Ideals generated by two elements in $mathbb{Z}[x]$
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Consider the following ideal $(2+x,x^2+5)$ in $mathbb{Z}[x]$. Then I showed that $(2+x,x^2+5)=(9,2+x)$.
But I am not able to do the same for ideals $(1-4x,x^2+5)$ and $(1+2x,x^2+5)$.
Is there some method?
EDIT: My goal is to show that these ideals can be written as $(c,ax+b)$ for some appropriate integers $a,b,c$.
Would be very grateful for help!
abstract-algebra polynomials ring-theory ideals
edited Dec 9 '18 at 0:27
Batominovski
33.1k33293
asked Dec 8 '18 at 23:44
ZFRZFR
5,17131440
$endgroup$
add a comment |
$begingroup$
Consider the following ideal $(2+x,x^2+5)$ in $mathbb{Z}[x]$. Then I showed that $(2+x,x^2+5)=(9,2+x)$.
But I am not able to do the same for ideals $(1-4x,x^2+5)$ and $(1+2x,x^2+5)$.
Is there some method?
EDIT: My goal is to show that these ideals can be written as $(c,ax+b)$ for some appropriate integers $a,b,c$.
Would be very grateful for help!
abstract-algebra polynomials ring-theory ideals
edited Dec 9 '18 at 0:27
Batominovski
33.1k33293
asked Dec 8 '18 at 23:44
ZFRZFR
5,17131440
$endgroup$
1
$begingroup$
what are you trying to do?
$endgroup$
– Andres Mejia
Dec 8 '18 at 23:50
$begingroup$
@AndresMejia, sorry, didn't understand your question?
$endgroup$
– ZFR
Dec 8 '18 at 23:51
$begingroup$
"I am not able to do the same..." what is "the same?"
$endgroup$
– Andres Mejia
Dec 8 '18 at 23:51
$begingroup$
@AndresMejia, I would like to show that these ideals is something like $(c,ax+b)$. Is it clear right now?
$endgroup$
– ZFR
Dec 8 '18 at 23:53
$begingroup$
yep! thanks :)${}{}$
$endgroup$
– Andres Mejia
Dec 8 '18 at 23:53
add a comment |
$begingroup$
Consider the following ideal $(2+x,x^2+5)$ in $mathbb{Z}[x]$. Then I showed that $(2+x,x^2+5)=(9,2+x)$.
But I am not able to do the same for ideals $(1-4x,x^2+5)$ and $(1+2x,x^2+5)$.
Is there some method?
EDIT: My goal is to show that these ideals can be written as $(c,ax+b)$ for some appropriate integers $a,b,c$.
Would be very grateful for help!
abstract-algebra polynomials ring-theory ideals
edited Dec 9 '18 at 0:27
Batominovski
33.1k33293
asked Dec 8 '18 at 23:44
ZFRZFR
5,17131440
$endgroup$
Consider the following ideal $(2+x,x^2+5)$ in $mathbb{Z}[x]$. Then I showed that $(2+x,x^2+5)=(9,2+x)$.
But I am not able to do the same for ideals $(1-4x,x^2+5)$ and $(1+2x,x^2+5)$.
Is there some method?
EDIT: My goal is to show that these ideals can be written as $(c,ax+b)$ for some appropriate integers $a,b,c$.
Would be very grateful for help!
abstract-algebra polynomials ring-theory ideals
abstract-algebra polynomials ring-theory ideals
edited Dec 9 '18 at 0:27
Batominovski
33.1k33293
asked Dec 8 '18 at 23:44
ZFRZFR
5,17131440
edited Dec 9 '18 at 0:27
Batominovski
33.1k33293
asked Dec 8 '18 at 23:44
ZFRZFR
5,17131440
edited Dec 9 '18 at 0:27
Batominovski
33.1k33293
edited Dec 9 '18 at 0:27
Batominovski
33.1k33293
edited Dec 9 '18 at 0:27
Batominovski
33.1k33293
33.1k33293
asked Dec 8 '18 at 23:44
ZFRZFR
5,17131440
asked Dec 8 '18 at 23:44
ZFRZFR
5,17131440
asked Dec 8 '18 at 23:44
ZFRZFR
5,17131440
5,17131440
1
$begingroup$
what are you trying to do?
$endgroup$
– Andres Mejia
Dec 8 '18 at 23:50
$begingroup$
@AndresMejia, sorry, didn't understand your question?
$endgroup$
– ZFR
Dec 8 '18 at 23:51
$begingroup$
"I am not able to do the same..." what is "the same?"
$endgroup$
– Andres Mejia
Dec 8 '18 at 23:51
$begingroup$
@AndresMejia, I would like to show that these ideals is something like $(c,ax+b)$. Is it clear right now?
$endgroup$
– ZFR
Dec 8 '18 at 23:53
$begingroup$
yep! thanks :)${}{}$
$endgroup$
– Andres Mejia
Dec 8 '18 at 23:53
add a comment |
1
$begingroup$
what are you trying to do?
