Mixing
Local ring of infinite dimension
Clash Royale CLAN TAG#URR8PPP
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Short version: Let $R$ be a commutative ring such that all chains of primes of $R$ with the same extremities have the same finite cardinality. Is $R$ locally finite-dimensional?
Longer version: Let $R$ be a commutative ring. Several slightly different definitions of the notion of catenarity of $R$ are in use, and here I would like to compare two of them.
A chain of primes of $R$ is a totally ordered set of primes of $R$ that has a minimum and a maximum. The minimum and the maximum of such a chain are called its extremities. A chain $C$ of primes of $R$ is called saturated if it is $subseteq$-maximal among the chains of primes of $R$ with the same extremities as $C$.
Now, we can give the two definitions of catenarity.
(A) All finite chains of primes of $R$ with the same extremities have the same cardinality.
(B) All chains of primes of $R$ with the same extremities have the same finite cardinality.
Clearly, (B) implies (A). Now, we consider the following two statements.
(1) $R$ fulfils (B).
(2) $R$ fulfils (A) and is locally finite-dimensional, i.e., all its primes have finite height.
It is easy to see that (2) implies (1). Moreover, if every prime of $R$ contains only finitely many minimal primes of $R$, then (1) and (2) are equivalent. In particular, they are equivalent if $R$ is noetherian or a domain.
My question is now the following:
Are (1) and (2) equivalent?
Of course, this comes down to asking whether (B) implies that $R$ is locally finite-dimemsional. I guess it does not and hence look for a counterexample. This amounts to finding a local ring $R$ fulfilling (B) with infinitely many minimal primes such that the (finite) dimensions $dim(R/mathfrakp)$ for the minimal primes $mathfrakp$ are unbounded. Does such a ring exist?
ac.commutative-algebra nonnoetherian krull-dimension
asked 3 hours ago
Fred Rohrer
3,71511432
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up vote
2
down vote
favorite
Short version: Let $R$ be a commutative ring such that all chains of primes of $R$ with the same extremities have the same finite cardinality. Is $R$ locally finite-dimensional?
Longer version: Let $R$ be a commutative ring. Several slightly different definitions of the notion of catenarity of $R$ are in use, and here I would like to compare two of them.
A chain of primes of $R$ is a totally ordered set of primes of $R$ that has a minimum and a maximum. The minimum and the maximum of such a chain are called its extremities. A chain $C$ of primes of $R$ is called saturated if it is $subseteq$-maximal among the chains of primes of $R$ with the same extremities as $C$.
Now, we can give the two definitions of catenarity.
(A) All finite chains of primes of $R$ with the same extremities have the same cardinality.
(B) All chains of primes of $R$ with the same extremities have the same finite cardinality.
Clearly, (B) implies (A). Now, we consider the following two statements.
(1) $R$ fulfils (B).
(2) $R$ fulfils (A) and is locally finite-dimensional, i.e., all its primes have finite height.
It is easy to see that (2) implies (1). Moreover, if every prime of $R$ contains only finitely many minimal primes of $R$, then (1) and (2) are equivalent. In particular, they are equivalent if $R$ is noetherian or a domain.
My question is now the following:
Are (1) and (2) equivalent?
Of course, this comes down to asking whether (B) implies that $R$ is locally finite-dimemsional. I guess it does not and hence look for a counterexample. This amounts to finding a local ring $R$ fulfilling (B) with infinitely many minimal primes such that the (finite) dimensions $dim(R/mathfrakp)$ for the minimal primes $mathfrakp$ are unbounded. Does such a ring exist?
ac.commutative-algebra nonnoetherian krull-dimension
asked 3 hours ago
Fred Rohrer
3,71511432
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Short version: Let $R$ be a commutative ring such that all chains of primes of $R$ with the same extremities have the same finite cardinality. Is $R$ locally finite-dimensional?
Longer version: Let $R$ be a commutative ring. Several slightly different definitions of the notion of catenarity of $R$ are in use, and here I would like to compare two of them.
A chain of primes of $R$ is a totally ordered set of primes of $R$ that has a minimum and a maximum. The minimum and the maximum of such a chain are called its extremities. A chain $C$ of primes of $R$ is called saturated if it is $subseteq$-maximal among the chains of primes of $R$ with the same extremities as $C$.
Now, we can give the two definitions of catenarity.
(A) All finite chains of primes of $R$ with the same extremities have the same cardinality.
(B) All chains of primes of $R$ with the same extremities have the same finite cardinality.
Clearly, (B) implies (A). Now, we consider the following two statements.
(1) $R$ fulfils (B).
(2) $R$ fulfils (A) and is locally finite-dimensional, i.e., all its primes have finite height.
It is easy to see that (2) implies (1). Moreover, if every prime of $R$ contains only finitely many minimal primes of $R$, then (1) and (2) are equivalent. In particular, they are equivalent if $R$ is noetherian or a domain.
My question is now the following:
Are (1) and (2) equivalent?
Of course, this comes down to asking whether (B) implies that $R$ is locally finite-dimemsional. I guess it does not and hence look for a counterexample. This amounts to finding a local ring $R$ fulfilling (B) with infinitely many minimal primes such that the (finite) dimensions $dim(R/mathfrakp)$ for the minimal primes $mathfrakp$ are unbounded. Does such a ring exist?
ac.commutative-algebra nonnoetherian krull-dimension
asked 3 hours ago
Fred Rohrer
3,71511432
Short version: Let $R$ be a commutative ring such that all chains of primes of $R$ with the same extremities have the same finite cardinality. Is $R$ locally finite-dimensional?
