Adapting the Chinese Remainder Theorem (CRT) for integers to polynomials












2












$begingroup$


I did a few examples using the CRT to solve congruences where everything was in terms of integers. I'm trying to use the same technique for polynomials over $mathbb{Q}$, but I'm getting stuck.




Here's an example with integers:



$begin{cases}x equiv 1 , (mathrm{mod} , 5) \
x equiv 2 , (mathrm{mod} , 7) \
x equiv 3 , (mathrm{mod} , 9) \
x equiv 4 , (mathrm{mod} , 11).
end{cases}$



Since all the moduli are pairwise relatively prime, we can use the CRT. Here's some notation I'm using:



$bullet , M$ denotes the product of the moduli (in this case, $M = 5 cdot7 cdot 9 cdot 11$)



$bullet , m_i $ denotes the modulus in the $i^{mathrm{th}}$ congruence



$bullet , M_i$ denotes $dfrac{M}{m_i}$



$bullet , y_i$ denotes the inverse of $M_i$ (mod $m_i$), i.e. $y_i$ satisfies $y_i M_i equiv 1$ (mod $m_i$).



Then $x = displaystyle sum_{i = 1}^n a_iM_iy_i$, and this solution is unique (mod $M$).




Now I want to apply the same technique to the following:



$begin{cases}
f(x) equiv 1 , (mathrm{ mod } , x^2 + 1) \
f(x) equiv x , (mathrm{mod} , x^4),
end{cases}$



where $f(x) in mathbb{Q}(x)$. Having checked that the moduli are relatively prime, we should be able to use the CRT. Using the notation above, I have the following:



$M = (x^4)(x^2 + 1)$



$M_1 = x^4$



$M_2 = x^2 + 1$



Here's where I run into a problem. I need to find $y_1, y_2$ such that



$begin{cases}
y_1 (x^4) equiv 1 , (mathrm{mod} , x^2 + 1) \
y_2 (x^2+1) equiv 1 , (mathrm{mod} , x^4).
end{cases}$



But how does one find $y_1, y_2$?






abstract-algebra ring-theory chinese-remainder-theorem







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    "Having checked that the moduli are relatively prime..." but that means precisely that there exist $p_1(x)$ and $p_2(x)$ such that $p_1(x)x^4 + p_2(x)(x^2+1)=1$.
    $endgroup$
    – kccu
    Apr 21 at 19:23










  • $begingroup$
    Are you saying that I can find $p_1(x)$ and $p_2(x)$ in general by using the extended Euclidean algorithm, and that $p_1(x)$ and $p_2(x)$ are precisely my $y_1$ and $y_2$?
    $endgroup$
    – Junglemath
    Apr 21 at 19:56
















2












$begingroup$


I did a few examples using the CRT to solve congruences where everything was in terms of integers. I'm trying to use the same technique for polynomials over $mathbb{Q}$, but I'm getting stuck.




Here's an example with integers:



$begin{cases}x equiv 1 , (mathrm{mod} , 5) \
x equiv 2 , (mathrm{mod} , 7) \
x equiv 3 , (mathrm{mod} , 9) \
x equiv 4 , (mathrm{mod} , 11).
end{cases}$



Since all the moduli are pairwise relatively prime, we can use the CRT. Here's some notation I'm using:



$bullet , M$ denotes the product of the moduli (in this case, $M = 5 cdot7 cdot 9 cdot 11$)



$bullet , m_i $ denotes the modulus in the $i^{mathrm{th}}$ congruence



$bullet , M_i$ denotes $dfrac{M}{m_i}$



$bullet , y_i$ denotes the inverse of $M_i$ (mod $m_i$), i.e. $y_i$ satisfies $y_i M_i equiv 1$ (mod $m_i$).



Then $x = displaystyle sum_{i = 1}^n a_iM_iy_i$, and this solution is unique (mod $M$).