$endgroup$
– Andres Mejia
Dec 8 '18 at 23:50
$begingroup$
@AndresMejia, sorry, didn't understand your question?
$endgroup$
– ZFR
Dec 8 '18 at 23:51
$begingroup$
"I am not able to do the same..." what is "the same?"
$endgroup$
– Andres Mejia
Dec 8 '18 at 23:51
$begingroup$
@AndresMejia, I would like to show that these ideals is something like $(c,ax+b)$. Is it clear right now?
$endgroup$
– ZFR
Dec 8 '18 at 23:53
$begingroup$
yep! thanks :)${}{}$
$endgroup$
– Andres Mejia
Dec 8 '18 at 23:53
1
1
$begingroup$
what are you trying to do?
$endgroup$
– Andres Mejia
Dec 8 '18 at 23:50
$begingroup$
what are you trying to do?
$endgroup$
– Andres Mejia
Dec 8 '18 at 23:50
$begingroup$
@AndresMejia, sorry, didn't understand your question?
$endgroup$
– ZFR
Dec 8 '18 at 23:51
$begingroup$
@AndresMejia, sorry, didn't understand your question?
$endgroup$
– ZFR
Dec 8 '18 at 23:51
$begingroup$
"I am not able to do the same..." what is "the same?"
$endgroup$
– Andres Mejia
Dec 8 '18 at 23:51
$begingroup$
"I am not able to do the same..." what is "the same?"
$endgroup$
– Andres Mejia
Dec 8 '18 at 23:51
$begingroup$
@AndresMejia, I would like to show that these ideals is something like $(c,ax+b)$. Is it clear right now?
$endgroup$
– ZFR
Dec 8 '18 at 23:53
$begingroup$
@AndresMejia, I would like to show that these ideals is something like $(c,ax+b)$. Is it clear right now?
$endgroup$
– ZFR
Dec 8 '18 at 23:53
$begingroup$
yep! thanks :)${}{}$
$endgroup$
– Andres Mejia
Dec 8 '18 at 23:53
$begingroup$
yep! thanks :)${}{}$
$endgroup$
– Andres Mejia
Dec 8 '18 at 23:53
add a comment |
2 Answers
2
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oldest
votes
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Hint $bmod I=(x^2!+!5,,1!-!4x)!:, color{#c00}{4xequiv 1},$ so $,0equiv 4(x^2!+!5)equiv (color{#c00}{4x})x!+!20equiv x!+!20$
Equivalently $I$ contains $,4(x^2!+!5)+x(1!-!4x) = 20!+!x$
Since $,x!+!20in I,$ we can further reduce mod $,x!+!20,$ too, i.e. use $,color{#0a0}{xequiv -20},$ to evaluate / reduce all other generators via $,f(color{#0a0}x)equiv f(color{#0a0}{-20}), $ i.e. we have
$$({x!-!a},, f,,g) = (x!-!a,, fbmod{x!-!color{#0a0}a},,gbmod{x!-!color{#0a0}a}) = (x!-!a,f(color{#0a0}a),g(color{#0a0}a))qquad$$
just like in the euclidean algorithm $, (a,b) = (a,,bbmod a). $ In summary it is simply:
$$(1!-!4x,,x^2!+!5) = (x!+!20,,underbrace{1!-!4x}_{large f(x)},,underbrace{x^2!+!5}_{large g(x)}) = (x!+!20,!!underbrace{81}_{!!!!large f(-20)},!underbrace{405}_{large g(-20)}!!) = (x!+!20,81)$$
Corollary $ fbmod I, =, f(-20)bmod 81, $ so $ fin Iiff 81mid f(-20)$
answered Dec 8 '18 at 23:56
Bill DubuqueBill Dubuque
210k29192645
$endgroup$
$begingroup$
Dear, Bill! Could you clarify it with words? I don't understand it :( Why $4x equiv 1$?
$endgroup$
– ZFR
Dec 8 '18 at 23:58
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the trick is to do it modulo $1-4x$. In the quotient $mathbb Z[x]/(1-4x)$, we have that $4x-1 equiv 0$ or $4x equiv 1$.
$endgroup$
– Andres Mejia
Dec 8 '18 at 23:59
$begingroup$
@ZFR $equiv$ is used to mean equality in a quotient ring.
$endgroup$
– Arthur
Dec 9 '18 at 0:02
$begingroup$
It is not clear how are you working with modulo? Why $4xequiv 1$ and it is not clear the rest of your answer.
$endgroup$
– ZFR
Dec 9 '18 at 0:02
1
$begingroup$
@ZFR Glad it was helpful. Alas, most textbooks don't explain things like this very well (if at all), since it's a bit too advanced for elementary books, but a bit too elementary for advanced books. So I try to highlight these ideas here to help fill the gaps.
