Position of $mathbb{Z}[sqrt{-d}]$ in class hierarchy for all integers $d$
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For any real number $d,$ define
$$mathbb{Z}[sqrt{-d}]: = {a+bsqrt{-d}: a,bin mathbb{Z}}.$$
It is well-known that $mathbb{Z}[sqrt{-2}]$ and $mathbb{Z}[sqrt{-1}]$ are ED but not fields,
$$mathbb{Z} bigg[ frac{1+sqrt{19}}{2} bigg]$$
is a PID but not ED, $mathbb{Z}[sqrt{-5}]$ and $mathbb{Z}[sqrt{-4}]$ are UFD but not PID.
What interests me is that different $d$ places $mathbb{Z}[sqrt{-d}]$ in different position in the class hierarchy.
Question: For all integer $d,$ do we know the position of $mathbb{Z}[sqrt{-d}]$ is the class hierarchy?
abstract-algebra ring-theory principal-ideal-domains unique-factorization-domains euclidean-domain
asked Nov 15 at 14:43
Idonknow
2,254746110
add a comment |
up vote
0
down vote
favorite
For any real number $d,$ define
$$mathbb{Z}[sqrt{-d}]: = {a+bsqrt{-d}: a,bin mathbb{Z}}.$$
It is well-known that $mathbb{Z}[sqrt{-2}]$ and $mathbb{Z}[sqrt{-1}]$ are ED but not fields,
$$mathbb{Z} bigg[ frac{1+sqrt{19}}{2} bigg]$$
is a PID but not ED, $mathbb{Z}[sqrt{-5}]$ and $mathbb{Z}[sqrt{-4}]$ are UFD but not PID.
What interests me is that different $d$ places $mathbb{Z}[sqrt{-d}]$ in different position in the class hierarchy.
Question: For all integer $d,$ do we know the position of $mathbb{Z}[sqrt{-d}]$ is the class hierarchy?
abstract-algebra ring-theory principal-ideal-domains unique-factorization-domains euclidean-domain
asked Nov 15 at 14:43
Idonknow
2,254746110
But more concretely; for $d=-14$ the ring has been known to be a UFD since the (early) 19th century, but has been shown to be Euclidean only in 2004. Which suggests that the answer for general $d$ is a firm no.
– Servaes
Nov 15 at 14:48
@Servaes So you are suggesting that $mathbb{Z}[sqrt{14}]$ is an ED? How does that gives a negative answer to my question?
– Idonknow
Nov 15 at 14:50
2
No, I am suggesting that determining where the ring is in the hierarchy is very difficult for most $d$.
– Servaes
Nov 15 at 14:51
@Servaes I see. However, I would be glad to see results for certain $d.$
– Idonknow
Nov 15 at 14:53
To see if and how the unique factorization fails look at the ideals with $ mathbb{Z}[sqrt{-d}]/( I cap mathbb{Z}[sqrt{-d}]) to O_{mathbb{Q}(sqrt{-d})} / I $ not being an isomorphism
– reuns
Nov 15 at 19:18
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
For any real number $d,$ define
$$mathbb{Z}[sqrt{-d}]: = {a+bsqrt{-d}: a,bin mathbb{Z}}.$$
It is well-known that $mathbb{Z}[sqrt{-2}]$ and $mathbb{Z}[sqrt{-1}]$ are ED but not fields,
$$mathbb{Z} bigg[ frac{1+sqrt{19}}{2} bigg]$$
is a PID but not ED, $mathbb{Z}[sqrt{-5}]$ and $mathbb{Z}[sqrt{-4}]$ are UFD but not PID.
What interests me is that different $d$ places $mathbb{Z}[sqrt{-d}]$ in different position in the class hierarchy.
Question: For all integer $d,$ do we know the position of $mathbb{Z}[sqrt{-d}]$ is the class hierarchy?
abstract-algebra ring-theory principal-ideal-domains unique-factorization-domains euclidean-domain
asked Nov 15 at 14:43
Idonknow
2,254746110
For any real number $d,$ define
$$mathbb{Z}[sqrt{-d}]: = {a+bsqrt{-d}: a,bin mathbb{Z}}.$$
It is well-known that $mathbb{Z}[sqrt{-2}]$ and $mathbb{Z}[sqrt{-1}]$ are ED but not fields,
$$mathbb{Z} bigg[ frac{1+sqrt{19}}{2} bigg]$$
is a PID but not ED, $mathbb{Z}[sqrt{-5}]$ and $mathbb{Z}[sqrt{-4}]$ are UFD but not PID.
What interests me is that different $d$ places $mathbb{Z}[sqrt{-d}]$ in different position in the class hierarchy.
