Mixing
Question about discrete valuation rings
$begingroup$
Suppose $R$ is a domain, not a field, and there exists an irreducible $t$ such that every non-zero element of $R$ can be written as $ut^n$, with $u$ a unit. Apparently $R$ is then Noetherian, a PID in fact. I'm trying to understand a proof of that. It says that, if $m$ is the maximal ideal generated by $t$, thus containing any given proper ideal $a$, and $m^k$ is the ideal generated by $t^k$, there is a maximum $k$ for which $a subset m^k$. Why is that?
abstract-algebra ring-theory
asked Feb 1 at 23:43
Romanda de GoreRomanda de Gore
3611
$endgroup$
|
show 1 more comment
$begingroup$
Suppose $R$ is a domain, not a field, and there exists an irreducible $t$ such that every non-zero element of $R$ can be written as $ut^n$, with $u$ a unit. Apparently $R$ is then Noetherian, a PID in fact. I'm trying to understand a proof of that. It says that, if $m$ is the maximal ideal generated by $t$, thus containing any given proper ideal $a$, and $m^k$ is the ideal generated by $t^k$, there is a maximum $k$ for which $a subset m^k$. Why is that?
abstract-algebra ring-theory
asked Feb 1 at 23:43
Romanda de GoreRomanda de Gore
3611
$endgroup$
$begingroup$
Is there no other hypothesis on $t$?
$endgroup$
– Bernard
Feb 1 at 23:58
$begingroup$
What, exactly, does this question have to do with discrete valuation rings? Cheers!
$endgroup$
– Robert Lewis
Feb 2 at 0:02
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@RobertLewis I would think that such a ring would necessarily be a DVR, no?
$endgroup$
– jgon
Feb 2 at 0:04
2
$begingroup$
@RobertLewis Yes, I think the assignment $v(x)=n$ if $x=ut^nne 0$, and $v(0)=infty$ should be a valuation on $R$? It's also been a little while for me since I last looked at DVRs though.
$endgroup$
– jgon
Feb 2 at 0:11
1
$begingroup$
@jgon: yes; I just wikied DVRs and what you said sounds right--or at least close! Thanks!
$endgroup$
– Robert Lewis
Feb 2 at 0:13
|
show 1 more comment
$begingroup$
Suppose $R$ is a domain, not a field, and there exists an irreducible $t$ such that every non-zero element of $R$ can be written as $ut^n$, with $u$ a unit. Apparently $R$ is then Noetherian, a PID in fact. I'm trying to understand a proof of that. It says that, if $m$ is the maximal ideal generated by $t$, thus containing any given proper ideal $a$, and $m^k$ is the ideal generated by $t^k$, there is a maximum $k$ for which $a subset m^k$. Why is that?
abstract-algebra ring-theory
asked Feb 1 at 23:43
Romanda de GoreRomanda de Gore
3611
$endgroup$
Suppose $R$ is a domain, not a field, and there exists an irreducible $t$ such that every non-zero element of $R$ can be written as $ut^n$, with $u$ a unit. Apparently $R$ is then Noetherian, a PID in fact. I'm trying to understand a proof of that. It says that, if $m$ is the maximal ideal generated by $t$, thus containing any given proper ideal $a$, and $m^k$ is the ideal generated by $t^k$, there is a maximum $k$ for which $a subset m^k$. Why is that?
abstract-algebra ring-theory
abstract-algebra ring-theory
asked Feb 1 at 23:43
Romanda de GoreRomanda de Gore
3611
asked Feb 1 at 23:43
Romanda de GoreRomanda de Gore
3611
asked Feb 1 at 23:43
Romanda de GoreRomanda de Gore
3611
asked Feb 1 at 23:43
Romanda de GoreRomanda de Gore
3611
asked Feb 1 at 23:43
Romanda de GoreRomanda de Gore
3611
3611
$begingroup$
Is there no other hypothesis on $t$?
$endgroup$
– Bernard
Feb 1 at 23:58
$begingroup$
What, exactly, does this question have to do with discrete valuation rings? Cheers!
$endgroup$
– Robert Lewis
Feb 2 at 0:02
$begingroup$
@RobertLewis I would think that such a ring would necessarily be a DVR, no?
$endgroup$
– jgon
Feb 2 at 0:04
2
$begingroup$
@RobertLewis Yes, I think the assignment $v(x)=n$ if $x=ut^nne 0$, and $v(0)=infty$ should be a valuation on $R$? It's also been a little while for me since I last looked at DVRs though.
$endgroup$
– jgon
Feb 2 at 0:11
1
$begingroup$
@jgon: yes; I just wikied DVRs and what you said sounds right--or at least close! Thanks!
