Are unique prime ideal factorization domains noetherian?
Let $A$ be a domain satisfying the following condition:
If $mathfrak p_1,dots,mathfrak p_k$ are distinct nonzero prime ideals of $A$, and if $m$ and $n$ are distinct elements of $mathbb N^k$, then we have
$$
mathfrak p_1^{m_1}cdotsmathfrak p_k^{m_k}nemathfrak p_1^{n_1}cdotsmathfrak p_k^{n_k}.
$$
Is $A$ necessarily noetherian?
This question is motivated by these outstanding answers of user26857 and Julian Rosen.
user26857's answer shows that noetherian domains do satisfy the above condition, whereas Julian's answer shows that many non-noetherian domains do not satisfy it.
commutative-algebra maximal-and-prime-ideals noetherian integral-domain
asked yesterday
Pierre-Yves Gaillard
13.2k23181
add a comment |
Let $A$ be a domain satisfying the following condition:
If $mathfrak p_1,dots,mathfrak p_k$ are distinct nonzero prime ideals of $A$, and if $m$ and $n$ are distinct elements of $mathbb N^k$, then we have
$$
mathfrak p_1^{m_1}cdotsmathfrak p_k^{m_k}nemathfrak p_1^{n_1}cdotsmathfrak p_k^{n_k}.
$$
Is $A$ necessarily noetherian?
This question is motivated by these outstanding answers of user26857 and Julian Rosen.
user26857's answer shows that noetherian domains do satisfy the above condition, whereas Julian's answer shows that many non-noetherian domains do not satisfy it.
commutative-algebra maximal-and-prime-ideals noetherian integral-domain
asked yesterday
Pierre-Yves Gaillard
13.2k23181
1
It's true for $1$-dimensional valuation domains. Is it true for valuation domains in general?
– Badam Baplan
yesterday
I think there are 1-dimensional valuation domains such that $m^2=m$.
– user26857
yesterday
Yes exactly. A $1$-dimensional valuation domain is Noetherian iff its max ideal is not idempotent iff it satisfies the property in question.
– Badam Baplan
yesterday
add a comment |
Let $A$ be a domain satisfying the following condition:
If $mathfrak p_1,dots,mathfrak p_k$ are distinct nonzero prime ideals of $A$, and if $m$ and $n$ are distinct elements of $mathbb N^k$, then we have
$$
mathfrak p_1^{m_1}cdotsmathfrak p_k^{m_k}nemathfrak p_1^{n_1}cdotsmathfrak p_k^{n_k}.
$$
Is $A$ necessarily noetherian?
This question is motivated by these outstanding answers of user26857 and Julian Rosen.
user26857's answer shows that noetherian domains do satisfy the above condition, whereas Julian's answer shows that many non-noetherian domains do not satisfy it.
commutative-algebra maximal-and-prime-ideals noetherian integral-domain
asked yesterday
Pierre-Yves Gaillard
13.2k23181
Let $A$ be a domain satisfying the following condition:
If $mathfrak p_1,dots,mathfrak p_k$ are distinct nonzero prime ideals of $A$, and if $m$ and $n$ are distinct elements of $mathbb N^k$, then we have
$$
mathfrak p_1^{m_1}cdotsmathfrak p_k^{m_k}nemathfrak p_1^{n_1}cdotsmathfrak p_k^{n_k}.
$$
Is $A$ necessarily noetherian?
This question is motivated by these outstanding answers of user26857 and Julian Rosen.
user26857's answer shows that noetherian domains do satisfy the above condition, whereas Julian's answer shows that many non-noetherian domains do not satisfy it.
commutative-algebra maximal-and-prime-ideals noetherian integral-domain
commutative-algebra maximal-and-prime-ideals noetherian integral-domain
asked yesterday
Pierre-Yves Gaillard
13.2k23181
asked yesterday
Pierre-Yves Gaillard
13.2k23181
edited yesterday
asked yesterday
Pierre-Yves Gaillard
13.2k23181
asked yesterday
Pierre-Yves Gaillard
13.2k23181
asked yesterday
Pierre-Yves Gaillard
13.2k23181
13.2k23181
1
It's true for $1$-dimensional valuation domains. Is it true for valuation domains in general?
