Is the ideal generated by ${4,x}$ a principal ideal in $Z[x]$? [duplicate]
$begingroup$
This question already has an answer here:
Show that $langle 2,x rangle$ is not a principal ideal in $mathbb Z [x]$
7 answers
I've : $I=<p,x>$ is not a principal ideal in $Z[x]$ where p is prime.
My question : Is $I=<p,x>$ a principal ideal in $Z[x]$ where p is not a prime?
More particularly, is the ideal generated by ${4,x}$ a principal ideal in $Z[x]$ ?
abstract-algebra ring-theory ideals principal-ideal-domains
asked Jan 12 at 15:46
BijanDattaBijanDatta
307113
$endgroup$
marked as duplicate by Saad, Dietrich Burde, user26857 abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra 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 13 at 20:06
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:
Show that $langle 2,x rangle$ is not a principal ideal in $mathbb Z [x]$
7 answers
I've : $I=<p,x>$ is not a principal ideal in $Z[x]$ where p is prime.
My question : Is $I=<p,x>$ a principal ideal in $Z[x]$ where p is not a prime?
More particularly, is the ideal generated by ${4,x}$ a principal ideal in $Z[x]$ ?
abstract-algebra ring-theory ideals principal-ideal-domains
asked Jan 12 at 15:46
BijanDattaBijanDatta
307113
$endgroup$
marked as duplicate by Saad, Dietrich Burde, user26857 abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra 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 13 at 20:06
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.
$begingroup$
If you have already that $I=(p,x)$ is not maximal, then where did you use there that $p$ is prime? Can you replace $2$ by $4$ in the standard proof?
$endgroup$
– Dietrich Burde
Jan 12 at 16:37
$begingroup$
You may want to check this answer of mine.
$endgroup$
– Andreas Caranti
Jan 12 at 17:38
add a comment |
$begingroup$
This question already has an answer here:
Show that $langle 2,x rangle$ is not a principal ideal in $mathbb Z [x]$
7 answers
I've : $I=<p,x>$ is not a principal ideal in $Z[x]$ where p is prime.
My question : Is $I=<p,x>$ a principal ideal in $Z[x]$ where p is not a prime?
More particularly, is the ideal generated by ${4,x}$ a principal ideal in $Z[x]$ ?
abstract-algebra ring-theory ideals principal-ideal-domains
asked Jan 12 at 15:46
BijanDattaBijanDatta
307113
$endgroup$
This question already has an answer here:
Show that $langle 2,x rangle$ is not a principal ideal in $mathbb Z [x]$
7 answers
I've : $I=<p,x>$ is not a principal ideal in $Z[x]$ where p is prime.
My question : Is $I=<p,x>$ a principal ideal in $Z[x]$ where p is not a prime?
More particularly, is the ideal generated by ${4,x}$ a principal ideal in $Z[x]$ ?
This question already has an answer here:
Show that $langle 2,x rangle$ is not a principal ideal in $mathbb Z [x]$
7 answers
abstract-algebra ring-theory ideals principal-ideal-domains
abstract-algebra ring-theory ideals principal-ideal-domains
asked Jan 12 at 15:46
BijanDattaBijanDatta
307113
asked Jan 12 at 15:46
BijanDattaBijanDatta
307113
asked Jan 12 at 15:46
BijanDattaBijanDatta
307113
asked Jan 12 at 15:46
BijanDattaBijanDatta
307113
asked Jan 12 at 15:46
BijanDattaBijanDatta
307113
307113
marked as duplicate by Saad, Dietrich Burde, user26857 abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra 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 13 at 20:06
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 Saad, Dietrich Burde, user26857 abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra 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 13 at 20:06
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.
$begingroup$
If you have already that $I=(p,x)$ is not maximal, then where did you use there that $p$ is prime? Can you replace $2$ by $4$ in the standard proof?
$endgroup$
– Dietrich Burde
Jan 12 at 16:37
$begingroup$
You may want to check this answer of mine.
$endgroup$
– Andreas Caranti
Jan 12 at 17:38
add a comment |
$begingroup$
If you have already that $I=(p,x)$ is not maximal, then where did you use there that $p$ is prime? Can you replace $2$ by $4$ in the standard proof?
$endgroup$
– Dietrich Burde
Jan 12 at 16:37
$begingroup$
You may want to check this answer of mine.
