Mixing
How can I show that the ideal $(2x, x^2+1)$ does not generate all of $mathbb Z[x]$?
$begingroup$
How can I show that the ideal $(2x, x^2+1)$ does not generate all of $mathbb Z[x]$?
Not sure of what to try first.
abstract-algebra polynomials ring-theory
asked Jan 31 at 23:27
Al JebrAl Jebr
4,42143478
$endgroup$
add a comment |
$begingroup$
How can I show that the ideal $(2x, x^2+1)$ does not generate all of $mathbb Z[x]$?
Not sure of what to try first.
abstract-algebra polynomials ring-theory
asked Jan 31 at 23:27
Al JebrAl Jebr
4,42143478
$endgroup$
add a comment |
$begingroup$
How can I show that the ideal $(2x, x^2+1)$ does not generate all of $mathbb Z[x]$?
Not sure of what to try first.
abstract-algebra polynomials ring-theory
asked Jan 31 at 23:27
Al JebrAl Jebr
4,42143478
$endgroup$
How can I show that the ideal $(2x, x^2+1)$ does not generate all of $mathbb Z[x]$?
Not sure of what to try first.
abstract-algebra polynomials ring-theory
abstract-algebra polynomials ring-theory
asked Jan 31 at 23:27
Al JebrAl Jebr
4,42143478
asked Jan 31 at 23:27
Al JebrAl Jebr
4,42143478
asked Jan 31 at 23:27
Al JebrAl Jebr
4,42143478
asked Jan 31 at 23:27
Al JebrAl Jebr
4,42143478
asked Jan 31 at 23:27
Al JebrAl Jebr
4,42143478
4,42143478
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint:
$(2x, x^2+1)subset (2, x^2+1)$ and
$$mathbf Z[x]/(2, x^2+1)simeq (mathbf Z/2mathbf Z)[x]big/(x^2+1)ne {0}.$$
edited Feb 2 at 18:14
Al Jebr
4,42143478
answered Jan 31 at 23:37
BernardBernard
124k741117
$endgroup$
$begingroup$
$(mathbf Z/2mathbf Z)[x]big/(x^2+1) = (mathbf Z/2mathbf Z)[x]$ since $1$ is a root of $x^2+1$ and $1 in mathbf Z/2mathbf Z[x]$. Is this correct?
$endgroup$
– Al Jebr
Feb 1 at 4:26
$begingroup$
No. It is the same as $mathbf Z[x]/bigl((x+1)^2bigr)$, which is isomorphic to $mathbf Z[x]/(x^2)$ (via the homomorphism $xmapsto x+1$).
$endgroup$
– Bernard
Feb 1 at 9:31
$begingroup$
But $1$ is a root of $x^2+1$ over $mathbf Z/2mathbf Z$. So, we have $(mathbf Z/2mathbf Z)[x]big/(x^2+1) = (mathbf Z/2mathbf Z)[1]=mathbf Z/2mathbf Z$. No?
$endgroup$
– Al Jebr
Feb 2 at 16:49
$begingroup$
Or is it because $x^2+1$ is reducible over $mathbf Z/2mathbf Z$ that I'm wrong?
$endgroup$
– Al Jebr
Feb 2 at 16:50
$begingroup$
No, because the ideal $bigl((x+1)^2bigr)$ is not the ideal $(x+1)$. Actually, $x+1$ is a non-zero nilpotent element.
$endgroup$
– Bernard
Feb 2 at 16:52
|
show 2 more comments
$begingroup$
First guess an element that is not inside, for example $x$.
Then prove that is not inside by contradiction.
