Determine the greatest common divisor of polynomials $x^2+1$ and $x^3+1$ in $Bbb Q[X]$.

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Exercise: Determine a gcd of the polynomials $x^2+1$ and $x^3+1$ in $Bbb Q[X]$.. Write the gcd as a combination of the given polynomials.



Is it correct that I keep using long division until the result is $0$, and then the previous result would be a gcd?



x^2+1 / x^3 + 1 x
x^3 + x
________-
1 - x

1 - x / x^2 + 1 -x-1
x^2 - x
________-
x + 1
x - 1
________-
2

2 / 1 - x -1/2x + 1/2
- x
______-
1
1
______-
0


Conclusion: A gcd is $2$?



I'm not sure what's meant with "Write the gcd as a combination of the given polynomials.", so it would be great if someone could point me in the right direction.










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    up vote
    2
    down vote

    favorite












    Exercise: Determine a gcd of the polynomials $x^2+1$ and $x^3+1$ in $Bbb Q[X]$.. Write the gcd as a combination of the given polynomials.



    Is it correct that I keep using long division until the result is $0$, and then the previous result would be a gcd?



    x^2+1 / x^3 + 1 x
    x^3 + x
    ________-
    1 - x

    1 - x / x^2 + 1 -x-1
    x^2 - x
    ________-
    x + 1
    x - 1
    ________-
    2

    2 / 1 - x -1/2x + 1/2
    - x
    ______-
    1
    1
    ______-
    0


    Conclusion: A gcd is $2$?



    I'm not sure what's meant with "Write the gcd as a combination of the given polynomials.", so it would be great if someone could point me in the right direction.










    share|cite|improve this question

























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Exercise: Determine a gcd of the polynomials $x^2+1$ and $x^3+1$ in $Bbb Q[X]$.. Write the gcd as a combination of the given polynomials.



      Is it correct that I keep using long division until the result is $0$, and then the previous result would be a gcd?



      x^2+1 / x^3 + 1 x
      x^3 + x
      ________-
      1 - x

      1 - x / x^2 + 1 -x-1
      x^2 - x
      ________-
      x + 1
      x - 1
      ________-
      2

      2 / 1 - x -1/2x + 1/2
      - x
      ______-
      1
      1
      ______-
      0


      Conclusion: A gcd is $2$?



      I'm not sure what's meant with "Write the gcd as a combination of the given polynomials.", so it would be great if someone could point me in the right direction.










      share|cite|improve this question















      Exercise: Determine a gcd of the polynomials $x^2+1$ and $x^3+1$ in $Bbb Q[X]$.. Write the gcd as a combination of the given polynomials.



      Is it correct that I keep using long division until the result is $0$, and then the previous result would be a gcd?



      x^2+1 / x^3 + 1 x
      x^3 + x
      ________-
      1 - x

      1 - x / x^2 + 1 -x-1
      x^2 - x
      ________-
      x + 1
      x - 1
      ________-
      2

      2 / 1 - x -1/2x + 1/2
      - x
      ______-
      1
      1
      ______-
      0


      Conclusion: A gcd is $2$?



      I'm not sure what's meant with "Write the gcd as a combination of the given polynomials.", so it would be great if someone could point me in the right direction.







      linear-algebra polynomials rational-numbers greatest-common-divisor






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      edited Sep 9 at 11:14









      thesmallprint

      2,3991617




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      asked Sep 9 at 10:24









      SJ19

      204




      204




















          3 Answers
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          The Euclidean algorithm steps are
          beginalign*
          x^3+1&=x (x^2+1) + (-x+1)\
          x^2+1&=-x (-x+1)+ (x+1)\
          -x+1&=-(x+1)+2
          endalign*



          So the GCD is $1$.



