How can i find a inverse of a polynomial in a quotient ring?












3












$begingroup$


I am asked to find the inverse of $widehat {x^3+1} $ in Q/[x]/I where I = $x^2-2$.



I am used to find the inverse by keep doing the division until a find a irreducible elements but the degree of the polynomial i asked to find the inverse is larger than the degree of polynomial in I.



In previous question, we have calculated the inverse of $widehat {X} $ which is $1/2$(X). And i am trying to breaking down $X^3 - 1 $ into $X(X^2-2)$ + $(2X+1)$. Is this bit some hints to approach this question or is it a new idea to find the inverse with a polynomial of a larger degree?



Thanks a lot!!






abstract-algebra polynomials ring-theory inverse







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$endgroup$








  • 1




    $begingroup$
    This is more or less the same as computing $1/((sqrt2)^3+1)$ in surds.
    $endgroup$
    – Lord Shark the Unknown
    Feb 4 at 19:24










  • $begingroup$
    Your calculation (breaking down $x^3+1$) shows that $widehat{x^3+1}=widehat{2x+1}$. Go from there.
    $endgroup$
    – Jyrki Lahtonen
    Feb 4 at 19:32


















3












$begingroup$


I am asked to find the inverse of $widehat {x^3+1} $ in Q/[x]/I where I = $x^2-2$.



I am used to find the inverse by keep doing the division until a find a irreducible elements but the degree of the polynomial i asked to find the inverse is larger than the degree of polynomial in I.



In previous question, we have calculated the inverse of $widehat {X} $ which is $1/2$(X). And i am trying to breaking down $X^3 - 1 $ into $X(X^2-2)$ + $(2X+1)$. Is this bit some hints to approach this question or is it a new idea to find the inverse with a polynomial of a larger degree?



Thanks a lot!!






abstract-algebra polynomials ring-theory inverse







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    This is more or less the same as computing $1/((sqrt2)^3+1)$ in surds.
    $endgroup$
    – Lord Shark the Unknown
    Feb 4 at 19:24










  • $begingroup$
    Your calculation (breaking down $x^3+1$) shows that $widehat{x^3+1}=widehat{2x+1}$. Go from there.
    $endgroup$
    – Jyrki Lahtonen
    Feb 4 at 19:32
















3












3








3





$begingroup$


I am asked to find the inverse of $widehat {x^3+1} $ in Q/[x]/I where I = $x^2-2$.



I am used to find the inverse by keep doing the division until a find a irreducible elements but the degree of the polynomial i asked to find the inverse is larger than the degree of polynomial in I.



In previous question, we have calculated the inverse of $widehat {X} $ which is $1/2$(X). And i am trying to breaking down $X^3 - 1 $ into $X(X^2-2)$ + $(2X+1)$. Is this bit some hints to approach this question or is it a new idea to find the inverse with a polynomial of a larger degree?



Thanks a lot!!






abstract-algebra polynomials ring-theory inverse







share|cite|improve this question











$endgroup$




I am asked to find the inverse of $widehat {x^3+1} $ in Q/[x]/I where I = $x^2-2$.



I am used to find the inverse by keep doing the division until a find a irreducible elements but the degree of the polynomial i asked to find the inverse is larger than the degree of polynomial in I.



In previous question, we have calculated the inverse of $widehat {X} $ which is $1/2$(X). And i am trying to breaking down $X^3 - 1 $ into $X(X^2-2)$ + $(2X+1)$. Is this bit some hints to approach this question or is it a new idea to find the inverse with a polynomial of a larger degree?



Thanks a lot!!






abstract-algebra polynomials ring-theory inverse




abstract-algebra polynomials ring-theory inverse






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 8 at 15:30







Thomas

















asked Feb 4 at 19:15









ThomasThomas

706




706








  • 1




    $begingroup$
    This is more or less the same as computing $1/((sqrt2)^3+1)$ in surds.
    $endgroup$
    – Lord Shark the Unknown
    Feb 4 at 19:24










  • $begingroup$
    Your calculation (breaking down $x^3+1$) shows that $widehat{x^3+1}=widehat{2x+1}$. Go from there.
    $endgroup$
    – Jyrki Lahtonen
    Feb 4 at 19:32
















  • 1




    $begingroup$
    This is more or less the same as computing $1/((sqrt2)^3+1)$ in surds.
    $endgroup$
    – Lord Shark the Unknown
    Feb 4 at 19:24










