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Is there a Euclidean ring that is not a domain?
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I have been reading about the equivalent definitions of PID. I learned that a domain $R$ is a PID if and only if $R$ has a Dedekind-Hasse norm. This followed from the fact that a ring $R$ is a principal ideal ring if $R$ admitted a Dedekind-Hasse norm.
$mathbbZ/nmathbbZ$ with $n$ composite is an example of a principal ideal ring that is not a domain. However, I was unable to think of a euclidean ring (a ring that admits an Euclidean function), but is not a domain.
- Is there a Euclidean ring that is not a domain?
- Is there a finite Euclidean ring? (It seems unplausible because it would be hard to give order.)
Added:
I was referring to the usual euclidean function as the definition does not seem to require the ring to be domain. If $R$ is a commutative unital ring,
a euclidean function is $N:R backslash 0 to mathbbZ_> 0$ such that for all nonzero $a,b in R$, there exists $q,r in R$ such that $a = bq +r$ and either $r=0$ or $N(r) <N(b)$.
commutative-algebra
asked Jul 16 at 21:06
libofmath
436
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up vote
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I have been reading about the equivalent definitions of PID. I learned that a domain $R$ is a PID if and only if $R$ has a Dedekind-Hasse norm. This followed from the fact that a ring $R$ is a principal ideal ring if $R$ admitted a Dedekind-Hasse norm.
$mathbbZ/nmathbbZ$ with $n$ composite is an example of a principal ideal ring that is not a domain. However, I was unable to think of a euclidean ring (a ring that admits an Euclidean function), but is not a domain.
- Is there a Euclidean ring that is not a domain?
- Is there a finite Euclidean ring? (It seems unplausible because it would be hard to give order.)
Added:
I was referring to the usual euclidean function as the definition does not seem to require the ring to be domain. If $R$ is a commutative unital ring,
a euclidean function is $N:R backslash 0 to mathbbZ_> 0$ such that for all nonzero $a,b in R$, there exists $q,r in R$ such that $a = bq +r$ and either $r=0$ or $N(r) <N(b)$.
commutative-algebra
asked Jul 16 at 21:06
libofmath
436
1
Please be more explicit about what you expect a Euclidean function to be on a non-domain.
– Mr. Chip
Jul 16 at 21:21
@Mr.Chip, I have added the definition. I read the first caveat in the following link where it says that the definition does not depend on $R$ being a domain.
– libofmath
Jul 16 at 21:37
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have been reading about the equivalent definitions of PID. I learned that a domain $R$ is a PID if and only if $R$ has a Dedekind-Hasse norm. This followed from the fact that a ring $R$ is a principal ideal ring if $R$ admitted a Dedekind-Hasse norm.
$mathbbZ/nmathbbZ$ with $n$ composite is an example of a principal ideal ring that is not a domain. However, I was unable to think of a euclidean ring (a ring that admits an Euclidean function), but is not a domain.
- Is there a Euclidean ring that is not a domain?
- Is there a finite Euclidean ring? (It seems unplausible because it would be hard to give order.)
Added:
I was referring to the usual euclidean function as the definition does not seem to require the ring to be domain. If $R$ is a commutative unital ring,
a euclidean function is $N:R backslash 0 to mathbbZ_> 0$ such that for all nonzero $a,b in R$, there exists $q,r in R$ such that $a = bq +r$ and either $r=0$ or $N(r) <N(b)$.
commutative-algebra
asked Jul 16 at 21:06
libofmath
436
I have been reading about the equivalent definitions of PID. I learned that a domain $R$ is a PID if and only if $R$ has a Dedekind-Hasse norm. This followed from the fact that a ring $R$ is a principal ideal ring if $R$ admitted a Dedekind-Hasse norm.
$mathbbZ/nmathbbZ$ with $n$ composite is an example of a principal ideal ring that is not a domain. However, I was unable to think of a euclidean ring (a ring that admits an Euclidean function), but is not a domain.
- Is there a Euclidean ring that is not a domain?
- Is there a finite Euclidean ring? (It seems unplausible because it would be hard to give order.)
Added:
I was referring to the usual euclidean function as the definition does not seem to require the ring to be domain. If $R$ is a commutative unital ring,
a euclidean function is $N:R backslash 0 to mathbbZ_> 0$ such that for all nonzero $a,b in R$, there exists $q,r in R$ such that $a = bq +r$ and either $r=0$ or $N(r) <N(b)$.
commutative-algebra
asked Jul 16 at 21:06
libofmath
436
edited Jul 16 at 21:34
asked Jul 16 at 21:06
libofmath
436
asked Jul 16 at 21:06
libofmath
436
asked Jul 16 at 21:06
libofmath
436
436
1
Please be more explicit about what you expect a Euclidean function to be on a non-domain.