          We can go back up
          beginalign*
          2 &= (-x+1) +(x+1)\
          &= (-x+1) + (x^2+1)+x (-x+1)\
          &= (-x+1) (x+1)+(x^2+1)\
          &= (x^3+1 - x(x^2+1))(x+1)+(x^2+1)\
          &= (x+1) (x^3+1) + (-x^2-x+1) (x^2+1)
          endalign*



          and so $1 = frac12 [(x+1) (x^3+1) + (-x^2-x+1) (x^2+1)]$






          share|cite|improve this answer






















          • Maybe I'm wrong but shouldn't gcd=1?? (It has to be a monic polynomial)
            – giannispapav
            Sep 9 at 11:16










          • Since the polynomials are members of $Bbb Q[x]$, wouldn't $$1=frac12(x+1) (x^3+1) + frac12(-x^2-x+1) (x^2+1)$$ work aswell? I fail to see how the gcd is even defined here.
            – cansomeonehelpmeout
            Sep 9 at 11:19











          • Thank you very much. Is it correct that 1 and 2 are both possible answers?
            – SJ19
            Sep 9 at 11:58










          • @SJ19: As a matter of convention, the greatest common divisor of polynomials is usually taken to mean the monic polynomial with the highest possible degree. So $1$ in this case.
            – TonyK
            Sep 9 at 12:05










          • @TonyK Alright. And how are you concluding GCD=1 from "-x+1 = -(x+1)+2"?
            – SJ19
            Sep 9 at 13:17

















          up vote
          1
          down vote













          Notice that $x^2+1$ is irreducible over $mathbbQ$. Let $d = gcd(x^2+1, x^3+1)$. Then $d mid x^2+1$ so $d$ is a nonzero constant, or $d$ is an associate of $x^2+1$.



          However, $x^2+1 notmid x^3+1$ so $d$ must be constant. We can choose $d = 1$ since $gcd$ is usually supposed to be monic.



          Finally, notice that



          $$1=left[frac12(x+1) right](x^3+1) + left[frac12(-x^2-x+1)right] (x^2+1)$$






          share|cite|improve this answer



























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



            $$ left( x^2 + 1 right) $$



            $$ left( x^3 + 1 right) = left( x^2 + 1 right) cdot colormagenta left( x right) + left( - x + 1 right) $$
            $$ left( x^2 + 1 right) = left( - x + 1 right) cdot colormagenta left( - x - 1 right) + left( 2 right) $$
            $$ left( - x + 1 right) = left( 2 right) cdot colormagenta left( frac - x + 1 2 right) + left( 0 right) $$
            $$ frac 01 $$
            $$ frac 10 $$
            $$ colormagenta left( x right) Longrightarrow Longrightarrow frac left( x right) left( 1 right) $$
            $$ colormagenta left( - x - 1 right) Longrightarrow Longrightarrow frac left( - x^2 - x + 1 right) left( - x - 1 right) $$
            $$ colormagenta left( frac - x + 1 2 right) Longrightarrow Longrightarrow frac left( frac x^3 + 1 2 right) left( frac x^2 + 1 2 right) $$
            $$ left( x^3 + 1 right) left( frac - x - 1 2 right) - left( x^2 + 1 right) left( frac - x^2 - x + 1 2 right) = left( -1 right) $$



            ..................................






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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              1
              down vote



              accepted










              The Euclidean algorithm steps are
              beginalign*
              x^3+1&=x (x^2+1) + (-x+1)\
              x^2+1&=-x (-x+1)+ (x+1)\
              -x+1&=-(x+1)+2
              endalign*



              So the GCD is $1$.



              We can go back up
              beginalign*
              2 &= (-x+1) +(x+1)\
              &= (-x+1) + (x^2+1)+x (-x+1)\
              &= (-x+1) (x+1)+(x^2+1)\
              &= (x^3+1 - x(x^2+1))(x+1)+(x^2+1)\
              &= (x+1) (x^3+1) + (-x^2-x+1) (x^2+1)
              endalign*



              and so $1 = frac12 [(x+1) (x^3+1) + (-x^2-x+1) (x^2+1)]$






              share|cite|improve this answer






















              • Maybe I'm wrong but shouldn't gcd=1?? (It has to be a monic polynomial)
                – giannispapav
                Sep 9 at 11:16