If $xin(2x,x^2+1)$ then there are polynomials $p(x)=a_0+a_1x+...+a_nx^n$, $q(x)=b_0+b_1x+...+b_kx^k$ such that $x=p(x)cdot 2x+q(x)cdot (x^2+1)$ so $$x=(a_0+a_1x+...+a_nx^n)2x+(b_0+b_1x+...+b_kx^k)(x^2+1)$$
Equating the coefficients at both sides
$1)$ $b_0=0$
$2)$ $2a_0+b_1=1$
$3)$ $2a_{r+1}+b_{r+2}+b_r=0$ for $rgeq 0$ or equivalently $b_r=-2a_{r+1}-b_{r+2}$
Using $3)$ lots of times on $2)$, as $b_r=0$ for $r$ big enough we get
$begin{align}
1&=2a_0+b_1 \
&= 2a_0-2a_2-b_3 \
& =2a_0-2a_2+2a_4+b_5 \
&dots \
&=2a_0-2a_2+2a_4+dots pm 2a_{lfloor n/2rfloor}
end{align}$
And as the left hand side is $1$ and the right hand side is even we get a contradiction. This shows $(2x,x^2+1)neq mathbb{Z}[x]$.
answered Jan 31 at 23:39
yamete kudasaiyamete kudasai
1,160818
$endgroup$
3
$begingroup$
As a remark, an ideal generate the entire ring iff 1 is inside. Hence you can skip the guess and use 1 from the beginning.
$endgroup$
– yamete kudasai
Jan 31 at 23:56
$begingroup$
The coefficients are not properly equated. We should have $b_0=0$ and $2a_0+b_1=1$ and $2a_1+b_0+b_2=0$.
$endgroup$
– Al Jebr
Feb 1 at 3:06
$begingroup$
I don't think this method is going to work.
$endgroup$
– Al Jebr
Feb 1 at 3:21
$begingroup$
@AlJebr I am sorry, I answered before without checking the computations. Now it should be good.
$endgroup$
– yamete kudasai
Feb 2 at 0:05
add a comment |
$begingroup$
Both generators evaluate to $2$ at $x=1$. It follows that $f(1)$ is an even integer for all $f(x)$ in the ideal. Therefore the ideal cannot be all of $Bbb{Z}[x]$.
answered Feb 2 at 18:30
Jyrki LahtonenJyrki Lahtonen
110k13172390
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
$(2x, x^2+1)subset (2, x^2+1)$ and
$$mathbf Z[x]/(2, x^2+1)simeq (mathbf Z/2mathbf Z)[x]big/(x^2+1)ne {0}.$$
edited Feb 2 at 18:14
Al Jebr
4,42143478
answered Jan 31 at 23:37
BernardBernard
124k741117
$endgroup$
$begingroup$
$(mathbf Z/2mathbf Z)[x]big/(x^2+1) = (mathbf Z/2mathbf Z)[x]$ since $1$ is a root of $x^2+1$ and $1 in mathbf Z/2mathbf Z[x]$. Is this correct?
$endgroup$
– Al Jebr
Feb 1 at 4:26
$begingroup$
No. It is the same as $mathbf Z[x]/bigl((x+1)^2bigr)$, which is isomorphic to $mathbf Z[x]/(x^2)$ (via the homomorphism $xmapsto x+1$).
$endgroup$
– Bernard
Feb 1 at 9:31
$begingroup$
But $1$ is a root of $x^2+1$ over $mathbf Z/2mathbf Z$. So, we have $(mathbf Z/2mathbf Z)[x]big/(x^2+1) = (mathbf Z/2mathbf Z)[1]=mathbf Z/2mathbf Z$. No?
$endgroup$
– Al Jebr
Feb 2 at 16:49
$begingroup$
Or is it because $x^2+1$ is reducible over $mathbf Z/2mathbf Z$ that I'm wrong?
$endgroup$
– Al Jebr
Feb 2 at 16:50
$begingroup$
No, because the ideal $bigl((x+1)^2bigr)$ is not the ideal $(x+1)$. Actually, $x+1$ is a non-zero nilpotent element.
$endgroup$
– Bernard
Feb 2 at 16:52
|
show 2 more comments
$begingroup$
Hint:
$(2x, x^2+1)subset (2, x^2+1)$ and
$$mathbf Z[x]/(2, x^2+1)simeq (mathbf Z/2mathbf Z)[x]big/(x^2+1)ne {0}.$$
edited Feb 2 at 18:14
Al Jebr
4,42143478
answered Jan 31 at 23:37
BernardBernard
124k741117
$endgroup$
$begingroup$
$(mathbf Z/2mathbf Z)[x]big/(x^2+1) = (mathbf Z/2mathbf Z)[x]$ since $1$ is a root of $x^2+1$ and $1 in mathbf Z/2mathbf Z[x]$. Is this correct?