$endgroup$
– Bill Dubuque
Dec 9 '18 at 1:47
|
show 13 more comments
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Note that $langle 1-4x,x^2+5rangle=langle x+20,81rangle$. This is because
$$x^2+5=(x-20)cdot(x+20)+5cdot 81,,$$
$$1-4x=(-4)cdot (x+20)+1cdot 81,,$$
$$x+20=4cdot(x^2+5)+xcdot(1-4x),,$$
and
$$81=16cdot(x^2+5)+(1+4x)cdot(1-4x),.$$
Similarly, $langle 1-2x,x^2+5rangle=langle x+10,21rangle$. This is because
$$x^2+5=(x-10)cdot(x+10)+5cdot 21,,$$
$$1-2x=(-2)cdot(x+10)+1cdot 21,,$$
$$x+10=2cdot(x^2+5)+xcdot(1-2x),,$$
and
$$21=4cdot(x^2+5)+(1+2x)cdot(1-2x),.$$
answered Dec 9 '18 at 0:25
BatominovskiBatominovski
33.1k33293
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1
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Good! but how you come up with those coefficients? The most difficult for me was finding the appropriate coefficients.
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– ZFR
Dec 9 '18 at 0:26
1
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My first goal was to find an integer $c$ that is definitely in the ideal. You want to eliminate $x^2$ from $x^2+5$. Say, your linear polynomial is $1-2x$. A multiple of $1-2x$ that only has $x^2$ and a constant term is $(1+2x)cdot(1-2x)$. This calculation leads to the last line of each case.
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– Batominovski
Dec 9 '18 at 0:34
1
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To find a linear polynomial $ax+b$ that is in the ideal is similar. You want to eliminate $x^2$ from $x^2+5$, but you want to leave the terms of order $1$ and $0$. A multiple of $1-2x$ that can eliminate $x^2$ from $x^2+5$ but leave a term of order $1$ is $xcdot(1-2x)$. This is the calculation in the second-to-last line of each case.
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– Batominovski
Dec 9 '18 at 0:38
1
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I want to note that you can't just use $1-2x$ for $ax+b$. It turns out that $langle 21,1-2xrangle$ is not the same as $langle 1-2x,x^2+5rangle$. If $|a|neq 1$, then you can never generate a monic polynomial $x^2+5$. Thus, the end result should have $a=1$ (or $a=-1$, but the sign doesn't really matter).
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– Batominovski
Dec 9 '18 at 0:43
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Hint $bmod I=(x^2!+!5,,1!-!4x)!:, color{#c00}{4xequiv 1},$ so $,0equiv 4(x^2!+!5)equiv (color{#c00}{4x})x!+!20equiv x!+!20$
Equivalently $I$ contains $,4(x^2!+!5)+x(1!-!4x) = 20!+!x$
Since $,x!+!20in I,$ we can further reduce mod $,x!+!20,$ too, i.e. use $,color{#0a0}{xequiv -20},$ to evaluate / reduce all other generators via $,f(color{#0a0}x)equiv f(color{#0a0}{-20}), $ i.e. we have
$$({x!-!a},, f,,g) = (x!-!a,, fbmod{x!-!color{#0a0}a},,gbmod{x!-!color{#0a0}a}) = (x!-!a,f(color{#0a0}a),g(color{#0a0}a))qquad$$
just like in the euclidean algorithm $, (a,b) = (a,,bbmod a). $ In summary it is simply:
$$(1!-!4x,,x^2!+!5) = (x!+!20,,underbrace{1!-!4x}_{large f(x)},,underbrace{x^2!+!5}_{large g(x)}) = (x!+!20,!!underbrace{81}_{!!!!large f(-20)},!underbrace{405}_{large g(-20)}!!) = (x!+!20,81)$$
Corollary $ fbmod I, =, f(-20)bmod 81, $ so $ fin Iiff 81mid f(-20)$
answered Dec 8 '18 at 23:56
Bill DubuqueBill Dubuque
210k29192645
$endgroup$
$begingroup$
Dear, Bill! Could you clarify it with words? I don't understand it :( Why $4x equiv 1$?
$endgroup$
– ZFR
Dec 8 '18 at 23:58
$begingroup$
the trick is to do it modulo $1-4x$. In the quotient $mathbb Z[x]/(1-4x)$, we have that $4x-1 equiv 0$ or $4x equiv 1$.
$endgroup$
– Andres Mejia
Dec 8 '18 at 23:59
$begingroup$
@ZFR $equiv$ is used to mean equality in a quotient ring.
$endgroup$
– Arthur
Dec 9 '18 at 0:02
$begingroup$
It is not clear how are you working with modulo? Why $4xequiv 1$ and it is not clear the rest of your answer.
$endgroup$
– ZFR
Dec 9 '18 at 0:02
1
$begingroup$
@ZFR Glad it was helpful. Alas, most textbooks don't explain things like this very well (if at all), since it's a bit too advanced for elementary books, but a bit too elementary for advanced books. So I try to highlight these ideas here to help fill the gaps.