              • Since the polynomials are members of $Bbb Q[x]$, wouldn't $$1=frac12(x+1) (x^3+1) + frac12(-x^2-x+1) (x^2+1)$$ work aswell? I fail to see how the gcd is even defined here.
                – cansomeonehelpmeout
                Sep 9 at 11:19











              • Thank you very much. Is it correct that 1 and 2 are both possible answers?
                – SJ19
                Sep 9 at 11:58










              • @SJ19: As a matter of convention, the greatest common divisor of polynomials is usually taken to mean the monic polynomial with the highest possible degree. So $1$ in this case.
                – TonyK
                Sep 9 at 12:05










              • @TonyK Alright. And how are you concluding GCD=1 from "-x+1 = -(x+1)+2"?
                – SJ19
                Sep 9 at 13:17














              up vote
              1
              down vote



              accepted










              The Euclidean algorithm steps are
              beginalign*
              x^3+1&=x (x^2+1) + (-x+1)\
              x^2+1&=-x (-x+1)+ (x+1)\
              -x+1&=-(x+1)+2
              endalign*



              So the GCD is $1$.



              We can go back up
              beginalign*
              2 &= (-x+1) +(x+1)\
              &= (-x+1) + (x^2+1)+x (-x+1)\
              &= (-x+1) (x+1)+(x^2+1)\
              &= (x^3+1 - x(x^2+1))(x+1)+(x^2+1)\
              &= (x+1) (x^3+1) + (-x^2-x+1) (x^2+1)
              endalign*



              and so $1 = frac12 [(x+1) (x^3+1) + (-x^2-x+1) (x^2+1)]$






              share|cite|improve this answer






















              • Maybe I'm wrong but shouldn't gcd=1?? (It has to be a monic polynomial)
                – giannispapav
                Sep 9 at 11:16










              • Since the polynomials are members of $Bbb Q[x]$, wouldn't $$1=frac12(x+1) (x^3+1) + frac12(-x^2-x+1) (x^2+1)$$ work aswell? I fail to see how the gcd is even defined here.
                – cansomeonehelpmeout
                Sep 9 at 11:19











              • Thank you very much. Is it correct that 1 and 2 are both possible answers?
                – SJ19
                Sep 9 at 11:58










              • @SJ19: As a matter of convention, the greatest common divisor of polynomials is usually taken to mean the monic polynomial with the highest possible degree. So $1$ in this case.
                – TonyK
                Sep 9 at 12:05










              • @TonyK Alright. And how are you concluding GCD=1 from "-x+1 = -(x+1)+2"?
                – SJ19
                Sep 9 at 13:17












              up vote
              1
              down vote



              accepted







              up vote
              1
              down vote



              accepted






              The Euclidean algorithm steps are
              beginalign*
              x^3+1&=x (x^2+1) + (-x+1)\
              x^2+1&=-x (-x+1)+ (x+1)\
              -x+1&=-(x+1)+2
              endalign*



              So the GCD is $1$.



              We can go back up
              beginalign*
              2 &= (-x+1) +(x+1)\
              &= (-x+1) + (x^2+1)+x (-x+1)\
              &= (-x+1) (x+1)+(x^2+1)\
              &= (x^3+1 - x(x^2+1))(x+1)+(x^2+1)\
              &= (x+1) (x^3+1) + (-x^2-x+1) (x^2+1)
              endalign*



              and so $1 = frac12 [(x+1) (x^3+1) + (-x^2-x+1) (x^2+1)]$






              share|cite|improve this answer














              The Euclidean algorithm steps are
              beginalign*
              x^3+1&=x (x^2+1) + (-x+1)\
              x^2+1&=-x (-x+1)+ (x+1)\
              -x+1&=-(x+1)+2
              endalign*



              So the GCD is $1$.



