Constructing a ring with a given spectrum
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A DVR necessarily has spectrum $ 0, mathfrakm $, but a DVR is also necessarily noetherian. Can we find an example of a non-noetherian (thus non DVR) ring, say $R$, with the spectrum $ 0, mathfrakm ?$
Clearly we need $R$ a domain since the zero ideal is in the spec. We also have $mathfrakm neq 0$ since then $R$ would be a field, thus noetherian but even if it didn’t follow that $mathfrakm neq 0$ I would require this so it’s interesting. Also, krull dimension is obviously 1 since we have one chain of prime ideals of height 1.
I have tried the standard non-noetherian things like starting with a polynomial ring in countably many variables and quotienting and localizing, but so far no luck. I always accidentally land on a field instead of a 2 point spectrum. I also haven’t been able to come up with a way to see this from an algebro-geometric side of things.
My commutative algebra skills are moderate, I went about halfway through Atiyah Macdonald. I’m sure someone with stronger comm alg intuition will see an easy construction?
algebraic-geometry ring-theory commutative-algebra
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A DVR necessarily has spectrum $ 0, mathfrakm $, but a DVR is also necessarily noetherian. Can we find an example of a non-noetherian (thus non DVR) ring, say $R$, with the spectrum $ 0, mathfrakm ?$
Clearly we need $R$ a domain since the zero ideal is in the spec. We also have $mathfrakm neq 0$ since then $R$ would be a field, thus noetherian but even if it didn’t follow that $mathfrakm neq 0$ I would require this so it’s interesting. Also, krull dimension is obviously 1 since we have one chain of prime ideals of height 1.
I have tried the standard non-noetherian things like starting with a polynomial ring in countably many variables and quotienting and localizing, but so far no luck. I always accidentally land on a field instead of a 2 point spectrum. I also haven’t been able to come up with a way to see this from an algebro-geometric side of things.
My commutative algebra skills are moderate, I went about halfway through Atiyah Macdonald. I’m sure someone with stronger comm alg intuition will see an easy construction?
algebraic-geometry ring-theory commutative-algebra
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
A DVR necessarily has spectrum $ 0, mathfrakm $, but a DVR is also necessarily noetherian. Can we find an example of a non-noetherian (thus non DVR) ring, say $R$, with the spectrum $ 0, mathfrakm ?$
Clearly we need $R$ a domain since the zero ideal is in the spec. We also have $mathfrakm neq 0$ since then $R$ would be a field, thus noetherian but even if it didn’t follow that $mathfrakm neq 0$ I would require this so it’s interesting. Also, krull dimension is obviously 1 since we have one chain of prime ideals of height 1.
I have tried the standard non-noetherian things like starting with a polynomial ring in countably many variables and quotienting and localizing, but so far no luck. I always accidentally land on a field instead of a 2 point spectrum. I also haven’t been able to come up with a way to see this from an algebro-geometric side of things.
My commutative algebra skills are moderate, I went about halfway through Atiyah Macdonald. I’m sure someone with stronger comm alg intuition will see an easy construction?
algebraic-geometry ring-theory commutative-algebra
A DVR necessarily has spectrum $ 0, mathfrakm $, but a DVR is also necessarily noetherian. Can we find an example of a non-noetherian (thus non DVR) ring, say $R$, with the spectrum $ 0, mathfrakm ?$
Clearly we need $R$ a domain since the zero ideal is in the spec. We also have $mathfrakm neq 0$ since then $R$ would be a field, thus noetherian but even if it didn’t follow that $mathfrakm neq 0$ I would require this so it’s interesting. Also, krull dimension is obviously 1 since we have one chain of prime ideals of height 1.
I have tried the standard non-noetherian things like starting with a polynomial ring in countably many variables and quotienting and localizing, but so far no luck. I always accidentally land on a field instead of a 2 point spectrum. I also haven’t been able to come up with a way to see this from an algebro-geometric side of things.
