Ideals generated by two elements in $mathbb{Z}[x]$












2












$begingroup$


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







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$endgroup$








  • 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
















2












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














2












2








2





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 0:27









Batominovski

33.1k33293




33.1k33293










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














  • 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










2 Answers
2






active

oldest

votes


















3












$begingroup$

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)$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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





















1












$begingroup$

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
$$21=4cdot(x^2+5)+(1+2x)cdot(1-2x),.$$






share|cite|improve this answer









$endgroup$









  • 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













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2 Answers
2






active

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2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

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)$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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


















3












$begingroup$

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)$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















3












3








3





$begingroup$

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)$






share|cite|improve this answer











$endgroup$



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)$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 9 '18 at 0:54

























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




















  • $begingroup$
    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 $4x equiv 1$.
$endgroup$
– Andres Mejia
Dec 8 '18 at 23:59




$begingroup$
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






$begingroup$
@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




$begingroup$
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




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






$begingroup$
@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













1












$begingroup$

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
$$21=4cdot(x^2+5)+(1+2x)cdot(1-2x),.$$






share|cite|improve this answer









$endgroup$









  • 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


















1












$begingroup$

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
$$21=4cdot(x^2+5)+(1+2x)cdot(1-2x),.$$






share|cite|improve this answer









$endgroup$









  • 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
















1












1








1





$begingroup$

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
$$21=4cdot(x^2+5)+(1+2x)cdot(1-2x),.$$






share|cite|improve this answer









$endgroup$



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
$$21=4cdot(x^2+5)+(1+2x)cdot(1-2x),.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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
















  • 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




















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