$endgroup$
– Al Jebr
Feb 1 at 4:26
$begingroup$
No. It is the same as $mathbf Z[x]/bigl((x+1)^2bigr)$, which is isomorphic to $mathbf Z[x]/(x^2)$ (via the homomorphism $xmapsto x+1$).
$endgroup$
– Bernard
Feb 1 at 9:31
$begingroup$
But $1$ is a root of $x^2+1$ over $mathbf Z/2mathbf Z$. So, we have $(mathbf Z/2mathbf Z)[x]big/(x^2+1) = (mathbf Z/2mathbf Z)[1]=mathbf Z/2mathbf Z$. No?
$endgroup$
– Al Jebr
Feb 2 at 16:49
$begingroup$
Or is it because $x^2+1$ is reducible over $mathbf Z/2mathbf Z$ that I'm wrong?
$endgroup$
– Al Jebr
Feb 2 at 16:50
$begingroup$
No, because the ideal $bigl((x+1)^2bigr)$ is not the ideal $(x+1)$. Actually, $x+1$ is a non-zero nilpotent element.
$endgroup$
– Bernard
Feb 2 at 16:52
|
show 2 more comments
$begingroup$
Hint:
$(2x, x^2+1)subset (2, x^2+1)$ and
$$mathbf Z[x]/(2, x^2+1)simeq (mathbf Z/2mathbf Z)[x]big/(x^2+1)ne {0}.$$
edited Feb 2 at 18:14
Al Jebr
4,42143478
answered Jan 31 at 23:37
BernardBernard
124k741117
$endgroup$
Hint:
$(2x, x^2+1)subset (2, x^2+1)$ and
$$mathbf Z[x]/(2, x^2+1)simeq (mathbf Z/2mathbf Z)[x]big/(x^2+1)ne {0}.$$
edited Feb 2 at 18:14
Al Jebr
4,42143478
answered Jan 31 at 23:37
BernardBernard
124k741117
edited Feb 2 at 18:14
Al Jebr
4,42143478
edited Feb 2 at 18:14
Al Jebr
4,42143478
edited Feb 2 at 18:14
Al Jebr
4,42143478
4,42143478
answered Jan 31 at 23:37
BernardBernard
124k741117
answered Jan 31 at 23:37
BernardBernard
124k741117
answered Jan 31 at 23:37
BernardBernard
124k741117
124k741117
$begingroup$
$(mathbf Z/2mathbf Z)[x]big/(x^2+1) = (mathbf Z/2mathbf Z)[x]$ since $1$ is a root of $x^2+1$ and $1 in mathbf Z/2mathbf Z[x]$. Is this correct?
$endgroup$
– Al Jebr
Feb 1 at 4:26
$begingroup$
No. It is the same as $mathbf Z[x]/bigl((x+1)^2bigr)$, which is isomorphic to $mathbf Z[x]/(x^2)$ (via the homomorphism $xmapsto x+1$).
$endgroup$
– Bernard
Feb 1 at 9:31
$begingroup$
But $1$ is a root of $x^2+1$ over $mathbf Z/2mathbf Z$. So, we have $(mathbf Z/2mathbf Z)[x]big/(x^2+1) = (mathbf Z/2mathbf Z)[1]=mathbf Z/2mathbf Z$. No?
$endgroup$
– Al Jebr
Feb 2 at 16:49
$begingroup$
Or is it because $x^2+1$ is reducible over $mathbf Z/2mathbf Z$ that I'm wrong?
$endgroup$
– Al Jebr
Feb 2 at 16:50
$begingroup$
No, because the ideal $bigl((x+1)^2bigr)$ is not the ideal $(x+1)$. Actually, $x+1$ is a non-zero nilpotent element.
