Discrete valuation arising from localizations of $R[[t]]$

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Let $R$ be a DVR and consider the ring of power series $R[[t]]$. Now let $mathfrak psubset R[[t]]$ be a prime ideal of height $1$.

Why the localization $R[[t]]_mathfrak p$ is again a DVR?

abstract-algebra commutative-algebra localization valuation-theory

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edited Aug 4 at 20:33

user26857

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up vote
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Let $R$ be a DVR and consider the ring of power series $R[[t]]$. Now let $mathfrak psubset R[[t]]$ be a prime ideal of height $1$.

Why the localization $R[[t]]_mathfrak p$ is again a DVR?

abstract-algebra commutative-algebra localization valuation-theory

share|cite|improve this question

edited Aug 4 at 20:33

user26857

38.7k123678

asked Jul 26 at 15:16

manifold

303213

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Let $R$ be a DVR and consider the ring of power series $R[[t]]$. Now let $mathfrak psubset R[[t]]$ be a prime ideal of height $1$.

Why the localization $R[[t]]_mathfrak p$ is again a DVR?

abstract-algebra commutative-algebra localization valuation-theory

share|cite|improve this question

edited Aug 4 at 20:33

user26857

38.7k123678

asked Jul 26 at 15:16

manifold

303213

Let $R$ be a DVR and consider the ring of power series $R[[t]]$. Now let $mathfrak psubset R[[t]]$ be a prime ideal of height $1$.

Why the localization $R[[t]]_mathfrak p$ is again a DVR?

abstract-algebra commutative-algebra localization valuation-theory

share|cite|improve this question

edited Aug 4 at 20:33

user26857

38.7k123678

asked Jul 26 at 15:16

manifold

303213

share|cite|improve this question

share|cite|improve this question

edited Aug 4 at 20:33

user26857

38.7k123678

edited Aug 4 at 20:33

user26857

38.7k123678

edited Aug 4 at 20:33

user26857

38.7k123678

38.7k123678

asked Jul 26 at 15:16

manifold

303213

asked Jul 26 at 15:16

manifold

303213

asked Jul 26 at 15:16

manifold

303213

303213

add a comment | 

add a comment | 

1 Answer
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The ring $R[[t]]$ is local of dimension $2$. Furthermore, its maximal ideal is generated by two elements, a uniformizing parameter of $R$ and $t$. Thus, $R[[t]]$ is regular. A localization of a regular ring is regular, and a regular local ring of dimension 1 is a DVR.

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answered Jul 27 at 10:37

Youngsu

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1 Answer
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1 Answer
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up vote
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down vote

The ring $R[[t]]$ is local of dimension $2$. Furthermore, its maximal ideal is generated by two elements, a uniformizing parameter of $R$ and $t$. Thus, $R[[t]]$ is regular. A localization of a regular ring is regular, and a regular local ring of dimension 1 is a DVR.

share|cite|improve this answer

answered Jul 27 at 10:37

Youngsu

1,693715

add a comment | 

up vote
3
down vote

The ring $R[[t]]$ is local of dimension $2$. Furthermore, its maximal ideal is generated by two elements, a uniformizing parameter of $R$ and $t$. Thus, $R[[t]]$ is regular. A localization of a regular ring is regular, and a regular local ring of dimension 1 is a DVR.

share|cite|improve this answer

answered Jul 27 at 10:37

Youngsu

1,693715

add a comment | 

up vote
3
down vote

up vote
3
down vote

The ring $R[[t]]$ is local of dimension $2$. Furthermore, its maximal ideal is generated by two elements, a uniformizing parameter of $R$ and $t$. Thus, $R[[t]]$ is regular. A localization of a regular ring is regular, and a regular local ring of dimension 1 is a DVR.

share|cite|improve this answer

answered Jul 27 at 10:37

Youngsu

1,693715

The ring $R[[t]]$ is local of dimension $2$. Furthermore, its maximal ideal is generated by two elements, a uniformizing parameter of $R$ and $t$. Thus, $R[[t]]$ is regular. A localization of a regular ring is regular, and a regular local ring of dimension 1 is a DVR.

share|cite|improve this answer

answered Jul 27 at 10:37

Youngsu

1,693715

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 10:37

Youngsu

1,693715

answered Jul 27 at 10:37

Youngsu

1,693715

answered Jul 27 at 10:37

Youngsu

1,693715

1,693715

add a comment | 

add a comment | 

 

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