Let $f = (x^4 - 1)(x + 2)$ and let $g = (x^2 - 4)(x + 1)$. Find a single polynomial that generates the ideal...












1












$begingroup$


The Problem:



Let $f = (x^4 - 1)(x + 2)$ and let $g = (x^2 - 4)(x + 1)$. Find a single polynomial that generates the ideal $I = (f,g)$ of $mathbb{R}[x]$.



My Approach:



This seemed a bit too easy, which made me think I'm misunderstanding the problem...



Since $f = (x+1)(x-1)(x^2 +1)(x+2)$ and $g= (x+2)(x-2)(x+1)$, it follows that $f$ and $g$ are both in the ideal generated by $p = x+1$ (or, for that matter, the ideal generated by $q = x-1$). Thus $p$ generates $I$.



Is that all there is to it?



EDIT: I had misdefined $g$.






abstract-algebra polynomials ring-theory







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$endgroup$

















    1












    $begingroup$


    The Problem:



    Let $f = (x^4 - 1)(x + 2)$ and let $g = (x^2 - 4)(x + 1)$. Find a single polynomial that generates the ideal $I = (f,g)$ of $mathbb{R}[x]$.



    My Approach:



    This seemed a bit too easy, which made me think I'm misunderstanding the problem...



    Since $f = (x+1)(x-1)(x^2 +1)(x+2)$ and $g= (x+2)(x-2)(x+1)$, it follows that $f$ and $g$ are both in the ideal generated by $p = x+1$ (or, for that matter, the ideal generated by $q = x-1$). Thus $p$ generates $I$.



    Is that all there is to it?



    EDIT: I had misdefined $g$.






    abstract-algebra polynomials ring-theory







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      The Problem:



      Let $f = (x^4 - 1)(x + 2)$ and let $g = (x^2 - 4)(x + 1)$. Find a single polynomial that generates the ideal $I = (f,g)$ of $mathbb{R}[x]$.



      My Approach:



      This seemed a bit too easy, which made me think I'm misunderstanding the problem...



      Since $f = (x+1)(x-1)(x^2 +1)(x+2)$ and $g= (x+2)(x-2)(x+1)$, it follows that $f$ and $g$ are both in the ideal generated by $p = x+1$ (or, for that matter, the ideal generated by $q = x-1$). Thus $p$ generates $I$.



      Is that all there is to it?



      EDIT: I had misdefined $g$.






      abstract-algebra polynomials ring-theory







      share|cite|improve this question











      $endgroup$




      The Problem:



      Let $f = (x^4 - 1)(x + 2)$ and let $g = (x^2 - 4)(x + 1)$. Find a single polynomial that generates the ideal $I = (f,g)$ of $mathbb{R}[x]$.



      My Approach:



      This seemed a bit too easy, which made me think I'm misunderstanding the problem...



      Since $f = (x+1)(x-1)(x^2 +1)(x+2)$ and $g= (x+2)(x-2)(x+1)$, it follows that $f$ and $g$ are both in the ideal generated by $p = x+1$ (or, for that matter, the ideal generated by $q = x-1$). Thus $p$ generates $I$.



      Is that all there is to it?



      EDIT: I had misdefined $g$.






      abstract-algebra polynomials ring-theory




      abstract-algebra polynomials ring-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 15 at 5:06







      thisisourconcerndude

















      asked Jan 15 at 4:58









      thisisourconcerndudethisisourconcerndude

      1,1271022




      1,1271022






















          2 Answers
          2






          active

          oldest

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          2












          $begingroup$

          It is well-known that $Bbb R[x]$ is a principal ideal domain; thus the ideal



          $(f(x), g(x)) = (d(x)) tag 1$



          for some



          $d(x) in Bbb R[x]; tag 2$



          it immediately follows that



          $d(x) = r(x)f(x) + s(x)g(x) tag 3$



          for some



          $r(x), s(x) in Bbb R[x]; tag 4$



          furthermore, it is clear from (1) that



          $d(x) mid f(x), ; d(x) mid g(x); tag 5$



          it is equally clear via (3) that any



          $p(x) in Bbb R[x] tag 6$



          such that



          $p(x) mid f(x), ; p(x) mid g(x) tag 7$



          will also satisfy



          $p(x) mid d(x); tag 8$



          (5) and (8) together show that $d(x)$ by definition is in fact given by



          $d(x) = gcd(f(x), g(x)); tag 9$



          that is, $d(x)$ is the greatest common divisor of $f(x)$, $g(x)$ in $Bbb R[x]$; this fact gives us a tool by which $d(x)$ may be had: calculating the irreducible, hence prime, factors of $f(x)$ and $g(x)$; we find that



