Show that $(x)$ is a prime ideal in $mathbb{Z}[x]$ (the polynomials with integer coefs) which is not maximal....
$begingroup$
This question already has an answer here:
ideal behind Maximal ideals are Prime ideals
1 answer
I showed that if $(x)$ is prime in $mathbb{Z}[x]$, then since
$mathbb{Z}[x] /(x)$ is isomorphic with $mathbb{Z}$ and $mathbb{Z}$ is not a field, so $(x)$ is not maximal. But when I want to show that $(x)$ is prime in $mathbb{Z}[x]$ I've got stuck... How could I show?
The old proofs doesn't give any isomorphic function! my question is a particular case, not a theorem (just for more info)
abstract-algebra ring-theory
edited Jan 2 at 12:17
stressed out
6,3501839
asked Jan 2 at 0:32
Arman_jrArman_jr
235
$endgroup$
marked as duplicate by Bill Dubuque ring-theory
Users with the ring-theory badge can single-handedly close ring-theory questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 2 at 1:00
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
ideal behind Maximal ideals are Prime ideals
1 answer
I showed that if $(x)$ is prime in $mathbb{Z}[x]$, then since
$mathbb{Z}[x] /(x)$ is isomorphic with $mathbb{Z}$ and $mathbb{Z}$ is not a field, so $(x)$ is not maximal. But when I want to show that $(x)$ is prime in $mathbb{Z}[x]$ I've got stuck... How could I show?
The old proofs doesn't give any isomorphic function! my question is a particular case, not a theorem (just for more info)
abstract-algebra ring-theory
edited Jan 2 at 12:17
stressed out
6,3501839
asked Jan 2 at 0:32
Arman_jrArman_jr
235
$endgroup$
marked as duplicate by Bill Dubuque ring-theory
Users with the ring-theory badge can single-handedly close ring-theory questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 2 at 1:00
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Please don't change your question after answers have been posted. Ask a new question instead if need be
$endgroup$
– Bill Dubuque
Jan 2 at 0:59
$begingroup$
You can find proofs in many prior answers, e.g. here and here
$endgroup$
– Bill Dubuque
Jan 2 at 1:15
$begingroup$
@bill Dubuque edited
$endgroup$
– Arman_jr
Jan 3 at 13:32
add a comment |
$begingroup$
This question already has an answer here:
ideal behind Maximal ideals are Prime ideals
1 answer
I showed that if $(x)$ is prime in $mathbb{Z}[x]$, then since
$mathbb{Z}[x] /(x)$ is isomorphic with $mathbb{Z}$ and $mathbb{Z}$ is not a field, so $(x)$ is not maximal. But when I want to show that $(x)$ is prime in $mathbb{Z}[x]$ I've got stuck... How could I show?
The old proofs doesn't give any isomorphic function! my question is a particular case, not a theorem (just for more info)
abstract-algebra ring-theory
edited Jan 2 at 12:17
stressed out
6,3501839
asked Jan 2 at 0:32
Arman_jrArman_jr
235
$endgroup$
This question already has an answer here:
ideal behind Maximal ideals are Prime ideals
1 answer
I showed that if $(x)$ is prime in $mathbb{Z}[x]$, then since
$mathbb{Z}[x] /(x)$ is isomorphic with $mathbb{Z}$ and $mathbb{Z}$ is not a field, so $(x)$ is not maximal. But when I want to show that $(x)$ is prime in $mathbb{Z}[x]$ I've got stuck... How could I show?