              We can go back up
              beginalign*
              2 &= (-x+1) +(x+1)\
              &= (-x+1) + (x^2+1)+x (-x+1)\
              &= (-x+1) (x+1)+(x^2+1)\
              &= (x^3+1 - x(x^2+1))(x+1)+(x^2+1)\
              &= (x+1) (x^3+1) + (-x^2-x+1) (x^2+1)
              endalign*



              and so $1 = frac12 [(x+1) (x^3+1) + (-x^2-x+1) (x^2+1)]$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Sep 9 at 11:44

























              answered Sep 9 at 10:47









              P. Quinton

              848111




              848111











              • Maybe I'm wrong but shouldn't gcd=1?? (It has to be a monic polynomial)
                – giannispapav
                Sep 9 at 11:16










              • Since the polynomials are members of $Bbb Q[x]$, wouldn't $$1=frac12(x+1) (x^3+1) + frac12(-x^2-x+1) (x^2+1)$$ work aswell? I fail to see how the gcd is even defined here.
                – cansomeonehelpmeout
                Sep 9 at 11:19











              • Thank you very much. Is it correct that 1 and 2 are both possible answers?
                – SJ19
                Sep 9 at 11:58










              • @SJ19: As a matter of convention, the greatest common divisor of polynomials is usually taken to mean the monic polynomial with the highest possible degree. So $1$ in this case.
                – TonyK
                Sep 9 at 12:05










              • @TonyK Alright. And how are you concluding GCD=1 from "-x+1 = -(x+1)+2"?
                – SJ19
                Sep 9 at 13:17
















              • Maybe I'm wrong but shouldn't gcd=1?? (It has to be a monic polynomial)
                – giannispapav
                Sep 9 at 11:16










              • Since the polynomials are members of $Bbb Q[x]$, wouldn't $$1=frac12(x+1) (x^3+1) + frac12(-x^2-x+1) (x^2+1)$$ work aswell? I fail to see how the gcd is even defined here.
                – cansomeonehelpmeout
                Sep 9 at 11:19











              • Thank you very much. Is it correct that 1 and 2 are both possible answers?
                – SJ19
                Sep 9 at 11:58










              • @SJ19: As a matter of convention, the greatest common divisor of polynomials is usually taken to mean the monic polynomial with the highest possible degree. So $1$ in this case.
                – TonyK
                Sep 9 at 12:05










              • @TonyK Alright. And how are you concluding GCD=1 from "-x+1 = -(x+1)+2"?
                – SJ19
                Sep 9 at 13:17















              Maybe I'm wrong but shouldn't gcd=1?? (It has to be a monic polynomial)
              – giannispapav
              Sep 9 at 11:16




              Maybe I'm wrong but shouldn't gcd=1?? (It has to be a monic polynomial)
              – giannispapav
              Sep 9 at 11:16












              Since the polynomials are members of $Bbb Q[x]$, wouldn't $$1=frac12(x+1) (x^3+1) + frac12(-x^2-x+1) (x^2+1)$$ work aswell? I fail to see how the gcd is even defined here.
              – cansomeonehelpmeout
              Sep 9 at 11:19





              Since the polynomials are members of $Bbb Q[x]$, wouldn't $$1=frac12(x+1) (x^3+1) + frac12(-x^2-x+1) (x^2+1)$$ work aswell? I fail to see how the gcd is even defined here.
              – cansomeonehelpmeout
              Sep 9 at 11:19













              Thank you very much. Is it correct that 1 and 2 are both possible answers?
              – SJ19
              Sep 9 at 11:58




              Thank you very much. Is it correct that 1 and 2 are both possible answers?
              – SJ19
              Sep 9 at 11:58












              @SJ19: As a matter of convention, the greatest common divisor of polynomials is usually taken to mean the monic polynomial with the highest possible degree. So $1$ in this case.
              – TonyK
              Sep 9 at 12:05




              @SJ19: As a matter of convention, the greatest common divisor of polynomials is usually taken to mean the monic polynomial with the highest possible degree. So $1$ in this case.
              – TonyK
              Sep 9 at 12:05












              @TonyK Alright. And how are you concluding GCD=1 from "-x+1 = -(x+1)+2"?
              – SJ19
              Sep 9 at 13:17




              @TonyK Alright. And how are you concluding GCD=1 from "-x+1 = -(x+1)+2"?
              – SJ19
              Sep 9 at 13:17










              up vote
              1
              down vote













              Notice that $x^2+1$ is irreducible over $mathbbQ$. Let $d = gcd(x^2+1, x^3+1)$. Then $d mid x^2+1$ so $d$ is a nonzero constant, or $d$ is an associate of $x^2+1$.