$endgroup$
– Robert Lewis
Feb 2 at 0:13
|
show 1 more comment
$begingroup$
Is there no other hypothesis on $t$?
$endgroup$
– Bernard
Feb 1 at 23:58
$begingroup$
What, exactly, does this question have to do with discrete valuation rings? Cheers!
$endgroup$
– Robert Lewis
Feb 2 at 0:02
$begingroup$
@RobertLewis I would think that such a ring would necessarily be a DVR, no?
$endgroup$
– jgon
Feb 2 at 0:04
2
$begingroup$
@RobertLewis Yes, I think the assignment $v(x)=n$ if $x=ut^nne 0$, and $v(0)=infty$ should be a valuation on $R$? It's also been a little while for me since I last looked at DVRs though.
$endgroup$
– jgon
Feb 2 at 0:11
1
$begingroup$
@jgon: yes; I just wikied DVRs and what you said sounds right--or at least close! Thanks!
$endgroup$
– Robert Lewis
Feb 2 at 0:13
$begingroup$
Is there no other hypothesis on $t$?
$endgroup$
– Bernard
Feb 1 at 23:58
$begingroup$
Is there no other hypothesis on $t$?
$endgroup$
– Bernard
Feb 1 at 23:58
$begingroup$
What, exactly, does this question have to do with discrete valuation rings? Cheers!
$endgroup$
– Robert Lewis
Feb 2 at 0:02
$begingroup$
What, exactly, does this question have to do with discrete valuation rings? Cheers!
$endgroup$
– Robert Lewis
Feb 2 at 0:02
$begingroup$
@RobertLewis I would think that such a ring would necessarily be a DVR, no?
$endgroup$
– jgon
Feb 2 at 0:04
$begingroup$
@RobertLewis I would think that such a ring would necessarily be a DVR, no?
$endgroup$
– jgon
Feb 2 at 0:04
2
2
$begingroup$
@RobertLewis Yes, I think the assignment $v(x)=n$ if $x=ut^nne 0$, and $v(0)=infty$ should be a valuation on $R$? It's also been a little while for me since I last looked at DVRs though.
$endgroup$
– jgon
Feb 2 at 0:11
$begingroup$
@RobertLewis Yes, I think the assignment $v(x)=n$ if $x=ut^nne 0$, and $v(0)=infty$ should be a valuation on $R$? It's also been a little while for me since I last looked at DVRs though.
$endgroup$
– jgon
Feb 2 at 0:11
1
1
$begingroup$
@jgon: yes; I just wikied DVRs and what you said sounds right--or at least close! Thanks!
$endgroup$
– Robert Lewis
Feb 2 at 0:13
$begingroup$
@jgon: yes; I just wikied DVRs and what you said sounds right--or at least close! Thanks!
$endgroup$
– Robert Lewis
Feb 2 at 0:13
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
The first step is to show that every element can be written uniquely as $ut^n$ (i.e. $u,n$ are unique).
Once you know that, you show that every (non-zero) ideal is some power of $mathfrak{m}$. Specifically, $mathfrak{a} = mathfrak{m}^k$ if $k$ is the smallest power such that there is some element $ut^k in mathfrak{m}$.
Then what's left is to show that $mathfrak{m}^{k} notsubseteq mathfrak{m}^{k+1}$.
None of the first steps are particularly hard, so I encourage you to try to prove them by yourself.
From the second and third parts, if $mathfrak{a} = mathfrak{m}^k$ then $mathfrak{m}^k$ is the largest power of $mathfrak{m}$ that contains $mathfrak{a}$.
answered Feb 2 at 0:13
Trevor GunnTrevor Gunn
14.9k32047
$endgroup$
add a comment |
$begingroup$
For $xne 0in R$, since we know that $x=ut^n$, with $u$ a unit, define $v(x)=n$. Define $v(0)=infty$.
Then for an ideal $a$, define $v(a)=min{v(x): xin a}$, which exists since the natural numbers union infinity is well ordered. Then for $k > v(a)$, $anotsubseteq m^k$, since if $v(a)=n$, then there is $xin a$ with $x=ut^n$, and $ut^nnotin m^k$ for $k > n$. Conversely, if $kle v(a)$, then $asubseteq m^k$, since every element of $a$ is divisible by $t^{v(a)}$, and thus divisible by $t^k$. Thus $v(a)$ is this maximum $k$ with $asubseteq m^k$.
answered Feb 2 at 0:10
jgonjgon
16.5k32143
$endgroup$
add a comment |
$begingroup$
Hint:
Note that $;bigl(At^nbigr)_{ninmathbf N}$ is a decreasing sequence, and show that $;displaystylebigcap_{ninmathbf N}At^n={0}$.
answered Feb 2 at 0:10
BernardBernard
124k741117
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The first step is to show that every element can be written uniquely as $ut^n$ (i.e. $u,n$ are unique).