$endgroup$
– Bernard
Feb 2 at 16:52
|
show 2 more comments
$begingroup$
$(mathbf Z/2mathbf Z)[x]big/(x^2+1) = (mathbf Z/2mathbf Z)[x]$ since $1$ is a root of $x^2+1$ and $1 in mathbf Z/2mathbf Z[x]$. Is this correct?
$endgroup$
– Al Jebr
Feb 1 at 4:26
$begingroup$
No. It is the same as $mathbf Z[x]/bigl((x+1)^2bigr)$, which is isomorphic to $mathbf Z[x]/(x^2)$ (via the homomorphism $xmapsto x+1$).
$endgroup$
– Bernard
Feb 1 at 9:31
$begingroup$
But $1$ is a root of $x^2+1$ over $mathbf Z/2mathbf Z$. So, we have $(mathbf Z/2mathbf Z)[x]big/(x^2+1) = (mathbf Z/2mathbf Z)[1]=mathbf Z/2mathbf Z$. No?
$endgroup$
– Al Jebr
Feb 2 at 16:49
$begingroup$
Or is it because $x^2+1$ is reducible over $mathbf Z/2mathbf Z$ that I'm wrong?
$endgroup$
– Al Jebr
Feb 2 at 16:50
$begingroup$
No, because the ideal $bigl((x+1)^2bigr)$ is not the ideal $(x+1)$. Actually, $x+1$ is a non-zero nilpotent element.
$endgroup$
– Bernard
Feb 2 at 16:52
$begingroup$
$(mathbf Z/2mathbf Z)[x]big/(x^2+1) = (mathbf Z/2mathbf Z)[x]$ since $1$ is a root of $x^2+1$ and $1 in mathbf Z/2mathbf Z[x]$. Is this correct?
$endgroup$
– Al Jebr
Feb 1 at 4:26
$begingroup$
$(mathbf Z/2mathbf Z)[x]big/(x^2+1) = (mathbf Z/2mathbf Z)[x]$ since $1$ is a root of $x^2+1$ and $1 in mathbf Z/2mathbf Z[x]$. Is this correct?
$endgroup$
– Al Jebr
Feb 1 at 4:26
$begingroup$
No. It is the same as $mathbf Z[x]/bigl((x+1)^2bigr)$, which is isomorphic to $mathbf Z[x]/(x^2)$ (via the homomorphism $xmapsto x+1$).
$endgroup$
– Bernard
Feb 1 at 9:31
$begingroup$
No. It is the same as $mathbf Z[x]/bigl((x+1)^2bigr)$, which is isomorphic to $mathbf Z[x]/(x^2)$ (via the homomorphism $xmapsto x+1$).
$endgroup$
– Bernard
Feb 1 at 9:31
$begingroup$
But $1$ is a root of $x^2+1$ over $mathbf Z/2mathbf Z$. So, we have $(mathbf Z/2mathbf Z)[x]big/(x^2+1) = (mathbf Z/2mathbf Z)[1]=mathbf Z/2mathbf Z$. No?
$endgroup$
– Al Jebr
Feb 2 at 16:49
$begingroup$
But $1$ is a root of $x^2+1$ over $mathbf Z/2mathbf Z$. So, we have $(mathbf Z/2mathbf Z)[x]big/(x^2+1) = (mathbf Z/2mathbf Z)[1]=mathbf Z/2mathbf Z$. No?
$endgroup$
– Al Jebr
Feb 2 at 16:49
$begingroup$
Or is it because $x^2+1$ is reducible over $mathbf Z/2mathbf Z$ that I'm wrong?
$endgroup$
– Al Jebr
Feb 2 at 16:50
$begingroup$
Or is it because $x^2+1$ is reducible over $mathbf Z/2mathbf Z$ that I'm wrong?
$endgroup$
– Al Jebr
Feb 2 at 16:50
$begingroup$
No, because the ideal $bigl((x+1)^2bigr)$ is not the ideal $(x+1)$. Actually, $x+1$ is a non-zero nilpotent element.
$endgroup$
– Bernard
Feb 2 at 16:52
$begingroup$
No, because the ideal $bigl((x+1)^2bigr)$ is not the ideal $(x+1)$. Actually, $x+1$ is a non-zero nilpotent element.