$endgroup$
– Andreas Caranti
Jan 12 at 17:38
$begingroup$
If you have already that $I=(p,x)$ is not maximal, then where did you use there that $p$ is prime? Can you replace $2$ by $4$ in the standard proof?
$endgroup$
– Dietrich Burde
Jan 12 at 16:37
$begingroup$
If you have already that $I=(p,x)$ is not maximal, then where did you use there that $p$ is prime? Can you replace $2$ by $4$ in the standard proof?
$endgroup$
– Dietrich Burde
Jan 12 at 16:37
$begingroup$
You may want to check this answer of mine.
$endgroup$
– Andreas Caranti
Jan 12 at 17:38
$begingroup$
You may want to check this answer of mine.
$endgroup$
– Andreas Caranti
Jan 12 at 17:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $langle4,xrangle$ was a principal ideal, then you would have $langle4,xrangle=langle p(x)rangle$, for some $p(x)inmathbb{Z}[x]$. Is this true? What can you tell about a polynomial $p(x)inmathbb{Z}[x]$ if you know that $p(x)mid4$ and that $p(x)mid x$?
answered Jan 12 at 15:50
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
$begingroup$
Ok. I've got this. Thanks.
$endgroup$
– BijanDatta
Jan 29 at 10:08
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $langle4,xrangle$ was a principal ideal, then you would have $langle4,xrangle=langle p(x)rangle$, for some $p(x)inmathbb{Z}[x]$. Is this true? What can you tell about a polynomial $p(x)inmathbb{Z}[x]$ if you know that $p(x)mid4$ and that $p(x)mid x$?
answered Jan 12 at 15:50
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
$begingroup$
Ok. I've got this. Thanks.
$endgroup$
– BijanDatta
Jan 29 at 10:08
add a comment |
$begingroup$
If $langle4,xrangle$ was a principal ideal, then you would have $langle4,xrangle=langle p(x)rangle$, for some $p(x)inmathbb{Z}[x]$. Is this true? What can you tell about a polynomial $p(x)inmathbb{Z}[x]$ if you know that $p(x)mid4$ and that $p(x)mid x$?
answered Jan 12 at 15:50
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
$begingroup$
Ok. I've got this. Thanks.
$endgroup$
– BijanDatta
Jan 29 at 10:08
add a comment |
$begingroup$
If $langle4,xrangle$ was a principal ideal, then you would have $langle4,xrangle=langle p(x)rangle$, for some $p(x)inmathbb{Z}[x]$. Is this true? What can you tell about a polynomial $p(x)inmathbb{Z}[x]$ if you know that $p(x)mid4$ and that $p(x)mid x$?
answered Jan 12 at 15:50
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
If $langle4,xrangle$ was a principal ideal, then you would have $langle4,xrangle=langle p(x)rangle$, for some $p(x)inmathbb{Z}[x]$. Is this true? What can you tell about a polynomial $p(x)inmathbb{Z}[x]$ if you know that $p(x)mid4$ and that $p(x)mid x$?
answered Jan 12 at 15:50
José Carlos SantosJosé Carlos Santos
167k22132235
answered Jan 12 at 15:50
José Carlos SantosJosé Carlos Santos
167k22132235
answered Jan 12 at 15:50
José Carlos SantosJosé Carlos Santos
167k22132235
answered Jan 12 at 15:50
José Carlos SantosJosé Carlos Santos
167k22132235
167k22132235
$begingroup$
Ok. I've got this. Thanks.
$endgroup$
– BijanDatta
Jan 29 at 10:08
add a comment |
$begingroup$
Ok. I've got this. Thanks.
$endgroup$
– BijanDatta
Jan 29 at 10:08
$begingroup$
Ok. I've got this. Thanks.
$endgroup$
– BijanDatta
Jan 29 at 10:08
$begingroup$
Ok. I've got this. Thanks.
$endgroup$
– BijanDatta
Jan 29 at 10:08
add a comment |
$begingroup$
If you have already that $I=(p,x)$ is not maximal, then where did you use there that $p$ is prime? Can you replace $2$ by $4$ in the standard proof?
$endgroup$
– Dietrich Burde
Jan 12 at 16:37
$begingroup$
You may want to check this answer of mine.
$endgroup$
– Andreas Caranti
Jan 12 at 17:38