– Badam Baplan
yesterday
I think there are 1-dimensional valuation domains such that $m^2=m$.
– user26857
yesterday
Yes exactly. A $1$-dimensional valuation domain is Noetherian iff its max ideal is not idempotent iff it satisfies the property in question.
– Badam Baplan
yesterday
add a comment |
1
It's true for $1$-dimensional valuation domains. Is it true for valuation domains in general?
– Badam Baplan
yesterday
I think there are 1-dimensional valuation domains such that $m^2=m$.
– user26857
yesterday
Yes exactly. A $1$-dimensional valuation domain is Noetherian iff its max ideal is not idempotent iff it satisfies the property in question.
– Badam Baplan
yesterday
1
1
It's true for $1$-dimensional valuation domains. Is it true for valuation domains in general?
– Badam Baplan
yesterday
It's true for $1$-dimensional valuation domains. Is it true for valuation domains in general?
– Badam Baplan
yesterday
I think there are 1-dimensional valuation domains such that $m^2=m$.
– user26857
yesterday
I think there are 1-dimensional valuation domains such that $m^2=m$.
– user26857
yesterday
Yes exactly. A $1$-dimensional valuation domain is Noetherian iff its max ideal is not idempotent iff it satisfies the property in question.
– Badam Baplan
yesterday
Yes exactly. A $1$-dimensional valuation domain is Noetherian iff its max ideal is not idempotent iff it satisfies the property in question.
– Badam Baplan
yesterday
add a comment |
1 Answer
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oldest
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Let's call domains having this property $UPIF$-domains, as in your first post.
Observe that a domain which is locally $UPIF$ is $UPIF$.
Indeed, given $$mathfrak p_1^{m_1}cdotsmathfrak p_k^{m_k}=mathfrak p_1^{n_1}cdotsmathfrak p_k^{n_k}$$
we can localize at each of the $mathfrak{p_i}$ one at a time and use the assumption that $D_{mathfrak{p_i}}$ is $UPIF$ to deduce that $m_i = n_i$.
From your linked Questions/Answers, we thus have that a locally Noetherian domain is locally UPIF, hence
A locally Noetherian domain is $UPIF$.
However, there are locally Noetherian domains that are not Noetherian.
A classic reference for such examples is section 2 of this paper of Heinzer and Ohm.
For a more recent reference, I'd suggest Loper's article on Almost Dedekind Domains Which Are Not Dedekind. In section 3, he discusses five distinct techniques for producing examples of non-Noetherian domains which are locally discrete valuation rings.
answered 7 hours ago
Badam Baplan
4,471722
add a comment |
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Let's call domains having this property $UPIF$-domains, as in your first post.
Observe that a domain which is locally $UPIF$ is $UPIF$.
Indeed, given $$mathfrak p_1^{m_1}cdotsmathfrak p_k^{m_k}=mathfrak p_1^{n_1}cdotsmathfrak p_k^{n_k}$$
we can localize at each of the $mathfrak{p_i}$ one at a time and use the assumption that $D_{mathfrak{p_i}}$ is $UPIF$ to deduce that $m_i = n_i$.
From your linked Questions/Answers, we thus have that a locally Noetherian domain is locally UPIF, hence
A locally Noetherian domain is $UPIF$.
However, there are locally Noetherian domains that are not Noetherian.
A classic reference for such examples is section 2 of this paper of Heinzer and Ohm.
For a more recent reference, I'd suggest Loper's article on Almost Dedekind Domains Which Are Not Dedekind. In section 3, he discusses five distinct techniques for producing examples of non-Noetherian domains which are locally discrete valuation rings.
answered 7 hours ago
Badam Baplan
4,471722
add a comment |
Let's call domains having this property $UPIF$-domains, as in your first post.