Now I want to apply the same technique to the following:



$begin{cases}
f(x) equiv 1 , (mathrm{ mod } , x^2 + 1) \
f(x) equiv x , (mathrm{mod} , x^4),
end{cases}$



where $f(x) in mathbb{Q}(x)$. Having checked that the moduli are relatively prime, we should be able to use the CRT. Using the notation above, I have the following:



$M = (x^4)(x^2 + 1)$



$M_1 = x^4$



$M_2 = x^2 + 1$



Here's where I run into a problem. I need to find $y_1, y_2$ such that



$begin{cases}
y_1 (x^4) equiv 1 , (mathrm{mod} , x^2 + 1) \
y_2 (x^2+1) equiv 1 , (mathrm{mod} , x^4).
end{cases}$



But how does one find $y_1, y_2$?






abstract-algebra ring-theory chinese-remainder-theorem







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    "Having checked that the moduli are relatively prime..." but that means precisely that there exist $p_1(x)$ and $p_2(x)$ such that $p_1(x)x^4 + p_2(x)(x^2+1)=1$.
    $endgroup$
    – kccu
    Apr 21 at 19:23










  • $begingroup$
    Are you saying that I can find $p_1(x)$ and $p_2(x)$ in general by using the extended Euclidean algorithm, and that $p_1(x)$ and $p_2(x)$ are precisely my $y_1$ and $y_2$?
    $endgroup$
    – Junglemath
    Apr 21 at 19:56














2












2








2





$begingroup$


I did a few examples using the CRT to solve congruences where everything was in terms of integers. I'm trying to use the same technique for polynomials over $mathbb{Q}$, but I'm getting stuck.




Here's an example with integers:



$begin{cases}x equiv 1 , (mathrm{mod} , 5) \
x equiv 2 , (mathrm{mod} , 7) \
x equiv 3 , (mathrm{mod} , 9) \
x equiv 4 , (mathrm{mod} , 11).
end{cases}$



Since all the moduli are pairwise relatively prime, we can use the CRT. Here's some notation I'm using:



$bullet , M$ denotes the product of the moduli (in this case, $M = 5 cdot7 cdot 9 cdot 11$)



$bullet , m_i $ denotes the modulus in the $i^{mathrm{th}}$ congruence



$bullet , M_i$ denotes $dfrac{M}{m_i}$



$bullet , y_i$ denotes the inverse of $M_i$ (mod $m_i$), i.e. $y_i$ satisfies $y_i M_i equiv 1$ (mod $m_i$).



Then $x = displaystyle sum_{i = 1}^n a_iM_iy_i$, and this solution is unique (mod $M$).




Now I want to apply the same technique to the following:



$begin{cases}
f(x) equiv 1 , (mathrm{ mod } , x^2 + 1) \
f(x) equiv x , (mathrm{mod} , x^4),
end{cases}$



where $f(x) in mathbb{Q}(x)$. Having checked that the moduli are relatively prime, we should be able to use the CRT. Using the notation above, I have the following:



$M = (x^4)(x^2 + 1)$



$M_1 = x^4$



$M_2 = x^2 + 1$



Here's where I run into a problem. I need to find $y_1, y_2$ such that



$begin{cases}
y_1 (x^4) equiv 1 , (mathrm{mod} , x^2 + 1) \
y_2 (x^2+1) equiv 1 , (mathrm{mod} , x^4).
end{cases}$



But how does one find $y_1, y_2$?






abstract-algebra ring-theory chinese-remainder-theorem







share|cite|improve this question









$endgroup$




I did a few examples using the CRT to solve congruences where everything was in terms of integers. I'm trying to use the same technique for polynomials over $mathbb{Q}$, but I'm getting stuck.




Here's an example with integers:



$begin{cases}x equiv 1 , (mathrm{mod} , 5) \
x equiv 2 , (mathrm{mod} , 7) \
x equiv 3 , (mathrm{mod} , 9) \
x equiv 4 , (mathrm{mod} , 11).
end{cases}$



Since all the moduli are pairwise relatively prime, we can use the CRT. Here's some notation I'm using:



$bullet , M$ denotes the product of the moduli (in this case, $M = 5 cdot7 cdot 9 cdot 11$)



$bullet , m_i $ denotes the modulus in the $i^{mathrm{th}}$ congruence



$bullet , M_i$ denotes $dfrac{M}{m_i}$



$bullet , y_i$ denotes the inverse of $M_i$ (mod $m_i$), i.e. $y_i$ satisfies $y_i M_i equiv 1$ (mod $m_i$).



Then $x = displaystyle sum_{i = 1}^n a_iM_iy_i$, and this solution is unique (mod $M$).




