Properties of rings of the form $mathbb{Z}+mathbb{Z}sqrt{p}i+mathbb{Z}sqrt{q}j+mathbb{Z}sqrt{r}k$
It is well-known that $mathbb{Z}+mathbb{Z}sqrt{5}i$ is not a unique factorization domain, since, for example, $6$ has two different factorizations into irreducibles: $(1+sqrt{5}i)(1-sqrt{5}i)=2cdot3$.
Let $p,q,r in mathbb{N}$ be odd primes (not necessarily three different).
Perhaps we should also require that each of ${p,q,r}$ is congruent to $1$ mod $4$ (perhaps no).
Let $R_{p,q,r}:=mathbb{Z}+mathbb{Z}sqrt{p}i+mathbb{Z}sqrt{q}j+mathbb{Z}sqrt{r}k$, where $i^2=j^2=k^2=-1$, $ij=k$ etc., as in the quaternions (replacing $mathbb{R}$ by $mathbb{Z}$).
Trivially, $R_{p,q,r}$ is an integral domain, because it is a subset of $mathbb{C}$.
Of course,$R_{p,q,r}$ is not a UFD, since, for example, $(1-sqrt{p}i)(1+sqrt{p}i)=1-pi^2=1+p=1+(1+4u)=2+4u=2(1+2u)$.
What else can be said about those integral domains?
For example: I guess that $R_{p,q,r}$ is a Euclidean domain, isn't it? Is there a known proof for this or one just needs to adjust the proof for $mathbb{Z}+mathbb{Z}sqrt{5}i$?
This question is relevant.
Any hints and comments are welcome!
commutative-algebra unique-factorization-domains euclidean-domain
asked Nov 30 '18 at 1:23
user237522user237522
2,1411617
add a comment |
It is well-known that $mathbb{Z}+mathbb{Z}sqrt{5}i$ is not a unique factorization domain, since, for example, $6$ has two different factorizations into irreducibles: $(1+sqrt{5}i)(1-sqrt{5}i)=2cdot3$.
Let $p,q,r in mathbb{N}$ be odd primes (not necessarily three different).
Perhaps we should also require that each of ${p,q,r}$ is congruent to $1$ mod $4$ (perhaps no).
Let $R_{p,q,r}:=mathbb{Z}+mathbb{Z}sqrt{p}i+mathbb{Z}sqrt{q}j+mathbb{Z}sqrt{r}k$, where $i^2=j^2=k^2=-1$, $ij=k$ etc., as in the quaternions (replacing $mathbb{R}$ by $mathbb{Z}$).
Trivially, $R_{p,q,r}$ is an integral domain, because it is a subset of $mathbb{C}$.
Of course,$R_{p,q,r}$ is not a UFD, since, for example, $(1-sqrt{p}i)(1+sqrt{p}i)=1-pi^2=1+p=1+(1+4u)=2+4u=2(1+2u)$.
What else can be said about those integral domains?
For example: I guess that $R_{p,q,r}$ is a Euclidean domain, isn't it? Is there a known proof for this or one just needs to adjust the proof for $mathbb{Z}+mathbb{Z}sqrt{5}i$?
This question is relevant.
Any hints and comments are welcome!
commutative-algebra unique-factorization-domains euclidean-domain
asked Nov 30 '18 at 1:23
user237522user237522
2,1411617
For the tensor product of this thing with a field see en.wikipedia.org/wiki/Quaternion_algebra .
– Qiaochu Yuan
Nov 30 '18 at 1:55
@QiaochuYuan, thank you very much for your comment.
– user237522
Nov 30 '18 at 2:02
add a comment |
It is well-known that $mathbb{Z}+mathbb{Z}sqrt{5}i$ is not a unique factorization domain, since, for example, $6$ has two different factorizations into irreducibles: $(1+sqrt{5}i)(1-sqrt{5}i)=2cdot3$.
Let $p,q,r in mathbb{N}$ be odd primes (not necessarily three different).
Perhaps we should also require that each of ${p,q,r}$ is congruent to $1$ mod $4$ (perhaps no).
Let $R_{p,q,r}:=mathbb{Z}+mathbb{Z}sqrt{p}i+mathbb{Z}sqrt{q}j+mathbb{Z}sqrt{r}k$, where $i^2=j^2=k^2=-1$, $ij=k$ etc., as in the quaternions (replacing $mathbb{R}$ by $mathbb{Z}$).