Question: For all integer $d,$ do we know the position of $mathbb{Z}[sqrt{-d}]$ is the class hierarchy?
abstract-algebra ring-theory principal-ideal-domains unique-factorization-domains euclidean-domain
abstract-algebra ring-theory principal-ideal-domains unique-factorization-domains euclidean-domain
asked Nov 15 at 14:43
Idonknow
2,254746110
asked Nov 15 at 14:43
Idonknow
2,254746110
edited Nov 15 at 14:50
asked Nov 15 at 14:43
Idonknow
2,254746110
asked Nov 15 at 14:43
Idonknow
2,254746110
asked Nov 15 at 14:43
Idonknow
2,254746110
2,254746110
But more concretely; for $d=-14$ the ring has been known to be a UFD since the (early) 19th century, but has been shown to be Euclidean only in 2004. Which suggests that the answer for general $d$ is a firm no.
– Servaes
Nov 15 at 14:48
@Servaes So you are suggesting that $mathbb{Z}[sqrt{14}]$ is an ED? How does that gives a negative answer to my question?
– Idonknow
Nov 15 at 14:50
2
No, I am suggesting that determining where the ring is in the hierarchy is very difficult for most $d$.
– Servaes
Nov 15 at 14:51
@Servaes I see. However, I would be glad to see results for certain $d.$
– Idonknow
Nov 15 at 14:53
To see if and how the unique factorization fails look at the ideals with $ mathbb{Z}[sqrt{-d}]/( I cap mathbb{Z}[sqrt{-d}]) to O_{mathbb{Q}(sqrt{-d})} / I $ not being an isomorphism
– reuns
Nov 15 at 19:18
add a comment |
But more concretely; for $d=-14$ the ring has been known to be a UFD since the (early) 19th century, but has been shown to be Euclidean only in 2004. Which suggests that the answer for general $d$ is a firm no.
– Servaes
Nov 15 at 14:48
@Servaes So you are suggesting that $mathbb{Z}[sqrt{14}]$ is an ED? How does that gives a negative answer to my question?
– Idonknow
Nov 15 at 14:50
2
No, I am suggesting that determining where the ring is in the hierarchy is very difficult for most $d$.
– Servaes
Nov 15 at 14:51
@Servaes I see. However, I would be glad to see results for certain $d.$
– Idonknow
Nov 15 at 14:53
To see if and how the unique factorization fails look at the ideals with $ mathbb{Z}[sqrt{-d}]/( I cap mathbb{Z}[sqrt{-d}]) to O_{mathbb{Q}(sqrt{-d})} / I $ not being an isomorphism
– reuns
Nov 15 at 19:18
But more concretely; for $d=-14$ the ring has been known to be a UFD since the (early) 19th century, but has been shown to be Euclidean only in 2004. Which suggests that the answer for general $d$ is a firm no.
– Servaes
Nov 15 at 14:48
But more concretely; for $d=-14$ the ring has been known to be a UFD since the (early) 19th century, but has been shown to be Euclidean only in 2004. Which suggests that the answer for general $d$ is a firm no.
– Servaes
Nov 15 at 14:48
@Servaes So you are suggesting that $mathbb{Z}[sqrt{14}]$ is an ED? How does that gives a negative answer to my question?
– Idonknow
Nov 15 at 14:50
@Servaes So you are suggesting that $mathbb{Z}[sqrt{14}]$ is an ED? How does that gives a negative answer to my question?
– Idonknow
Nov 15 at 14:50
2
2
No, I am suggesting that determining where the ring is in the hierarchy is very difficult for most $d$.
– Servaes
Nov 15 at 14:51
No, I am suggesting that determining where the ring is in the hierarchy is very difficult for most $d$.
– Servaes
Nov 15 at 14:51
@Servaes I see. However, I would be glad to see results for certain $d.$
– Idonknow
Nov 15 at 14:53
@Servaes I see. However, I would be glad to see results for certain $d.$
– Idonknow
Nov 15 at 14:53
To see if and how the unique factorization fails look at the ideals with $ mathbb{Z}[sqrt{-d}]/( I cap mathbb{Z}[sqrt{-d}]) to O_{mathbb{Q}(sqrt{-d})} / I $ not being an isomorphism
– reuns
Nov 15 at 19:18
To see if and how the unique factorization fails look at the ideals with $ mathbb{Z}[sqrt{-d}]/( I cap mathbb{Z}[sqrt{-d}]) to O_{mathbb{Q}(sqrt{-d})} / I $ not being an isomorphism
– reuns
Nov 15 at 19:18
add a comment |
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But more concretely; for $d=-14$ the ring has been known to be a UFD since the (early) 19th century, but has been shown to be Euclidean only in 2004. Which suggests that the answer for general $d$ is a firm no.
– Servaes
Nov 15 at 14:48
@Servaes So you are suggesting that $mathbb{Z}[sqrt{14}]$ is an ED? How does that gives a negative answer to my question?
– Idonknow
Nov 15 at 14:50
2
No, I am suggesting that determining where the ring is in the hierarchy is very difficult for most $d$.
– Servaes
Nov 15 at 14:51
@Servaes I see. However, I would be glad to see results for certain $d.$
– Idonknow
Nov 15 at 14:53
To see if and how the unique factorization fails look at the ideals with $ mathbb{Z}[sqrt{-d}]/( I cap mathbb{Z}[sqrt{-d}]) to O_{mathbb{Q}(sqrt{-d})} / I $ not being an isomorphism
– reuns
Nov 15 at 19:18