Mixing
Does {0} ideal prime imply the ring is an integral domain? [closed]
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If P={0} is a prime ideal of a ring R, is R an Integral domain?
abstract-algebra ring-theory maximal-and-prime-ideals
asked Jan 8 at 21:09
Amrita DeyAmrita Dey
113
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closed as off-topic by Davide Giraudo, egreg, rschwieb, jgon, Alexander Gruber♦ Jan 8 at 23:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Davide Giraudo, egreg, rschwieb, jgon, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
If P={0} is a prime ideal of a ring R, is R an Integral domain?
abstract-algebra ring-theory maximal-and-prime-ideals
asked Jan 8 at 21:09
Amrita DeyAmrita Dey
113
$endgroup$
closed as off-topic by Davide Giraudo, egreg, rschwieb, jgon, Alexander Gruber♦ Jan 8 at 23:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Davide Giraudo, egreg, rschwieb, jgon, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
4
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Definitions are your friends. Look at the definition of ${0}$ being a prime ideal, and then compare to the definition of an integral domain. It is true...
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– hardmath
Jan 8 at 21:13
1
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Your statement is certainly true if $R$ is commutative. Some interesting things can happen if $R$ is not commutative. You might want to consult the $Prime Ring$ entry on Wiki-pedia.
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– Chris Leary
Jan 8 at 21:18
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It is actually equivalent.
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– Bernard
Jan 8 at 21:19
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@Bernard Well, since it’s not clear what things your statement was meant to apply to, let’s get specific. $R/I$ is a prime ring off $I$ is a (noncommutative definition) prime ideal. $R/I$ is a (possibly noncommutative) domain iff $I$ is a “completely prime” ideal, that is, one satisfying the commutative definition of prime ideal. “Prime” and “completely prime” are equivalent for commutative rings. There. That ought to clarify things.
$endgroup$
– rschwieb
Jan 8 at 22:59
$begingroup$
To the OP : are you assuming commutativity? You should say so one way or the other.
$endgroup$
– rschwieb
Jan 8 at 23:01
add a comment |
$begingroup$
If P={0} is a prime ideal of a ring R, is R an Integral domain?
abstract-algebra ring-theory maximal-and-prime-ideals
asked Jan 8 at 21:09
Amrita DeyAmrita Dey
113
$endgroup$
If P={0} is a prime ideal of a ring R, is R an Integral domain?
abstract-algebra ring-theory maximal-and-prime-ideals
abstract-algebra ring-theory maximal-and-prime-ideals
asked Jan 8 at 21:09
Amrita DeyAmrita Dey
113
asked Jan 8 at 21:09
Amrita DeyAmrita Dey
113
edited Jan 9 at 7:51
Amrita Dey
asked Jan 8 at 21:09
Amrita DeyAmrita Dey
113
asked Jan 8 at 21:09
Amrita DeyAmrita Dey
113
asked Jan 8 at 21:09
Amrita DeyAmrita Dey
113
113
closed as off-topic by Davide Giraudo, egreg, rschwieb, jgon, Alexander Gruber♦ Jan 8 at 23:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Davide Giraudo, egreg, rschwieb, jgon, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Davide Giraudo, egreg, rschwieb, jgon, Alexander Gruber♦ Jan 8 at 23:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Davide Giraudo, egreg, rschwieb, jgon, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
4
$begingroup$
Definitions are your friends. Look at the definition of ${0}$ being a prime ideal, and then compare to the definition of an integral domain. It is true...
$endgroup$
– hardmath
Jan 8 at 21:13
1
$begingroup$
Your statement is certainly true if $R$ is commutative. Some interesting things can happen if $R$ is not commutative. You might want to consult the $Prime Ring$ entry on Wiki-pedia.
$endgroup$
– Chris Leary
Jan 8 at 21:18
$begingroup$
It is actually equivalent.
$endgroup$
– Bernard
Jan 8 at 21:19
$begingroup$
@Bernard Well, since it’s not clear what things your statement was meant to apply to, let’s get specific. $R/I$ is a prime ring off $I$ is a (noncommutative definition) prime ideal. $R/I$ is a (possibly noncommutative) domain iff $I$ is a “completely prime” ideal, that is, one satisfying the commutative definition of prime ideal. “Prime” and “completely prime” are equivalent for commutative rings. There. That ought to clarify things.
$endgroup$
– rschwieb
Jan 8 at 22:59
$begingroup$
To the OP : are you assuming commutativity? You should say so one way or the other.
$endgroup$
– rschwieb
Jan 8 at 23:01
add a comment |
4
$begingroup$
Definitions are your friends. Look at the definition of ${0}$ being a prime ideal, and then compare to the definition of an integral domain. It is true...