– Mr. Chip
Jul 16 at 21:21
@Mr.Chip, I have added the definition. I read the first caveat in the following link where it says that the definition does not depend on $R$ being a domain.
– libofmath
Jul 16 at 21:37
add a comment |Â
1
Please be more explicit about what you expect a Euclidean function to be on a non-domain.
– Mr. Chip
Jul 16 at 21:21
@Mr.Chip, I have added the definition. I read the first caveat in the following link where it says that the definition does not depend on $R$ being a domain.
– libofmath
Jul 16 at 21:37
1
1
Please be more explicit about what you expect a Euclidean function to be on a non-domain.
– Mr. Chip
Jul 16 at 21:21
Please be more explicit about what you expect a Euclidean function to be on a non-domain.
– Mr. Chip
Jul 16 at 21:21
@Mr.Chip, I have added the definition. I read the first caveat in the following link where it says that the definition does not depend on $R$ being a domain.
– libofmath
Jul 16 at 21:37
@Mr.Chip, I have added the definition. I read the first caveat in the following link where it says that the definition does not depend on $R$ being a domain.
– libofmath
Jul 16 at 21:37
add a comment |Â
1 Answer
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I think $R = mathbbF_2[x]/(x^2)$ with $(N(1),N(x),N(1+x)) = (1,2,1)$ works.
Notice that $1, 1+x$ are units, while $x$ is nilpotent and so a zero divisor. We have equations like $$1 = 0x + 1 = (1 + x)(1+x), quad 1+x = 0x + (1 + x) = 1(1+x).$$
Also $x = (1+x)x = 1x.$ I think that covers it.
answered Jul 16 at 21:46
Mr. Chip
3,2031028
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I think $R = mathbbF_2[x]/(x^2)$ with $(N(1),N(x),N(1+x)) = (1,2,1)$ works.
Notice that $1, 1+x$ are units, while $x$ is nilpotent and so a zero divisor. We have equations like $$1 = 0x + 1 = (1 + x)(1+x), quad 1+x = 0x + (1 + x) = 1(1+x).$$
Also $x = (1+x)x = 1x.$ I think that covers it.
answered Jul 16 at 21:46
Mr. Chip
3,2031028
add a comment |Â
up vote
0
down vote
I think $R = mathbbF_2[x]/(x^2)$ with $(N(1),N(x),N(1+x)) = (1,2,1)$ works.
Notice that $1, 1+x$ are units, while $x$ is nilpotent and so a zero divisor. We have equations like $$1 = 0x + 1 = (1 + x)(1+x), quad 1+x = 0x + (1 + x) = 1(1+x).$$
Also $x = (1+x)x = 1x.$ I think that covers it.
answered Jul 16 at 21:46
Mr. Chip
3,2031028
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I think $R = mathbbF_2[x]/(x^2)$ with $(N(1),N(x),N(1+x)) = (1,2,1)$ works.
Notice that $1, 1+x$ are units, while $x$ is nilpotent and so a zero divisor. We have equations like $$1 = 0x + 1 = (1 + x)(1+x), quad 1+x = 0x + (1 + x) = 1(1+x).$$
Also $x = (1+x)x = 1x.$ I think that covers it.
answered Jul 16 at 21:46
Mr. Chip
3,2031028
I think $R = mathbbF_2[x]/(x^2)$ with $(N(1),N(x),N(1+x)) = (1,2,1)$ works.
Notice that $1, 1+x$ are units, while $x$ is nilpotent and so a zero divisor. We have equations like $$1 = 0x + 1 = (1 + x)(1+x), quad 1+x = 0x + (1 + x) = 1(1+x).$$
Also $x = (1+x)x = 1x.$ I think that covers it.
answered Jul 16 at 21:46
Mr. Chip
3,2031028
edited Jul 16 at 21:51
answered Jul 16 at 21:46
Mr. Chip
3,2031028
answered Jul 16 at 21:46
Mr. Chip
3,2031028
answered Jul 16 at 21:46
Mr. Chip
3,2031028
3,2031028
add a comment |Â
add a comment |Â
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1
Please be more explicit about what you expect a Euclidean function to be on a non-domain.
– Mr. Chip
Jul 16 at 21:21
@Mr.Chip, I have added the definition. I read the first caveat in the following link where it says that the definition does not depend on $R$ being a domain.
– libofmath
Jul 16 at 21:37