Mixing
Is every ideal of Quotient Ring a PID?
$begingroup$
We are given a principal ideal domain $R$ and an ideal of this domain, $I$. Is it true that every proper ideal of $displaystylefrac{R}{I}$ a principal ideal domain ?
I have proven that every proper ideal, say, $overline{K}$ of $displaystylefrac{R}{I}$ is an ideal generated by an element of the $displaystylefrac{R}{I}$. Is it true that $overline{K}$ is an integral domain ?
abstract-algebra principal-ideal-domains
asked Jan 23 at 16:26
Minto PMinto P
667
$endgroup$
add a comment |
$begingroup$
We are given a principal ideal domain $R$ and an ideal of this domain, $I$. Is it true that every proper ideal of $displaystylefrac{R}{I}$ a principal ideal domain ?
I have proven that every proper ideal, say, $overline{K}$ of $displaystylefrac{R}{I}$ is an ideal generated by an element of the $displaystylefrac{R}{I}$. Is it true that $overline{K}$ is an integral domain ?
abstract-algebra principal-ideal-domains
asked Jan 23 at 16:26
Minto PMinto P
667
$endgroup$
add a comment |
$begingroup$
We are given a principal ideal domain $R$ and an ideal of this domain, $I$. Is it true that every proper ideal of $displaystylefrac{R}{I}$ a principal ideal domain ?
I have proven that every proper ideal, say, $overline{K}$ of $displaystylefrac{R}{I}$ is an ideal generated by an element of the $displaystylefrac{R}{I}$. Is it true that $overline{K}$ is an integral domain ?
abstract-algebra principal-ideal-domains
asked Jan 23 at 16:26
Minto PMinto P
667
$endgroup$
We are given a principal ideal domain $R$ and an ideal of this domain, $I$. Is it true that every proper ideal of $displaystylefrac{R}{I}$ a principal ideal domain ?
I have proven that every proper ideal, say, $overline{K}$ of $displaystylefrac{R}{I}$ is an ideal generated by an element of the $displaystylefrac{R}{I}$. Is it true that $overline{K}$ is an integral domain ?
abstract-algebra principal-ideal-domains
abstract-algebra principal-ideal-domains
asked Jan 23 at 16:26
Minto PMinto P
667
asked Jan 23 at 16:26
Minto PMinto P
667
asked Jan 23 at 16:26
Minto PMinto P
667
asked Jan 23 at 16:26
Minto PMinto P
667
asked Jan 23 at 16:26
Minto PMinto P
667
667
add a comment |
add a comment |
2 Answers
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No, in general $K$ can have zero divisors, and it will not always have an identity.
For example: $mathbb Z/12mathbb Z$ and its ideal $6mathbb Z/12mathbb Z$ are examples for both.
$K$ will not have zero divisors if $I$ is prime, but in a PID this would mean that $I={0}$ or else a maximal ideal, which are both pretty uninteresting cases.
The best you can say is, as you concluded, that $R/I$ is always a principal ideal ring.
answered Jan 23 at 16:52
rschwiebrschwieb
107k12102251
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add a comment |
$begingroup$
Consider $mathbb{Z}$ and $p,q,r$, three prime integers. $mathbb{Z}/pqrmathbb{Z}$ is isomorphic to $mathbb{Z}/ptimesmathbb{Z}/qtimes mathbb{Z}/r$. $mathbb{Z}/ptimesmathbb{Z}/qtimes 0$ is a proper ideal of $mathbb{Z}/ptimesmathbb{Z}/qtimes mathbb{Z}/r$ and is not integral.
edited Jan 23 at 19:19
J. W. Tanner
3,2401320
answered Jan 23 at 16:52
Tsemo AristideTsemo Aristide
59.4k11445
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
oldest
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$begingroup$
No, in general $K$ can have zero divisors, and it will not always have an identity.
For example: $mathbb Z/12mathbb Z$ and its ideal $6mathbb Z/12mathbb Z$ are examples for both.
$K$ will not have zero divisors if $I$ is prime, but in a PID this would mean that $I={0}$ or else a maximal ideal, which are both pretty uninteresting cases.
The best you can say is, as you concluded, that $R/I$ is always a principal ideal ring.
answered Jan 23 at 16:52
rschwiebrschwieb
107k12102251
$endgroup$
add a comment |
$begingroup$
No, in general $K$ can have zero divisors, and it will not always have an identity.
For example: $mathbb Z/12mathbb Z$ and its ideal $6mathbb Z/12mathbb Z$ are examples for both.
$K$ will not have zero divisors if $I$ is prime, but in a PID this would mean that $I={0}$ or else a maximal ideal, which are both pretty uninteresting cases.
