Valuation ring and localizations [closed]
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Theorem. Let $D$ be an integral domain with identity. The following conditions are equivalent.
(1) $D_P$ is a valuation ring for each proper prime $P$ in $D$.
(2) $D_M$ is a valuation ring for each maximal ideal $M$ in $D$.
How do I prove this? I tried working with the definitions, but it didn't turn out to be helpful. Help.
abstract-algebra maximal-and-prime-ideals localization valuation-theory
edited Dec 18 '18 at 18:28
user26857
39.3k124183
asked Dec 3 '18 at 17:46
GentianaGentiana
244
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closed as off-topic by user26857, Leucippus, Tianlalu, The Chaz 2.0, Brahadeesh Dec 19 '18 at 3:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user26857, Leucippus, Tianlalu, The Chaz 2.0, Brahadeesh
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$begingroup$
Theorem. Let $D$ be an integral domain with identity. The following conditions are equivalent.
(1) $D_P$ is a valuation ring for each proper prime $P$ in $D$.
(2) $D_M$ is a valuation ring for each maximal ideal $M$ in $D$.
How do I prove this? I tried working with the definitions, but it didn't turn out to be helpful. Help.
abstract-algebra maximal-and-prime-ideals localization valuation-theory
edited Dec 18 '18 at 18:28
user26857
39.3k124183
asked Dec 3 '18 at 17:46
GentianaGentiana
244
$endgroup$
closed as off-topic by user26857, Leucippus, Tianlalu, The Chaz 2.0, Brahadeesh Dec 19 '18 at 3:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user26857, Leucippus, Tianlalu, The Chaz 2.0, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Theorem. Let $D$ be an integral domain with identity. The following conditions are equivalent.
(1) $D_P$ is a valuation ring for each proper prime $P$ in $D$.
(2) $D_M$ is a valuation ring for each maximal ideal $M$ in $D$.
How do I prove this? I tried working with the definitions, but it didn't turn out to be helpful. Help.
abstract-algebra maximal-and-prime-ideals localization valuation-theory
edited Dec 18 '18 at 18:28
user26857
39.3k124183
asked Dec 3 '18 at 17:46
GentianaGentiana
244
$endgroup$
Theorem. Let $D$ be an integral domain with identity. The following conditions are equivalent.
(1) $D_P$ is a valuation ring for each proper prime $P$ in $D$.
(2) $D_M$ is a valuation ring for each maximal ideal $M$ in $D$.
How do I prove this? I tried working with the definitions, but it didn't turn out to be helpful. Help.
abstract-algebra maximal-and-prime-ideals localization valuation-theory
abstract-algebra maximal-and-prime-ideals localization valuation-theory
edited Dec 18 '18 at 18:28
user26857
39.3k124183
asked Dec 3 '18 at 17:46
GentianaGentiana
244
edited Dec 18 '18 at 18:28
user26857
39.3k124183
asked Dec 3 '18 at 17:46
GentianaGentiana
244
edited Dec 18 '18 at 18:28
user26857
39.3k124183
edited Dec 18 '18 at 18:28
user26857
39.3k124183
edited Dec 18 '18 at 18:28
user26857
39.3k124183
39.3k124183
asked Dec 3 '18 at 17:46
GentianaGentiana
244
asked Dec 3 '18 at 17:46
GentianaGentiana
244
asked Dec 3 '18 at 17:46
GentianaGentiana
244
244
closed as off-topic by user26857, Leucippus, Tianlalu, The Chaz 2.0, Brahadeesh Dec 19 '18 at 3:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user26857, Leucippus, Tianlalu, The Chaz 2.0, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by user26857, Leucippus, Tianlalu, The Chaz 2.0, Brahadeesh Dec 19 '18 at 3:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user26857, Leucippus, Tianlalu, The Chaz 2.0, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
active
oldest
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Let's take as definition of a valuation ring $V$ with field of fractions $K$ that for each $u in K$, either $u in V$ or $u^{-1} in V$.
Fact: Let $V$ be a valuation ring with field of fractions $K$. Let $R$ be a ring $V subseteq R subseteq K$. Then $R$ is a valuation ring.
Proof of fact is immediate.
(1) -> (2) is obvious since maximal ideals are prime.