              However, $x^2+1 notmid x^3+1$ so $d$ must be constant. We can choose $d = 1$ since $gcd$ is usually supposed to be monic.



              Finally, notice that



              $$1=left[frac12(x+1) right](x^3+1) + left[frac12(-x^2-x+1)right] (x^2+1)$$






              share|cite|improve this answer
























                up vote
                1
                down vote













                Notice that $x^2+1$ is irreducible over $mathbbQ$. Let $d = gcd(x^2+1, x^3+1)$. Then $d mid x^2+1$ so $d$ is a nonzero constant, or $d$ is an associate of $x^2+1$.



                However, $x^2+1 notmid x^3+1$ so $d$ must be constant. We can choose $d = 1$ since $gcd$ is usually supposed to be monic.



                Finally, notice that



                $$1=left[frac12(x+1) right](x^3+1) + left[frac12(-x^2-x+1)right] (x^2+1)$$






                share|cite|improve this answer






















                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  Notice that $x^2+1$ is irreducible over $mathbbQ$. Let $d = gcd(x^2+1, x^3+1)$. Then $d mid x^2+1$ so $d$ is a nonzero constant, or $d$ is an associate of $x^2+1$.



                  However, $x^2+1 notmid x^3+1$ so $d$ must be constant. We can choose $d = 1$ since $gcd$ is usually supposed to be monic.



                  Finally, notice that



                  $$1=left[frac12(x+1) right](x^3+1) + left[frac12(-x^2-x+1)right] (x^2+1)$$






                  share|cite|improve this answer












                  Notice that $x^2+1$ is irreducible over $mathbbQ$. Let $d = gcd(x^2+1, x^3+1)$. Then $d mid x^2+1$ so $d$ is a nonzero constant, or $d$ is an associate of $x^2+1$.



                  However, $x^2+1 notmid x^3+1$ so $d$ must be constant. We can choose $d = 1$ since $gcd$ is usually supposed to be monic.



                  Finally, notice that



                  $$1=left[frac12(x+1) right](x^3+1) + left[frac12(-x^2-x+1)right] (x^2+1)$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Sep 9 at 11:50









                  mechanodroid

                  24.7k62245




                  24.7k62245




















                      up vote
                      1
                      down vote













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



                      $$ left( x^2 + 1 right) $$



                      $$ left( x^3 + 1 right) = left( x^2 + 1 right) cdot colormagenta left( x right) + left( - x + 1 right) $$
                      $$ left( x^2 + 1 right) = left( - x + 1 right) cdot colormagenta left( - x - 1 right) + left( 2 right) $$
                      $$ left( - x + 1 right) = left( 2 right) cdot colormagenta left( frac - x + 1 2 right) + left( 0 right) $$
                      $$ frac 01 $$
                      $$ frac 10 $$
                      $$ colormagenta left( x right) Longrightarrow Longrightarrow frac left( x right) left( 1 right) $$
                      $$ colormagenta left( - x - 1 right) Longrightarrow Longrightarrow frac left( - x^2 - x + 1 right) left( - x - 1 right) $$
                      $$ colormagenta left( frac - x + 1 2 right) Longrightarrow Longrightarrow frac left( frac x^3 + 1 2 right) left( frac x^2 + 1 2 right) $$
                      $$ left( x^3 + 1 right) left( frac - x - 1 2 right) - left( x^2 + 1 right) left( frac - x^2 - x + 1 2 right) = left( -1 right) $$