$endgroup$
– Bill Dubuque
Dec 9 '18 at 1:47
|
show 13 more comments
$begingroup$
Hint $bmod I=(x^2!+!5,,1!-!4x)!:, color{#c00}{4xequiv 1},$ so $,0equiv 4(x^2!+!5)equiv (color{#c00}{4x})x!+!20equiv x!+!20$
Equivalently $I$ contains $,4(x^2!+!5)+x(1!-!4x) = 20!+!x$
Since $,x!+!20in I,$ we can further reduce mod $,x!+!20,$ too, i.e. use $,color{#0a0}{xequiv -20},$ to evaluate / reduce all other generators via $,f(color{#0a0}x)equiv f(color{#0a0}{-20}), $ i.e. we have
$$({x!-!a},, f,,g) = (x!-!a,, fbmod{x!-!color{#0a0}a},,gbmod{x!-!color{#0a0}a}) = (x!-!a,f(color{#0a0}a),g(color{#0a0}a))qquad$$
just like in the euclidean algorithm $, (a,b) = (a,,bbmod a). $ In summary it is simply:
$$(1!-!4x,,x^2!+!5) = (x!+!20,,underbrace{1!-!4x}_{large f(x)},,underbrace{x^2!+!5}_{large g(x)}) = (x!+!20,!!underbrace{81}_{!!!!large f(-20)},!underbrace{405}_{large g(-20)}!!) = (x!+!20,81)$$
Corollary $ fbmod I, =, f(-20)bmod 81, $ so $ fin Iiff 81mid f(-20)$
answered Dec 8 '18 at 23:56
Bill DubuqueBill Dubuque
210k29192645
$endgroup$
$begingroup$
Dear, Bill! Could you clarify it with words? I don't understand it :( Why $4x equiv 1$?
$endgroup$
– ZFR
Dec 8 '18 at 23:58
$begingroup$
the trick is to do it modulo $1-4x$. In the quotient $mathbb Z[x]/(1-4x)$, we have that $4x-1 equiv 0$ or $4x equiv 1$.
$endgroup$
– Andres Mejia
Dec 8 '18 at 23:59
$begingroup$
@ZFR $equiv$ is used to mean equality in a quotient ring.
$endgroup$
– Arthur
Dec 9 '18 at 0:02
$begingroup$
It is not clear how are you working with modulo? Why $4xequiv 1$ and it is not clear the rest of your answer.
$endgroup$
– ZFR
Dec 9 '18 at 0:02
1
$begingroup$
@ZFR Glad it was helpful. Alas, most textbooks don't explain things like this very well (if at all), since it's a bit too advanced for elementary books, but a bit too elementary for advanced books. So I try to highlight these ideas here to help fill the gaps.
$endgroup$
– Bill Dubuque
Dec 9 '18 at 1:47
|
show 13 more comments
$begingroup$
Hint $bmod I=(x^2!+!5,,1!-!4x)!:, color{#c00}{4xequiv 1},$ so $,0equiv 4(x^2!+!5)equiv (color{#c00}{4x})x!+!20equiv x!+!20$
Equivalently $I$ contains $,4(x^2!+!5)+x(1!-!4x) = 20!+!x$
Since $,x!+!20in I,$ we can further reduce mod $,x!+!20,$ too, i.e. use $,color{#0a0}{xequiv -20},$ to evaluate / reduce all other generators via $,f(color{#0a0}x)equiv f(color{#0a0}{-20}), $ i.e. we have
$$({x!-!a},, f,,g) = (x!-!a,, fbmod{x!-!color{#0a0}a},,gbmod{x!-!color{#0a0}a}) = (x!-!a,f(color{#0a0}a),g(color{#0a0}a))qquad$$
just like in the euclidean algorithm $, (a,b) = (a,,bbmod a). $ In summary it is simply:
$$(1!-!4x,,x^2!+!5) = (x!+!20,,underbrace{1!-!4x}_{large f(x)},,underbrace{x^2!+!5}_{large g(x)}) = (x!+!20,!!underbrace{81}_{!!!!large f(-20)},!underbrace{405}_{large g(-20)}!!) = (x!+!20,81)$$
Corollary $ fbmod I, =, f(-20)bmod 81, $ so $ fin Iiff 81mid f(-20)$
answered Dec 8 '18 at 23:56
Bill DubuqueBill Dubuque
210k29192645
$endgroup$
Hint $bmod I=(x^2!+!5,,1!-!4x)!:, color{#c00}{4xequiv 1},$ so $,0equiv 4(x^2!+!5)equiv (color{#c00}{4x})x!+!20equiv x!+!20$
Equivalently $I$ contains $,4(x^2!+!5)+x(1!-!4x) = 20!+!x$
Since $,x!+!20in I,$ we can further reduce mod $,x!+!20,$ too, i.e. use $,color{#0a0}{xequiv -20},$ to evaluate / reduce all other generators via $,f(color{#0a0}x)equiv f(color{#0a0}{-20}), $ i.e. we have
$$({x!-!a},, f,,g) = (x!-!a,, fbmod{x!-!color{#0a0}a},,gbmod{x!-!color{#0a0}a}) = (x!-!a,f(color{#0a0}a),g(color{#0a0}a))qquad$$
just like in the euclidean algorithm $, (a,b) = (a,,bbmod a). $ In summary it is simply:
$$(1!-!4x,,x^2!+!5) = (x!+!20,,underbrace{1!-!4x}_{large f(x)},,underbrace{x^2!+!5}_{large g(x)}) = (x!+!20,!!underbrace{81}_{!!!!large f(-20)},!underbrace{405}_{large g(-20)}!!) = (x!+!20,81)$$
Corollary $ fbmod I, =, f(-20)bmod 81, $ so $ fin Iiff 81mid f(-20)$
answered Dec 8 '18 at 23:56
Bill DubuqueBill Dubuque
210k29192645
edited Dec 9 '18 at 0:54
answered Dec 8 '18 at 23:56
Bill DubuqueBill Dubuque
210k29192645
answered Dec 8 '18 at 23:56
Bill DubuqueBill Dubuque
210k29192645
answered Dec 8 '18 at 23:56
Bill DubuqueBill Dubuque
210k29192645
210k29192645
$begingroup$
Dear, Bill! Could you clarify it with words? I don't understand it :( Why $4x equiv 1$?
$endgroup$
– ZFR
Dec 8 '18 at 23:58
$begingroup$
the trick is to do it modulo $1-4x$. In the quotient $mathbb Z[x]/(1-4x)$, we have that $4x-1 equiv 0$ or $4x equiv 1$.
$endgroup$
– Andres Mejia
Dec 8 '18 at 23:59
$begingroup$
@ZFR $equiv$ is used to mean equality in a quotient ring.
$endgroup$
– Arthur
Dec 9 '18 at 0:02
$begingroup$
It is not clear how are you working with modulo? Why $4xequiv 1$ and it is not clear the rest of your answer.
$endgroup$
– ZFR
Dec 9 '18 at 0:02
1
$begingroup$
@ZFR Glad it was helpful. Alas, most textbooks don't explain things like this very well (if at all), since it's a bit too advanced for elementary books, but a bit too elementary for advanced books. So I try to highlight these ideas here to help fill the gaps.
$endgroup$
– Bill Dubuque
Dec 9 '18 at 1:47
|
show 13 more comments
$begingroup$
Dear, Bill! Could you clarify it with words? I don't understand it :( Why $4x equiv 1$?
$endgroup$
– ZFR
Dec 8 '18 at 23:58
$begingroup$
the trick is to do it modulo $1-4x$. In the quotient $mathbb Z[x]/(1-4x)$, we have that $4x-1 equiv 0$ or $4x equiv 1$.
$endgroup$
– Andres Mejia
Dec 8 '18 at 23:59
$begingroup$
@ZFR $equiv$ is used to mean equality in a quotient ring.
$endgroup$
– Arthur
Dec 9 '18 at 0:02
$begingroup$
It is not clear how are you working with modulo? Why $4xequiv 1$ and it is not clear the rest of your answer.
$endgroup$
– ZFR
Dec 9 '18 at 0:02
1
$begingroup$
@ZFR Glad it was helpful. Alas, most textbooks don't explain things like this very well (if at all), since it's a bit too advanced for elementary books, but a bit too elementary for advanced books. So I try to highlight these ideas here to help fill the gaps.
$endgroup$
– Bill Dubuque
Dec 9 '18 at 1:47
$begingroup$
Dear, Bill! Could you clarify it with words? I don't understand it :( Why $4x equiv 1$?
$endgroup$
– ZFR
Dec 8 '18 at 23:58
$begingroup$
Dear, Bill! Could you clarify it with words? I don't understand it :( Why $4x equiv 1$?
$endgroup$
– ZFR
Dec 8 '18 at 23:58
$begingroup$
the trick is to do it modulo $1-4x$. In the quotient $mathbb Z[x]/(1-4x)$, we have that $4x-1 equiv 0$ or $4x equiv 1$.
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– Andres Mejia
Dec 8 '18 at 23:59
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the trick is to do it modulo $1-4x$. In the quotient $mathbb Z[x]/(1-4x)$, we have that $4x-1 equiv 0$ or $4x equiv 1$.
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– Andres Mejia
Dec 8 '18 at 23:59
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@ZFR $equiv$ is used to mean equality in a quotient ring.
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– Arthur
Dec 9 '18 at 0:02
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@ZFR $equiv$ is used to mean equality in a quotient ring.
