Application Chinese Remainder Theorem to Dedekind Domains
I have a question about the application of CRT in a proof of following thread:
Dedekind domain with a finite number of prime ideals is principal
The claim is that a Dedekind domain with a finite number of prime ideals is already principal:
In his answer @pki uses following argument:
Let $R$ be a Dedekind ring and assume that the prime ideals are $mathfrak{p}_1,ldots,mathfrak{p}_n$. Then $mathfrak{p}_1^2,mathfrak{p}_2,ldots,mathfrak{p}_n$ are coprime. Pick an element $pi in mathfrak{p}_1setminus mathfrak{p}_1^2$ and by CRT you can find an $xin R$ s.t.
$$ xequiv pi,(textrm{mod } mathfrak{p}_1^2),;; xequiv 1,(textrm{mod } mathfrak{p}_k),; k=2,ldots,n $$
Factoring we must have $(x)=mathfrak{p}_1$ (???)
Indeed the CRT provides a $x$ such that $ xequiv pi,(textrm{mod } mathfrak{p}_1^2),;; xequiv 1,(textrm{mod } mathfrak{p}_k),; k=2,ldots,n $ holds.
But why we get $(x)=mathfrak{p}_1$?
Factoring just implies $(bar{x}) = mathfrak{p}_1/ mathfrak{p}_1^2$. How to deduce $(x)=mathfrak{p}_1$?
commutative-algebra dedekind-domain
add a comment |
I have a question about the application of CRT in a proof of following thread:
Dedekind domain with a finite number of prime ideals is principal
The claim is that a Dedekind domain with a finite number of prime ideals is already principal:
In his answer @pki uses following argument:
Let $R$ be a Dedekind ring and assume that the prime ideals are $mathfrak{p}_1,ldots,mathfrak{p}_n$. Then $mathfrak{p}_1^2,mathfrak{p}_2,ldots,mathfrak{p}_n$ are coprime. Pick an element $pi in mathfrak{p}_1setminus mathfrak{p}_1^2$ and by CRT you can find an $xin R$ s.t.
$$ xequiv pi,(textrm{mod } mathfrak{p}_1^2),;; xequiv 1,(textrm{mod } mathfrak{p}_k),; k=2,ldots,n $$
Factoring we must have $(x)=mathfrak{p}_1$ (???)
Indeed the CRT provides a $x$ such that $ xequiv pi,(textrm{mod } mathfrak{p}_1^2),;; xequiv 1,(textrm{mod } mathfrak{p}_k),; k=2,ldots,n $ holds.
But why we get $(x)=mathfrak{p}_1$?
Factoring just implies $(bar{x}) = mathfrak{p}_1/ mathfrak{p}_1^2$. How to deduce $(x)=mathfrak{p}_1$?
commutative-algebra dedekind-domain
add a comment |
I have a question about the application of CRT in a proof of following thread:
Dedekind domain with a finite number of prime ideals is principal
The claim is that a Dedekind domain with a finite number of prime ideals is already principal:
In his answer @pki uses following argument:
Let $R$ be a Dedekind ring and assume that the prime ideals are $mathfrak{p}_1,ldots,mathfrak{p}_n$. Then $mathfrak{p}_1^2,mathfrak{p}_2,ldots,mathfrak{p}_n$ are coprime. Pick an element $pi in mathfrak{p}_1setminus mathfrak{p}_1^2$ and by CRT you can find an $xin R$ s.t.
$$ xequiv pi,(textrm{mod } mathfrak{p}_1^2),;; xequiv 1,(textrm{mod } mathfrak{p}_k),; k=2,ldots,n $$
Factoring we must have $(x)=mathfrak{p}_1$ (???)
Indeed the CRT provides a $x$ such that $ xequiv pi,(textrm{mod } mathfrak{p}_1^2),;; xequiv 1,(textrm{mod } mathfrak{p}_k),; k=2,ldots,n $ holds.
But why we get $(x)=mathfrak{p}_1$?
Factoring just implies $(bar{x}) = mathfrak{p}_1/ mathfrak{p}_1^2$. How to deduce $(x)=mathfrak{p}_1$?
commutative-algebra dedekind-domain
I have a question about the application of CRT in a proof of following thread:
Dedekind domain with a finite number of prime ideals is principal
The claim is that a Dedekind domain with a finite number of prime ideals is already principal:
In his answer @pki uses following argument:
Let $R$ be a Dedekind ring and assume that the prime ideals are $mathfrak{p}_1,ldots,mathfrak{p}_n$. Then $mathfrak{p}_1^2,mathfrak{p}_2,ldots,mathfrak{p}_n$ are coprime. Pick an element $pi in mathfrak{p}_1setminus mathfrak{p}_1^2$ and by CRT you can find an $xin R$ s.t.
$$ xequiv pi,(textrm{mod } mathfrak{p}_1^2),;; xequiv 1,(textrm{mod } mathfrak{p}_k),; k=2,ldots,n $$
Factoring we must have $(x)=mathfrak{p}_1$ (???)
