If $dmid e$, does the ideal $(x^d-1, 1+x+cdots+x^{e-1})subsetmathbb{Z}[x]$ contain $1+x+cdots x^{d-1}$?
up vote
2
down vote
favorite
If $dmid e$, does the ideal $(x^d-1, 1+x+cdots+x^{e-1})subsetmathbb{Z}[x]$ contain $1+x+cdots+x^{d-1}$?
For an integer $n$, let $chi_n(X) := 1+X+cdots+X^{n-1}$.
Certainly $chi_d(x)$ divides both $x^d-1$ and $chi_e(x)$. Moreover, the quotient $chi_e(x)/chi_d(x)$ is coprime to $(x^d-1)/chi_d(x)$.
Since the polynomials involved are all primitive, I feel like some form of Gauss's lemma should imply the result, but I'm not sure.
abstract-algebra polynomials
edited Nov 18 at 23:32
user26857
39.1k123882
asked Nov 15 at 23:52
stupid_question_bot
1,888415
add a comment |
up vote
2
down vote
favorite
If $dmid e$, does the ideal $(x^d-1, 1+x+cdots+x^{e-1})subsetmathbb{Z}[x]$ contain $1+x+cdots+x^{d-1}$?
For an integer $n$, let $chi_n(X) := 1+X+cdots+X^{n-1}$.
Certainly $chi_d(x)$ divides both $x^d-1$ and $chi_e(x)$. Moreover, the quotient $chi_e(x)/chi_d(x)$ is coprime to $(x^d-1)/chi_d(x)$.
Since the polynomials involved are all primitive, I feel like some form of Gauss's lemma should imply the result, but I'm not sure.
abstract-algebra polynomials
edited Nov 18 at 23:32
user26857
39.1k123882
asked Nov 15 at 23:52
stupid_question_bot
1,888415
1
For example if $d=2,e=4$, then the ideal $langle x^2-1, x^3+x^2+x+1 rangle$ is contained in the kernel of the morphism $mathbb{Z}[x] to mathbb{Z} / 4mathbb{Z}, x mapsto 1$ whereas $x+1$ is not in this kernel.
– Daniel Schepler
Nov 16 at 0:08
Generalizing the above. Just note that $chi_e=chi_{e-d}(X^d-1)+chi_{e-d}+chi_d$. Now, let $e=qd$, and one can show, through induction, $chi_e=(chi_{(q-1)d}+cdots+chi_{d})(X^d-1)+qchi_d$. So one sees that in general $(chi_{qd},X^d-1)=(qchi_d,X^d-1)=(chi_d)(q,X+1)$, which can only equal $(chi_d)$ if $q=1$.
– Josué Tonelli-Cueto
Nov 16 at 0:43
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
If $dmid e$, does the ideal $(x^d-1, 1+x+cdots+x^{e-1})subsetmathbb{Z}[x]$ contain $1+x+cdots+x^{d-1}$?
For an integer $n$, let $chi_n(X) := 1+X+cdots+X^{n-1}$.
Certainly $chi_d(x)$ divides both $x^d-1$ and $chi_e(x)$. Moreover, the quotient $chi_e(x)/chi_d(x)$ is coprime to $(x^d-1)/chi_d(x)$.
Since the polynomials involved are all primitive, I feel like some form of Gauss's lemma should imply the result, but I'm not sure.
abstract-algebra polynomials
edited Nov 18 at 23:32
user26857
39.1k123882
asked Nov 15 at 23:52
stupid_question_bot
1,888415
If $dmid e$, does the ideal $(x^d-1, 1+x+cdots+x^{e-1})subsetmathbb{Z}[x]$ contain $1+x+cdots+x^{d-1}$?
For an integer $n$, let $chi_n(X) := 1+X+cdots+X^{n-1}$.
Certainly $chi_d(x)$ divides both $x^d-1$ and $chi_e(x)$. Moreover, the quotient $chi_e(x)/chi_d(x)$ is coprime to $(x^d-1)/chi_d(x)$.