  • $begingroup$
    Your calculation (breaking down $x^3+1$) shows that $widehat{x^3+1}=widehat{2x+1}$. Go from there.
    $endgroup$
    – Jyrki Lahtonen
    Feb 4 at 19:32










1




1




$begingroup$
This is more or less the same as computing $1/((sqrt2)^3+1)$ in surds.
$endgroup$
– Lord Shark the Unknown
Feb 4 at 19:24




$begingroup$
This is more or less the same as computing $1/((sqrt2)^3+1)$ in surds.
$endgroup$
– Lord Shark the Unknown
Feb 4 at 19:24












$begingroup$
Your calculation (breaking down $x^3+1$) shows that $widehat{x^3+1}=widehat{2x+1}$. Go from there.
$endgroup$
– Jyrki Lahtonen
Feb 4 at 19:32






$begingroup$
Your calculation (breaking down $x^3+1$) shows that $widehat{x^3+1}=widehat{2x+1}$. Go from there.
$endgroup$
– Jyrki Lahtonen
Feb 4 at 19:32












3 Answers
3






active

oldest

votes


















3












$begingroup$

Hint $ color{#c00}{x^2 = 2} Rightarrow dfrac{1}{1+color{#c00}{x^2} x} ,=, dfrac{1}{1+color{#c00}2x} = dfrac{1}{1+2color{#c00}x}, dfrac{1-2x}{1-2color{#c00}x} = dfrac{1,-,2x }{1-4(color{#c00}2)}$



i.e. $ $ rationalize the denom of $, dfrac{1}{1+2sqrt{2}}.,$ More generally, use the Extended Euclidean Algorithm.



For much further discussion see here and its links.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    extended Euc:



    $$ left( x^{3} + 1 right) $$



    $$ left( x^{2} - 2 right) $$



    $$ left( x^{3} + 1 right) = left( x^{2} - 2 right) cdot color{magenta}{ left( x right) } + left( 2 x + 1 right) $$
    $$ left( x^{2} - 2 right) = left( 2 x + 1 right) cdot color{magenta}{ left( frac{ 2 x - 1 }{ 4 } right) } + left( frac{ -7}{4 } right) $$
    $$ left( 2 x + 1 right) = left( frac{ -7}{4 } right) cdot color{magenta}{ left( frac{ - 8 x - 4 }{ 7 } right) } + left( 0 right) $$
    $$ frac{ 0}{1} $$
    $$ frac{ 1}{0} $$
    $$ color{magenta}{ left( x right) } Longrightarrow Longrightarrow frac{ left( x right) }{ left( 1 right) } $$
    $$ color{magenta}{ left( frac{ 2 x - 1 }{ 4 } right) } Longrightarrow Longrightarrow frac{ left( frac{ 2 x^{2} - x + 4 }{ 4 } right) }{ left( frac{ 2 x - 1 }{ 4 } right) } $$
    $$ color{magenta}{ left( frac{ - 8 x - 4 }{ 7 } right) } Longrightarrow Longrightarrow frac{ left( frac{ - 4 x^{3} - 4 }{ 7 } right) }{ left( frac{ - 4 x^{2} + 8 }{ 7 } right) } $$
    $$ left( x^{3} + 1 right) left( frac{ 2 x - 1 }{ 7 } right) - left( x^{2} - 2 right) left( frac{ 2 x^{2} - x + 4 }{ 7 } right) = left( 1 right) $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      If one insists on using the longer way of the extended Euclidean algorithm, then it is much clearer & simpler to use it in the augmented matrix form.
      $endgroup$
      – Bill Dubuque
      Feb 4 at 20:46












    • $begingroup$
      @BillDubuque I am very fond of continued fractions, and was delighted to learn Bezout for (rational) polynomials could also be done that way. Matter of taste, I guess, or of effort invested.
      $endgroup$
      – Will Jagy
      Feb 4 at 20:53










    • $begingroup$
      Certainly CFs are useful but - as in this critique of a similar prior answer - it will likely prove incomprehensible to most readers without a link to a careful explanation of the method. I always link my extended Euclidean algorithm answers to a detailed exposition and this seems to work well in this regard (I get very few questions).
      $endgroup$
      – Bill Dubuque
      Feb 11 at 16:24





















