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A Noetherian integral domain is a UFD iff $(f):(g)$ is principal

A Noetherian domain $R$ is a UFD iff $I_{fg}={hin R mid hgin(f)}$ is principal for all $f,gin R$.Is any UFD also a PID?In a Noetherian integral domain, a principal prime ideal can't have proper non-zero prime idealsIf in a UFD every maximal ideal is principal then it is a PIDIs there a constructive proof that a Euclidean domain is a UFD?Noetherian domain with a unique non-zero prime ideal is a UFD?Example for Noetherian domain, non-UFD $R$ of dimension one such that every localization $R_P$ is UFD.Integral domains $R$ such that any ring $R subset S subset K$ is NoetherianNoetherian domain R is a UFD if every prime ideal of height 1 in R is principal.The relation between Noetherian valuation domain, principal ideal domain, factor ring and Dedekind$R$ a $PID Rightarrow R[[X]]$ a $UFD$

5

1

$begingroup$

Let $R$ be a Noetherian integral domain. For $f, g in R$, define $(f):(g)={h in R mid hg in (f) }$. Sow that $R$ is a UFD if and only if $(f):(g)$ is principal for all $f,g in R$.

It is easy to show that $(f):(g)$ is an ideal. For the forward direction, I suspect that I need to use the fact that $R$ is Noetherian to show that $(f):(g)$ is principal. For the reverse direction, I know that if every $(f):(g)$ is principal then the ring $R$ is a PID. But I don't know how to proceed. Any ideas?

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

share|cite|improve this question

edited yesterday

user26857

39.3k124183

asked Mar 5 '16 at 23:47

user112358user112358

885518

$endgroup$

add a comment | 

5

1

$begingroup$

Let $R$ be a Noetherian integral domain. For $f, g in R$, define $(f):(g)={h in R mid hg in (f) }$. Sow that $R$ is a UFD if and only if $(f):(g)$ is principal for all $f,g in R$.

It is easy to show that $(f):(g)$ is an ideal. For the forward direction, I suspect that I need to use the fact that $R$ is Noetherian to show that $(f):(g)$ is principal. For the reverse direction, I know that if every $(f):(g)$ is principal then the ring $R$ is a PID. But I don't know how to proceed. Any ideas?

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

share|cite|improve this question

edited yesterday

user26857

39.3k124183

asked Mar 5 '16 at 23:47

user112358user112358

885518

$endgroup$

add a comment | 

$begingroup$

Let $R$ be a Noetherian integral domain. For $f, g in R$, define $(f):(g)={h in R mid hg in (f) }$. Sow that $R$ is a UFD if and only if $(f):(g)$ is principal for all $f,g in R$.

It is easy to show that $(f):(g)$ is an ideal. For the forward direction, I suspect that I need to use the fact that $R$ is Noetherian to show that $(f):(g)$ is principal. For the reverse direction, I know that if every $(f):(g)$ is principal then the ring $R$ is a PID. But I don't know how to proceed. Any ideas?

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

share|cite|improve this question

edited yesterday

user26857

39.3k124183

asked Mar 5 '16 at 23:47

user112358user112358

885518

$endgroup$

share|cite|improve this question

edited yesterday

user26857

39.3k124183

asked Mar 5 '16 at 23:47

user112358user112358

885518

share|cite|improve this question

edited yesterday

user26857

39.3k124183

asked Mar 5 '16 at 23:47

user112358user112358

885518

edited yesterday

user26857

39.3k124183

edited yesterday

user26857

39.3k124183

asked Mar 5 '16 at 23:47

user112358user112358

885518

asked Mar 5 '16 at 23:47

user112358user112358

885518

1 Answer
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2

$begingroup$

I don't really understand what you mean by "I know that if every $(f):(g)$ is principal then the ring $R$ is a PID" : you seem to imply that every noetherian UFD is a PID, which is false.

The first implication (assuming $R$ is a UFD) is actually true even if $R$ is not noetherian. Indeed, if $R$ is a UFD, then write $f = uprod p_i^{a_i}$ and $g=vprod p_i^{b_i}$. Then for any $h = wprod p_i^{c_i}$, you get $hgin (f)$ iff $forall i,$ $b_i + c_i geqslant a_i$. Then putting $d_i = max(0 ; a_i-b_i)$ you get $(f):(g) = (prod p_i^{d_i})$.

As for the second implication, you only need that $R$ admits an irreducible factor decomposition (which of course is always true when $R$ is noetherian). In this case it is well-known that $R$ is a UFD (meaning that the decomposition is unique) if and only if irreducible elements satisfy the Gauss (or Euclid, or whatever) lemma : $pmid ab$ implies $pmid a$ or $pmid b$.

First observe that if $(f):(g) = (x)$ then since $fg in (f)$ you get $fin (f):(g) = (x)$ and hence $xmid f$. So if $p$ is irreducible, and $(p):(a) = (x)$, you get $xmid p$, and thus $(x) = (1)$ or $(x) = (p)$. The first case happens if and only if $pmid a$ by definition.

Now assume $pmid ab$ and $pnmid b$. Then $abin (p)$ so by definition $ain (p):(b)$, but $(p):(b)=(p)$ by the previous observation, so $pmid a$. This is the Gauss lemma.

share|cite|improve this answer

edited yesterday

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39.3k124183

answered Mar 6 '16 at 3:05

Captain LamaCaptain Lama

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$endgroup$

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2

$begingroup$

I don't really understand what you mean by "I know that if every $(f):(g)$ is principal then the ring $R$ is a PID" : you seem to imply that every noetherian UFD is a PID, which is false.