Now I want to apply the same technique to the following:



$begin{cases}
f(x) equiv 1 , (mathrm{ mod } , x^2 + 1) \
f(x) equiv x , (mathrm{mod} , x^4),
end{cases}$



where $f(x) in mathbb{Q}(x)$. Having checked that the moduli are relatively prime, we should be able to use the CRT. Using the notation above, I have the following:



$M = (x^4)(x^2 + 1)$



$M_1 = x^4$



$M_2 = x^2 + 1$



Here's where I run into a problem. I need to find $y_1, y_2$ such that



$begin{cases}
y_1 (x^4) equiv 1 , (mathrm{mod} , x^2 + 1) \
y_2 (x^2+1) equiv 1 , (mathrm{mod} , x^4).
end{cases}$



But how does one find $y_1, y_2$?






abstract-algebra ring-theory chinese-remainder-theorem




abstract-algebra ring-theory chinese-remainder-theorem






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Apr 21 at 19:16









JunglemathJunglemath

6017




6017








  • 1




    $begingroup$
    "Having checked that the moduli are relatively prime..." but that means precisely that there exist $p_1(x)$ and $p_2(x)$ such that $p_1(x)x^4 + p_2(x)(x^2+1)=1$.
    $endgroup$
    – kccu
    Apr 21 at 19:23










  • $begingroup$
    Are you saying that I can find $p_1(x)$ and $p_2(x)$ in general by using the extended Euclidean algorithm, and that $p_1(x)$ and $p_2(x)$ are precisely my $y_1$ and $y_2$?
    $endgroup$
    – Junglemath
    Apr 21 at 19:56














  • 1




    $begingroup$
    "Having checked that the moduli are relatively prime..." but that means precisely that there exist $p_1(x)$ and $p_2(x)$ such that $p_1(x)x^4 + p_2(x)(x^2+1)=1$.
    $endgroup$
    – kccu
    Apr 21 at 19:23










  • $begingroup$
    Are you saying that I can find $p_1(x)$ and $p_2(x)$ in general by using the extended Euclidean algorithm, and that $p_1(x)$ and $p_2(x)$ are precisely my $y_1$ and $y_2$?
    $endgroup$
    – Junglemath
    Apr 21 at 19:56








1




1




$begingroup$
"Having checked that the moduli are relatively prime..." but that means precisely that there exist $p_1(x)$ and $p_2(x)$ such that $p_1(x)x^4 + p_2(x)(x^2+1)=1$.
$endgroup$
– kccu
Apr 21 at 19:23




$begingroup$
"Having checked that the moduli are relatively prime..." but that means precisely that there exist $p_1(x)$ and $p_2(x)$ such that $p_1(x)x^4 + p_2(x)(x^2+1)=1$.
$endgroup$
– kccu
Apr 21 at 19:23












$begingroup$
Are you saying that I can find $p_1(x)$ and $p_2(x)$ in general by using the extended Euclidean algorithm, and that $p_1(x)$ and $p_2(x)$ are precisely my $y_1$ and $y_2$?
$endgroup$
– Junglemath
Apr 21 at 19:56




$begingroup$
Are you saying that I can find $p_1(x)$ and $p_2(x)$ in general by using the extended Euclidean algorithm, and that $p_1(x)$ and $p_2(x)$ are precisely my $y_1$ and $y_2$?
$endgroup$
– Junglemath
Apr 21 at 19:56










2 Answers
2






active

oldest

votes


















3












$begingroup$

To find $y_1$ and $y_2$ consider solving the problem
$$y_1x^4+y_2(x^2+1)=1.$$
This is not always easy to solve, but in this case a solution comes to mind. Note that by difference of squares
$$(x^2-1)(x^2+1)=x^4-1,$$
hence
$$x^4+[(-1)(x^2-1)](x^2+1)=1.$$
This tells us that we can choose
$$y_1=1,$$
$$y_2=(1-x^2).$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Is there an algorithmic way of solving these, rather than relying on intuition?
    $endgroup$
    – Junglemath
    Apr 21 at 19:59










  • $begingroup$
    @Junglemath By Euclidean algorithm you can find two polynomials $p(x), q(x) in mathbb{Q}[x]$ such that $p(x) x^4 + q(x) (x^2 + 1) = 1$.
    $endgroup$
    – Paolo
    Apr 21 at 20:09