$endgroup$
– hardmath
Jan 8 at 21:13
1
$begingroup$
Your statement is certainly true if $R$ is commutative. Some interesting things can happen if $R$ is not commutative. You might want to consult the $Prime Ring$ entry on Wiki-pedia.
$endgroup$
– Chris Leary
Jan 8 at 21:18
$begingroup$
It is actually equivalent.
$endgroup$
– Bernard
Jan 8 at 21:19
$begingroup$
@Bernard Well, since it’s not clear what things your statement was meant to apply to, let’s get specific. $R/I$ is a prime ring off $I$ is a (noncommutative definition) prime ideal. $R/I$ is a (possibly noncommutative) domain iff $I$ is a “completely prime” ideal, that is, one satisfying the commutative definition of prime ideal. “Prime” and “completely prime” are equivalent for commutative rings. There. That ought to clarify things.
$endgroup$
– rschwieb
Jan 8 at 22:59
$begingroup$
To the OP : are you assuming commutativity? You should say so one way or the other.
$endgroup$
– rschwieb
Jan 8 at 23:01
4
4
$begingroup$
Definitions are your friends. Look at the definition of ${0}$ being a prime ideal, and then compare to the definition of an integral domain. It is true...
$endgroup$
– hardmath
Jan 8 at 21:13
$begingroup$
Definitions are your friends. Look at the definition of ${0}$ being a prime ideal, and then compare to the definition of an integral domain. It is true...
$endgroup$
– hardmath
Jan 8 at 21:13
1
1
$begingroup$
Your statement is certainly true if $R$ is commutative. Some interesting things can happen if $R$ is not commutative. You might want to consult the $Prime Ring$ entry on Wiki-pedia.
$endgroup$
– Chris Leary
Jan 8 at 21:18
$begingroup$
Your statement is certainly true if $R$ is commutative. Some interesting things can happen if $R$ is not commutative. You might want to consult the $Prime Ring$ entry on Wiki-pedia.
$endgroup$
– Chris Leary
Jan 8 at 21:18
$begingroup$
It is actually equivalent.
$endgroup$
– Bernard
Jan 8 at 21:19
$begingroup$
It is actually equivalent.
$endgroup$
– Bernard
Jan 8 at 21:19
$begingroup$
@Bernard Well, since it’s not clear what things your statement was meant to apply to, let’s get specific. $R/I$ is a prime ring off $I$ is a (noncommutative definition) prime ideal. $R/I$ is a (possibly noncommutative) domain iff $I$ is a “completely prime” ideal, that is, one satisfying the commutative definition of prime ideal. “Prime” and “completely prime” are equivalent for commutative rings. There. That ought to clarify things.
$endgroup$
– rschwieb
Jan 8 at 22:59
$begingroup$
@Bernard Well, since it’s not clear what things your statement was meant to apply to, let’s get specific. $R/I$ is a prime ring off $I$ is a (noncommutative definition) prime ideal. $R/I$ is a (possibly noncommutative) domain iff $I$ is a “completely prime” ideal, that is, one satisfying the commutative definition of prime ideal. “Prime” and “completely prime” are equivalent for commutative rings. There. That ought to clarify things.
$endgroup$
– rschwieb
Jan 8 at 22:59
$begingroup$
To the OP : are you assuming commutativity? You should say so one way or the other.
$endgroup$
– rschwieb
Jan 8 at 23:01
$begingroup$
To the OP : are you assuming commutativity? You should say so one way or the other.
$endgroup$
– rschwieb
Jan 8 at 23:01
add a comment |
2 Answers
2
active
oldest
votes
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Yes this is true, why don't you try showing this? What needs to be shown? By definition, we need that if $a,bin R$, then $ab=0$ implies $a=0$ or $b=0$. Assuming $ab=0$, we have $abin P={0}$, and now use the definition of $P$ being a prime ideal.
Alternatively, the fact follows from the following proposition (since $Rcong R/{0}$ for any ring $R$):
Proposition: An ideal $P$ of a ring $R$ is prime if and only if $R/P$ is an integral domain.
answered Jan 8 at 21:14
Alex MathersAlex Mathers
10.9k21344
$endgroup$
$begingroup$
The proposition you are using holds good only when R is a commutative ring with unity. But in my question R is considered to be any ring.. That is where my doubt lies.
$endgroup$
– Amrita Dey
Jan 9 at 7:53
add a comment |
$begingroup$
Assuming a commutative ring (there is a generalized notion of prime ideal for a noncommutative ring).