                      ..................................






                      share|cite|improve this answer
























                        up vote
                        1
                        down vote













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



                        $$ left( x^2 + 1 right) $$



                        $$ left( x^3 + 1 right) = left( x^2 + 1 right) cdot colormagenta left( x right) + left( - x + 1 right) $$
                        $$ left( x^2 + 1 right) = left( - x + 1 right) cdot colormagenta left( - x - 1 right) + left( 2 right) $$
                        $$ left( - x + 1 right) = left( 2 right) cdot colormagenta left( frac - x + 1 2 right) + left( 0 right) $$
                        $$ frac 01 $$
                        $$ frac 10 $$
                        $$ colormagenta left( x right) Longrightarrow Longrightarrow frac left( x right) left( 1 right) $$
                        $$ colormagenta left( - x - 1 right) Longrightarrow Longrightarrow frac left( - x^2 - x + 1 right) left( - x - 1 right) $$
                        $$ colormagenta left( frac - x + 1 2 right) Longrightarrow Longrightarrow frac left( frac x^3 + 1 2 right) left( frac x^2 + 1 2 right) $$
                        $$ left( x^3 + 1 right) left( frac - x - 1 2 right) - left( x^2 + 1 right) left( frac - x^2 - x + 1 2 right) = left( -1 right) $$



                        ..................................






                        share|cite|improve this answer






















                          up vote
                          1
                          down vote










                          up vote
                          1
                          down vote









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



                          $$ left( x^2 + 1 right) $$



                          $$ left( x^3 + 1 right) = left( x^2 + 1 right) cdot colormagenta left( x right) + left( - x + 1 right) $$
                          $$ left( x^2 + 1 right) = left( - x + 1 right) cdot colormagenta left( - x - 1 right) + left( 2 right) $$
                          $$ left( - x + 1 right) = left( 2 right) cdot colormagenta left( frac - x + 1 2 right) + left( 0 right) $$
                          $$ frac 01 $$
                          $$ frac 10 $$
                          $$ colormagenta left( x right) Longrightarrow Longrightarrow frac left( x right) left( 1 right) $$
                          $$ colormagenta left( - x - 1 right) Longrightarrow Longrightarrow frac left( - x^2 - x + 1 right) left( - x - 1 right) $$
                          $$ colormagenta left( frac - x + 1 2 right) Longrightarrow Longrightarrow frac left( frac x^3 + 1 2 right) left( frac x^2 + 1 2 right) $$
                          $$ left( x^3 + 1 right) left( frac - x - 1 2 right) - left( x^2 + 1 right) left( frac - x^2 - x + 1 2 right) = left( -1 right) $$



                          ..................................






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



                          $$ left( x^2 + 1 right) $$



                          $$ left( x^3 + 1 right) = left( x^2 + 1 right) cdot colormagenta left( x right) + left( - x + 1 right) $$
                          $$ left( x^2 + 1 right) = left( - x + 1 right) cdot colormagenta left( - x - 1 right) + left( 2 right) $$
                          $$ left( - x + 1 right) = left( 2 right) cdot colormagenta left( frac - x + 1 2 right) + left( 0 right) $$
                          $$ frac 01 $$
                          $$ frac 10 $$
                          $$ colormagenta left( x right) Longrightarrow Longrightarrow frac left( x right) left( 1 right) $$
                          $$ colormagenta left( - x - 1 right) Longrightarrow Longrightarrow frac left( - x^2 - x + 1 right) left( - x - 1 right) $$
                          $$ colormagenta left( frac - x + 1 2 right) Longrightarrow Longrightarrow frac left( frac x^3 + 1 2 right) left( frac x^2 + 1 2 right) $$
                          $$ left( x^3 + 1 right) left( frac - x - 1 2 right) - left( x^2 + 1 right) left( frac - x^2 - x + 1 2 right) = left( -1 right) $$



                          ..................................







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                          share|cite|improve this answer










                          answered Sep 9 at 14:33









                          Will Jagy

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