Trivially, $R_{p,q,r}$ is an integral domain, because it is a subset of $mathbb{C}$.
Of course,$R_{p,q,r}$ is not a UFD, since, for example, $(1-sqrt{p}i)(1+sqrt{p}i)=1-pi^2=1+p=1+(1+4u)=2+4u=2(1+2u)$.
What else can be said about those integral domains?
For example: I guess that $R_{p,q,r}$ is a Euclidean domain, isn't it? Is there a known proof for this or one just needs to adjust the proof for $mathbb{Z}+mathbb{Z}sqrt{5}i$?
This question is relevant.
Any hints and comments are welcome!
commutative-algebra unique-factorization-domains euclidean-domain
asked Nov 30 '18 at 1:23
user237522user237522
2,1411617
It is well-known that $mathbb{Z}+mathbb{Z}sqrt{5}i$ is not a unique factorization domain, since, for example, $6$ has two different factorizations into irreducibles: $(1+sqrt{5}i)(1-sqrt{5}i)=2cdot3$.
Let $p,q,r in mathbb{N}$ be odd primes (not necessarily three different).
Perhaps we should also require that each of ${p,q,r}$ is congruent to $1$ mod $4$ (perhaps no).
Let $R_{p,q,r}:=mathbb{Z}+mathbb{Z}sqrt{p}i+mathbb{Z}sqrt{q}j+mathbb{Z}sqrt{r}k$, where $i^2=j^2=k^2=-1$, $ij=k$ etc., as in the quaternions (replacing $mathbb{R}$ by $mathbb{Z}$).
Trivially, $R_{p,q,r}$ is an integral domain, because it is a subset of $mathbb{C}$.
Of course,$R_{p,q,r}$ is not a UFD, since, for example, $(1-sqrt{p}i)(1+sqrt{p}i)=1-pi^2=1+p=1+(1+4u)=2+4u=2(1+2u)$.
What else can be said about those integral domains?
For example: I guess that $R_{p,q,r}$ is a Euclidean domain, isn't it? Is there a known proof for this or one just needs to adjust the proof for $mathbb{Z}+mathbb{Z}sqrt{5}i$?
This question is relevant.
Any hints and comments are welcome!
commutative-algebra unique-factorization-domains euclidean-domain
commutative-algebra unique-factorization-domains euclidean-domain
asked Nov 30 '18 at 1:23
user237522user237522
2,1411617
asked Nov 30 '18 at 1:23
user237522user237522
2,1411617
edited Nov 30 '18 at 1:29
user237522
asked Nov 30 '18 at 1:23
user237522user237522
2,1411617
asked Nov 30 '18 at 1:23
user237522user237522
2,1411617
asked Nov 30 '18 at 1:23
user237522user237522
2,1411617
2,1411617
For the tensor product of this thing with a field see en.wikipedia.org/wiki/Quaternion_algebra .
– Qiaochu Yuan
Nov 30 '18 at 1:55
@QiaochuYuan, thank you very much for your comment.
– user237522
Nov 30 '18 at 2:02
add a comment |
For the tensor product of this thing with a field see en.wikipedia.org/wiki/Quaternion_algebra .
– Qiaochu Yuan
Nov 30 '18 at 1:55
@QiaochuYuan, thank you very much for your comment.
– user237522
Nov 30 '18 at 2:02
For the tensor product of this thing with a field see en.wikipedia.org/wiki/Quaternion_algebra .
– Qiaochu Yuan
Nov 30 '18 at 1:55
For the tensor product of this thing with a field see en.wikipedia.org/wiki/Quaternion_algebra .
– Qiaochu Yuan
Nov 30 '18 at 1:55
@QiaochuYuan, thank you very much for your comment.
– user237522
Nov 30 '18 at 2:02
@QiaochuYuan, thank you very much for your comment.
– user237522
Nov 30 '18 at 2:02
add a comment |
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For the tensor product of this thing with a field see en.wikipedia.org/wiki/Quaternion_algebra .
– Qiaochu Yuan
Nov 30 '18 at 1:55
@QiaochuYuan, thank you very much for your comment.
– user237522
Nov 30 '18 at 2:02