The best you can say is, as you concluded, that $R/I$ is always a principal ideal ring.
answered Jan 23 at 16:52
rschwiebrschwieb
107k12102251
$endgroup$
add a comment |
$begingroup$
No, in general $K$ can have zero divisors, and it will not always have an identity.
For example: $mathbb Z/12mathbb Z$ and its ideal $6mathbb Z/12mathbb Z$ are examples for both.
$K$ will not have zero divisors if $I$ is prime, but in a PID this would mean that $I={0}$ or else a maximal ideal, which are both pretty uninteresting cases.
The best you can say is, as you concluded, that $R/I$ is always a principal ideal ring.
answered Jan 23 at 16:52
rschwiebrschwieb
107k12102251
$endgroup$
No, in general $K$ can have zero divisors, and it will not always have an identity.
For example: $mathbb Z/12mathbb Z$ and its ideal $6mathbb Z/12mathbb Z$ are examples for both.
$K$ will not have zero divisors if $I$ is prime, but in a PID this would mean that $I={0}$ or else a maximal ideal, which are both pretty uninteresting cases.
The best you can say is, as you concluded, that $R/I$ is always a principal ideal ring.
answered Jan 23 at 16:52
rschwiebrschwieb
107k12102251
edited Jan 23 at 16:58
answered Jan 23 at 16:52
rschwiebrschwieb
107k12102251
answered Jan 23 at 16:52
rschwiebrschwieb
107k12102251
answered Jan 23 at 16:52
rschwiebrschwieb
107k12102251
107k12102251
add a comment |
add a comment |
$begingroup$
Consider $mathbb{Z}$ and $p,q,r$, three prime integers. $mathbb{Z}/pqrmathbb{Z}$ is isomorphic to $mathbb{Z}/ptimesmathbb{Z}/qtimes mathbb{Z}/r$. $mathbb{Z}/ptimesmathbb{Z}/qtimes 0$ is a proper ideal of $mathbb{Z}/ptimesmathbb{Z}/qtimes mathbb{Z}/r$ and is not integral.
edited Jan 23 at 19:19
J. W. Tanner
3,2401320
answered Jan 23 at 16:52
Tsemo AristideTsemo Aristide
59.4k11445
$endgroup$
add a comment |
$begingroup$
Consider $mathbb{Z}$ and $p,q,r$, three prime integers. $mathbb{Z}/pqrmathbb{Z}$ is isomorphic to $mathbb{Z}/ptimesmathbb{Z}/qtimes mathbb{Z}/r$. $mathbb{Z}/ptimesmathbb{Z}/qtimes 0$ is a proper ideal of $mathbb{Z}/ptimesmathbb{Z}/qtimes mathbb{Z}/r$ and is not integral.
edited Jan 23 at 19:19
J. W. Tanner
3,2401320
answered Jan 23 at 16:52
Tsemo AristideTsemo Aristide
59.4k11445
$endgroup$
add a comment |
$begingroup$
Consider $mathbb{Z}$ and $p,q,r$, three prime integers. $mathbb{Z}/pqrmathbb{Z}$ is isomorphic to $mathbb{Z}/ptimesmathbb{Z}/qtimes mathbb{Z}/r$. $mathbb{Z}/ptimesmathbb{Z}/qtimes 0$ is a proper ideal of $mathbb{Z}/ptimesmathbb{Z}/qtimes mathbb{Z}/r$ and is not integral.
edited Jan 23 at 19:19
J. W. Tanner
3,2401320
answered Jan 23 at 16:52
Tsemo AristideTsemo Aristide
59.4k11445
$endgroup$
Consider $mathbb{Z}$ and $p,q,r$, three prime integers. $mathbb{Z}/pqrmathbb{Z}$ is isomorphic to $mathbb{Z}/ptimesmathbb{Z}/qtimes mathbb{Z}/r$. $mathbb{Z}/ptimesmathbb{Z}/qtimes 0$ is a proper ideal of $mathbb{Z}/ptimesmathbb{Z}/qtimes mathbb{Z}/r$ and is not integral.
edited Jan 23 at 19:19
J. W. Tanner
3,2401320
answered Jan 23 at 16:52
Tsemo AristideTsemo Aristide
59.4k11445
edited Jan 23 at 19:19
J. W. Tanner
3,2401320
edited Jan 23 at 19:19
J. W. Tanner
3,2401320
edited Jan 23 at 19:19
J. W. Tanner
3,2401320
3,2401320
answered Jan 23 at 16:52
Tsemo AristideTsemo Aristide
59.4k11445
answered Jan 23 at 16:52
Tsemo AristideTsemo Aristide
59.4k11445
answered Jan 23 at 16:52
Tsemo AristideTsemo Aristide
59.4k11445
59.4k11445
add a comment |
add a comment |
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