The first implication (assuming $R$ is a UFD) is actually true even if $R$ is not noetherian. Indeed, if $R$ is a UFD, then write $f = uprod p_i^{a_i}$ and $g=vprod p_i^{b_i}$. Then for any $h = wprod p_i^{c_i}$, you get $hgin (f)$ iff $forall i,$ $b_i + c_i geqslant a_i$. Then putting $d_i = max(0 ; a_i-b_i)$ you get $(f):(g) = (prod p_i^{d_i})$.

As for the second implication, you only need that $R$ admits an irreducible factor decomposition (which of course is always true when $R$ is noetherian). In this case it is well-known that $R$ is a UFD (meaning that the decomposition is unique) if and only if irreducible elements satisfy the Gauss (or Euclid, or whatever) lemma : $pmid ab$ implies $pmid a$ or $pmid b$.

First observe that if $(f):(g) = (x)$ then since $fg in (f)$ you get $fin (f):(g) = (x)$ and hence $xmid f$. So if $p$ is irreducible, and $(p):(a) = (x)$, you get $xmid p$, and thus $(x) = (1)$ or $(x) = (p)$. The first case happens if and only if $pmid a$ by definition.

Now assume $pmid ab$ and $pnmid b$. Then $abin (p)$ so by definition $ain (p):(b)$, but $(p):(b)=(p)$ by the previous observation, so $pmid a$. This is the Gauss lemma.

share|cite|improve this answer

edited yesterday

user26857

39.3k124183

answered Mar 6 '16 at 3:05

Captain LamaCaptain Lama

10.1k1030

$endgroup$

add a comment | 

2

$begingroup$

I don't really understand what you mean by "I know that if every $(f):(g)$ is principal then the ring $R$ is a PID" : you seem to imply that every noetherian UFD is a PID, which is false.

The first implication (assuming $R$ is a UFD) is actually true even if $R$ is not noetherian. Indeed, if $R$ is a UFD, then write $f = uprod p_i^{a_i}$ and $g=vprod p_i^{b_i}$. Then for any $h = wprod p_i^{c_i}$, you get $hgin (f)$ iff $forall i,$ $b_i + c_i geqslant a_i$. Then putting $d_i = max(0 ; a_i-b_i)$ you get $(f):(g) = (prod p_i^{d_i})$.

As for the second implication, you only need that $R$ admits an irreducible factor decomposition (which of course is always true when $R$ is noetherian). In this case it is well-known that $R$ is a UFD (meaning that the decomposition is unique) if and only if irreducible elements satisfy the Gauss (or Euclid, or whatever) lemma : $pmid ab$ implies $pmid a$ or $pmid b$.

First observe that if $(f):(g) = (x)$ then since $fg in (f)$ you get $fin (f):(g) = (x)$ and hence $xmid f$. So if $p$ is irreducible, and $(p):(a) = (x)$, you get $xmid p$, and thus $(x) = (1)$ or $(x) = (p)$. The first case happens if and only if $pmid a$ by definition.

Now assume $pmid ab$ and $pnmid b$. Then $abin (p)$ so by definition $ain (p):(b)$, but $(p):(b)=(p)$ by the previous observation, so $pmid a$. This is the Gauss lemma.

share|cite|improve this answer

edited yesterday

user26857

39.3k124183

answered Mar 6 '16 at 3:05

Captain LamaCaptain Lama

10.1k1030

$endgroup$

add a comment | 

$begingroup$

I don't really understand what you mean by "I know that if every $(f):(g)$ is principal then the ring $R$ is a PID" : you seem to imply that every noetherian UFD is a PID, which is false.

The first implication (assuming $R$ is a UFD) is actually true even if $R$ is not noetherian. Indeed, if $R$ is a UFD, then write $f = uprod p_i^{a_i}$ and $g=vprod p_i^{b_i}$. Then for any $h = wprod p_i^{c_i}$, you get $hgin (f)$ iff $forall i,$ $b_i + c_i geqslant a_i$. Then putting $d_i = max(0 ; a_i-b_i)$ you get $(f):(g) = (prod p_i^{d_i})$.

As for the second implication, you only need that $R$ admits an irreducible factor decomposition (which of course is always true when $R$ is noetherian). In this case it is well-known that $R$ is a UFD (meaning that the decomposition is unique) if and only if irreducible elements satisfy the Gauss (or Euclid, or whatever) lemma : $pmid ab$ implies $pmid a$ or $pmid b$.

First observe that if $(f):(g) = (x)$ then since $fg in (f)$ you get $fin (f):(g) = (x)$ and hence $xmid f$. So if $p$ is irreducible, and $(p):(a) = (x)$, you get $xmid p$, and thus $(x) = (1)$ or $(x) = (p)$. The first case happens if and only if $pmid a$ by definition.

Now assume $pmid ab$ and $pnmid b$. Then $abin (p)$ so by definition $ain (p):(b)$, but $(p):(b)=(p)$ by the previous observation, so $pmid a$. This is the Gauss lemma.

share|cite|improve this answer

edited yesterday

user26857

39.3k124183

answered Mar 6 '16 at 3:05

Captain LamaCaptain Lama

10.1k1030

$endgroup$

share|cite|improve this answer

edited yesterday

user26857

39.3k124183

answered Mar 6 '16 at 3:05

Captain LamaCaptain Lama

10.1k1030

edited yesterday

user26857

39.3k124183

edited yesterday

user26857

39.3k124183

answered Mar 6 '16 at 3:05

Captain LamaCaptain Lama

10.1k1030

answered Mar 6 '16 at 3:05

Captain LamaCaptain Lama

10.1k1030