  • 2




    $begingroup$
    @Junglemath The polynomials over a field form a Euclidean Domain, so yes, there is. If the gcd of $f_1,f_2inmathbb{Q}[x]$ is a unit, then you can perform the Euclidean Algorithm to find their gcd. Now, you can reverse the algorithm to write them as a linear combination of their gcd the same way you would for integers. I said not easy, because the process can be time consuming and very tedious.
    $endgroup$
    – Melody
    Apr 21 at 20:10








  • 2




    $begingroup$
    @Junglemath I didn't really convert it. I knew in advanced the single equation had a solution. This is because $x^4$ and $x^2+1$ have no common roots, hence no common irreducible factors. This means they are relatively prime, in which case we can write $1$ as a linear combination. Knowing that, I knew solving the single equation would give rise to a solution to the congruence equations. This is completely analogous to how you can solve everything over the integers.
    $endgroup$
    – Melody
    Apr 21 at 20:22








  • 2




    $begingroup$
    @Junglemath I describe here at length this method of scaling the Bezout equation into a CRT solution.
    $endgroup$
    – Bill Dubuque
    Apr 21 at 21:28





















2












$begingroup$

Bu applying $ abbmod ac, =, a(bbmod c) $ [Mod Distributive Law] $ $ it is a bit simpler:



$ f-x,bmod, {x^{large 4}(x^{large 2}!+!1)}, =, x^{large 4}underbrace{{left[dfrac{color{#c00}f-x}{color{#0a0}{x^{large 4}}}bmod {x^{large 2}!+!1}right]}}_{large color{#0a0}{x^{Large 4}} equiv 1 {rm by} x^{Large 2} equiv -1 } =, x^{large 4}[1-x], $ by $,color{#c00}fequiv 1pmod{!x^{large 2}!+!1}$



Remark $ $ Here are further examples done using MDL (an operational form of CRT).



You can find further details here on transforming the Bezout equation into a CRT solution (the method sketched in Melody's answer).






share|cite|improve this answer











$endgroup$














    Your Answer








    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3196235%2fadapting-the-chinese-remainder-theorem-crt-for-integers-to-polynomials%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    To find $y_1$ and $y_2$ consider solving the problem
    $$y_1x^4+y_2(x^2+1)=1.$$
    This is not always easy to solve, but in this case a solution comes to mind. Note that by difference of squares
    $$(x^2-1)(x^2+1)=x^4-1,$$
    hence
    $$x^4+[(-1)(x^2-1)](x^2+1)=1.$$
    This tells us that we can choose
    $$y_1=1,$$
    $$y_2=(1-x^2).$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Is there an algorithmic way of solving these, rather than relying on intuition?
      $endgroup$
      – Junglemath
      Apr 21 at 19:59










    • $begingroup$
      @Junglemath By Euclidean algorithm you can find two polynomials $p(x), q(x) in mathbb{Q}[x]$ such that $p(x) x^4 + q(x) (x^2 + 1) = 1$.
      $endgroup$
      – Paolo
      Apr 21 at 20:09








    • 2




      $begingroup$
      @Junglemath The polynomials over a field form a Euclidean Domain, so yes, there is. If the gcd of $f_1,f_2inmathbb{Q}[x]$ is a unit, then you can perform the Euclidean Algorithm to find their gcd. Now, you can reverse the algorithm to write them as a linear combination of their gcd the same way you would for integers. I said not easy, because the process can be time consuming and very tedious.
      $endgroup$
      – Melody
      Apr 21 at 20:10








    • 2




      $begingroup$
      @Junglemath I didn't really convert it. I knew in advanced the single equation had a solution. This is because $x^4$ and $x^2+1$ have no common roots, hence no common irreducible factors. This means they are relatively prime, in which case we can write $1$ as a linear combination. Knowing that, I knew solving the single equation would give rise to a solution to the congruence equations. This is completely analogous to how you can solve everything over the integers.
      $endgroup$
      – Melody
      Apr 21 at 20:22








    • 2




      $begingroup$
      @Junglemath I describe here at length this method of scaling the Bezout equation into a CRT solution.
      $endgroup$
      – Bill Dubuque
      Apr 21 at 21:28


















    3












    $begingroup$

    To find $y_1$ and $y_2$ consider solving the problem
    $$y_1x^4+y_2(x^2+1)=1.$$
    This is not always easy to solve, but in this case a solution comes to mind. Note that by difference of squares
    $$(x^2-1)(x^2+1)=x^4-1,$$
    hence
    $$x^4+[(-1)(x^2-1)](x^2+1)=1.$$
    This tells us that we can choose
    $$y_1=1,$$
    $$y_2=(1-x^2).$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Is there an algorithmic way of solving these, rather than relying on intuition?
      $endgroup$
      – Junglemath
      Apr 21 at 19:59