My commutative algebra skills are moderate, I went about halfway through Atiyah Macdonald. I’m sure someone with stronger comm alg intuition will see an easy construction?
algebraic-geometry ring-theory commutative-algebra
algebraic-geometry ring-theory commutative-algebra
edited 2 days ago
asked 2 days ago
Prince M
1,7231520
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2 Answers
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To expand on Bernard's answer, let $k$ be a field and consider the ring $R = k[[x^mathbbR_ge 0]]$ of formal power series of the form $sum_r in S c_r x^r$ where $r in mathbbR_ge 0$ and $S subseteq mathbbR_ge 0$ is well-ordered (this guarantees that multiplication is well-defined). This is a valuation ring.
For $f in R$ a nonzero power series, write $nu(f) in mathbbR_ge 0$ for the smallest exponent of a nonzero term in $f$. If $I$ is any nonzero ideal of $R$, then if some nonzero $f in I$ has valuation $nu(f)$, then multiplying by every element of $R$, and observing that nonzero every $g in R$ can be written as $x^nu(g)$ times a unit, we see that $I$ contains every element of $R$ with valuation $ge nu(f)$. So the ideals of $R$ come in two infinite families, namely
$$I_r = f in R : nu(f) ge r , r in mathbbR_ge 0$$
and
$$J_r = f in R : nu(f) > r , r in mathbbR_ge 0$$
(together with the zero ideal, which you can think of as $I_infty$). These ideals are totally ordered and in particular show that $R$ is very far from Noetherian.
The corresponding quotients $R/I_r$ and $R/J_r$ contain nontrivial nilpotents, and hence $I_r$ and $J_r$ are not prime, except for $J_0$, which is the unique maximal ideal, and $I_0$, which is the unit ideal. So $R$ has exactly two prime ideals, $(0)$ and $J_0$, as desired.
Great answer, explicit and detailed!
– Prince M
yesterday
add a comment |
up vote
3
down vote
A non-discrete valuation ring of height $1$ is such an example.
Ok give me an example verifying the existence of a non discrete valuation ring of height 1
– Prince M
2 days ago
@Prince: I construct exactly such a thing in my answer.
– Qiaochu Yuan
2 days ago
I see that! I upvoted yours and will probably accept it. I tried to comment on it and say thank you, but it said my comment was too short.
– Prince M
2 days ago
Too be honest I always avoid rings of formal power series because I never learned them so they feel uncomfortable.. I should probably get over that!
– Prince M
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
To expand on Bernard's answer, let $k$ be a field and consider the ring $R = k[[x^mathbbR_ge 0]]$ of formal power series of the form $sum_r in S c_r x^r$ where $r in mathbbR_ge 0$ and $S subseteq mathbbR_ge 0$ is well-ordered (this guarantees that multiplication is well-defined). This is a valuation ring.
For $f in R$ a nonzero power series, write $nu(f) in mathbbR_ge 0$ for the smallest exponent of a nonzero term in $f$. If $I$ is any nonzero ideal of $R$, then if some nonzero $f in I$ has valuation $nu(f)$, then multiplying by every element of $R$, and observing that nonzero every $g in R$ can be written as $x^nu(g)$ times a unit, we see that $I$ contains every element of $R$ with valuation $ge nu(f)$. So the ideals of $R$ come in two infinite families, namely
$$I_r = f in R : nu(f) ge r , r in mathbbR_ge 0$$
and
$$J_r = f in R : nu(f) > r , r in mathbbR_ge 0$$
(together with the zero ideal, which you can think of as $I_infty$). These ideals are totally ordered and in particular show that $R$ is very far from Noetherian.
The corresponding quotients $R/I_r$ and $R/J_r$ contain nontrivial nilpotents, and hence $I_r$ and $J_r$ are not prime, except for $J_0$, which is the unique maximal ideal, and $I_0$, which is the unit ideal. So $R$ has exactly two prime ideals, $(0)$ and $J_0$, as desired.
Great answer, explicit and detailed!
– Prince M
yesterday
add a comment |
up vote
2
down vote
accepted
To expand on Bernard's answer, let $k$ be a field and consider the ring $R = k[[x^mathbbR_ge 0]]$ of formal power series of the form $sum_r in S c_r x^r$ where $r in mathbbR_ge 0$ and $S subseteq mathbbR_ge 0$ is well-ordered (this guarantees that multiplication is well-defined). This is a valuation ring.