    1












    $begingroup$

    Applying the generalized Euclidean algorithm to $x^3+1$ and to $x^2-2$, you get that$$-frac74=left(frac{x^2}2-frac x4+1right)(x^2-2)-left(frac x2-frac14right)(x^3+1).$$Therefore,$$left(-frac{2x^2}7+frac x7-frac47right)(x^2-2)+left(frac{2x}7-frac17right)(x^3+1)=1$$and so the inverse of $x^3+1$ in $mathbb{Q}[x]/langle x^2-2rangle$ is $dfrac{2x}7-dfrac17$.






    share|cite|improve this answer











    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Hint $ color{#c00}{x^2 = 2} Rightarrow dfrac{1}{1+color{#c00}{x^2} x} ,=, dfrac{1}{1+color{#c00}2x} = dfrac{1}{1+2color{#c00}x}, dfrac{1-2x}{1-2color{#c00}x} = dfrac{1,-,2x }{1-4(color{#c00}2)}$



      i.e. $ $ rationalize the denom of $, dfrac{1}{1+2sqrt{2}}.,$ More generally, use the Extended Euclidean Algorithm.



      For much further discussion see here and its links.






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        Hint $ color{#c00}{x^2 = 2} Rightarrow dfrac{1}{1+color{#c00}{x^2} x} ,=, dfrac{1}{1+color{#c00}2x} = dfrac{1}{1+2color{#c00}x}, dfrac{1-2x}{1-2color{#c00}x} = dfrac{1,-,2x }{1-4(color{#c00}2)}$



        i.e. $ $ rationalize the denom of $, dfrac{1}{1+2sqrt{2}}.,$ More generally, use the Extended Euclidean Algorithm.



        For much further discussion see here and its links.






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          Hint $ color{#c00}{x^2 = 2} Rightarrow dfrac{1}{1+color{#c00}{x^2} x} ,=, dfrac{1}{1+color{#c00}2x} = dfrac{1}{1+2color{#c00}x}, dfrac{1-2x}{1-2color{#c00}x} = dfrac{1,-,2x }{1-4(color{#c00}2)}$



          i.e. $ $ rationalize the denom of $, dfrac{1}{1+2sqrt{2}}.,$ More generally, use the Extended Euclidean Algorithm.



          For much further discussion see here and its links.






          share|cite|improve this answer











          $endgroup$



          Hint $ color{#c00}{x^2 = 2} Rightarrow dfrac{1}{1+color{#c00}{x^2} x} ,=, dfrac{1}{1+color{#c00}2x} = dfrac{1}{1+2color{#c00}x}, dfrac{1-2x}{1-2color{#c00}x} = dfrac{1,-,2x }{1-4(color{#c00}2)}$



          i.e. $ $ rationalize the denom of $, dfrac{1}{1+2sqrt{2}}.,$ More generally, use the Extended Euclidean Algorithm.



          For much further discussion see here and its links.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Feb 11 at 16:36

























          answered Feb 4 at 19:44









          Bill DubuqueBill Dubuque

          213k29195654




          213k29195654























              1












              $begingroup$

              extended Euc:



              $$ left( x^{3} + 1 right) $$



              $$ left( x^{2} - 2 right) $$



              $$ left( x^{3} + 1 right) = left( x^{2} - 2 right) cdot color{magenta}{ left( x right) } + left( 2 x + 1 right) $$
              $$ left( x^{2} - 2 right) = left( 2 x + 1 right) cdot color{magenta}{ left( frac{ 2 x - 1 }{ 4 } right) } + left( frac{ -7}{4 } right) $$
              $$ left( 2 x + 1 right) = left( frac{ -7}{4 } right) cdot color{magenta}{ left( frac{ - 8 x - 4 }{ 7 } right) } + left( 0 right) $$
              $$ frac{ 0}{1} $$
              $$ frac{ 1}{0} $$
              $$ color{magenta}{ left( x right) } Longrightarrow Longrightarrow frac{ left( x right) }{ left( 1 right) } $$
              $$ color{magenta}{ left( frac{ 2 x - 1 }{ 4 } right) } Longrightarrow Longrightarrow frac{ left( frac{ 2 x^{2} - x + 4 }{ 4 } right) }{ left( frac{ 2 x - 1 }{ 4 } right) } $$
              $$ color{magenta}{ left( frac{ - 8 x - 4 }{ 7 } right) } Longrightarrow Longrightarrow frac{ left( frac{ - 4 x^{3} - 4 }{ 7 } right) }{ left( frac{ - 4 x^{2} + 8 }{ 7 } right) } $$
              $$ left( x^{3} + 1 right) left( frac{ 2 x - 1 }{ 7 } right) - left( x^{2} - 2 right) left( frac{ 2 x^{2} - x + 4 }{ 7 } right) = left( 1 right) $$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                If one insists on using the longer way of the extended Euclidean algorithm, then it is much clearer & simpler to use it in the augmented matrix form.
                $endgroup$
                – Bill Dubuque
                Feb 4 at 20:46












