Mixing
An infinite dimensional local domain whose chains of primes are finite
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
Does there exist a local domain of infinite dimension in which every chain of prime ideals is finite?
Of course, such a ring must be neither noetherian nor catenary.
(This question arose while trying to compare different definitions of catenarity, and more precisely while trying to understand what may cause codimensions in topological spaces to be infinite.)
ag.algebraic-geometry ac.commutative-algebra nonnoetherian krull-dimension
asked 4 hours ago
Fred Rohrer
3,74211432
add a comment |Â
up vote
3
down vote
favorite
Does there exist a local domain of infinite dimension in which every chain of prime ideals is finite?
Of course, such a ring must be neither noetherian nor catenary.
(This question arose while trying to compare different definitions of catenarity, and more precisely while trying to understand what may cause codimensions in topological spaces to be infinite.)
ag.algebraic-geometry ac.commutative-algebra nonnoetherian krull-dimension
asked 4 hours ago
Fred Rohrer
3,74211432
It's not completely obvious to me that it can't be noetherian. If there's a noetherian example $A$, then we can suppose that $A/P$ has finite dimension for every nonzero prime $P$. How does one get a contradiction?
– YCor
3 hours ago
@YCor: Local noetherian rings have finite dimension, since primes in noetherian rings have finite height.
– Fred Rohrer
3 hours ago
Oh sorry, I hadn't seen the local assumption. I was thinking of a nonlocal noetherian ring, for which it sounds not obvious to me.
– YCor
3 hours ago
@YCor: Actually, my original problem was whether there is a spectral topological space with closed irreducible subsets $Ysubseteq Z$ such that $rm codim(Y,Z)=infty$ and that every chain of closed irreducible subsets with extremities $Y$ and $Z$ is finite. This can be reduced to the question given here.
– Fred Rohrer
3 hours ago
Just an obvious comment ... the ring you are looking for can't be Valuation ring either ...
– user521337
1 hour ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Does there exist a local domain of infinite dimension in which every chain of prime ideals is finite?
Of course, such a ring must be neither noetherian nor catenary.
(This question arose while trying to compare different definitions of catenarity, and more precisely while trying to understand what may cause codimensions in topological spaces to be infinite.)
ag.algebraic-geometry ac.commutative-algebra nonnoetherian krull-dimension
asked 4 hours ago
Fred Rohrer
3,74211432
Does there exist a local domain of infinite dimension in which every chain of prime ideals is finite?
Of course, such a ring must be neither noetherian nor catenary.
(This question arose while trying to compare different definitions of catenarity, and more precisely while trying to understand what may cause codimensions in topological spaces to be infinite.)
ag.algebraic-geometry ac.commutative-algebra nonnoetherian krull-dimension
ag.algebraic-geometry ac.commutative-algebra nonnoetherian krull-dimension
asked 4 hours ago
Fred Rohrer
3,74211432
asked 4 hours ago
Fred Rohrer
3,74211432
asked 4 hours ago
Fred Rohrer
3,74211432
asked 4 hours ago
Fred Rohrer
3,74211432
asked 4 hours ago
Fred Rohrer
3,74211432
3,74211432
It's not completely obvious to me that it can't be noetherian. If there's a noetherian example $A$, then we can suppose that $A/P$ has finite dimension for every nonzero prime $P$. How does one get a contradiction?
– YCor
3 hours ago
@YCor: Local noetherian rings have finite dimension, since primes in noetherian rings have finite height.
– Fred Rohrer
3 hours ago
Oh sorry, I hadn't seen the local assumption. I was thinking of a nonlocal noetherian ring, for which it sounds not obvious to me.
– YCor
3 hours ago
@YCor: Actually, my original problem was whether there is a spectral topological space with closed irreducible subsets $Ysubseteq Z$ such that $rm codim(Y,Z)=infty$ and that every chain of closed irreducible subsets with extremities $Y$ and $Z$ is finite. This can be reduced to the question given here.
– Fred Rohrer
3 hours ago
Just an obvious comment ... the ring you are looking for can't be Valuation ring either ...
– user521337
1 hour ago
add a comment |Â
It's not completely obvious to me that it can't be noetherian. If there's a noetherian example $A$, then we can suppose that $A/P$ has finite dimension for every nonzero prime $P$. How does one get a contradiction?
– YCor
3 hours ago
@YCor: Local noetherian rings have finite dimension, since primes in noetherian rings have finite height.
– Fred Rohrer
3 hours ago
Oh sorry, I hadn't seen the local assumption. I was thinking of a nonlocal noetherian ring, for which it sounds not obvious to me.