Indeed the CRT provides a $x$ such that $ xequiv pi,(textrm{mod } mathfrak{p}_1^2),;; xequiv 1,(textrm{mod } mathfrak{p}_k),; k=2,ldots,n $ holds.
But why we get $(x)=mathfrak{p}_1$?
Factoring just implies $(bar{x}) = mathfrak{p}_1/ mathfrak{p}_1^2$. How to deduce $(x)=mathfrak{p}_1$?
commutative-algebra dedekind-domain
commutative-algebra dedekind-domain
asked Mar 30 '18 at 18:27
KarlPeterKarlPeter
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First $newcommand{pp}{mathfrak{p}}newcommand{mm}{mathfrak{m}}(x)=pp_1^{e_1}cdots pp_n^{e_n}$ for some $e_iinnewcommand{NN}{mathbb{N}}NN$. If $e_i ge 1$ for $ige 2$, then $xin(x)subset pp_i$, however, $xequiv 1 pmod{pp_i}$, so this is impossible. Hence $(x)=pp_1^{e_1}$, for some $e_1ge 1$ (since $(x)subset pp_1$ by assumption, so $(x)ne (1)$). However, $xnotin pp_1^2$ either, so $xin(x)=pp_1^{e_1}$ forces $e_1=1$. Thus $(x)=pp_1$ as desired.
Hi, thank you for the answer. One step stays unclear: How do you get $newcommand{pp}{mathfrak{p}}newcommand{mm}{mathfrak{m}}(x)=pp_1^{e_1}cdots pp_n^{e_n}$?
– KarlPeter
Mar 30 '18 at 18:43
@KarlPeter Ah, that's a definition of Dedekind domain, every nonzero ideal factors into primes. See Wikipedia: en.wikipedia.org/wiki/Dedekind_domain#Alternative_definitions
– jgon
Mar 30 '18 at 18:45
@KarlPeter For a proof you'll want to look at a resource on algebraic number theory, e.g. Milne's course notes here: jmilne.org/math/CourseNotes/ant.html It's a little too involved for an answer here.
– jgon
Mar 30 '18 at 18:46
add a comment |
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First $newcommand{pp}{mathfrak{p}}newcommand{mm}{mathfrak{m}}(x)=pp_1^{e_1}cdots pp_n^{e_n}$ for some $e_iinnewcommand{NN}{mathbb{N}}NN$. If $e_i ge 1$ for $ige 2$, then $xin(x)subset pp_i$, however, $xequiv 1 pmod{pp_i}$, so this is impossible. Hence $(x)=pp_1^{e_1}$, for some $e_1ge 1$ (since $(x)subset pp_1$ by assumption, so $(x)ne (1)$). However, $xnotin pp_1^2$ either, so $xin(x)=pp_1^{e_1}$ forces $e_1=1$. Thus $(x)=pp_1$ as desired.
Hi, thank you for the answer. One step stays unclear: How do you get $newcommand{pp}{mathfrak{p}}newcommand{mm}{mathfrak{m}}(x)=pp_1^{e_1}cdots pp_n^{e_n}$?
– KarlPeter
Mar 30 '18 at 18:43
@KarlPeter Ah, that's a definition of Dedekind domain, every nonzero ideal factors into primes. See Wikipedia: en.wikipedia.org/wiki/Dedekind_domain#Alternative_definitions
– jgon
Mar 30 '18 at 18:45
@KarlPeter For a proof you'll want to look at a resource on algebraic number theory, e.g. Milne's course notes here: jmilne.org/math/CourseNotes/ant.html It's a little too involved for an answer here.
– jgon
Mar 30 '18 at 18:46
add a comment |
First $newcommand{pp}{mathfrak{p}}newcommand{mm}{mathfrak{m}}(x)=pp_1^{e_1}cdots pp_n^{e_n}$ for some $e_iinnewcommand{NN}{mathbb{N}}NN$. If $e_i ge 1$ for $ige 2$, then $xin(x)subset pp_i$, however, $xequiv 1 pmod{pp_i}$, so this is impossible. Hence $(x)=pp_1^{e_1}$, for some $e_1ge 1$ (since $(x)subset pp_1$ by assumption, so $(x)ne (1)$). However, $xnotin pp_1^2$ either, so $xin(x)=pp_1^{e_1}$ forces $e_1=1$. Thus $(x)=pp_1$ as desired.
Hi, thank you for the answer. One step stays unclear: How do you get $newcommand{pp}{mathfrak{p}}newcommand{mm}{mathfrak{m}}(x)=pp_1^{e_1}cdots pp_n^{e_n}$?
– KarlPeter
Mar 30 '18 at 18:43
@KarlPeter Ah, that's a definition of Dedekind domain, every nonzero ideal factors into primes. See Wikipedia: en.wikipedia.org/wiki/Dedekind_domain#Alternative_definitions
– jgon
Mar 30 '18 at 18:45
@KarlPeter For a proof you'll want to look at a resource on algebraic number theory, e.g. Milne's course notes here: jmilne.org/math/CourseNotes/ant.html It's a little too involved for an answer here.