(2) -> (1) Let $mathfrak{p}$ be a prime ideal. It is contained in some maximal ideal $mathfrak{m}$. We have naturally that $D subseteq D_mathfrak{m} subseteq D_mathfrak{p} subseteq K$. So $D_mathfrak{p}$ is a valuation ring by the above fact and the assumption that $D_mathfrak{m}$ is a valuation ring.
answered Dec 11 '18 at 8:11
Badam BaplanBadam Baplan
4,611722
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's take as definition of a valuation ring $V$ with field of fractions $K$ that for each $u in K$, either $u in V$ or $u^{-1} in V$.
Fact: Let $V$ be a valuation ring with field of fractions $K$. Let $R$ be a ring $V subseteq R subseteq K$. Then $R$ is a valuation ring.
Proof of fact is immediate.
(1) -> (2) is obvious since maximal ideals are prime.
(2) -> (1) Let $mathfrak{p}$ be a prime ideal. It is contained in some maximal ideal $mathfrak{m}$. We have naturally that $D subseteq D_mathfrak{m} subseteq D_mathfrak{p} subseteq K$. So $D_mathfrak{p}$ is a valuation ring by the above fact and the assumption that $D_mathfrak{m}$ is a valuation ring.
answered Dec 11 '18 at 8:11
Badam BaplanBadam Baplan
4,611722
$endgroup$
add a comment |
$begingroup$
Let's take as definition of a valuation ring $V$ with field of fractions $K$ that for each $u in K$, either $u in V$ or $u^{-1} in V$.
Fact: Let $V$ be a valuation ring with field of fractions $K$. Let $R$ be a ring $V subseteq R subseteq K$. Then $R$ is a valuation ring.
Proof of fact is immediate.
(1) -> (2) is obvious since maximal ideals are prime.
(2) -> (1) Let $mathfrak{p}$ be a prime ideal. It is contained in some maximal ideal $mathfrak{m}$. We have naturally that $D subseteq D_mathfrak{m} subseteq D_mathfrak{p} subseteq K$. So $D_mathfrak{p}$ is a valuation ring by the above fact and the assumption that $D_mathfrak{m}$ is a valuation ring.
answered Dec 11 '18 at 8:11
Badam BaplanBadam Baplan
4,611722
$endgroup$
add a comment |
$begingroup$
Let's take as definition of a valuation ring $V$ with field of fractions $K$ that for each $u in K$, either $u in V$ or $u^{-1} in V$.
Fact: Let $V$ be a valuation ring with field of fractions $K$. Let $R$ be a ring $V subseteq R subseteq K$. Then $R$ is a valuation ring.
Proof of fact is immediate.
(1) -> (2) is obvious since maximal ideals are prime.
(2) -> (1) Let $mathfrak{p}$ be a prime ideal. It is contained in some maximal ideal $mathfrak{m}$. We have naturally that $D subseteq D_mathfrak{m} subseteq D_mathfrak{p} subseteq K$. So $D_mathfrak{p}$ is a valuation ring by the above fact and the assumption that $D_mathfrak{m}$ is a valuation ring.
answered Dec 11 '18 at 8:11
Badam BaplanBadam Baplan
4,611722
$endgroup$
Let's take as definition of a valuation ring $V$ with field of fractions $K$ that for each $u in K$, either $u in V$ or $u^{-1} in V$.
Fact: Let $V$ be a valuation ring with field of fractions $K$. Let $R$ be a ring $V subseteq R subseteq K$. Then $R$ is a valuation ring.
Proof of fact is immediate.
(1) -> (2) is obvious since maximal ideals are prime.
(2) -> (1) Let $mathfrak{p}$ be a prime ideal. It is contained in some maximal ideal $mathfrak{m}$. We have naturally that $D subseteq D_mathfrak{m} subseteq D_mathfrak{p} subseteq K$. So $D_mathfrak{p}$ is a valuation ring by the above fact and the assumption that $D_mathfrak{m}$ is a valuation ring.
answered Dec 11 '18 at 8:11
Badam BaplanBadam Baplan
4,611722
answered Dec 11 '18 at 8:11
Badam BaplanBadam Baplan
4,611722
answered Dec 11 '18 at 8:11
Badam BaplanBadam Baplan
4,611722
answered Dec 11 '18 at 8:11
Badam BaplanBadam Baplan
4,611722
4,611722
add a comment |
add a comment |