$endgroup$
– Bernard
Feb 2 at 16:52
|
show 2 more comments
$begingroup$
First guess an element that is not inside, for example $x$.
Then prove that is not inside by contradiction.
If $xin(2x,x^2+1)$ then there are polynomials $p(x)=a_0+a_1x+...+a_nx^n$, $q(x)=b_0+b_1x+...+b_kx^k$ such that $x=p(x)cdot 2x+q(x)cdot (x^2+1)$ so $$x=(a_0+a_1x+...+a_nx^n)2x+(b_0+b_1x+...+b_kx^k)(x^2+1)$$
Equating the coefficients at both sides
$1)$ $b_0=0$
$2)$ $2a_0+b_1=1$
$3)$ $2a_{r+1}+b_{r+2}+b_r=0$ for $rgeq 0$ or equivalently $b_r=-2a_{r+1}-b_{r+2}$
Using $3)$ lots of times on $2)$, as $b_r=0$ for $r$ big enough we get
$begin{align}
1&=2a_0+b_1 \
&= 2a_0-2a_2-b_3 \
& =2a_0-2a_2+2a_4+b_5 \
&dots \
&=2a_0-2a_2+2a_4+dots pm 2a_{lfloor n/2rfloor}
end{align}$
And as the left hand side is $1$ and the right hand side is even we get a contradiction. This shows $(2x,x^2+1)neq mathbb{Z}[x]$.
answered Jan 31 at 23:39
yamete kudasaiyamete kudasai
1,160818
$endgroup$
3
$begingroup$
As a remark, an ideal generate the entire ring iff 1 is inside. Hence you can skip the guess and use 1 from the beginning.
$endgroup$
– yamete kudasai
Jan 31 at 23:56
$begingroup$
The coefficients are not properly equated. We should have $b_0=0$ and $2a_0+b_1=1$ and $2a_1+b_0+b_2=0$.
$endgroup$
– Al Jebr
Feb 1 at 3:06
$begingroup$
I don't think this method is going to work.
$endgroup$
– Al Jebr
Feb 1 at 3:21
$begingroup$
@AlJebr I am sorry, I answered before without checking the computations. Now it should be good.
$endgroup$
– yamete kudasai
Feb 2 at 0:05
add a comment |
$begingroup$
First guess an element that is not inside, for example $x$.
Then prove that is not inside by contradiction.
If $xin(2x,x^2+1)$ then there are polynomials $p(x)=a_0+a_1x+...+a_nx^n$, $q(x)=b_0+b_1x+...+b_kx^k$ such that $x=p(x)cdot 2x+q(x)cdot (x^2+1)$ so $$x=(a_0+a_1x+...+a_nx^n)2x+(b_0+b_1x+...+b_kx^k)(x^2+1)$$
Equating the coefficients at both sides
$1)$ $b_0=0$
$2)$ $2a_0+b_1=1$
$3)$ $2a_{r+1}+b_{r+2}+b_r=0$ for $rgeq 0$ or equivalently $b_r=-2a_{r+1}-b_{r+2}$
Using $3)$ lots of times on $2)$, as $b_r=0$ for $r$ big enough we get
$begin{align}
1&=2a_0+b_1 \
&= 2a_0-2a_2-b_3 \
& =2a_0-2a_2+2a_4+b_5 \
&dots \
&=2a_0-2a_2+2a_4+dots pm 2a_{lfloor n/2rfloor}
end{align}$
And as the left hand side is $1$ and the right hand side is even we get a contradiction. This shows $(2x,x^2+1)neq mathbb{Z}[x]$.
answered Jan 31 at 23:39
yamete kudasaiyamete kudasai
1,160818
$endgroup$
3
$begingroup$
As a remark, an ideal generate the entire ring iff 1 is inside. Hence you can skip the guess and use 1 from the beginning.
$endgroup$
– yamete kudasai
Jan 31 at 23:56
$begingroup$
The coefficients are not properly equated. We should have $b_0=0$ and $2a_0+b_1=1$ and $2a_1+b_0+b_2=0$.