Since the polynomials involved are all primitive, I feel like some form of Gauss's lemma should imply the result, but I'm not sure.
abstract-algebra polynomials
abstract-algebra polynomials
edited Nov 18 at 23:32
user26857
39.1k123882
asked Nov 15 at 23:52
stupid_question_bot
1,888415
edited Nov 18 at 23:32
user26857
39.1k123882
asked Nov 15 at 23:52
stupid_question_bot
1,888415
edited Nov 18 at 23:32
user26857
39.1k123882
edited Nov 18 at 23:32
user26857
39.1k123882
edited Nov 18 at 23:32
user26857
39.1k123882
39.1k123882
asked Nov 15 at 23:52
stupid_question_bot
1,888415
asked Nov 15 at 23:52
stupid_question_bot
1,888415
asked Nov 15 at 23:52
stupid_question_bot
1,888415
1,888415
1
For example if $d=2,e=4$, then the ideal $langle x^2-1, x^3+x^2+x+1 rangle$ is contained in the kernel of the morphism $mathbb{Z}[x] to mathbb{Z} / 4mathbb{Z}, x mapsto 1$ whereas $x+1$ is not in this kernel.
– Daniel Schepler
Nov 16 at 0:08
Generalizing the above. Just note that $chi_e=chi_{e-d}(X^d-1)+chi_{e-d}+chi_d$. Now, let $e=qd$, and one can show, through induction, $chi_e=(chi_{(q-1)d}+cdots+chi_{d})(X^d-1)+qchi_d$. So one sees that in general $(chi_{qd},X^d-1)=(qchi_d,X^d-1)=(chi_d)(q,X+1)$, which can only equal $(chi_d)$ if $q=1$.
– Josué Tonelli-Cueto
Nov 16 at 0:43
add a comment |
1
For example if $d=2,e=4$, then the ideal $langle x^2-1, x^3+x^2+x+1 rangle$ is contained in the kernel of the morphism $mathbb{Z}[x] to mathbb{Z} / 4mathbb{Z}, x mapsto 1$ whereas $x+1$ is not in this kernel.
– Daniel Schepler
Nov 16 at 0:08
Generalizing the above. Just note that $chi_e=chi_{e-d}(X^d-1)+chi_{e-d}+chi_d$. Now, let $e=qd$, and one can show, through induction, $chi_e=(chi_{(q-1)d}+cdots+chi_{d})(X^d-1)+qchi_d$. So one sees that in general $(chi_{qd},X^d-1)=(qchi_d,X^d-1)=(chi_d)(q,X+1)$, which can only equal $(chi_d)$ if $q=1$.
– Josué Tonelli-Cueto
Nov 16 at 0:43
1
1
For example if $d=2,e=4$, then the ideal $langle x^2-1, x^3+x^2+x+1 rangle$ is contained in the kernel of the morphism $mathbb{Z}[x] to mathbb{Z} / 4mathbb{Z}, x mapsto 1$ whereas $x+1$ is not in this kernel.
– Daniel Schepler
Nov 16 at 0:08
For example if $d=2,e=4$, then the ideal $langle x^2-1, x^3+x^2+x+1 rangle$ is contained in the kernel of the morphism $mathbb{Z}[x] to mathbb{Z} / 4mathbb{Z}, x mapsto 1$ whereas $x+1$ is not in this kernel.
– Daniel Schepler
Nov 16 at 0:08
Generalizing the above. Just note that $chi_e=chi_{e-d}(X^d-1)+chi_{e-d}+chi_d$. Now, let $e=qd$, and one can show, through induction, $chi_e=(chi_{(q-1)d}+cdots+chi_{d})(X^d-1)+qchi_d$. So one sees that in general $(chi_{qd},X^d-1)=(qchi_d,X^d-1)=(chi_d)(q,X+1)$, which can only equal $(chi_d)$ if $q=1$.