The old proofs doesn't give any isomorphic function! my question is a particular case, not a theorem (just for more info)
This question already has an answer here:
ideal behind Maximal ideals are Prime ideals
1 answer
abstract-algebra ring-theory
abstract-algebra ring-theory
edited Jan 2 at 12:17
stressed out
6,3501839
asked Jan 2 at 0:32
Arman_jrArman_jr
235
edited Jan 2 at 12:17
stressed out
6,3501839
asked Jan 2 at 0:32
Arman_jrArman_jr
235
edited Jan 2 at 12:17
stressed out
6,3501839
edited Jan 2 at 12:17
stressed out
6,3501839
edited Jan 2 at 12:17
stressed out
6,3501839
6,3501839
asked Jan 2 at 0:32
Arman_jrArman_jr
235
asked Jan 2 at 0:32
Arman_jrArman_jr
235
asked Jan 2 at 0:32
Arman_jrArman_jr
235
235
marked as duplicate by Bill Dubuque ring-theory
Users with the ring-theory badge can single-handedly close ring-theory questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 2 at 1:00
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Bill Dubuque ring-theory
Users with the ring-theory badge can single-handedly close ring-theory questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 2 at 1:00
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Please don't change your question after answers have been posted. Ask a new question instead if need be
$endgroup$
– Bill Dubuque
Jan 2 at 0:59
$begingroup$
You can find proofs in many prior answers, e.g. here and here
$endgroup$
– Bill Dubuque
Jan 2 at 1:15
$begingroup$
@bill Dubuque edited
$endgroup$
– Arman_jr
Jan 3 at 13:32
add a comment |
$begingroup$
Please don't change your question after answers have been posted. Ask a new question instead if need be
$endgroup$
– Bill Dubuque
Jan 2 at 0:59
$begingroup$
You can find proofs in many prior answers, e.g. here and here
$endgroup$
– Bill Dubuque
Jan 2 at 1:15
$begingroup$
@bill Dubuque edited
$endgroup$
– Arman_jr
Jan 3 at 13:32
$begingroup$
Please don't change your question after answers have been posted. Ask a new question instead if need be
$endgroup$
– Bill Dubuque
Jan 2 at 0:59
$begingroup$
Please don't change your question after answers have been posted. Ask a new question instead if need be
$endgroup$
– Bill Dubuque
Jan 2 at 0:59
$begingroup$
You can find proofs in many prior answers, e.g. here and here
$endgroup$
– Bill Dubuque
Jan 2 at 1:15
$begingroup$
You can find proofs in many prior answers, e.g. here and here
$endgroup$
– Bill Dubuque
Jan 2 at 1:15
$begingroup$
@bill Dubuque edited
$endgroup$
– Arman_jr
Jan 3 at 13:32
$begingroup$
@bill Dubuque edited
$endgroup$
– Arman_jr
Jan 3 at 13:32
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$mathbb{Z}$ is a domain. And $A/I$ is a domain if and only if $I$ is a prime ideal.
answered Jan 2 at 0:40
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$mathbb{Z}$ is a domain. And $A/I$ is a domain if and only if $I$ is a prime ideal.
answered Jan 2 at 0:40
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
$endgroup$
add a comment |
$begingroup$
$mathbb{Z}$ is a domain. And $A/I$ is a domain if and only if $I$ is a prime ideal.
answered Jan 2 at 0:40
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
$endgroup$
add a comment |
$begingroup$
$mathbb{Z}$ is a domain. And $A/I$ is a domain if and only if $I$ is a prime ideal.
answered Jan 2 at 0:40
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
$endgroup$
$mathbb{Z}$ is a domain. And $A/I$ is a domain if and only if $I$ is a prime ideal.
answered Jan 2 at 0:40
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
answered Jan 2 at 0:40
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
answered Jan 2 at 0:40
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
answered Jan 2 at 0:40
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
802110
add a comment |
add a comment |
$begingroup$
Please don't change your question after answers have been posted. Ask a new question instead if need be
$endgroup$
– Bill Dubuque
Jan 2 at 0:59
$begingroup$
You can find proofs in many prior answers, e.g. here and here
$endgroup$
– Bill Dubuque
Jan 2 at 1:15
$begingroup$
@bill Dubuque edited
$endgroup$
– Arman_jr
Jan 3 at 13:32