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– Arthur
Dec 9 '18 at 0:02
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It is not clear how are you working with modulo? Why $4xequiv 1$ and it is not clear the rest of your answer.
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– ZFR
Dec 9 '18 at 0:02
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It is not clear how are you working with modulo? Why $4xequiv 1$ and it is not clear the rest of your answer.
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– ZFR
Dec 9 '18 at 0:02
1
1
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@ZFR Glad it was helpful. Alas, most textbooks don't explain things like this very well (if at all), since it's a bit too advanced for elementary books, but a bit too elementary for advanced books. So I try to highlight these ideas here to help fill the gaps.
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– Bill Dubuque
Dec 9 '18 at 1:47
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@ZFR Glad it was helpful. Alas, most textbooks don't explain things like this very well (if at all), since it's a bit too advanced for elementary books, but a bit too elementary for advanced books. So I try to highlight these ideas here to help fill the gaps.
$endgroup$
– Bill Dubuque
Dec 9 '18 at 1:47
|
show 13 more comments
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Note that $langle 1-4x,x^2+5rangle=langle x+20,81rangle$. This is because
$$x^2+5=(x-20)cdot(x+20)+5cdot 81,,$$
$$1-4x=(-4)cdot (x+20)+1cdot 81,,$$
$$x+20=4cdot(x^2+5)+xcdot(1-4x),,$$
and
$$81=16cdot(x^2+5)+(1+4x)cdot(1-4x),.$$
Similarly, $langle 1-2x,x^2+5rangle=langle x+10,21rangle$. This is because
$$x^2+5=(x-10)cdot(x+10)+5cdot 21,,$$
$$1-2x=(-2)cdot(x+10)+1cdot 21,,$$
$$x+10=2cdot(x^2+5)+xcdot(1-2x),,$$
and
$$21=4cdot(x^2+5)+(1+2x)cdot(1-2x),.$$
answered Dec 9 '18 at 0:25
BatominovskiBatominovski
33.1k33293
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1
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Good! but how you come up with those coefficients? The most difficult for me was finding the appropriate coefficients.
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– ZFR
Dec 9 '18 at 0:26
1
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My first goal was to find an integer $c$ that is definitely in the ideal. You want to eliminate $x^2$ from $x^2+5$. Say, your linear polynomial is $1-2x$. A multiple of $1-2x$ that only has $x^2$ and a constant term is $(1+2x)cdot(1-2x)$. This calculation leads to the last line of each case.
$endgroup$
– Batominovski
Dec 9 '18 at 0:34
1
$begingroup$
To find a linear polynomial $ax+b$ that is in the ideal is similar. You want to eliminate $x^2$ from $x^2+5$, but you want to leave the terms of order $1$ and $0$. A multiple of $1-2x$ that can eliminate $x^2$ from $x^2+5$ but leave a term of order $1$ is $xcdot(1-2x)$. This is the calculation in the second-to-last line of each case.
$endgroup$
– Batominovski
Dec 9 '18 at 0:38
1
$begingroup$
I want to note that you can't just use $1-2x$ for $ax+b$. It turns out that $langle 21,1-2xrangle$ is not the same as $langle 1-2x,x^2+5rangle$. If $|a|neq 1$, then you can never generate a monic polynomial $x^2+5$. Thus, the end result should have $a=1$ (or $a=-1$, but the sign doesn't really matter).
$endgroup$
– Batominovski
Dec 9 '18 at 0:43
add a comment |
$begingroup$
Note that $langle 1-4x,x^2+5rangle=langle x+20,81rangle$. This is because
$$x^2+5=(x-20)cdot(x+20)+5cdot 81,,$$
$$1-4x=(-4)cdot (x+20)+1cdot 81,,$$
$$x+20=4cdot(x^2+5)+xcdot(1-4x),,$$
and
$$81=16cdot(x^2+5)+(1+4x)cdot(1-4x),.$$
Similarly, $langle 1-2x,x^2+5rangle=langle x+10,21rangle$. This is because
$$x^2+5=(x-10)cdot(x+10)+5cdot 21,,$$
$$1-2x=(-2)cdot(x+10)+1cdot 21,,$$
$$x+10=2cdot(x^2+5)+xcdot(1-2x),,$$
and
$$21=4cdot(x^2+5)+(1+2x)cdot(1-2x),.$$
answered Dec 9 '18 at 0:25
BatominovskiBatominovski
33.1k33293
$endgroup$
1
$begingroup$
Good! but how you come up with those coefficients? The most difficult for me was finding the appropriate coefficients.
$endgroup$
– ZFR
Dec 9 '18 at 0:26
1
$begingroup$
My first goal was to find an integer $c$ that is definitely in the ideal. You want to eliminate $x^2$ from $x^2+5$. Say, your linear polynomial is $1-2x$. A multiple of $1-2x$ that only has $x^2$ and a constant term is $(1+2x)cdot(1-2x)$. This calculation leads to the last line of each case.