              • $begingroup$
                @BillDubuque I am very fond of continued fractions, and was delighted to learn Bezout for (rational) polynomials could also be done that way. Matter of taste, I guess, or of effort invested.
                $endgroup$
                – Will Jagy
                Feb 4 at 20:53










              • $begingroup$
                Certainly CFs are useful but - as in this critique of a similar prior answer - it will likely prove incomprehensible to most readers without a link to a careful explanation of the method. I always link my extended Euclidean algorithm answers to a detailed exposition and this seems to work well in this regard (I get very few questions).
                $endgroup$
                – Bill Dubuque
                Feb 11 at 16:24


















              1












              $begingroup$

              extended Euc:



              $$ left( x^{3} + 1 right) $$



              $$ left( x^{2} - 2 right) $$



              $$ left( x^{3} + 1 right) = left( x^{2} - 2 right) cdot color{magenta}{ left( x right) } + left( 2 x + 1 right) $$
              $$ left( x^{2} - 2 right) = left( 2 x + 1 right) cdot color{magenta}{ left( frac{ 2 x - 1 }{ 4 } right) } + left( frac{ -7}{4 } right) $$
              $$ left( 2 x + 1 right) = left( frac{ -7}{4 } right) cdot color{magenta}{ left( frac{ - 8 x - 4 }{ 7 } right) } + left( 0 right) $$
              $$ frac{ 0}{1} $$
              $$ frac{ 1}{0} $$
              $$ color{magenta}{ left( x right) } Longrightarrow Longrightarrow frac{ left( x right) }{ left( 1 right) } $$
              $$ color{magenta}{ left( frac{ 2 x - 1 }{ 4 } right) } Longrightarrow Longrightarrow frac{ left( frac{ 2 x^{2} - x + 4 }{ 4 } right) }{ left( frac{ 2 x - 1 }{ 4 } right) } $$
              $$ color{magenta}{ left( frac{ - 8 x - 4 }{ 7 } right) } Longrightarrow Longrightarrow frac{ left( frac{ - 4 x^{3} - 4 }{ 7 } right) }{ left( frac{ - 4 x^{2} + 8 }{ 7 } right) } $$
              $$ left( x^{3} + 1 right) left( frac{ 2 x - 1 }{ 7 } right) - left( x^{2} - 2 right) left( frac{ 2 x^{2} - x + 4 }{ 7 } right) = left( 1 right) $$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                If one insists on using the longer way of the extended Euclidean algorithm, then it is much clearer & simpler to use it in the augmented matrix form.
                $endgroup$
                – Bill Dubuque
                Feb 4 at 20:46












              • $begingroup$
                @BillDubuque I am very fond of continued fractions, and was delighted to learn Bezout for (rational) polynomials could also be done that way. Matter of taste, I guess, or of effort invested.
                $endgroup$
                – Will Jagy
                Feb 4 at 20:53










              • $begingroup$
                Certainly CFs are useful but - as in this critique of a similar prior answer - it will likely prove incomprehensible to most readers without a link to a careful explanation of the method. I always link my extended Euclidean algorithm answers to a detailed exposition and this seems to work well in this regard (I get very few questions).
                $endgroup$
                – Bill Dubuque
                Feb 11 at 16:24
















              1












              1








              1





              $begingroup$

              extended Euc:



              $$ left( x^{3} + 1 right) $$



              $$ left( x^{2} - 2 right) $$



              $$ left( x^{3} + 1 right) = left( x^{2} - 2 right) cdot color{magenta}{ left( x right) } + left( 2 x + 1 right) $$
              $$ left( x^{2} - 2 right) = left( 2 x + 1 right) cdot color{magenta}{ left( frac{ 2 x - 1 }{ 4 } right) } + left( frac{ -7}{4 } right) $$
              $$ left( 2 x + 1 right) = left( frac{ -7}{4 } right) cdot color{magenta}{ left( frac{ - 8 x - 4 }{ 7 } right) } + left( 0 right) $$
              $$ frac{ 0}{1} $$
              $$ frac{ 1}{0} $$
              $$ color{magenta}{ left( x right) } Longrightarrow Longrightarrow frac{ left( x right) }{ left( 1 right) } $$
              $$ color{magenta}{ left( frac{ 2 x - 1 }{ 4 } right) } Longrightarrow Longrightarrow frac{ left( frac{ 2 x^{2} - x + 4 }{ 4 } right) }{ left( frac{ 2 x - 1 }{ 4 } right) } $$
              $$ color{magenta}{ left( frac{ - 8 x - 4 }{ 7 } right) } Longrightarrow Longrightarrow frac{ left( frac{ - 4 x^{3} - 4 }{ 7 } right) }{ left( frac{ - 4 x^{2} + 8 }{ 7 } right) } $$
              $$ left( x^{3} + 1 right) left( frac{ 2 x - 1 }{ 7 } right) - left( x^{2} - 2 right) left( frac{ 2 x^{2} - x + 4 }{ 7 } right) = left( 1 right) $$






              share|cite|improve this answer









              $endgroup$



              extended Euc:



              $$ left( x^{3} + 1 right) $$



              $$ left( x^{2} - 2 right) $$



              $$ left( x^{3} + 1 right) = left( x^{2} - 2 right) cdot color{magenta}{ left( x right) } + left( 2 x + 1 right) $$
              $$ left( x^{2} - 2 right) = left( 2 x + 1 right) cdot color{magenta}{ left( frac{ 2 x - 1 }{ 4 } right) } + left( frac{ -7}{4 } right) $$
              $$ left( 2 x + 1 right) = left( frac{ -7}{4 } right) cdot color{magenta}{ left( frac{ - 8 x - 4 }{ 7 } right) } + left( 0 right) $$
              $$ frac{ 0}{1} $$
              $$ frac{ 1}{0} $$
              $$ color{magenta}{ left( x right) } Longrightarrow Longrightarrow frac{ left( x right) }{ left( 1 right) } $$
              $$ color{magenta}{ left( frac{ 2 x - 1 }{ 4 } right) } Longrightarrow Longrightarrow frac{ left( frac{ 2 x^{2} - x + 4 }{ 4 } right) }{ left( frac{ 2 x - 1 }{ 4 } right) } $$
              $$ color{magenta}{ left( frac{ - 8 x - 4 }{ 7 } right) } Longrightarrow Longrightarrow frac{ left( frac{ - 4 x^{3} - 4 }{ 7 } right) }{ left( frac{ - 4 x^{2} + 8 }{ 7 } right) } $$
              $$ left( x^{3} + 1 right) left( frac{ 2 x - 1 }{ 7 } right) - left( x^{2} - 2 right) left( frac{ 2 x^{2} - x + 4 }{ 7 } right) = left( 1 right) $$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Feb 4 at 20:28









              Will JagyWill Jagy

              104k5102201




              104k5102201












              • $begingroup$
                If one insists on using the longer way of the extended Euclidean algorithm, then it is much clearer & simpler to use it in the augmented matrix form.
                $endgroup$
                – Bill Dubuque
                Feb 4 at 20:46












              • $begingroup$
                @BillDubuque I am very fond of continued fractions, and was delighted to learn Bezout for (rational) polynomials could also be done that way. Matter of taste, I guess, or of effort invested.
                $endgroup$
                – Will Jagy
                Feb 4 at 20:53










              • $begingroup$
                Certainly CFs are useful but - as in this critique of a similar prior answer - it will likely prove incomprehensible to most readers without a link to a careful explanation of the method. I always link my extended Euclidean algorithm answers to a detailed exposition and this seems to work well in this regard (I get very few questions).
                $endgroup$
                – Bill Dubuque
                Feb 11 at 16:24




















              • $begingroup$
                If one insists on using the longer way of the extended Euclidean algorithm, then it is much clearer & simpler to use it in the augmented matrix form.
                $endgroup$
                – Bill Dubuque
                Feb 4 at 20:46












              • $begingroup$
                @BillDubuque I am very fond of continued fractions, and was delighted to learn Bezout for (rational) polynomials could also be done that way. Matter of taste, I guess, or of effort invested.
                $endgroup$
                – Will Jagy
                Feb 4 at 20:53