Once you know that, you show that every (non-zero) ideal is some power of $mathfrak{m}$. Specifically, $mathfrak{a} = mathfrak{m}^k$ if $k$ is the smallest power such that there is some element $ut^k in mathfrak{m}$.
Then what's left is to show that $mathfrak{m}^{k} notsubseteq mathfrak{m}^{k+1}$.
None of the first steps are particularly hard, so I encourage you to try to prove them by yourself.
From the second and third parts, if $mathfrak{a} = mathfrak{m}^k$ then $mathfrak{m}^k$ is the largest power of $mathfrak{m}$ that contains $mathfrak{a}$.
answered Feb 2 at 0:13
Trevor GunnTrevor Gunn
14.9k32047
$endgroup$
add a comment |
$begingroup$
The first step is to show that every element can be written uniquely as $ut^n$ (i.e. $u,n$ are unique).
Once you know that, you show that every (non-zero) ideal is some power of $mathfrak{m}$. Specifically, $mathfrak{a} = mathfrak{m}^k$ if $k$ is the smallest power such that there is some element $ut^k in mathfrak{m}$.
Then what's left is to show that $mathfrak{m}^{k} notsubseteq mathfrak{m}^{k+1}$.
None of the first steps are particularly hard, so I encourage you to try to prove them by yourself.
From the second and third parts, if $mathfrak{a} = mathfrak{m}^k$ then $mathfrak{m}^k$ is the largest power of $mathfrak{m}$ that contains $mathfrak{a}$.
answered Feb 2 at 0:13
Trevor GunnTrevor Gunn
14.9k32047
$endgroup$
add a comment |
$begingroup$
The first step is to show that every element can be written uniquely as $ut^n$ (i.e. $u,n$ are unique).
Once you know that, you show that every (non-zero) ideal is some power of $mathfrak{m}$. Specifically, $mathfrak{a} = mathfrak{m}^k$ if $k$ is the smallest power such that there is some element $ut^k in mathfrak{m}$.
Then what's left is to show that $mathfrak{m}^{k} notsubseteq mathfrak{m}^{k+1}$.
None of the first steps are particularly hard, so I encourage you to try to prove them by yourself.
From the second and third parts, if $mathfrak{a} = mathfrak{m}^k$ then $mathfrak{m}^k$ is the largest power of $mathfrak{m}$ that contains $mathfrak{a}$.
answered Feb 2 at 0:13
Trevor GunnTrevor Gunn
14.9k32047
$endgroup$
The first step is to show that every element can be written uniquely as $ut^n$ (i.e. $u,n$ are unique).
Once you know that, you show that every (non-zero) ideal is some power of $mathfrak{m}$. Specifically, $mathfrak{a} = mathfrak{m}^k$ if $k$ is the smallest power such that there is some element $ut^k in mathfrak{m}$.
Then what's left is to show that $mathfrak{m}^{k} notsubseteq mathfrak{m}^{k+1}$.
None of the first steps are particularly hard, so I encourage you to try to prove them by yourself.
From the second and third parts, if $mathfrak{a} = mathfrak{m}^k$ then $mathfrak{m}^k$ is the largest power of $mathfrak{m}$ that contains $mathfrak{a}$.
answered Feb 2 at 0:13
Trevor GunnTrevor Gunn
14.9k32047
answered Feb 2 at 0:13
Trevor GunnTrevor Gunn
14.9k32047
answered Feb 2 at 0:13
Trevor GunnTrevor Gunn
14.9k32047
answered Feb 2 at 0:13
Trevor GunnTrevor Gunn
14.9k32047
14.9k32047
add a comment |
add a comment |
$begingroup$
For $xne 0in R$, since we know that $x=ut^n$, with $u$ a unit, define $v(x)=n$. Define $v(0)=infty$.
Then for an ideal $a$, define $v(a)=min{v(x): xin a}$, which exists since the natural numbers union infinity is well ordered. Then for $k > v(a)$, $anotsubseteq m^k$, since if $v(a)=n$, then there is $xin a$ with $x=ut^n$, and $ut^nnotin m^k$ for $k > n$. Conversely, if $kle v(a)$, then $asubseteq m^k$, since every element of $a$ is divisible by $t^{v(a)}$, and thus divisible by $t^k$. Thus $v(a)$ is this maximum $k$ with $asubseteq m^k$.
answered Feb 2 at 0:10
jgonjgon
16.5k32143
$endgroup$
add a comment |
$begingroup$
For $xne 0in R$, since we know that $x=ut^n$, with $u$ a unit, define $v(x)=n$. Define $v(0)=infty$.