$endgroup$
– Batominovski
Dec 9 '18 at 0:34
1
$begingroup$
To find a linear polynomial $ax+b$ that is in the ideal is similar. You want to eliminate $x^2$ from $x^2+5$, but you want to leave the terms of order $1$ and $0$. A multiple of $1-2x$ that can eliminate $x^2$ from $x^2+5$ but leave a term of order $1$ is $xcdot(1-2x)$. This is the calculation in the second-to-last line of each case.
$endgroup$
– Batominovski
Dec 9 '18 at 0:38
1
$begingroup$
I want to note that you can't just use $1-2x$ for $ax+b$. It turns out that $langle 21,1-2xrangle$ is not the same as $langle 1-2x,x^2+5rangle$. If $|a|neq 1$, then you can never generate a monic polynomial $x^2+5$. Thus, the end result should have $a=1$ (or $a=-1$, but the sign doesn't really matter).
$endgroup$
– Batominovski
Dec 9 '18 at 0:43
add a comment |
$begingroup$
Note that $langle 1-4x,x^2+5rangle=langle x+20,81rangle$. This is because
$$x^2+5=(x-20)cdot(x+20)+5cdot 81,,$$
$$1-4x=(-4)cdot (x+20)+1cdot 81,,$$
$$x+20=4cdot(x^2+5)+xcdot(1-4x),,$$
and
$$81=16cdot(x^2+5)+(1+4x)cdot(1-4x),.$$
Similarly, $langle 1-2x,x^2+5rangle=langle x+10,21rangle$. This is because
$$x^2+5=(x-10)cdot(x+10)+5cdot 21,,$$
$$1-2x=(-2)cdot(x+10)+1cdot 21,,$$
$$x+10=2cdot(x^2+5)+xcdot(1-2x),,$$
and
$$21=4cdot(x^2+5)+(1+2x)cdot(1-2x),.$$
answered Dec 9 '18 at 0:25
BatominovskiBatominovski
33.1k33293
$endgroup$
Note that $langle 1-4x,x^2+5rangle=langle x+20,81rangle$. This is because
$$x^2+5=(x-20)cdot(x+20)+5cdot 81,,$$
$$1-4x=(-4)cdot (x+20)+1cdot 81,,$$
$$x+20=4cdot(x^2+5)+xcdot(1-4x),,$$
and
$$81=16cdot(x^2+5)+(1+4x)cdot(1-4x),.$$
Similarly, $langle 1-2x,x^2+5rangle=langle x+10,21rangle$. This is because
$$x^2+5=(x-10)cdot(x+10)+5cdot 21,,$$
$$1-2x=(-2)cdot(x+10)+1cdot 21,,$$
$$x+10=2cdot(x^2+5)+xcdot(1-2x),,$$
and
$$21=4cdot(x^2+5)+(1+2x)cdot(1-2x),.$$
answered Dec 9 '18 at 0:25
BatominovskiBatominovski
33.1k33293
answered Dec 9 '18 at 0:25
BatominovskiBatominovski
33.1k33293
answered Dec 9 '18 at 0:25
BatominovskiBatominovski
33.1k33293
answered Dec 9 '18 at 0:25
BatominovskiBatominovski
33.1k33293
33.1k33293
1
$begingroup$
Good! but how you come up with those coefficients? The most difficult for me was finding the appropriate coefficients.
$endgroup$
– ZFR
Dec 9 '18 at 0:26
1
$begingroup$
My first goal was to find an integer $c$ that is definitely in the ideal. You want to eliminate $x^2$ from $x^2+5$. Say, your linear polynomial is $1-2x$. A multiple of $1-2x$ that only has $x^2$ and a constant term is $(1+2x)cdot(1-2x)$. This calculation leads to the last line of each case.
$endgroup$
– Batominovski
Dec 9 '18 at 0:34
1
$begingroup$
To find a linear polynomial $ax+b$ that is in the ideal is similar. You want to eliminate $x^2$ from $x^2+5$, but you want to leave the terms of order $1$ and $0$. A multiple of $1-2x$ that can eliminate $x^2$ from $x^2+5$ but leave a term of order $1$ is $xcdot(1-2x)$. This is the calculation in the second-to-last line of each case.
$endgroup$
– Batominovski
Dec 9 '18 at 0:38
1
$begingroup$
I want to note that you can't just use $1-2x$ for $ax+b$. It turns out that $langle 21,1-2xrangle$ is not the same as $langle 1-2x,x^2+5rangle$. If $|a|neq 1$, then you can never generate a monic polynomial $x^2+5$. Thus, the end result should have $a=1$ (or $a=-1$, but the sign doesn't really matter).
$endgroup$
– Batominovski
Dec 9 '18 at 0:43
add a comment |
1
$begingroup$
Good! but how you come up with those coefficients? The most difficult for me was finding the appropriate coefficients.