Longer version: Let $R$ be a commutative ring. Several slightly different definitions of the notion of catenarity of $R$ are in use, and here I would like to compare two of them.
A chain of primes of $R$ is a totally ordered set of primes of $R$ that has a minimum and a maximum. The minimum and the maximum of such a chain are called its extremities. A chain $C$ of primes of $R$ is called saturated if it is $subseteq$-maximal among the chains of primes of $R$ with the same extremities as $C$.
Now, we can give the two definitions of catenarity.
(A) All finite chains of primes of $R$ with the same extremities have the same cardinality.
(B) All chains of primes of $R$ with the same extremities have the same finite cardinality.
Clearly, (B) implies (A). Now, we consider the following two statements.
(1) $R$ fulfils (B).
(2) $R$ fulfils (A) and is locally finite-dimensional, i.e., all its primes have finite height.
It is easy to see that (2) implies (1). Moreover, if every prime of $R$ contains only finitely many minimal primes of $R$, then (1) and (2) are equivalent. In particular, they are equivalent if $R$ is noetherian or a domain.
My question is now the following:
Are (1) and (2) equivalent?
Of course, this comes down to asking whether (B) implies that $R$ is locally finite-dimemsional. I guess it does not and hence look for a counterexample. This amounts to finding a local ring $R$ fulfilling (B) with infinitely many minimal primes such that the (finite) dimensions $dim(R/mathfrakp)$ for the minimal primes $mathfrakp$ are unbounded. Does such a ring exist?
ac.commutative-algebra nonnoetherian krull-dimension
ac.commutative-algebra nonnoetherian krull-dimension
asked 3 hours ago
Fred Rohrer
3,71511432
asked 3 hours ago
Fred Rohrer
3,71511432
asked 3 hours ago
Fred Rohrer
3,71511432
asked 3 hours ago
Fred Rohrer
3,71511432
asked 3 hours ago
Fred Rohrer
3,71511432
3,71511432
add a comment |Â
add a comment |Â
1 Answer
1
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up vote
4
down vote
Let $k$ be a field. Let $x_i, j$, $1 leq j leq i in mathbfN$ be variables. Consider the ring
$$
R = k[x_i, j]/(x_i, j x_i', j', i not = i')
$$
Let $mathfrak m$ be the maximal ideal generated by all $x_i, j$. For any prime ideal $mathfrak p not = mathfrak m$ there exists a unique $i$ such that $x_i', j' in mathfrak p$ for $i' not = i$. Hence chains happen in the finite polynomial rings $k[x_i, 1, ldots, x_i, i]$. Using this it is easy to see that $R$ is a counter example.
answered 1 hour ago
darx
411
New contributor
darx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Let $k$ be a field. Let $x_i, j$, $1 leq j leq i in mathbfN$ be variables. Consider the ring
$$
R = k[x_i, j]/(x_i, j x_i', j', i not = i')
$$
Let $mathfrak m$ be the maximal ideal generated by all $x_i, j$. For any prime ideal $mathfrak p not = mathfrak m$ there exists a unique $i$ such that $x_i', j' in mathfrak p$ for $i' not = i$. Hence chains happen in the finite polynomial rings $k[x_i, 1, ldots, x_i, i]$. Using this it is easy to see that $R$ is a counter example.
answered 1 hour ago
darx
411
New contributor
darx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
4
down vote
Let $k$ be a field. Let $x_i, j$, $1 leq j leq i in mathbfN$ be variables. Consider the ring
$$
R = k[x_i, j]/(x_i, j x_i', j', i not = i')
$$
Let $mathfrak m$ be the maximal ideal generated by all $x_i, j$. For any prime ideal $mathfrak p not = mathfrak m$ there exists a unique $i$ such that $x_i', j' in mathfrak p$ for $i' not = i$. Hence chains happen in the finite polynomial rings $k[x_i, 1, ldots, x_i, i]$. Using this it is easy to see that $R$ is a counter example.
answered 1 hour ago
darx
411
New contributor
darx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Let $k$ be a field. Let $x_i, j$, $1 leq j leq i in mathbfN$ be variables. Consider the ring
$$
R = k[x_i, j]/(x_i, j x_i', j', i not = i')
$$
Let $mathfrak m$ be the maximal ideal generated by all $x_i, j$. For any prime ideal $mathfrak p not = mathfrak m$ there exists a unique $i$ such that $x_i', j' in mathfrak p$ for $i' not = i$. Hence chains happen in the finite polynomial rings $k[x_i, 1, ldots, x_i, i]$. Using this it is easy to see that $R$ is a counter example.
answered 1 hour ago
darx
411
New contributor
darx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Let $k$ be a field. Let $x_i, j$, $1 leq j leq i in mathbfN$ be variables. Consider the ring
$$
R = k[x_i, j]/(x_i, j x_i', j', i not = i')
$$
Let $mathfrak m$ be the maximal ideal generated by all $x_i, j$. For any prime ideal $mathfrak p not = mathfrak m$ there exists a unique $i$ such that $x_i', j' in mathfrak p$ for $i' not = i$. Hence chains happen in the finite polynomial rings $k[x_i, 1, ldots, x_i, i]$. Using this it is easy to see that $R$ is a counter example.
answered 1 hour ago
darx
411
New contributor
darx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
darx
411
New contributor
darx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
darx
411
answered 1 hour ago
darx
411
411
New contributor
darx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
darx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
darx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
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