    • $begingroup$
      @Junglemath By Euclidean algorithm you can find two polynomials $p(x), q(x) in mathbb{Q}[x]$ such that $p(x) x^4 + q(x) (x^2 + 1) = 1$.
      $endgroup$
      – Paolo
      Apr 21 at 20:09








    • 2




      $begingroup$
      @Junglemath The polynomials over a field form a Euclidean Domain, so yes, there is. If the gcd of $f_1,f_2inmathbb{Q}[x]$ is a unit, then you can perform the Euclidean Algorithm to find their gcd. Now, you can reverse the algorithm to write them as a linear combination of their gcd the same way you would for integers. I said not easy, because the process can be time consuming and very tedious.
      $endgroup$
      – Melody
      Apr 21 at 20:10








    • 2




      $begingroup$
      @Junglemath I didn't really convert it. I knew in advanced the single equation had a solution. This is because $x^4$ and $x^2+1$ have no common roots, hence no common irreducible factors. This means they are relatively prime, in which case we can write $1$ as a linear combination. Knowing that, I knew solving the single equation would give rise to a solution to the congruence equations. This is completely analogous to how you can solve everything over the integers.
      $endgroup$
      – Melody
      Apr 21 at 20:22








    • 2




      $begingroup$
      @Junglemath I describe here at length this method of scaling the Bezout equation into a CRT solution.
      $endgroup$
      – Bill Dubuque
      Apr 21 at 21:28
















    3












    3








    3





    $begingroup$

    To find $y_1$ and $y_2$ consider solving the problem
    $$y_1x^4+y_2(x^2+1)=1.$$
    This is not always easy to solve, but in this case a solution comes to mind. Note that by difference of squares
    $$(x^2-1)(x^2+1)=x^4-1,$$
    hence
    $$x^4+[(-1)(x^2-1)](x^2+1)=1.$$
    This tells us that we can choose
    $$y_1=1,$$
    $$y_2=(1-x^2).$$






    share|cite|improve this answer











    $endgroup$



    To find $y_1$ and $y_2$ consider solving the problem
    $$y_1x^4+y_2(x^2+1)=1.$$
    This is not always easy to solve, but in this case a solution comes to mind. Note that by difference of squares
    $$(x^2-1)(x^2+1)=x^4-1,$$
    hence
    $$x^4+[(-1)(x^2-1)](x^2+1)=1.$$
    This tells us that we can choose
    $$y_1=1,$$
    $$y_2=(1-x^2).$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Apr 21 at 19:53

























    answered Apr 21 at 19:26









    MelodyMelody

    1,46112




    1,46112












    • $begingroup$
      Is there an algorithmic way of solving these, rather than relying on intuition?
      $endgroup$
      – Junglemath
      Apr 21 at 19:59










    • $begingroup$
      @Junglemath By Euclidean algorithm you can find two polynomials $p(x), q(x) in mathbb{Q}[x]$ such that $p(x) x^4 + q(x) (x^2 + 1) = 1$.
      $endgroup$
      – Paolo
      Apr 21 at 20:09








    • 2




      $begingroup$
      @Junglemath The polynomials over a field form a Euclidean Domain, so yes, there is. If the gcd of $f_1,f_2inmathbb{Q}[x]$ is a unit, then you can perform the Euclidean Algorithm to find their gcd. Now, you can reverse the algorithm to write them as a linear combination of their gcd the same way you would for integers. I said not easy, because the process can be time consuming and very tedious.
      $endgroup$
      – Melody
      Apr 21 at 20:10








    • 2




      $begingroup$
      @Junglemath I didn't really convert it. I knew in advanced the single equation had a solution. This is because $x^4$ and $x^2+1$ have no common roots, hence no common irreducible factors. This means they are relatively prime, in which case we can write $1$ as a linear combination. Knowing that, I knew solving the single equation would give rise to a solution to the congruence equations. This is completely analogous to how you can solve everything over the integers.
      $endgroup$
      – Melody
      Apr 21 at 20:22