– Josué Tonelli-Cueto
Nov 16 at 0:43
Generalizing the above. Just note that $chi_e=chi_{e-d}(X^d-1)+chi_{e-d}+chi_d$. Now, let $e=qd$, and one can show, through induction, $chi_e=(chi_{(q-1)d}+cdots+chi_{d})(X^d-1)+qchi_d$. So one sees that in general $(chi_{qd},X^d-1)=(qchi_d,X^d-1)=(chi_d)(q,X+1)$, which can only equal $(chi_d)$ if $q=1$.
– Josué Tonelli-Cueto
Nov 16 at 0:43
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
As a simple counterexample, let $d=1$ and $e=2$.
Note that when $d=1$, the expression
$$1+x+cdots+x^{d-1}$$
is presumably intended to be equal to $1$, not $2$.
Then in $mathbb{Z}[x]$, the ideal $(x-1,x+1)$ does not contain $1$.
Suppose instead that
$$a(x)(x-1)+b(x)(x+1)=1$$
for some $a,bin mathbb{Z}[x]$.
Now plug in $x=1$, and note that the $text{LHS}$ is divisible by $2$, while the $text{RHS}$ is equal to $1$.
$qquadoverline{
qquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquad
}
$
More generally, let $d,e$ be any positive integers such that $d < e$.
If we suppose that in $mathbb{Z}[x]$, the polynomial
$$1+x + cdots + x^{d-1}$$
is an element of the ideal
$$(x^d-1,1+x + cdots + x^{e-1})$$
we would have
$$a(x)(x^d-1)+b(x)(1+x + cdots + x^{e-1})=1+x + cdots + x^{d-1}$$
for some $a,bin mathbb{Z}[x]$.
But then if we plug in $x=1$, the $text{LHS}$ is divisible by $e$, while the $text{RHS}$ is equal to $d$.
answered Nov 16 at 0:08
quasi
35.9k22562
2
4 doesn't divide 2.
– Seewoo Lee
Nov 16 at 0:21
I haven't been following closely, but it seems you have changed earlier incorrect answers into the answer given long ago in a comment. At the least you should acknowledge DS's prior comment
– Bill Dubuque
Nov 16 at 0:35
Actually, I didn't look at the comments to the OP.
– quasi
Nov 16 at 0:37
Fixed! In my original comments I miscomputed $chi_1$ as $0$. That was my mistake. So $+1$.
– Josué Tonelli-Cueto
Nov 16 at 0:45
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
As a simple counterexample, let $d=1$ and $e=2$.
Note that when $d=1$, the expression
$$1+x+cdots+x^{d-1}$$
is presumably intended to be equal to $1$, not $2$.
Then in $mathbb{Z}[x]$, the ideal $(x-1,x+1)$ does not contain $1$.
Suppose instead that
$$a(x)(x-1)+b(x)(x+1)=1$$
for some $a,bin mathbb{Z}[x]$.
Now plug in $x=1$, and note that the $text{LHS}$ is divisible by $2$, while the $text{RHS}$ is equal to $1$.
$qquadoverline{
qquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquad
}
$
More generally, let $d,e$ be any positive integers such that $d < e$.
If we suppose that in $mathbb{Z}[x]$, the polynomial
$$1+x + cdots + x^{d-1}$$
is an element of the ideal
$$(x^d-1,1+x + cdots + x^{e-1})$$
we would have
$$a(x)(x^d-1)+b(x)(1+x + cdots + x^{e-1})=1+x + cdots + x^{d-1}$$
for some $a,bin mathbb{Z}[x]$.
But then if we plug in $x=1$, the $text{LHS}$ is divisible by $e$, while the $text{RHS}$ is equal to $d$.
answered Nov 16 at 0:08
quasi
35.9k22562
2
4 doesn't divide 2.
– Seewoo Lee
Nov 16 at 0:21
I haven't been following closely, but it seems you have changed earlier incorrect answers into the answer given long ago in a comment. At the least you should acknowledge DS's prior comment
– Bill Dubuque
Nov 16 at 0:35
Actually, I didn't look at the comments to the OP.
– quasi
Nov 16 at 0:37
Fixed! In my original comments I miscomputed $chi_1$ as $0$. That was my mistake. So $+1$.