Yes. This follows directly from the definitions, since ${0}$ a prime ideal implies $xy=0implies xyin{0}implies xin{0}lor yin{0}implies x=0lor y=0$.
answered Jan 8 at 21:20
Chris CusterChris Custer
12.3k3825
$endgroup$
$begingroup$
What if the ring is not commutative?
$endgroup$
– Amrita Dey
Jan 9 at 20:08
$begingroup$
Then the condition is called being a domain. But the definition of prime ideal is different.
$endgroup$
– Chris Custer
Jan 9 at 20:12
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes this is true, why don't you try showing this? What needs to be shown? By definition, we need that if $a,bin R$, then $ab=0$ implies $a=0$ or $b=0$. Assuming $ab=0$, we have $abin P={0}$, and now use the definition of $P$ being a prime ideal.
Alternatively, the fact follows from the following proposition (since $Rcong R/{0}$ for any ring $R$):
Proposition: An ideal $P$ of a ring $R$ is prime if and only if $R/P$ is an integral domain.
answered Jan 8 at 21:14
Alex MathersAlex Mathers
10.9k21344
$endgroup$
$begingroup$
The proposition you are using holds good only when R is a commutative ring with unity. But in my question R is considered to be any ring.. That is where my doubt lies.
$endgroup$
– Amrita Dey
Jan 9 at 7:53
add a comment |
$begingroup$
Yes this is true, why don't you try showing this? What needs to be shown? By definition, we need that if $a,bin R$, then $ab=0$ implies $a=0$ or $b=0$. Assuming $ab=0$, we have $abin P={0}$, and now use the definition of $P$ being a prime ideal.
Alternatively, the fact follows from the following proposition (since $Rcong R/{0}$ for any ring $R$):
Proposition: An ideal $P$ of a ring $R$ is prime if and only if $R/P$ is an integral domain.
answered Jan 8 at 21:14
Alex MathersAlex Mathers
10.9k21344
$endgroup$
$begingroup$
The proposition you are using holds good only when R is a commutative ring with unity. But in my question R is considered to be any ring.. That is where my doubt lies.
$endgroup$
– Amrita Dey
Jan 9 at 7:53
add a comment |
$begingroup$
Yes this is true, why don't you try showing this? What needs to be shown? By definition, we need that if $a,bin R$, then $ab=0$ implies $a=0$ or $b=0$. Assuming $ab=0$, we have $abin P={0}$, and now use the definition of $P$ being a prime ideal.
Alternatively, the fact follows from the following proposition (since $Rcong R/{0}$ for any ring $R$):
Proposition: An ideal $P$ of a ring $R$ is prime if and only if $R/P$ is an integral domain.
answered Jan 8 at 21:14
Alex MathersAlex Mathers
10.9k21344
$endgroup$
Yes this is true, why don't you try showing this? What needs to be shown? By definition, we need that if $a,bin R$, then $ab=0$ implies $a=0$ or $b=0$. Assuming $ab=0$, we have $abin P={0}$, and now use the definition of $P$ being a prime ideal.
Alternatively, the fact follows from the following proposition (since $Rcong R/{0}$ for any ring $R$):
Proposition: An ideal $P$ of a ring $R$ is prime if and only if $R/P$ is an integral domain.
answered Jan 8 at 21:14
Alex MathersAlex Mathers
10.9k21344
answered Jan 8 at 21:14
Alex MathersAlex Mathers
10.9k21344
answered Jan 8 at 21:14
Alex MathersAlex Mathers
10.9k21344
answered Jan 8 at 21:14
Alex MathersAlex Mathers
10.9k21344
10.9k21344
$begingroup$
The proposition you are using holds good only when R is a commutative ring with unity. But in my question R is considered to be any ring.. That is where my doubt lies.
$endgroup$
– Amrita Dey
Jan 9 at 7:53
add a comment |
$begingroup$
The proposition you are using holds good only when R is a commutative ring with unity. But in my question R is considered to be any ring.. That is where my doubt lies.
$endgroup$
– Amrita Dey
Jan 9 at 7:53
$begingroup$
The proposition you are using holds good only when R is a commutative ring with unity. But in my question R is considered to be any ring.. That is where my doubt lies.
$endgroup$
– Amrita Dey
Jan 9 at 7:53
$begingroup$
The proposition you are using holds good only when R is a commutative ring with unity. But in my question R is considered to be any ring.. That is where my doubt lies.
$endgroup$
– Amrita Dey
Jan 9 at 7:53
add a comment |
$begingroup$
Assuming a commutative ring (there is a generalized notion of prime ideal for a noncommutative ring).