    • 2




      $begingroup$
      @Junglemath I describe here at length this method of scaling the Bezout equation into a CRT solution.
      $endgroup$
      – Bill Dubuque
      Apr 21 at 21:28




















    • $begingroup$
      Is there an algorithmic way of solving these, rather than relying on intuition?
      $endgroup$
      – Junglemath
      Apr 21 at 19:59










    • $begingroup$
      @Junglemath By Euclidean algorithm you can find two polynomials $p(x), q(x) in mathbb{Q}[x]$ such that $p(x) x^4 + q(x) (x^2 + 1) = 1$.
      $endgroup$
      – Paolo
      Apr 21 at 20:09








    • 2




      $begingroup$
      @Junglemath The polynomials over a field form a Euclidean Domain, so yes, there is. If the gcd of $f_1,f_2inmathbb{Q}[x]$ is a unit, then you can perform the Euclidean Algorithm to find their gcd. Now, you can reverse the algorithm to write them as a linear combination of their gcd the same way you would for integers. I said not easy, because the process can be time consuming and very tedious.
      $endgroup$
      – Melody
      Apr 21 at 20:10








    • 2




      $begingroup$
      @Junglemath I didn't really convert it. I knew in advanced the single equation had a solution. This is because $x^4$ and $x^2+1$ have no common roots, hence no common irreducible factors. This means they are relatively prime, in which case we can write $1$ as a linear combination. Knowing that, I knew solving the single equation would give rise to a solution to the congruence equations. This is completely analogous to how you can solve everything over the integers.
      $endgroup$
      – Melody
      Apr 21 at 20:22








    • 2




      $begingroup$
      @Junglemath I describe here at length this method of scaling the Bezout equation into a CRT solution.
      $endgroup$
      – Bill Dubuque
      Apr 21 at 21:28


















    $begingroup$
    Is there an algorithmic way of solving these, rather than relying on intuition?
    $endgroup$
    – Junglemath
    Apr 21 at 19:59




    $begingroup$
    Is there an algorithmic way of solving these, rather than relying on intuition?
    $endgroup$
    – Junglemath
    Apr 21 at 19:59












    $begingroup$
    @Junglemath By Euclidean algorithm you can find two polynomials $p(x), q(x) in mathbb{Q}[x]$ such that $p(x) x^4 + q(x) (x^2 + 1) = 1$.
    $endgroup$
    – Paolo
    Apr 21 at 20:09






    $begingroup$
    @Junglemath By Euclidean algorithm you can find two polynomials $p(x), q(x) in mathbb{Q}[x]$ such that $p(x) x^4 + q(x) (x^2 + 1) = 1$.
    $endgroup$
    – Paolo
    Apr 21 at 20:09






    2




    2




    $begingroup$
    @Junglemath The polynomials over a field form a Euclidean Domain, so yes, there is. If the gcd of $f_1,f_2inmathbb{Q}[x]$ is a unit, then you can perform the Euclidean Algorithm to find their gcd. Now, you can reverse the algorithm to write them as a linear combination of their gcd the same way you would for integers. I said not easy, because the process can be time consuming and very tedious.
    $endgroup$
    – Melody
    Apr 21 at 20:10






    $begingroup$
    @Junglemath The polynomials over a field form a Euclidean Domain, so yes, there is. If the gcd of $f_1,f_2inmathbb{Q}[x]$ is a unit, then you can perform the Euclidean Algorithm to find their gcd. Now, you can reverse the algorithm to write them as a linear combination of their gcd the same way you would for integers. I said not easy, because the process can be time consuming and very tedious.
    $endgroup$
    – Melody
    Apr 21 at 20:10






    2




    2




    $begingroup$
    @Junglemath I didn't really convert it. I knew in advanced the single equation had a solution. This is because $x^4$ and $x^2+1$ have no common roots, hence no common irreducible factors. This means they are relatively prime, in which case we can write $1$ as a linear combination. Knowing that, I knew solving the single equation would give rise to a solution to the congruence equations. This is completely analogous to how you can solve everything over the integers.
    $endgroup$
    – Melody
    Apr 21 at 20:22






    $begingroup$
    @Junglemath I didn't really convert it. I knew in advanced the single equation had a solution. This is because $x^4$ and $x^2+1$ have no common roots, hence no common irreducible factors. This means they are relatively prime, in which case we can write $1$ as a linear combination. Knowing that, I knew solving the single equation would give rise to a solution to the congruence equations. This is completely analogous to how you can solve everything over the integers.
    $endgroup$
    – Melody
    Apr 21 at 20:22






