units of some ring, under what conditions on $G$, can we say that $U(S)cong Hom_{mathbb Z} (G,mathbb Z)$, for some ring $S$ ?
My thoughts : Given any torsion free abelian group $G$, I can show that the group of units of the Group ring $(mathbb Z/(2) )[G]$ is isomorphic to $G$ . Since the dual group of a torsion free abelian group is again torsion free, so I'm done in the case of torsion free abelian groups.
I don't know what happens if the group has a torsion.
Are there any other general sufficient or necessary conditions ?
ring-theory commutative-algebra modules homological-algebra abelian-groups
$endgroup$
1
$begingroup$
Since there is no functoriality needed, why not look first at "Let $G$ be an abelian group (... of a very special kind, after looking at torsion...), find a ring $S$ having $U(S)=G$."
$endgroup$
– dan_fulea
Jan 31 at 18:05
add a comment |
$begingroup$
All rings below are commutative with unity.
For a ring $R$, let $U(R)$ denote its group of units, which is in particular abelian.
Now, let $R$ be a ring. Consider the dual group of $U(R)$ namely $hat {U(R)} := Hom_{mathbb Z} (U(R),mathbb Z)$.
Under what conditions on $ U(R)$ , can we say that there exists a ring $S$ such that $U(S) cong hat {U(R)}$ ?
In other words : Given an abelian group $G$, which is the group of units of some ring, under what conditions on $G$, can we say that $U(S)cong Hom_{mathbb Z} (G,mathbb Z)$, for some ring $S$ ?
My thoughts : Given any torsion free abelian group $G$, I can show that the group of units of the Group ring $(mathbb Z/(2) )[G]$ is isomorphic to $G$ . Since the dual group of a torsion free abelian group is again torsion free, so I'm done in the case of torsion free abelian groups.
I don't know what happens if the group has a torsion.
Are there any other general sufficient or necessary conditions ?
ring-theory commutative-algebra modules homological-algebra abelian-groups
$endgroup$
All rings below are commutative with unity.
For a ring $R$, let $U(R)$ denote its group of units, which is in particular abelian.
Now, let $R$ be a ring. Consider the dual group of $U(R)$ namely $hat {U(R)} := Hom_{mathbb Z} (U(R),mathbb Z)$.
Under what conditions on $ U(R)$ , can we say that there exists a ring $S$ such that $U(S) cong hat {U(R)}$ ?
In other words : Given an abelian group $G$, which is the group of units of some ring, under what conditions on $G$, can we say that $U(S)cong Hom_{mathbb Z} (G,mathbb Z)$, for some ring $S$ ?
My thoughts : Given any torsion free abelian group $G$, I can show that the group of units of the Group ring $(mathbb Z/(2) )[G]$ is isomorphic to $G$ . Since the dual group of a torsion free abelian group is again torsion free, so I'm done in the case of torsion free abelian groups.
I don't know what happens if the group has a torsion.
Are there any other general sufficient or necessary conditions ?
ring-theory commutative-algebra modules homological-algebra abelian-groups
ring-theory commutative-algebra modules homological-algebra abelian-groups
asked Jan 31 at 17:54
user640299
1
$begingroup$
Since there is no functoriality needed, why not look first at "Let $G$ be an abelian group (... of a very special kind, after looking at torsion...), find a ring $S$ having $U(S)=G$."
$endgroup$
– dan_fulea
Jan 31 at 18:05
add a comment |
1
$begingroup$
Since there is no functoriality needed, why not look first at "Let $G$ be an abelian group (... of a very special kind, after looking at torsion...), find a ring $S$ having $U(S)=G$."
$endgroup$
– dan_fulea
Jan 31 at 18:05
1
1
$begingroup$
Since there is no functoriality needed, why not look first at "Let $G$ be an abelian group (... of a very special kind, after looking at torsion...), find a ring $S$ having $U(S)=G$."
$endgroup$
– dan_fulea
Jan 31 at 18:05
$begingroup$
Since there is no functoriality needed, why not look first at "Let $G$ be an abelian group (... of a very special kind, after looking at torsion...), find a ring $S$ having $U(S)=G$."