– YCor
3 hours ago
@YCor: Actually, my original problem was whether there is a spectral topological space with closed irreducible subsets $Ysubseteq Z$ such that $rm codim(Y,Z)=infty$ and that every chain of closed irreducible subsets with extremities $Y$ and $Z$ is finite. This can be reduced to the question given here.
– Fred Rohrer
3 hours ago
Just an obvious comment ... the ring you are looking for can't be Valuation ring either ...
– user521337
1 hour ago
It's not completely obvious to me that it can't be noetherian. If there's a noetherian example $A$, then we can suppose that $A/P$ has finite dimension for every nonzero prime $P$. How does one get a contradiction?
– YCor
3 hours ago
It's not completely obvious to me that it can't be noetherian. If there's a noetherian example $A$, then we can suppose that $A/P$ has finite dimension for every nonzero prime $P$. How does one get a contradiction?
– YCor
3 hours ago
@YCor: Local noetherian rings have finite dimension, since primes in noetherian rings have finite height.
– Fred Rohrer
3 hours ago
@YCor: Local noetherian rings have finite dimension, since primes in noetherian rings have finite height.
– Fred Rohrer
3 hours ago
Oh sorry, I hadn't seen the local assumption. I was thinking of a nonlocal noetherian ring, for which it sounds not obvious to me.
– YCor
3 hours ago
Oh sorry, I hadn't seen the local assumption. I was thinking of a nonlocal noetherian ring, for which it sounds not obvious to me.
– YCor
3 hours ago
@YCor: Actually, my original problem was whether there is a spectral topological space with closed irreducible subsets $Ysubseteq Z$ such that $rm codim(Y,Z)=infty$ and that every chain of closed irreducible subsets with extremities $Y$ and $Z$ is finite. This can be reduced to the question given here.
– Fred Rohrer
3 hours ago
@YCor: Actually, my original problem was whether there is a spectral topological space with closed irreducible subsets $Ysubseteq Z$ such that $rm codim(Y,Z)=infty$ and that every chain of closed irreducible subsets with extremities $Y$ and $Z$ is finite. This can be reduced to the question given here.
– Fred Rohrer
3 hours ago
Just an obvious comment ... the ring you are looking for can't be Valuation ring either ...
– user521337
1 hour ago
Just an obvious comment ... the ring you are looking for can't be Valuation ring either ...
– user521337
1 hour ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
Choose a field $k$ and a Noetherian $k$-algebra $R$ of infinite dimension. (I know you know such a thing exists.) Let $R' subset R[x]$ be the set of polynomials $f = sum a_i x^i$ whose constant term is constant, i.e., $a_0 in k subset R$. Then we have
$$
textSpec(R') = textSpec(R[x]) amalg_textSpec(R) textSpec(k)
$$
by Tag 0B7J. Observe that $R'[1/x] = R[x, 1/x]$ is Noetherian. Observe that for any prime $mathfrak p subset R$ the prime $mathfrak p' = R' cap mathfrak p R[x]$ is contained in the maximal ideal $mathfrak m = textKer(R' to k) = sqrtxR'$ (small detail omitted). Thus the local ring of $R'$ and $mathfrak m$ is an example. Namely, its punctured spectrum is Noetherian of infinite dimension (as the localization of $R'$ at $mathfrak p'$ is the same as the localization of $R[x]$ at $mathfrak p R[x]$ which is flat over $R_mathfrak p$ and hence has dimension $geq dim(R_mathfrak p)$. Enjoy!
answered 1 hour ago
darx
311
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
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Choose a field $k$ and a Noetherian $k$-algebra $R$ of infinite dimension. (I know you know such a thing exists.) Let $R' subset R[x]$ be the set of polynomials $f = sum a_i x^i$ whose constant term is constant, i.e., $a_0 in k subset R$. Then we have
$$
textSpec(R') = textSpec(R[x]) amalg_textSpec(R) textSpec(k)
$$
by Tag 0B7J. Observe that $R'[1/x] = R[x, 1/x]$ is Noetherian. Observe that for any prime $mathfrak p subset R$ the prime $mathfrak p' = R' cap mathfrak p R[x]$ is contained in the maximal ideal $mathfrak m = textKer(R' to k) = sqrtxR'$ (small detail omitted). Thus the local ring of $R'$ and $mathfrak m$ is an example. Namely, its punctured spectrum is Noetherian of infinite dimension (as the localization of $R'$ at $mathfrak p'$ is the same as the localization of $R[x]$ at $mathfrak p R[x]$ which is flat over $R_mathfrak p$ and hence has dimension $geq dim(R_mathfrak p)$. Enjoy!