    2




    2




    $begingroup$
    @Junglemath I describe here at length this method of scaling the Bezout equation into a CRT solution.
    $endgroup$
    – Bill Dubuque
    Apr 21 at 21:28






    $begingroup$
    @Junglemath I describe here at length this method of scaling the Bezout equation into a CRT solution.
    $endgroup$
    – Bill Dubuque
    Apr 21 at 21:28













    2












    $begingroup$

    Bu applying $ abbmod ac, =, a(bbmod c) $ [Mod Distributive Law] $ $ it is a bit simpler:



    $ f-x,bmod, {x^{large 4}(x^{large 2}!+!1)}, =, x^{large 4}underbrace{{left[dfrac{color{#c00}f-x}{color{#0a0}{x^{large 4}}}bmod {x^{large 2}!+!1}right]}}_{large color{#0a0}{x^{Large 4}} equiv 1 {rm by} x^{Large 2} equiv -1 } =, x^{large 4}[1-x], $ by $,color{#c00}fequiv 1pmod{!x^{large 2}!+!1}$



    Remark $ $ Here are further examples done using MDL (an operational form of CRT).



    You can find further details here on transforming the Bezout equation into a CRT solution (the method sketched in Melody's answer).






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      Bu applying $ abbmod ac, =, a(bbmod c) $ [Mod Distributive Law] $ $ it is a bit simpler:



      $ f-x,bmod, {x^{large 4}(x^{large 2}!+!1)}, =, x^{large 4}underbrace{{left[dfrac{color{#c00}f-x}{color{#0a0}{x^{large 4}}}bmod {x^{large 2}!+!1}right]}}_{large color{#0a0}{x^{Large 4}} equiv 1 {rm by} x^{Large 2} equiv -1 } =, x^{large 4}[1-x], $ by $,color{#c00}fequiv 1pmod{!x^{large 2}!+!1}$



      Remark $ $ Here are further examples done using MDL (an operational form of CRT).



      You can find further details here on transforming the Bezout equation into a CRT solution (the method sketched in Melody's answer).






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Bu applying $ abbmod ac, =, a(bbmod c) $ [Mod Distributive Law] $ $ it is a bit simpler:



        $ f-x,bmod, {x^{large 4}(x^{large 2}!+!1)}, =, x^{large 4}underbrace{{left[dfrac{color{#c00}f-x}{color{#0a0}{x^{large 4}}}bmod {x^{large 2}!+!1}right]}}_{large color{#0a0}{x^{Large 4}} equiv 1 {rm by} x^{Large 2} equiv -1 } =, x^{large 4}[1-x], $ by $,color{#c00}fequiv 1pmod{!x^{large 2}!+!1}$



        Remark $ $ Here are further examples done using MDL (an operational form of CRT).



        You can find further details here on transforming the Bezout equation into a CRT solution (the method sketched in Melody's answer).






        share|cite|improve this answer











        $endgroup$



        Bu applying $ abbmod ac, =, a(bbmod c) $ [Mod Distributive Law] $ $ it is a bit simpler:



        $ f-x,bmod, {x^{large 4}(x^{large 2}!+!1)}, =, x^{large 4}underbrace{{left[dfrac{color{#c00}f-x}{color{#0a0}{x^{large 4}}}bmod {x^{large 2}!+!1}right]}}_{large color{#0a0}{x^{Large 4}} equiv 1 {rm by} x^{Large 2} equiv -1 } =, x^{large 4}[1-x], $ by $,color{#c00}fequiv 1pmod{!x^{large 2}!+!1}$



        Remark $ $ Here are further examples done using MDL (an operational form of CRT).



        You can find further details here on transforming the Bezout equation into a CRT solution (the method sketched in Melody's answer).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Apr 21 at 21:41

























        answered Apr 21 at 21:25









        Bill DubuqueBill Dubuque

        214k29198660




        214k29198660






























            draft saved

            draft discarded




















































            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.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3196235%2fadapting-the-chinese-remainder-theorem-crt-for-integers-to-polynomials%23new-answer', 'question_page');
            }
            );

            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







            -

            Popular posts from this blog

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
            Read more

            In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
            Read more

            How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

            Image
            1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
            Read more