Mixing
On Noetherian domains whose proper submodules of the fraction field are projective
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Let $R$ be a Noetherian domain with fraction field $K$. If every proper $R$-submodule of $K$, containing $R$ (i.e. all $R$-submodules $M$ of $K$ such that $R subseteq M subsetneq K$) is projective over $R$, then is $R$ integrally closed in $K$ ?
If this is not true in general, what if we strengthen the hypothesis to say every proper $R$-submodule of $K$ is projective over $R$ ?
ring-theory commutative-algebra modules projective-module integral-extensions
edited Jan 30 at 16:30
user26857
39.5k124283
asked Jan 29 at 23:31
user102248user102248
30619
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add a comment |
$begingroup$
Let $R$ be a Noetherian domain with fraction field $K$. If every proper $R$-submodule of $K$, containing $R$ (i.e. all $R$-submodules $M$ of $K$ such that $R subseteq M subsetneq K$) is projective over $R$, then is $R$ integrally closed in $K$ ?
If this is not true in general, what if we strengthen the hypothesis to say every proper $R$-submodule of $K$ is projective over $R$ ?
ring-theory commutative-algebra modules projective-module integral-extensions
edited Jan 30 at 16:30
user26857
39.5k124283
asked Jan 29 at 23:31
user102248user102248
30619
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Please don't self-delete your question after receiving an answer, so that the community can evaluate its value to this site.
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– GNUSupporter 8964民主女神 地下教會
Feb 6 at 21:16
add a comment |
$begingroup$
Let $R$ be a Noetherian domain with fraction field $K$. If every proper $R$-submodule of $K$, containing $R$ (i.e. all $R$-submodules $M$ of $K$ such that $R subseteq M subsetneq K$) is projective over $R$, then is $R$ integrally closed in $K$ ?
If this is not true in general, what if we strengthen the hypothesis to say every proper $R$-submodule of $K$ is projective over $R$ ?
ring-theory commutative-algebra modules projective-module integral-extensions
edited Jan 30 at 16:30
user26857
39.5k124283
asked Jan 29 at 23:31
user102248user102248
30619
$endgroup$
Let $R$ be a Noetherian domain with fraction field $K$. If every proper $R$-submodule of $K$, containing $R$ (i.e. all $R$-submodules $M$ of $K$ such that $R subseteq M subsetneq K$) is projective over $R$, then is $R$ integrally closed in $K$ ?
If this is not true in general, what if we strengthen the hypothesis to say every proper $R$-submodule of $K$ is projective over $R$ ?
ring-theory commutative-algebra modules projective-module integral-extensions
ring-theory commutative-algebra modules projective-module integral-extensions
edited Jan 30 at 16:30
user26857
39.5k124283
asked Jan 29 at 23:31
user102248user102248
30619
edited Jan 30 at 16:30
user26857
39.5k124283
asked Jan 29 at 23:31
user102248user102248
30619
edited Jan 30 at 16:30
user26857
39.5k124283
edited Jan 30 at 16:30
user26857
39.5k124283
edited Jan 30 at 16:30
user26857
39.5k124283
39.5k124283
asked Jan 29 at 23:31
user102248user102248
30619
asked Jan 29 at 23:31
user102248user102248
30619
asked Jan 29 at 23:31
user102248user102248
30619
30619
$begingroup$
Please don't self-delete your question after receiving an answer, so that the community can evaluate its value to this site.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 6 at 21:16
add a comment |
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Please don't self-delete your question after receiving an answer, so that the community can evaluate its value to this site.
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– GNUSupporter 8964民主女神 地下教會
Feb 6 at 21:16
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Please don't self-delete your question after receiving an answer, so that the community can evaluate its value to this site.
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– GNUSupporter 8964民主女神 地下教會
Feb 6 at 21:16
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Please don't self-delete your question after receiving an answer, so that the community can evaluate its value to this site.
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– GNUSupporter 8964民主女神 地下教會
Feb 6 at 21:16
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1 Answer
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This condition is extremely strong: it is equivalent to $R$ being a DVR or a field. In particular, it does imply that $R$ is integrally closed.
Indeed, suppose $R$ is not a field. Let $rin R$ be a nonzero nonunit element. Then $R[r^{-1}]$ is a submodule of $K$ containing $R$ which is not projective. By hypothesis, this is only possible if $R[r^{-1}]=K$. This means $r$ is in every nonzero prime ideal of $R$. Since $rin R$ was an arbitrary nonzero nonunit element, this implies that $R$ has a unique nonzero prime ideal $P$, consisting of all nonunits.
To conclude that $R$ is a DVR, it suffices to show that $P$ is principal. To prove this, let $ain P$ be nonzero and let $M=a^{-1}Psubset K$. Then $M$ is a proper submodule of $K$ which contains $R$, and thus is projective. Since $R$ is local, this means $M$ is free. But multiplication by $a$ is an isomorphism $Mto P$, and thus $P$ is free. Since $Psubset R$ it cannot be free on more than one generator, and so $P$ is free on one generator and in particular is principal.
answered Jan 30 at 0:04
Eric WofseyEric Wofsey
192k14216350
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1 Answer
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$begingroup$
This condition is extremely strong: it is equivalent to $R$ being a DVR or a field. In particular, it does imply that $R$ is integrally closed.