– Josué Tonelli-Cueto
Nov 16 at 0:45
add a comment |
up vote
2
down vote
accepted
As a simple counterexample, let $d=1$ and $e=2$.
Note that when $d=1$, the expression
$$1+x+cdots+x^{d-1}$$
is presumably intended to be equal to $1$, not $2$.
Then in $mathbb{Z}[x]$, the ideal $(x-1,x+1)$ does not contain $1$.
Suppose instead that
$$a(x)(x-1)+b(x)(x+1)=1$$
for some $a,bin mathbb{Z}[x]$.
Now plug in $x=1$, and note that the $text{LHS}$ is divisible by $2$, while the $text{RHS}$ is equal to $1$.
$qquadoverline{
qquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquad
}
$
More generally, let $d,e$ be any positive integers such that $d < e$.
If we suppose that in $mathbb{Z}[x]$, the polynomial
$$1+x + cdots + x^{d-1}$$
is an element of the ideal
$$(x^d-1,1+x + cdots + x^{e-1})$$
we would have
$$a(x)(x^d-1)+b(x)(1+x + cdots + x^{e-1})=1+x + cdots + x^{d-1}$$
for some $a,bin mathbb{Z}[x]$.
But then if we plug in $x=1$, the $text{LHS}$ is divisible by $e$, while the $text{RHS}$ is equal to $d$.
answered Nov 16 at 0:08
quasi
35.9k22562
2
4 doesn't divide 2.
– Seewoo Lee
Nov 16 at 0:21
I haven't been following closely, but it seems you have changed earlier incorrect answers into the answer given long ago in a comment. At the least you should acknowledge DS's prior comment
– Bill Dubuque
Nov 16 at 0:35
Actually, I didn't look at the comments to the OP.
– quasi
Nov 16 at 0:37
Fixed! In my original comments I miscomputed $chi_1$ as $0$. That was my mistake. So $+1$.
– Josué Tonelli-Cueto
Nov 16 at 0:45
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
As a simple counterexample, let $d=1$ and $e=2$.
Note that when $d=1$, the expression
$$1+x+cdots+x^{d-1}$$
is presumably intended to be equal to $1$, not $2$.
Then in $mathbb{Z}[x]$, the ideal $(x-1,x+1)$ does not contain $1$.
Suppose instead that
$$a(x)(x-1)+b(x)(x+1)=1$$
for some $a,bin mathbb{Z}[x]$.
Now plug in $x=1$, and note that the $text{LHS}$ is divisible by $2$, while the $text{RHS}$ is equal to $1$.
$qquadoverline{
qquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquad
}
$
More generally, let $d,e$ be any positive integers such that $d < e$.
If we suppose that in $mathbb{Z}[x]$, the polynomial
$$1+x + cdots + x^{d-1}$$
is an element of the ideal
$$(x^d-1,1+x + cdots + x^{e-1})$$
we would have
$$a(x)(x^d-1)+b(x)(1+x + cdots + x^{e-1})=1+x + cdots + x^{d-1}$$
for some $a,bin mathbb{Z}[x]$.
But then if we plug in $x=1$, the $text{LHS}$ is divisible by $e$, while the $text{RHS}$ is equal to $d$.
answered Nov 16 at 0:08
quasi
35.9k22562
As a simple counterexample, let $d=1$ and $e=2$.
Note that when $d=1$, the expression
$$1+x+cdots+x^{d-1}$$
is presumably intended to be equal to $1$, not $2$.
Then in $mathbb{Z}[x]$, the ideal $(x-1,x+1)$ does not contain $1$.
Suppose instead that
$$a(x)(x-1)+b(x)(x+1)=1$$
for some $a,bin mathbb{Z}[x]$.
Now plug in $x=1$, and note that the $text{LHS}$ is divisible by $2$, while the $text{RHS}$ is equal to $1$.
$qquadoverline{
qquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquad
}
$
More generally, let $d,e$ be any positive integers such that $d < e$.