– jgon
Mar 30 '18 at 18:46
add a comment |
First $newcommand{pp}{mathfrak{p}}newcommand{mm}{mathfrak{m}}(x)=pp_1^{e_1}cdots pp_n^{e_n}$ for some $e_iinnewcommand{NN}{mathbb{N}}NN$. If $e_i ge 1$ for $ige 2$, then $xin(x)subset pp_i$, however, $xequiv 1 pmod{pp_i}$, so this is impossible. Hence $(x)=pp_1^{e_1}$, for some $e_1ge 1$ (since $(x)subset pp_1$ by assumption, so $(x)ne (1)$). However, $xnotin pp_1^2$ either, so $xin(x)=pp_1^{e_1}$ forces $e_1=1$. Thus $(x)=pp_1$ as desired.
First $newcommand{pp}{mathfrak{p}}newcommand{mm}{mathfrak{m}}(x)=pp_1^{e_1}cdots pp_n^{e_n}$ for some $e_iinnewcommand{NN}{mathbb{N}}NN$. If $e_i ge 1$ for $ige 2$, then $xin(x)subset pp_i$, however, $xequiv 1 pmod{pp_i}$, so this is impossible. Hence $(x)=pp_1^{e_1}$, for some $e_1ge 1$ (since $(x)subset pp_1$ by assumption, so $(x)ne (1)$). However, $xnotin pp_1^2$ either, so $xin(x)=pp_1^{e_1}$ forces $e_1=1$. Thus $(x)=pp_1$ as desired.
answered Mar 30 '18 at 18:39
jgonjgon
13.4k21941
13.4k21941
Hi, thank you for the answer. One step stays unclear: How do you get $newcommand{pp}{mathfrak{p}}newcommand{mm}{mathfrak{m}}(x)=pp_1^{e_1}cdots pp_n^{e_n}$?
– KarlPeter
Mar 30 '18 at 18:43
@KarlPeter Ah, that's a definition of Dedekind domain, every nonzero ideal factors into primes. See Wikipedia: en.wikipedia.org/wiki/Dedekind_domain#Alternative_definitions
– jgon
Mar 30 '18 at 18:45
@KarlPeter For a proof you'll want to look at a resource on algebraic number theory, e.g. Milne's course notes here: jmilne.org/math/CourseNotes/ant.html It's a little too involved for an answer here.
– jgon
Mar 30 '18 at 18:46
add a comment |
Hi, thank you for the answer. One step stays unclear: How do you get $newcommand{pp}{mathfrak{p}}newcommand{mm}{mathfrak{m}}(x)=pp_1^{e_1}cdots pp_n^{e_n}$?
– KarlPeter
Mar 30 '18 at 18:43
@KarlPeter Ah, that's a definition of Dedekind domain, every nonzero ideal factors into primes. See Wikipedia: en.wikipedia.org/wiki/Dedekind_domain#Alternative_definitions
– jgon
Mar 30 '18 at 18:45
@KarlPeter For a proof you'll want to look at a resource on algebraic number theory, e.g. Milne's course notes here: jmilne.org/math/CourseNotes/ant.html It's a little too involved for an answer here.
– jgon
Mar 30 '18 at 18:46
Hi, thank you for the answer. One step stays unclear: How do you get $newcommand{pp}{mathfrak{p}}newcommand{mm}{mathfrak{m}}(x)=pp_1^{e_1}cdots pp_n^{e_n}$?
– KarlPeter
Mar 30 '18 at 18:43
Hi, thank you for the answer. One step stays unclear: How do you get $newcommand{pp}{mathfrak{p}}newcommand{mm}{mathfrak{m}}(x)=pp_1^{e_1}cdots pp_n^{e_n}$?
– KarlPeter
Mar 30 '18 at 18:43
@KarlPeter Ah, that's a definition of Dedekind domain, every nonzero ideal factors into primes. See Wikipedia: en.wikipedia.org/wiki/Dedekind_domain#Alternative_definitions
– jgon
Mar 30 '18 at 18:45
@KarlPeter Ah, that's a definition of Dedekind domain, every nonzero ideal factors into primes. See Wikipedia: en.wikipedia.org/wiki/Dedekind_domain#Alternative_definitions
– jgon
Mar 30 '18 at 18:45
@KarlPeter For a proof you'll want to look at a resource on algebraic number theory, e.g. Milne's course notes here: jmilne.org/math/CourseNotes/ant.html It's a little too involved for an answer here.
– jgon
Mar 30 '18 at 18:46
@KarlPeter For a proof you'll want to look at a resource on algebraic number theory, e.g. Milne's course notes here: jmilne.org/math/CourseNotes/ant.html It's a little too involved for an answer here.
– jgon
Mar 30 '18 at 18:46
add a comment |
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