Indeed, suppose $R$ is not a field. Let $rin R$ be a nonzero nonunit element. Then $R[r^{-1}]$ is a submodule of $K$ containing $R$ which is not projective. By hypothesis, this is only possible if $R[r^{-1}]=K$. This means $r$ is in every nonzero prime ideal of $R$. Since $rin R$ was an arbitrary nonzero nonunit element, this implies that $R$ has a unique nonzero prime ideal $P$, consisting of all nonunits.
To conclude that $R$ is a DVR, it suffices to show that $P$ is principal. To prove this, let $ain P$ be nonzero and let $M=a^{-1}Psubset K$. Then $M$ is a proper submodule of $K$ which contains $R$, and thus is projective. Since $R$ is local, this means $M$ is free. But multiplication by $a$ is an isomorphism $Mto P$, and thus $P$ is free. Since $Psubset R$ it cannot be free on more than one generator, and so $P$ is free on one generator and in particular is principal.
answered Jan 30 at 0:04
Eric WofseyEric Wofsey
192k14216350
$endgroup$
add a comment |
$begingroup$
This condition is extremely strong: it is equivalent to $R$ being a DVR or a field. In particular, it does imply that $R$ is integrally closed.
Indeed, suppose $R$ is not a field. Let $rin R$ be a nonzero nonunit element. Then $R[r^{-1}]$ is a submodule of $K$ containing $R$ which is not projective. By hypothesis, this is only possible if $R[r^{-1}]=K$. This means $r$ is in every nonzero prime ideal of $R$. Since $rin R$ was an arbitrary nonzero nonunit element, this implies that $R$ has a unique nonzero prime ideal $P$, consisting of all nonunits.
To conclude that $R$ is a DVR, it suffices to show that $P$ is principal. To prove this, let $ain P$ be nonzero and let $M=a^{-1}Psubset K$. Then $M$ is a proper submodule of $K$ which contains $R$, and thus is projective. Since $R$ is local, this means $M$ is free. But multiplication by $a$ is an isomorphism $Mto P$, and thus $P$ is free. Since $Psubset R$ it cannot be free on more than one generator, and so $P$ is free on one generator and in particular is principal.
answered Jan 30 at 0:04
Eric WofseyEric Wofsey
192k14216350
$endgroup$
add a comment |
$begingroup$
This condition is extremely strong: it is equivalent to $R$ being a DVR or a field. In particular, it does imply that $R$ is integrally closed.
Indeed, suppose $R$ is not a field. Let $rin R$ be a nonzero nonunit element. Then $R[r^{-1}]$ is a submodule of $K$ containing $R$ which is not projective. By hypothesis, this is only possible if $R[r^{-1}]=K$. This means $r$ is in every nonzero prime ideal of $R$. Since $rin R$ was an arbitrary nonzero nonunit element, this implies that $R$ has a unique nonzero prime ideal $P$, consisting of all nonunits.
To conclude that $R$ is a DVR, it suffices to show that $P$ is principal. To prove this, let $ain P$ be nonzero and let $M=a^{-1}Psubset K$. Then $M$ is a proper submodule of $K$ which contains $R$, and thus is projective. Since $R$ is local, this means $M$ is free. But multiplication by $a$ is an isomorphism $Mto P$, and thus $P$ is free. Since $Psubset R$ it cannot be free on more than one generator, and so $P$ is free on one generator and in particular is principal.
answered Jan 30 at 0:04
Eric WofseyEric Wofsey
192k14216350
$endgroup$
This condition is extremely strong: it is equivalent to $R$ being a DVR or a field. In particular, it does imply that $R$ is integrally closed.
Indeed, suppose $R$ is not a field. Let $rin R$ be a nonzero nonunit element. Then $R[r^{-1}]$ is a submodule of $K$ containing $R$ which is not projective. By hypothesis, this is only possible if $R[r^{-1}]=K$. This means $r$ is in every nonzero prime ideal of $R$. Since $rin R$ was an arbitrary nonzero nonunit element, this implies that $R$ has a unique nonzero prime ideal $P$, consisting of all nonunits.
To conclude that $R$ is a DVR, it suffices to show that $P$ is principal. To prove this, let $ain P$ be nonzero and let $M=a^{-1}Psubset K$. Then $M$ is a proper submodule of $K$ which contains $R$, and thus is projective. Since $R$ is local, this means $M$ is free. But multiplication by $a$ is an isomorphism $Mto P$, and thus $P$ is free. Since $Psubset R$ it cannot be free on more than one generator, and so $P$ is free on one generator and in particular is principal.
answered Jan 30 at 0:04
Eric WofseyEric Wofsey
192k14216350
answered Jan 30 at 0:04
Eric WofseyEric Wofsey
192k14216350
answered Jan 30 at 0:04
Eric WofseyEric Wofsey
192k14216350
answered Jan 30 at 0:04
Eric WofseyEric Wofsey
192k14216350
192k14216350
add a comment |
add a comment |
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– GNUSupporter 8964民主女神 地下教會
Feb 6 at 21:16