answered 1 hour ago
darx
311
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
3
down vote
Choose a field $k$ and a Noetherian $k$-algebra $R$ of infinite dimension. (I know you know such a thing exists.) Let $R' subset R[x]$ be the set of polynomials $f = sum a_i x^i$ whose constant term is constant, i.e., $a_0 in k subset R$. Then we have
$$
textSpec(R') = textSpec(R[x]) amalg_textSpec(R) textSpec(k)
$$
by Tag 0B7J. Observe that $R'[1/x] = R[x, 1/x]$ is Noetherian. Observe that for any prime $mathfrak p subset R$ the prime $mathfrak p' = R' cap mathfrak p R[x]$ is contained in the maximal ideal $mathfrak m = textKer(R' to k) = sqrtxR'$ (small detail omitted). Thus the local ring of $R'$ and $mathfrak m$ is an example. Namely, its punctured spectrum is Noetherian of infinite dimension (as the localization of $R'$ at $mathfrak p'$ is the same as the localization of $R[x]$ at $mathfrak p R[x]$ which is flat over $R_mathfrak p$ and hence has dimension $geq dim(R_mathfrak p)$. Enjoy!
answered 1 hour ago
darx
311
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
3
down vote
up vote
3
down vote
Choose a field $k$ and a Noetherian $k$-algebra $R$ of infinite dimension. (I know you know such a thing exists.) Let $R' subset R[x]$ be the set of polynomials $f = sum a_i x^i$ whose constant term is constant, i.e., $a_0 in k subset R$. Then we have
$$
textSpec(R') = textSpec(R[x]) amalg_textSpec(R) textSpec(k)
$$
by Tag 0B7J. Observe that $R'[1/x] = R[x, 1/x]$ is Noetherian. Observe that for any prime $mathfrak p subset R$ the prime $mathfrak p' = R' cap mathfrak p R[x]$ is contained in the maximal ideal $mathfrak m = textKer(R' to k) = sqrtxR'$ (small detail omitted). Thus the local ring of $R'$ and $mathfrak m$ is an example. Namely, its punctured spectrum is Noetherian of infinite dimension (as the localization of $R'$ at $mathfrak p'$ is the same as the localization of $R[x]$ at $mathfrak p R[x]$ which is flat over $R_mathfrak p$ and hence has dimension $geq dim(R_mathfrak p)$. Enjoy!
answered 1 hour ago
darx
311
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.
Choose a field $k$ and a Noetherian $k$-algebra $R$ of infinite dimension. (I know you know such a thing exists.) Let $R' subset R[x]$ be the set of polynomials $f = sum a_i x^i$ whose constant term is constant, i.e., $a_0 in k subset R$. Then we have
$$
textSpec(R') = textSpec(R[x]) amalg_textSpec(R) textSpec(k)
$$
by Tag 0B7J. Observe that $R'[1/x] = R[x, 1/x]$ is Noetherian. Observe that for any prime $mathfrak p subset R$ the prime $mathfrak p' = R' cap mathfrak p R[x]$ is contained in the maximal ideal $mathfrak m = textKer(R' to k) = sqrtxR'$ (small detail omitted). Thus the local ring of $R'$ and $mathfrak m$ is an example. Namely, its punctured spectrum is Noetherian of infinite dimension (as the localization of $R'$ at $mathfrak p'$ is the same as the localization of $R[x]$ at $mathfrak p R[x]$ which is flat over $R_mathfrak p$ and hence has dimension $geq dim(R_mathfrak p)$. Enjoy!
answered 1 hour ago
darx
311
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
311
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
311
answered 1 hour ago
darx
311
311
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
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f311907%2fan-infinite-dimensional-local-domain-whose-chains-of-primes-are-finite%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
It's not completely obvious to me that it can't be noetherian. If there's a noetherian example $A$, then we can suppose that $A/P$ has finite dimension for every nonzero prime $P$. How does one get a contradiction?
– YCor
3 hours ago
@YCor: Local noetherian rings have finite dimension, since primes in noetherian rings have finite height.
– Fred Rohrer
3 hours ago
Oh sorry, I hadn't seen the local assumption. I was thinking of a nonlocal noetherian ring, for which it sounds not obvious to me.
– YCor
3 hours ago
@YCor: Actually, my original problem was whether there is a spectral topological space with closed irreducible subsets $Ysubseteq Z$ such that $rm codim(Y,Z)=infty$ and that every chain of closed irreducible subsets with extremities $Y$ and $Z$ is finite. This can be reduced to the question given here.
– Fred Rohrer
3 hours ago
Just an obvious comment ... the ring you are looking for can't be Valuation ring either ...
– user521337
1 hour ago