For $f in R$ a nonzero power series, write $nu(f) in mathbbR_ge 0$ for the smallest exponent of a nonzero term in $f$. If $I$ is any nonzero ideal of $R$, then if some nonzero $f in I$ has valuation $nu(f)$, then multiplying by every element of $R$, and observing that nonzero every $g in R$ can be written as $x^nu(g)$ times a unit, we see that $I$ contains every element of $R$ with valuation $ge nu(f)$. So the ideals of $R$ come in two infinite families, namely
$$I_r = f in R : nu(f) ge r , r in mathbbR_ge 0$$
and
$$J_r = f in R : nu(f) > r , r in mathbbR_ge 0$$
(together with the zero ideal, which you can think of as $I_infty$). These ideals are totally ordered and in particular show that $R$ is very far from Noetherian.
The corresponding quotients $R/I_r$ and $R/J_r$ contain nontrivial nilpotents, and hence $I_r$ and $J_r$ are not prime, except for $J_0$, which is the unique maximal ideal, and $I_0$, which is the unit ideal. So $R$ has exactly two prime ideals, $(0)$ and $J_0$, as desired.
Great answer, explicit and detailed!
– Prince M
yesterday
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
To expand on Bernard's answer, let $k$ be a field and consider the ring $R = k[[x^mathbbR_ge 0]]$ of formal power series of the form $sum_r in S c_r x^r$ where $r in mathbbR_ge 0$ and $S subseteq mathbbR_ge 0$ is well-ordered (this guarantees that multiplication is well-defined). This is a valuation ring.
For $f in R$ a nonzero power series, write $nu(f) in mathbbR_ge 0$ for the smallest exponent of a nonzero term in $f$. If $I$ is any nonzero ideal of $R$, then if some nonzero $f in I$ has valuation $nu(f)$, then multiplying by every element of $R$, and observing that nonzero every $g in R$ can be written as $x^nu(g)$ times a unit, we see that $I$ contains every element of $R$ with valuation $ge nu(f)$. So the ideals of $R$ come in two infinite families, namely
$$I_r = f in R : nu(f) ge r , r in mathbbR_ge 0$$
and
$$J_r = f in R : nu(f) > r , r in mathbbR_ge 0$$
(together with the zero ideal, which you can think of as $I_infty$). These ideals are totally ordered and in particular show that $R$ is very far from Noetherian.
The corresponding quotients $R/I_r$ and $R/J_r$ contain nontrivial nilpotents, and hence $I_r$ and $J_r$ are not prime, except for $J_0$, which is the unique maximal ideal, and $I_0$, which is the unit ideal. So $R$ has exactly two prime ideals, $(0)$ and $J_0$, as desired.
To expand on Bernard's answer, let $k$ be a field and consider the ring $R = k[[x^mathbbR_ge 0]]$ of formal power series of the form $sum_r in S c_r x^r$ where $r in mathbbR_ge 0$ and $S subseteq mathbbR_ge 0$ is well-ordered (this guarantees that multiplication is well-defined). This is a valuation ring.
For $f in R$ a nonzero power series, write $nu(f) in mathbbR_ge 0$ for the smallest exponent of a nonzero term in $f$. If $I$ is any nonzero ideal of $R$, then if some nonzero $f in I$ has valuation $nu(f)$, then multiplying by every element of $R$, and observing that nonzero every $g in R$ can be written as $x^nu(g)$ times a unit, we see that $I$ contains every element of $R$ with valuation $ge nu(f)$. So the ideals of $R$ come in two infinite families, namely
$$I_r = f in R : nu(f) ge r , r in mathbbR_ge 0$$
and
$$J_r = f in R : nu(f) > r , r in mathbbR_ge 0$$
(together with the zero ideal, which you can think of as $I_infty$). These ideals are totally ordered and in particular show that $R$ is very far from Noetherian.