              • $begingroup$
                Certainly CFs are useful but - as in this critique of a similar prior answer - it will likely prove incomprehensible to most readers without a link to a careful explanation of the method. I always link my extended Euclidean algorithm answers to a detailed exposition and this seems to work well in this regard (I get very few questions).
                $endgroup$
                – Bill Dubuque
                Feb 11 at 16:24


















              $begingroup$
              If one insists on using the longer way of the extended Euclidean algorithm, then it is much clearer & simpler to use it in the augmented matrix form.
              $endgroup$
              – Bill Dubuque
              Feb 4 at 20:46






              $begingroup$
              If one insists on using the longer way of the extended Euclidean algorithm, then it is much clearer & simpler to use it in the augmented matrix form.
              $endgroup$
              – Bill Dubuque
              Feb 4 at 20:46














              $begingroup$
              @BillDubuque I am very fond of continued fractions, and was delighted to learn Bezout for (rational) polynomials could also be done that way. Matter of taste, I guess, or of effort invested.
              $endgroup$
              – Will Jagy
              Feb 4 at 20:53




              $begingroup$
              @BillDubuque I am very fond of continued fractions, and was delighted to learn Bezout for (rational) polynomials could also be done that way. Matter of taste, I guess, or of effort invested.
              $endgroup$
              – Will Jagy
              Feb 4 at 20:53












              $begingroup$
              Certainly CFs are useful but - as in this critique of a similar prior answer - it will likely prove incomprehensible to most readers without a link to a careful explanation of the method. I always link my extended Euclidean algorithm answers to a detailed exposition and this seems to work well in this regard (I get very few questions).
              $endgroup$
              – Bill Dubuque
              Feb 11 at 16:24






              $begingroup$
              Certainly CFs are useful but - as in this critique of a similar prior answer - it will likely prove incomprehensible to most readers without a link to a careful explanation of the method. I always link my extended Euclidean algorithm answers to a detailed exposition and this seems to work well in this regard (I get very few questions).
              $endgroup$
              – Bill Dubuque
              Feb 11 at 16:24













              1












              $begingroup$

              Applying the generalized Euclidean algorithm to $x^3+1$ and to $x^2-2$, you get that$$-frac74=left(frac{x^2}2-frac x4+1right)(x^2-2)-left(frac x2-frac14right)(x^3+1).$$Therefore,$$left(-frac{2x^2}7+frac x7-frac47right)(x^2-2)+left(frac{2x}7-frac17right)(x^3+1)=1$$and so the inverse of $x^3+1$ in $mathbb{Q}[x]/langle x^2-2rangle$ is $dfrac{2x}7-dfrac17$.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                Applying the generalized Euclidean algorithm to $x^3+1$ and to $x^2-2$, you get that$$-frac74=left(frac{x^2}2-frac x4+1right)(x^2-2)-left(frac x2-frac14right)(x^3+1).$$Therefore,$$left(-frac{2x^2}7+frac x7-frac47right)(x^2-2)+left(frac{2x}7-frac17right)(x^3+1)=1$$and so the inverse of $x^3+1$ in $mathbb{Q}[x]/langle x^2-2rangle$ is $dfrac{2x}7-dfrac17$.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Applying the generalized Euclidean algorithm to $x^3+1$ and to $x^2-2$, you get that$$-frac74=left(frac{x^2}2-frac x4+1right)(x^2-2)-left(frac x2-frac14right)(x^3+1).$$Therefore,$$left(-frac{2x^2}7+frac x7-frac47right)(x^2-2)+left(frac{2x}7-frac17right)(x^3+1)=1$$and so the inverse of $x^3+1$ in $mathbb{Q}[x]/langle x^2-2rangle$ is $dfrac{2x}7-dfrac17$.






                  share|cite|improve this answer











                  $endgroup$



                  Applying the generalized Euclidean algorithm to $x^3+1$ and to $x^2-2$, you get that$$-frac74=left(frac{x^2}2-frac x4+1right)(x^2-2)-left(frac x2-frac14right)(x^3+1).$$Therefore,$$left(-frac{2x^2}7+frac x7-frac47right)(x^2-2)+left(frac{2x}7-frac17right)(x^3+1)=1$$and so the inverse of $x^3+1$ in $mathbb{Q}[x]/langle x^2-2rangle$ is $dfrac{2x}7-dfrac17$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Feb 6 at 0:06









                  user26857

                  39.4k124183




                  39.4k124183










                  answered Feb 4 at 19:33









                  José Carlos SantosJosé Carlos Santos

                  170k23132238




                  170k23132238






























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