Yes. This follows directly from the definitions, since ${0}$ a prime ideal implies $xy=0implies xyin{0}implies xin{0}lor yin{0}implies x=0lor y=0$.
answered Jan 8 at 21:20
Chris CusterChris Custer
12.3k3825
$endgroup$
$begingroup$
What if the ring is not commutative?
$endgroup$
– Amrita Dey
Jan 9 at 20:08
$begingroup$
Then the condition is called being a domain. But the definition of prime ideal is different.
$endgroup$
– Chris Custer
Jan 9 at 20:12
add a comment |
$begingroup$
Assuming a commutative ring (there is a generalized notion of prime ideal for a noncommutative ring).
Yes. This follows directly from the definitions, since ${0}$ a prime ideal implies $xy=0implies xyin{0}implies xin{0}lor yin{0}implies x=0lor y=0$.
answered Jan 8 at 21:20
Chris CusterChris Custer
12.3k3825
$endgroup$
$begingroup$
What if the ring is not commutative?
$endgroup$
– Amrita Dey
Jan 9 at 20:08
$begingroup$
Then the condition is called being a domain. But the definition of prime ideal is different.
$endgroup$
– Chris Custer
Jan 9 at 20:12
add a comment |
$begingroup$
Assuming a commutative ring (there is a generalized notion of prime ideal for a noncommutative ring).
Yes. This follows directly from the definitions, since ${0}$ a prime ideal implies $xy=0implies xyin{0}implies xin{0}lor yin{0}implies x=0lor y=0$.
answered Jan 8 at 21:20
Chris CusterChris Custer
12.3k3825
$endgroup$
Assuming a commutative ring (there is a generalized notion of prime ideal for a noncommutative ring).
Yes. This follows directly from the definitions, since ${0}$ a prime ideal implies $xy=0implies xyin{0}implies xin{0}lor yin{0}implies x=0lor y=0$.
answered Jan 8 at 21:20
Chris CusterChris Custer
12.3k3825
edited Jan 9 at 15:29
answered Jan 8 at 21:20
Chris CusterChris Custer
12.3k3825
answered Jan 8 at 21:20
Chris CusterChris Custer
12.3k3825
answered Jan 8 at 21:20
Chris CusterChris Custer
12.3k3825
12.3k3825
$begingroup$
What if the ring is not commutative?
$endgroup$
– Amrita Dey
Jan 9 at 20:08
$begingroup$
Then the condition is called being a domain. But the definition of prime ideal is different.
$endgroup$
– Chris Custer
Jan 9 at 20:12
add a comment |
$begingroup$
What if the ring is not commutative?
$endgroup$
– Amrita Dey
Jan 9 at 20:08
$begingroup$
Then the condition is called being a domain. But the definition of prime ideal is different.
$endgroup$
– Chris Custer
Jan 9 at 20:12
$begingroup$
What if the ring is not commutative?
$endgroup$
– Amrita Dey
Jan 9 at 20:08
$begingroup$
What if the ring is not commutative?
$endgroup$
– Amrita Dey
Jan 9 at 20:08
$begingroup$
Then the condition is called being a domain. But the definition of prime ideal is different.
$endgroup$
– Chris Custer
Jan 9 at 20:12
$begingroup$
Then the condition is called being a domain. But the definition of prime ideal is different.
$endgroup$
– Chris Custer
Jan 9 at 20:12
add a comment |
4
$begingroup$
Definitions are your friends. Look at the definition of ${0}$ being a prime ideal, and then compare to the definition of an integral domain. It is true...
$endgroup$
– hardmath
Jan 8 at 21:13
1
$begingroup$
Your statement is certainly true if $R$ is commutative. Some interesting things can happen if $R$ is not commutative. You might want to consult the $Prime Ring$ entry on Wiki-pedia.
$endgroup$
– Chris Leary
Jan 8 at 21:18
$begingroup$
It is actually equivalent.
$endgroup$
– Bernard
Jan 8 at 21:19
$begingroup$
@Bernard Well, since it’s not clear what things your statement was meant to apply to, let’s get specific. $R/I$ is a prime ring off $I$ is a (noncommutative definition) prime ideal. $R/I$ is a (possibly noncommutative) domain iff $I$ is a “completely prime” ideal, that is, one satisfying the commutative definition of prime ideal. “Prime” and “completely prime” are equivalent for commutative rings. There. That ought to clarify things.
$endgroup$
– rschwieb
Jan 8 at 22:59
$begingroup$
To the OP : are you assuming commutativity? You should say so one way or the other.
$endgroup$
– rschwieb
Jan 8 at 23:01