If we suppose that in $mathbb{Z}[x]$, the polynomial
$$1+x + cdots + x^{d-1}$$
is an element of the ideal
$$(x^d-1,1+x + cdots + x^{e-1})$$
we would have
$$a(x)(x^d-1)+b(x)(1+x + cdots + x^{e-1})=1+x + cdots + x^{d-1}$$
for some $a,bin mathbb{Z}[x]$.
But then if we plug in $x=1$, the $text{LHS}$ is divisible by $e$, while the $text{RHS}$ is equal to $d$.
answered Nov 16 at 0:08
quasi
35.9k22562
edited Nov 16 at 3:42
answered Nov 16 at 0:08
quasi
35.9k22562
answered Nov 16 at 0:08
quasi
35.9k22562
answered Nov 16 at 0:08
quasi
35.9k22562
35.9k22562
2
4 doesn't divide 2.
– Seewoo Lee
Nov 16 at 0:21
I haven't been following closely, but it seems you have changed earlier incorrect answers into the answer given long ago in a comment. At the least you should acknowledge DS's prior comment
– Bill Dubuque
Nov 16 at 0:35
Actually, I didn't look at the comments to the OP.
– quasi
Nov 16 at 0:37
Fixed! In my original comments I miscomputed $chi_1$ as $0$. That was my mistake. So $+1$.
– Josué Tonelli-Cueto
Nov 16 at 0:45
add a comment |
2
4 doesn't divide 2.
– Seewoo Lee
Nov 16 at 0:21
I haven't been following closely, but it seems you have changed earlier incorrect answers into the answer given long ago in a comment. At the least you should acknowledge DS's prior comment
– Bill Dubuque
Nov 16 at 0:35
Actually, I didn't look at the comments to the OP.
– quasi
Nov 16 at 0:37
Fixed! In my original comments I miscomputed $chi_1$ as $0$. That was my mistake. So $+1$.
– Josué Tonelli-Cueto
Nov 16 at 0:45
2
2
4 doesn't divide 2.
– Seewoo Lee
Nov 16 at 0:21
4 doesn't divide 2.
– Seewoo Lee
Nov 16 at 0:21
I haven't been following closely, but it seems you have changed earlier incorrect answers into the answer given long ago in a comment. At the least you should acknowledge DS's prior comment
– Bill Dubuque
Nov 16 at 0:35
I haven't been following closely, but it seems you have changed earlier incorrect answers into the answer given long ago in a comment. At the least you should acknowledge DS's prior comment
– Bill Dubuque
Nov 16 at 0:35
Actually, I didn't look at the comments to the OP.
– quasi
Nov 16 at 0:37
Actually, I didn't look at the comments to the OP.
– quasi
Nov 16 at 0:37
Fixed! In my original comments I miscomputed $chi_1$ as $0$. That was my mistake. So $+1$.
– Josué Tonelli-Cueto
Nov 16 at 0:45
Fixed! In my original comments I miscomputed $chi_1$ as $0$. That was my mistake. So $+1$.
– Josué Tonelli-Cueto
Nov 16 at 0:45
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3000512%2fif-d-mid-e-does-the-ideal-xd-1-1x-cdotsxe-1-subset-mathbbzx-c%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
For example if $d=2,e=4$, then the ideal $langle x^2-1, x^3+x^2+x+1 rangle$ is contained in the kernel of the morphism $mathbb{Z}[x] to mathbb{Z} / 4mathbb{Z}, x mapsto 1$ whereas $x+1$ is not in this kernel.
– Daniel Schepler
Nov 16 at 0:08
Generalizing the above. Just note that $chi_e=chi_{e-d}(X^d-1)+chi_{e-d}+chi_d$. Now, let $e=qd$, and one can show, through induction, $chi_e=(chi_{(q-1)d}+cdots+chi_{d})(X^d-1)+qchi_d$. So one sees that in general $(chi_{qd},X^d-1)=(qchi_d,X^d-1)=(chi_d)(q,X+1)$, which can only equal $(chi_d)$ if $q=1$.
– Josué Tonelli-Cueto
Nov 16 at 0:43