The corresponding quotients $R/I_r$ and $R/J_r$ contain nontrivial nilpotents, and hence $I_r$ and $J_r$ are not prime, except for $J_0$, which is the unique maximal ideal, and $I_0$, which is the unit ideal. So $R$ has exactly two prime ideals, $(0)$ and $J_0$, as desired.
edited yesterday
answered 2 days ago
Qiaochu Yuan
274k32578914
274k32578914
Great answer, explicit and detailed!
– Prince M
yesterday
add a comment |
Great answer, explicit and detailed!
– Prince M
yesterday
Great answer, explicit and detailed!
– Prince M
yesterday
Great answer, explicit and detailed!
– Prince M
yesterday
add a comment |
up vote
3
down vote
A non-discrete valuation ring of height $1$ is such an example.
Ok give me an example verifying the existence of a non discrete valuation ring of height 1
– Prince M
2 days ago
@Prince: I construct exactly such a thing in my answer.
– Qiaochu Yuan
2 days ago
I see that! I upvoted yours and will probably accept it. I tried to comment on it and say thank you, but it said my comment was too short.
– Prince M
2 days ago
Too be honest I always avoid rings of formal power series because I never learned them so they feel uncomfortable.. I should probably get over that!
– Prince M
2 days ago
add a comment |
up vote
3
down vote
A non-discrete valuation ring of height $1$ is such an example.
Ok give me an example verifying the existence of a non discrete valuation ring of height 1
– Prince M
2 days ago
@Prince: I construct exactly such a thing in my answer.
– Qiaochu Yuan
2 days ago
I see that! I upvoted yours and will probably accept it. I tried to comment on it and say thank you, but it said my comment was too short.
– Prince M
2 days ago
Too be honest I always avoid rings of formal power series because I never learned them so they feel uncomfortable.. I should probably get over that!
– Prince M
2 days ago
add a comment |
up vote
3
down vote
up vote
3
down vote
A non-discrete valuation ring of height $1$ is such an example.
A non-discrete valuation ring of height $1$ is such an example.
answered 2 days ago
Bernard
115k637108
115k637108
Ok give me an example verifying the existence of a non discrete valuation ring of height 1
– Prince M
2 days ago
@Prince: I construct exactly such a thing in my answer.
– Qiaochu Yuan
2 days ago
I see that! I upvoted yours and will probably accept it. I tried to comment on it and say thank you, but it said my comment was too short.
– Prince M
2 days ago
Too be honest I always avoid rings of formal power series because I never learned them so they feel uncomfortable.. I should probably get over that!
– Prince M
2 days ago
add a comment |
Ok give me an example verifying the existence of a non discrete valuation ring of height 1
– Prince M
2 days ago
@Prince: I construct exactly such a thing in my answer.
– Qiaochu Yuan
2 days ago
I see that! I upvoted yours and will probably accept it. I tried to comment on it and say thank you, but it said my comment was too short.
– Prince M
2 days ago
Too be honest I always avoid rings of formal power series because I never learned them so they feel uncomfortable.. I should probably get over that!
– Prince M
2 days ago
Ok give me an example verifying the existence of a non discrete valuation ring of height 1
– Prince M
2 days ago
Ok give me an example verifying the existence of a non discrete valuation ring of height 1
– Prince M
2 days ago
@Prince: I construct exactly such a thing in my answer.
– Qiaochu Yuan
2 days ago
@Prince: I construct exactly such a thing in my answer.
– Qiaochu Yuan
2 days ago
I see that! I upvoted yours and will probably accept it. I tried to comment on it and say thank you, but it said my comment was too short.
– Prince M
2 days ago
I see that! I upvoted yours and will probably accept it. I tried to comment on it and say thank you, but it said my comment was too short.
– Prince M
2 days ago
Too be honest I always avoid rings of formal power series because I never learned them so they feel uncomfortable.. I should probably get over that!
– Prince M
2 days ago
Too be honest I always avoid rings of formal power series because I never learned them so they feel uncomfortable.. I should probably get over that!
– Prince M
2 days ago
add a comment |
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