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Multiplicative group of the fraction field of a UFD

2

$begingroup$

Multiplicative group of rational $mathbb Q^*cong mathbb Z_2 times bigoplus_i=1^infty mathbb Z$.

My question:

In general, how to characterize the multiplicative group $K^∗$ of the fraction field $K$ of a UFD $R$ ?

Can we use the same method which was used in $mathbb Q^*$?

And is there any reference about this?

Thank you for your time and help!

abstract-algebra group-theory ring-theory unique-factorization-domains

share|cite|improve this question

edited Mar 25 at 7:42

Andrews

asked Dec 3 '18 at 20:50

AndrewsAndrews

1,2962423

$endgroup$

add a comment | 

2

$begingroup$

Multiplicative group of rational $mathbb Q^*cong mathbb Z_2 times bigoplus_i=1^infty mathbb Z$.

My question:

In general, how to characterize the multiplicative group $K^∗$ of the fraction field $K$ of a UFD $R$ ?

Can we use the same method which was used in $mathbb Q^*$?

And is there any reference about this?

Thank you for your time and help!

abstract-algebra group-theory ring-theory unique-factorization-domains

share|cite|improve this question

edited Mar 25 at 7:42

Andrews

asked Dec 3 '18 at 20:50

AndrewsAndrews

1,2962423

$endgroup$

add a comment | 

$begingroup$

Multiplicative group of rational $mathbb Q^*cong mathbb Z_2 times bigoplus_i=1^infty mathbb Z$.

My question:

In general, how to characterize the multiplicative group $K^∗$ of the fraction field $K$ of a UFD $R$ ?

Can we use the same method which was used in $mathbb Q^*$?

And is there any reference about this?

Thank you for your time and help!

abstract-algebra group-theory ring-theory unique-factorization-domains

share|cite|improve this question

edited Mar 25 at 7:42

Andrews

asked Dec 3 '18 at 20:50

AndrewsAndrews

1,2962423

$endgroup$

share|cite|improve this question

edited Mar 25 at 7:42

Andrews

asked Dec 3 '18 at 20:50

AndrewsAndrews

1,2962423

share|cite|improve this question

edited Mar 25 at 7:42

Andrews

asked Dec 3 '18 at 20:50

AndrewsAndrews

1,2962423

asked Dec 3 '18 at 20:50

AndrewsAndrews

1,2962423

asked Dec 3 '18 at 20:50

AndrewsAndrews

1,2962423

1 Answer
1

active

oldest

votes

3

$begingroup$

Your general statement is totally false (note that the answer you linked stated that description for a number field, not for the field of fractions of an arbitrary UFD). For one thing, the description you gave would say that $K^*$ is always countable, which is certainly wrong. It's not even correct if you remove the restriction that the free part has countable rank: for instance, $mathbbC$ is a UFD and is its own fraction field, and its multiplicative group has no nontrivial cyclic group as a direct summand (since it is divisible).

In general, if $R$ is a UFD, then the multiplicative group of the field of fractions of $R$ is isomorphic to $R^timestimes bigoplus_I mathbbZ$ where $I$ is the set of irreducible elements of $R$ (modulo units). The proof is exactly the same as in the case of $mathbbQ$. Note though that the group $R^times$ of units in $R$ can be much more complicated than just a cyclic group and $I$ could have any cardinality.

share|cite|improve this answer

edited Dec 3 '18 at 21:15

answered Dec 3 '18 at 21:09

Eric WofseyEric Wofsey

193k14221352

$endgroup$

add a comment | 

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3

$begingroup$

Your general statement is totally false (note that the answer you linked stated that description for a number field, not for the field of fractions of an arbitrary UFD). For one thing, the description you gave would say that $K^*$ is always countable, which is certainly wrong. It's not even correct if you remove the restriction that the free part has countable rank: for instance, $mathbbC$ is a UFD and is its own fraction field, and its multiplicative group has no nontrivial cyclic group as a direct summand (since it is divisible).

In general, if $R$ is a UFD, then the multiplicative group of the field of fractions of $R$ is isomorphic to $R^timestimes bigoplus_I mathbbZ$ where $I$ is the set of irreducible elements of $R$ (modulo units). The proof is exactly the same as in the case of $mathbbQ$. Note though that the group $R^times$ of units in $R$ can be much more complicated than just a cyclic group and $I$ could have any cardinality.

share|cite|improve this answer

edited Dec 3 '18 at 21:15

answered Dec 3 '18 at 21:09

Eric WofseyEric Wofsey

193k14221352

$endgroup$

add a comment | 

3

$begingroup$

Your general statement is totally false (note that the answer you linked stated that description for a number field, not for the field of fractions of an arbitrary UFD). For one thing, the description you gave would say that $K^*$ is always countable, which is certainly wrong. It's not even correct if you remove the restriction that the free part has countable rank: for instance, $mathbbC$ is a UFD and is its own fraction field, and its multiplicative group has no nontrivial cyclic group as a direct summand (since it is divisible).

In general, if $R$ is a UFD, then the multiplicative group of the field of fractions of $R$ is isomorphic to $R^timestimes bigoplus_I mathbbZ$ where $I$ is the set of irreducible elements of $R$ (modulo units). The proof is exactly the same as in the case of $mathbbQ$. Note though that the group $R^times$ of units in $R$ can be much more complicated than just a cyclic group and $I$ could have any cardinality.

share|cite|improve this answer

edited Dec 3 '18 at 21:15

answered Dec 3 '18 at 21:09

Eric WofseyEric Wofsey

193k14221352

$endgroup$

add a comment | 

$begingroup$

Your general statement is totally false (note that the answer you linked stated that description for a number field, not for the field of fractions of an arbitrary UFD). For one thing, the description you gave would say that $K^*$ is always countable, which is certainly wrong. It's not even correct if you remove the restriction that the free part has countable rank: for instance, $mathbbC$ is a UFD and is its own fraction field, and its multiplicative group has no nontrivial cyclic group as a direct summand (since it is divisible).

In general, if $R$ is a UFD, then the multiplicative group of the field of fractions of $R$ is isomorphic to $R^timestimes bigoplus_I mathbbZ$ where $I$ is the set of irreducible elements of $R$ (modulo units). The proof is exactly the same as in the case of $mathbbQ$. Note though that the group $R^times$ of units in $R$ can be much more complicated than just a cyclic group and $I$ could have any cardinality.

share|cite|improve this answer

edited Dec 3 '18 at 21:15

answered Dec 3 '18 at 21:09

Eric WofseyEric Wofsey

193k14221352

$endgroup$

share|cite|improve this answer

edited Dec 3 '18 at 21:15

answered Dec 3 '18 at 21:09

Eric WofseyEric Wofsey

193k14221352

answered Dec 3 '18 at 21:09

Eric WofseyEric Wofsey

193k14221352

answered Dec 3 '18 at 21:09

Eric WofseyEric Wofsey

193k14221352