$endgroup$
– dan_fulea
Jan 31 at 18:05
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The dual group of any abelian group is torsion-free. Indeed, this is immediate from the fact that $mathbb{Z}$ is torsion-free: if $finoperatorname{Hom}(G,mathbb{Z})$ is torsion, then $f(x)$ is a torsion element of $mathbb{Z}$ for all $xin G$ so $f=0$. So, your argument works for any $G$, not just torsion-free groups.
answered Feb 1 at 4:21
Eric WofseyEric Wofsey
192k14220352
$endgroup$
$begingroup$
Ah indeed ... thanks ... in fact if $R$ is a domain and $M,N$ are $R$ modules with $N$ torsion free then so is $Hom_R (M,N)$
$endgroup$
– user640299
Feb 1 at 13:53
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
The dual group of any abelian group is torsion-free. Indeed, this is immediate from the fact that $mathbb{Z}$ is torsion-free: if $finoperatorname{Hom}(G,mathbb{Z})$ is torsion, then $f(x)$ is a torsion element of $mathbb{Z}$ for all $xin G$ so $f=0$. So, your argument works for any $G$, not just torsion-free groups.
answered Feb 1 at 4:21
Eric WofseyEric Wofsey
192k14220352
$endgroup$
$begingroup$
Ah indeed ... thanks ... in fact if $R$ is a domain and $M,N$ are $R$ modules with $N$ torsion free then so is $Hom_R (M,N)$
$endgroup$
– user640299
Feb 1 at 13:53
add a comment |
$begingroup$
The dual group of any abelian group is torsion-free. Indeed, this is immediate from the fact that $mathbb{Z}$ is torsion-free: if $finoperatorname{Hom}(G,mathbb{Z})$ is torsion, then $f(x)$ is a torsion element of $mathbb{Z}$ for all $xin G$ so $f=0$. So, your argument works for any $G$, not just torsion-free groups.
answered Feb 1 at 4:21
Eric WofseyEric Wofsey
192k14220352
$endgroup$
$begingroup$
Ah indeed ... thanks ... in fact if $R$ is a domain and $M,N$ are $R$ modules with $N$ torsion free then so is $Hom_R (M,N)$
$endgroup$
– user640299
Feb 1 at 13:53
add a comment |
$begingroup$
The dual group of any abelian group is torsion-free. Indeed, this is immediate from the fact that $mathbb{Z}$ is torsion-free: if $finoperatorname{Hom}(G,mathbb{Z})$ is torsion, then $f(x)$ is a torsion element of $mathbb{Z}$ for all $xin G$ so $f=0$. So, your argument works for any $G$, not just torsion-free groups.
answered Feb 1 at 4:21
Eric WofseyEric Wofsey
192k14220352
$endgroup$
The dual group of any abelian group is torsion-free. Indeed, this is immediate from the fact that $mathbb{Z}$ is torsion-free: if $finoperatorname{Hom}(G,mathbb{Z})$ is torsion, then $f(x)$ is a torsion element of $mathbb{Z}$ for all $xin G$ so $f=0$. So, your argument works for any $G$, not just torsion-free groups.
answered Feb 1 at 4:21
Eric WofseyEric Wofsey
192k14220352
answered Feb 1 at 4:21
Eric WofseyEric Wofsey
192k14220352
answered Feb 1 at 4:21
Eric WofseyEric Wofsey
192k14220352
answered Feb 1 at 4:21
Eric WofseyEric Wofsey
192k14220352
192k14220352
$begingroup$
Ah indeed ... thanks ... in fact if $R$ is a domain and $M,N$ are $R$ modules with $N$ torsion free then so is $Hom_R (M,N)$
$endgroup$
– user640299
Feb 1 at 13:53
add a comment |
$begingroup$
Ah indeed ... thanks ... in fact if $R$ is a domain and $M,N$ are $R$ modules with $N$ torsion free then so is $Hom_R (M,N)$
$endgroup$
– user640299
Feb 1 at 13:53
$begingroup$
Ah indeed ... thanks ... in fact if $R$ is a domain and $M,N$ are $R$ modules with $N$ torsion free then so is $Hom_R (M,N)$
$endgroup$
– user640299
Feb 1 at 13:53
$begingroup$
Ah indeed ... thanks ... in fact if $R$ is a domain and $M,N$ are $R$ modules with $N$ torsion free then so is $Hom_R (M,N)$
$endgroup$
– user640299
Feb 1 at 13:53
add a comment |
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1
$begingroup$
Since there is no functoriality needed, why not look first at "Let $G$ be an abelian group (... of a very special kind, after looking at torsion...), find a ring $S$ having $U(S)=G$."
$endgroup$
– dan_fulea
Jan 31 at 18:05