$endgroup$
– Al Jebr
Feb 1 at 3:06
$begingroup$
I don't think this method is going to work.
$endgroup$
– Al Jebr
Feb 1 at 3:21
$begingroup$
@AlJebr I am sorry, I answered before without checking the computations. Now it should be good.
$endgroup$
– yamete kudasai
Feb 2 at 0:05
add a comment |
$begingroup$
First guess an element that is not inside, for example $x$.
Then prove that is not inside by contradiction.
If $xin(2x,x^2+1)$ then there are polynomials $p(x)=a_0+a_1x+...+a_nx^n$, $q(x)=b_0+b_1x+...+b_kx^k$ such that $x=p(x)cdot 2x+q(x)cdot (x^2+1)$ so $$x=(a_0+a_1x+...+a_nx^n)2x+(b_0+b_1x+...+b_kx^k)(x^2+1)$$
Equating the coefficients at both sides
$1)$ $b_0=0$
$2)$ $2a_0+b_1=1$
$3)$ $2a_{r+1}+b_{r+2}+b_r=0$ for $rgeq 0$ or equivalently $b_r=-2a_{r+1}-b_{r+2}$
Using $3)$ lots of times on $2)$, as $b_r=0$ for $r$ big enough we get
$begin{align}
1&=2a_0+b_1 \
&= 2a_0-2a_2-b_3 \
& =2a_0-2a_2+2a_4+b_5 \
&dots \
&=2a_0-2a_2+2a_4+dots pm 2a_{lfloor n/2rfloor}
end{align}$
And as the left hand side is $1$ and the right hand side is even we get a contradiction. This shows $(2x,x^2+1)neq mathbb{Z}[x]$.
answered Jan 31 at 23:39
yamete kudasaiyamete kudasai
1,160818
$endgroup$
First guess an element that is not inside, for example $x$.
Then prove that is not inside by contradiction.
If $xin(2x,x^2+1)$ then there are polynomials $p(x)=a_0+a_1x+...+a_nx^n$, $q(x)=b_0+b_1x+...+b_kx^k$ such that $x=p(x)cdot 2x+q(x)cdot (x^2+1)$ so $$x=(a_0+a_1x+...+a_nx^n)2x+(b_0+b_1x+...+b_kx^k)(x^2+1)$$
Equating the coefficients at both sides
$1)$ $b_0=0$
$2)$ $2a_0+b_1=1$
$3)$ $2a_{r+1}+b_{r+2}+b_r=0$ for $rgeq 0$ or equivalently $b_r=-2a_{r+1}-b_{r+2}$
Using $3)$ lots of times on $2)$, as $b_r=0$ for $r$ big enough we get
$begin{align}
1&=2a_0+b_1 \
&= 2a_0-2a_2-b_3 \
& =2a_0-2a_2+2a_4+b_5 \
&dots \
&=2a_0-2a_2+2a_4+dots pm 2a_{lfloor n/2rfloor}
end{align}$
And as the left hand side is $1$ and the right hand side is even we get a contradiction. This shows $(2x,x^2+1)neq mathbb{Z}[x]$.
answered Jan 31 at 23:39
yamete kudasaiyamete kudasai
1,160818
edited Feb 6 at 22:59
answered Jan 31 at 23:39
yamete kudasaiyamete kudasai
1,160818
answered Jan 31 at 23:39
yamete kudasaiyamete kudasai
1,160818
answered Jan 31 at 23:39
yamete kudasaiyamete kudasai
1,160818
1,160818
3
$begingroup$
As a remark, an ideal generate the entire ring iff 1 is inside. Hence you can skip the guess and use 1 from the beginning.
$endgroup$
– yamete kudasai
Jan 31 at 23:56
$begingroup$
The coefficients are not properly equated. We should have $b_0=0$ and $2a_0+b_1=1$ and $2a_1+b_0+b_2=0$.
$endgroup$
– Al Jebr
Feb 1 at 3:06
$begingroup$
I don't think this method is going to work.
$endgroup$
– Al Jebr
Feb 1 at 3:21
$begingroup$
@AlJebr I am sorry, I answered before without checking the computations. Now it should be good.