$endgroup$
– ZFR
Dec 9 '18 at 0:26
1
$begingroup$
My first goal was to find an integer $c$ that is definitely in the ideal. You want to eliminate $x^2$ from $x^2+5$. Say, your linear polynomial is $1-2x$. A multiple of $1-2x$ that only has $x^2$ and a constant term is $(1+2x)cdot(1-2x)$. This calculation leads to the last line of each case.
$endgroup$
– Batominovski
Dec 9 '18 at 0:34
1
$begingroup$
To find a linear polynomial $ax+b$ that is in the ideal is similar. You want to eliminate $x^2$ from $x^2+5$, but you want to leave the terms of order $1$ and $0$. A multiple of $1-2x$ that can eliminate $x^2$ from $x^2+5$ but leave a term of order $1$ is $xcdot(1-2x)$. This is the calculation in the second-to-last line of each case.
$endgroup$
– Batominovski
Dec 9 '18 at 0:38
1
$begingroup$
I want to note that you can't just use $1-2x$ for $ax+b$. It turns out that $langle 21,1-2xrangle$ is not the same as $langle 1-2x,x^2+5rangle$. If $|a|neq 1$, then you can never generate a monic polynomial $x^2+5$. Thus, the end result should have $a=1$ (or $a=-1$, but the sign doesn't really matter).
$endgroup$
– Batominovski
Dec 9 '18 at 0:43
1
1
$begingroup$
Good! but how you come up with those coefficients? The most difficult for me was finding the appropriate coefficients.
$endgroup$
– ZFR
Dec 9 '18 at 0:26
$begingroup$
Good! but how you come up with those coefficients? The most difficult for me was finding the appropriate coefficients.
$endgroup$
– ZFR
Dec 9 '18 at 0:26
1
1
$begingroup$
My first goal was to find an integer $c$ that is definitely in the ideal. You want to eliminate $x^2$ from $x^2+5$. Say, your linear polynomial is $1-2x$. A multiple of $1-2x$ that only has $x^2$ and a constant term is $(1+2x)cdot(1-2x)$. This calculation leads to the last line of each case.
$endgroup$
– Batominovski
Dec 9 '18 at 0:34
$begingroup$
My first goal was to find an integer $c$ that is definitely in the ideal. You want to eliminate $x^2$ from $x^2+5$. Say, your linear polynomial is $1-2x$. A multiple of $1-2x$ that only has $x^2$ and a constant term is $(1+2x)cdot(1-2x)$. This calculation leads to the last line of each case.
$endgroup$
– Batominovski
Dec 9 '18 at 0:34
1
1
$begingroup$
To find a linear polynomial $ax+b$ that is in the ideal is similar. You want to eliminate $x^2$ from $x^2+5$, but you want to leave the terms of order $1$ and $0$. A multiple of $1-2x$ that can eliminate $x^2$ from $x^2+5$ but leave a term of order $1$ is $xcdot(1-2x)$. This is the calculation in the second-to-last line of each case.
$endgroup$
– Batominovski
Dec 9 '18 at 0:38
$begingroup$
To find a linear polynomial $ax+b$ that is in the ideal is similar. You want to eliminate $x^2$ from $x^2+5$, but you want to leave the terms of order $1$ and $0$. A multiple of $1-2x$ that can eliminate $x^2$ from $x^2+5$ but leave a term of order $1$ is $xcdot(1-2x)$. This is the calculation in the second-to-last line of each case.
$endgroup$
– Batominovski
Dec 9 '18 at 0:38
1
1
$begingroup$
I want to note that you can't just use $1-2x$ for $ax+b$. It turns out that $langle 21,1-2xrangle$ is not the same as $langle 1-2x,x^2+5rangle$. If $|a|neq 1$, then you can never generate a monic polynomial $x^2+5$. Thus, the end result should have $a=1$ (or $a=-1$, but the sign doesn't really matter).
$endgroup$
– Batominovski
Dec 9 '18 at 0:43
$begingroup$
I want to note that you can't just use $1-2x$ for $ax+b$. It turns out that $langle 21,1-2xrangle$ is not the same as $langle 1-2x,x^2+5rangle$. If $|a|neq 1$, then you can never generate a monic polynomial $x^2+5$. Thus, the end result should have $a=1$ (or $a=-1$, but the sign doesn't really matter).
$endgroup$
– Batominovski
Dec 9 '18 at 0:43
add a comment |
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what are you trying to do?
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– Andres Mejia
Dec 8 '18 at 23:50
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@AndresMejia, sorry, didn't understand your question?
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– ZFR
Dec 8 '18 at 23:51
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"I am not able to do the same..." what is "the same?"
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– Andres Mejia
Dec 8 '18 at 23:51
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@AndresMejia, I would like to show that these ideals is something like $(c,ax+b)$. Is it clear right now?
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– ZFR
Dec 8 '18 at 23:53
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yep! thanks :)${}{}$
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– Andres Mejia
Dec 8 '18 at 23:53