Then for an ideal $a$, define $v(a)=min{v(x): xin a}$, which exists since the natural numbers union infinity is well ordered. Then for $k > v(a)$, $anotsubseteq m^k$, since if $v(a)=n$, then there is $xin a$ with $x=ut^n$, and $ut^nnotin m^k$ for $k > n$. Conversely, if $kle v(a)$, then $asubseteq m^k$, since every element of $a$ is divisible by $t^{v(a)}$, and thus divisible by $t^k$. Thus $v(a)$ is this maximum $k$ with $asubseteq m^k$.
answered Feb 2 at 0:10
jgonjgon
16.5k32143
$endgroup$
add a comment |
$begingroup$
For $xne 0in R$, since we know that $x=ut^n$, with $u$ a unit, define $v(x)=n$. Define $v(0)=infty$.
Then for an ideal $a$, define $v(a)=min{v(x): xin a}$, which exists since the natural numbers union infinity is well ordered. Then for $k > v(a)$, $anotsubseteq m^k$, since if $v(a)=n$, then there is $xin a$ with $x=ut^n$, and $ut^nnotin m^k$ for $k > n$. Conversely, if $kle v(a)$, then $asubseteq m^k$, since every element of $a$ is divisible by $t^{v(a)}$, and thus divisible by $t^k$. Thus $v(a)$ is this maximum $k$ with $asubseteq m^k$.
answered Feb 2 at 0:10
jgonjgon
16.5k32143
$endgroup$
For $xne 0in R$, since we know that $x=ut^n$, with $u$ a unit, define $v(x)=n$. Define $v(0)=infty$.
Then for an ideal $a$, define $v(a)=min{v(x): xin a}$, which exists since the natural numbers union infinity is well ordered. Then for $k > v(a)$, $anotsubseteq m^k$, since if $v(a)=n$, then there is $xin a$ with $x=ut^n$, and $ut^nnotin m^k$ for $k > n$. Conversely, if $kle v(a)$, then $asubseteq m^k$, since every element of $a$ is divisible by $t^{v(a)}$, and thus divisible by $t^k$. Thus $v(a)$ is this maximum $k$ with $asubseteq m^k$.
answered Feb 2 at 0:10
jgonjgon
16.5k32143
answered Feb 2 at 0:10
jgonjgon
16.5k32143
answered Feb 2 at 0:10
jgonjgon
16.5k32143
answered Feb 2 at 0:10
jgonjgon
16.5k32143
16.5k32143
add a comment |
add a comment |
$begingroup$
Hint:
Note that $;bigl(At^nbigr)_{ninmathbf N}$ is a decreasing sequence, and show that $;displaystylebigcap_{ninmathbf N}At^n={0}$.
answered Feb 2 at 0:10
BernardBernard
124k741117
$endgroup$
add a comment |
$begingroup$
Hint:
Note that $;bigl(At^nbigr)_{ninmathbf N}$ is a decreasing sequence, and show that $;displaystylebigcap_{ninmathbf N}At^n={0}$.
answered Feb 2 at 0:10
BernardBernard
124k741117
$endgroup$
add a comment |
$begingroup$
Hint:
Note that $;bigl(At^nbigr)_{ninmathbf N}$ is a decreasing sequence, and show that $;displaystylebigcap_{ninmathbf N}At^n={0}$.
answered Feb 2 at 0:10
BernardBernard
124k741117
$endgroup$
Hint:
Note that $;bigl(At^nbigr)_{ninmathbf N}$ is a decreasing sequence, and show that $;displaystylebigcap_{ninmathbf N}At^n={0}$.
answered Feb 2 at 0:10
BernardBernard
124k741117
answered Feb 2 at 0:10
BernardBernard
124k741117
answered Feb 2 at 0:10
BernardBernard
124k741117
answered Feb 2 at 0:10
BernardBernard
124k741117
124k741117
add a comment |
add a comment |
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$begingroup$
Is there no other hypothesis on $t$?
$endgroup$
– Bernard
Feb 1 at 23:58
$begingroup$
What, exactly, does this question have to do with discrete valuation rings? Cheers!
$endgroup$
– Robert Lewis
Feb 2 at 0:02
$begingroup$
@RobertLewis I would think that such a ring would necessarily be a DVR, no?
$endgroup$
– jgon
Feb 2 at 0:04
2
$begingroup$
@RobertLewis Yes, I think the assignment $v(x)=n$ if $x=ut^nne 0$, and $v(0)=infty$ should be a valuation on $R$? It's also been a little while for me since I last looked at DVRs though.
$endgroup$
– jgon
Feb 2 at 0:11
1
$begingroup$
@jgon: yes; I just wikied DVRs and what you said sounds right--or at least close! Thanks!
$endgroup$
– Robert Lewis
Feb 2 at 0:13