$endgroup$
– yamete kudasai
Feb 2 at 0:05
add a comment |
3
$begingroup$
As a remark, an ideal generate the entire ring iff 1 is inside. Hence you can skip the guess and use 1 from the beginning.
$endgroup$
– yamete kudasai
Jan 31 at 23:56
$begingroup$
The coefficients are not properly equated. We should have $b_0=0$ and $2a_0+b_1=1$ and $2a_1+b_0+b_2=0$.
$endgroup$
– Al Jebr
Feb 1 at 3:06
$begingroup$
I don't think this method is going to work.
$endgroup$
– Al Jebr
Feb 1 at 3:21
$begingroup$
@AlJebr I am sorry, I answered before without checking the computations. Now it should be good.
$endgroup$
– yamete kudasai
Feb 2 at 0:05
3
3
$begingroup$
As a remark, an ideal generate the entire ring iff 1 is inside. Hence you can skip the guess and use 1 from the beginning.
$endgroup$
– yamete kudasai
Jan 31 at 23:56
$begingroup$
As a remark, an ideal generate the entire ring iff 1 is inside. Hence you can skip the guess and use 1 from the beginning.
$endgroup$
– yamete kudasai
Jan 31 at 23:56
$begingroup$
The coefficients are not properly equated. We should have $b_0=0$ and $2a_0+b_1=1$ and $2a_1+b_0+b_2=0$.
$endgroup$
– Al Jebr
Feb 1 at 3:06
$begingroup$
The coefficients are not properly equated. We should have $b_0=0$ and $2a_0+b_1=1$ and $2a_1+b_0+b_2=0$.
$endgroup$
– Al Jebr
Feb 1 at 3:06
$begingroup$
I don't think this method is going to work.
$endgroup$
– Al Jebr
Feb 1 at 3:21
$begingroup$
I don't think this method is going to work.
$endgroup$
– Al Jebr
Feb 1 at 3:21
$begingroup$
@AlJebr I am sorry, I answered before without checking the computations. Now it should be good.
$endgroup$
– yamete kudasai
Feb 2 at 0:05
$begingroup$
@AlJebr I am sorry, I answered before without checking the computations. Now it should be good.
$endgroup$
– yamete kudasai
Feb 2 at 0:05
add a comment |
$begingroup$
Both generators evaluate to $2$ at $x=1$. It follows that $f(1)$ is an even integer for all $f(x)$ in the ideal. Therefore the ideal cannot be all of $Bbb{Z}[x]$.
answered Feb 2 at 18:30
Jyrki LahtonenJyrki Lahtonen
110k13172390
$endgroup$
add a comment |
$begingroup$
Both generators evaluate to $2$ at $x=1$. It follows that $f(1)$ is an even integer for all $f(x)$ in the ideal. Therefore the ideal cannot be all of $Bbb{Z}[x]$.
answered Feb 2 at 18:30
Jyrki LahtonenJyrki Lahtonen
110k13172390
$endgroup$
add a comment |
$begingroup$
Both generators evaluate to $2$ at $x=1$. It follows that $f(1)$ is an even integer for all $f(x)$ in the ideal. Therefore the ideal cannot be all of $Bbb{Z}[x]$.
answered Feb 2 at 18:30
Jyrki LahtonenJyrki Lahtonen
110k13172390
$endgroup$
Both generators evaluate to $2$ at $x=1$. It follows that $f(1)$ is an even integer for all $f(x)$ in the ideal. Therefore the ideal cannot be all of $Bbb{Z}[x]$.
answered Feb 2 at 18:30
Jyrki LahtonenJyrki Lahtonen
110k13172390
edited Feb 5 at 20:39
answered Feb 2 at 18:30
Jyrki LahtonenJyrki Lahtonen
110k13172390
answered Feb 2 at 18:30
Jyrki LahtonenJyrki Lahtonen
110k13172390
answered Feb 2 at 18:30
Jyrki LahtonenJyrki Lahtonen
110k13172390
110k13172390
add a comment |
add a comment |
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