Observe that a domain which is locally $UPIF$ is $UPIF$.
Indeed, given $$mathfrak p_1^{m_1}cdotsmathfrak p_k^{m_k}=mathfrak p_1^{n_1}cdotsmathfrak p_k^{n_k}$$
we can localize at each of the $mathfrak{p_i}$ one at a time and use the assumption that $D_{mathfrak{p_i}}$ is $UPIF$ to deduce that $m_i = n_i$.
From your linked Questions/Answers, we thus have that a locally Noetherian domain is locally UPIF, hence
A locally Noetherian domain is $UPIF$.
However, there are locally Noetherian domains that are not Noetherian.
A classic reference for such examples is section 2 of this paper of Heinzer and Ohm.
For a more recent reference, I'd suggest Loper's article on Almost Dedekind Domains Which Are Not Dedekind. In section 3, he discusses five distinct techniques for producing examples of non-Noetherian domains which are locally discrete valuation rings.
answered 7 hours ago
Badam Baplan
4,471722
add a comment |
Let's call domains having this property $UPIF$-domains, as in your first post.
Observe that a domain which is locally $UPIF$ is $UPIF$.
Indeed, given $$mathfrak p_1^{m_1}cdotsmathfrak p_k^{m_k}=mathfrak p_1^{n_1}cdotsmathfrak p_k^{n_k}$$
we can localize at each of the $mathfrak{p_i}$ one at a time and use the assumption that $D_{mathfrak{p_i}}$ is $UPIF$ to deduce that $m_i = n_i$.
From your linked Questions/Answers, we thus have that a locally Noetherian domain is locally UPIF, hence
A locally Noetherian domain is $UPIF$.
However, there are locally Noetherian domains that are not Noetherian.
A classic reference for such examples is section 2 of this paper of Heinzer and Ohm.
For a more recent reference, I'd suggest Loper's article on Almost Dedekind Domains Which Are Not Dedekind. In section 3, he discusses five distinct techniques for producing examples of non-Noetherian domains which are locally discrete valuation rings.
answered 7 hours ago
Badam Baplan
4,471722
Let's call domains having this property $UPIF$-domains, as in your first post.
Observe that a domain which is locally $UPIF$ is $UPIF$.
Indeed, given $$mathfrak p_1^{m_1}cdotsmathfrak p_k^{m_k}=mathfrak p_1^{n_1}cdotsmathfrak p_k^{n_k}$$
we can localize at each of the $mathfrak{p_i}$ one at a time and use the assumption that $D_{mathfrak{p_i}}$ is $UPIF$ to deduce that $m_i = n_i$.
From your linked Questions/Answers, we thus have that a locally Noetherian domain is locally UPIF, hence
A locally Noetherian domain is $UPIF$.
However, there are locally Noetherian domains that are not Noetherian.
A classic reference for such examples is section 2 of this paper of Heinzer and Ohm.
For a more recent reference, I'd suggest Loper's article on Almost Dedekind Domains Which Are Not Dedekind. In section 3, he discusses five distinct techniques for producing examples of non-Noetherian domains which are locally discrete valuation rings.
answered 7 hours ago
Badam Baplan
4,471722
edited 5 hours ago
answered 7 hours ago
Badam Baplan
4,471722
answered 7 hours ago
Badam Baplan
4,471722
answered 7 hours ago
Badam Baplan
4,471722
4,471722
add a comment |
add a comment |
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1
It's true for $1$-dimensional valuation domains. Is it true for valuation domains in general?
– Badam Baplan
yesterday
I think there are 1-dimensional valuation domains such that $m^2=m$.
– user26857
yesterday
Yes exactly. A $1$-dimensional valuation domain is Noetherian iff its max ideal is not idempotent iff it satisfies the property in question.
– Badam Baplan
yesterday