If $R$ is a principal ideal domain and $s$ is a non-zero prime ideal, then $frac{R}{s}$ has finitely many...
$begingroup$
If $R$ is a principal ideal domain and $s$ is a non-zero prime ideal, then $frac{R}{s}$ has finitely many prime ideals .
How can I prove it?
My attempt:
As $R$ is a PID then $s$ will be maximal ideal so $frac {R}{s}$ will be field which has precisely two prime ideal . One is {$0$} and another one is $frac {R}{s}$ .
Can anyone correct me If I have gone wrong anywhere?
abstract-algebra ring-theory principal-ideal-domains
asked Dec 14 '18 at 7:01
cmicmi
1,121312
$endgroup$
add a comment |
$begingroup$
If $R$ is a principal ideal domain and $s$ is a non-zero prime ideal, then $frac{R}{s}$ has finitely many prime ideals .
How can I prove it?
My attempt:
As $R$ is a PID then $s$ will be maximal ideal so $frac {R}{s}$ will be field which has precisely two prime ideal . One is {$0$} and another one is $frac {R}{s}$ .
Can anyone correct me If I have gone wrong anywhere?
abstract-algebra ring-theory principal-ideal-domains
asked Dec 14 '18 at 7:01
cmicmi
1,121312
$endgroup$
add a comment |
$begingroup$
If $R$ is a principal ideal domain and $s$ is a non-zero prime ideal, then $frac{R}{s}$ has finitely many prime ideals .
How can I prove it?
My attempt:
As $R$ is a PID then $s$ will be maximal ideal so $frac {R}{s}$ will be field which has precisely two prime ideal . One is {$0$} and another one is $frac {R}{s}$ .
Can anyone correct me If I have gone wrong anywhere?
abstract-algebra ring-theory principal-ideal-domains
asked Dec 14 '18 at 7:01
cmicmi
1,121312
$endgroup$
If $R$ is a principal ideal domain and $s$ is a non-zero prime ideal, then $frac{R}{s}$ has finitely many prime ideals .
How can I prove it?
My attempt:
As $R$ is a PID then $s$ will be maximal ideal so $frac {R}{s}$ will be field which has precisely two prime ideal . One is {$0$} and another one is $frac {R}{s}$ .
Can anyone correct me If I have gone wrong anywhere?
abstract-algebra ring-theory principal-ideal-domains
abstract-algebra ring-theory principal-ideal-domains
asked Dec 14 '18 at 7:01
cmicmi
1,121312
asked Dec 14 '18 at 7:01
cmicmi
1,121312
asked Dec 14 '18 at 7:01
cmicmi
1,121312
asked Dec 14 '18 at 7:01
cmicmi
1,121312
asked Dec 14 '18 at 7:01
cmicmi
1,121312
1,121312
add a comment |
add a comment |
1 Answer
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$begingroup$
The whole ring is not a prime ideal, by convention. It's the same way that $1$ is not a prime number. So the only prime ideal in a field is ${0}$.
Apart from that, the proof looks good (assuming you are allowed to use the theorem that nonzero prime ideals are maximal in PIDs).
edited Dec 14 '18 at 15:53
Bill Dubuque
212k29195649
answered Dec 14 '18 at 7:11
ArthurArthur
116k7116199
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
The whole ring is not a prime ideal, by convention. It's the same way that $1$ is not a prime number. So the only prime ideal in a field is ${0}$.
Apart from that, the proof looks good (assuming you are allowed to use the theorem that nonzero prime ideals are maximal in PIDs).
edited Dec 14 '18 at 15:53
Bill Dubuque
212k29195649
answered Dec 14 '18 at 7:11
ArthurArthur
116k7116199
$endgroup$
add a comment |
$begingroup$
The whole ring is not a prime ideal, by convention. It's the same way that $1$ is not a prime number. So the only prime ideal in a field is ${0}$.
Apart from that, the proof looks good (assuming you are allowed to use the theorem that nonzero prime ideals are maximal in PIDs).
edited Dec 14 '18 at 15:53
Bill Dubuque
212k29195649
answered Dec 14 '18 at 7:11
ArthurArthur
116k7116199
$endgroup$
add a comment |
$begingroup$
The whole ring is not a prime ideal, by convention. It's the same way that $1$ is not a prime number. So the only prime ideal in a field is ${0}$.
Apart from that, the proof looks good (assuming you are allowed to use the theorem that nonzero prime ideals are maximal in PIDs).
edited Dec 14 '18 at 15:53
Bill Dubuque
212k29195649
answered Dec 14 '18 at 7:11
ArthurArthur
116k7116199
$endgroup$
The whole ring is not a prime ideal, by convention. It's the same way that $1$ is not a prime number. So the only prime ideal in a field is ${0}$.
Apart from that, the proof looks good (assuming you are allowed to use the theorem that nonzero prime ideals are maximal in PIDs).
edited Dec 14 '18 at 15:53
Bill Dubuque
212k29195649
answered Dec 14 '18 at 7:11
ArthurArthur
116k7116199
edited Dec 14 '18 at 15:53
Bill Dubuque
212k29195649
edited Dec 14 '18 at 15:53
Bill Dubuque
212k29195649
edited Dec 14 '18 at 15:53
Bill Dubuque
212k29195649
212k29195649
answered Dec 14 '18 at 7:11
ArthurArthur
116k7116199
answered Dec 14 '18 at 7:11
ArthurArthur
116k7116199
answered Dec 14 '18 at 7:11
ArthurArthur
116k7116199
116k7116199
add a comment |
add a comment |
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