          $f(x) = (x^4 - 1)(x + 2) = (x^2 + 1)(x + 1)(x - 1)(x + 2), tag{10}$



          $g(x) = (x^2 - 4)(x + 1) = (x - 2)(x + 2)( x + 1); tag{11}$



          inspection of (10), (11) reveals that every factor on the right-hand sides of these equations is a prime (irreducible) in $Bbb R[x]$, being either linear or the quadratic $x^2 + 1$ which has no roots in $Bbb R$; the factors they have in common are $x + 1$ and $x + 2$; therefore, since $Bbb R[x]$ is a unique factorization domain (as is any principal ideal domain), the unique prime factors of $d(x)$ must be these two; that is



          $d(x) = (x + 1)(x + 2). tag{12}$



          $OEDelta$.






          share|cite|improve this answer











          $endgroup$





















            3












            $begingroup$

            No: you have shown that $(f,g)$ is contained in $(p)$ (or $(q)$), but you need to show $(f,g)$ is equal to the ideal generated by some single polynomial. In other words, you would need to show the reverse inclusion as well, that $(f,g)$ contains $(p)$. You have not proved that, and in fact it is not actually true for the $p$ you have chosen.






            share|cite|improve this answer









            $endgroup$













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              2 Answers
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              active

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              2 Answers
              2






              active

              oldest

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              active

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              active

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              2












              $begingroup$

              It is well-known that $Bbb R[x]$ is a principal ideal domain; thus the ideal



              $(f(x), g(x)) = (d(x)) tag 1$



              for some



              $d(x) in Bbb R[x]; tag 2$



              it immediately follows that



              $d(x) = r(x)f(x) + s(x)g(x) tag 3$



              for some



              $r(x), s(x) in Bbb R[x]; tag 4$



              furthermore, it is clear from (1) that



              $d(x) mid f(x), ; d(x) mid g(x); tag 5$



              it is equally clear via (3) that any



              $p(x) in Bbb R[x] tag 6$



              such that



              $p(x) mid f(x), ; p(x) mid g(x) tag 7$



              will also satisfy



              $p(x) mid d(x); tag 8$



              (5) and (8) together show that $d(x)$ by definition is in fact given by



              $d(x) = gcd(f(x), g(x)); tag 9$



              that is, $d(x)$ is the greatest common divisor of $f(x)$, $g(x)$ in $Bbb R[x]$; this fact gives us a tool by which $d(x)$ may be had: calculating the irreducible, hence prime, factors of $f(x)$ and $g(x)$; we find that



              $f(x) = (x^4 - 1)(x + 2) = (x^2 + 1)(x + 1)(x - 1)(x + 2), tag{10}$



              $g(x) = (x^2 - 4)(x + 1) = (x - 2)(x + 2)( x + 1); tag{11}$



              inspection of (10), (11) reveals that every factor on the right-hand sides of these equations is a prime (irreducible) in $Bbb R[x]$, being either linear or the quadratic $x^2 + 1$ which has no roots in $Bbb R$; the factors they have in common are $x + 1$ and $x + 2$; therefore, since $Bbb R[x]$ is a unique factorization domain (as is any principal ideal domain), the unique prime factors of $d(x)$ must be these two; that is



              $d(x) = (x + 1)(x + 2). tag{12}$



              $OEDelta$.






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                It is well-known that $Bbb R[x]$ is a principal ideal domain; thus the ideal



                $(f(x), g(x)) = (d(x)) tag 1$



                for some



                $d(x) in Bbb R[x]; tag 2$



                it immediately follows that



                $d(x) = r(x)f(x) + s(x)g(x) tag 3$



                for some



                $r(x), s(x) in Bbb R[x]; tag 4$



                furthermore, it is clear from (1) that



                $d(x) mid f(x), ; d(x) mid g(x); tag 5$



                it is equally clear via (3) that any



                $p(x) in Bbb R[x] tag 6$



                such that



                $p(x) mid f(x), ; p(x) mid g(x) tag 7$



                will also satisfy



                $p(x) mid d(x); tag 8$



                (5) and (8) together show that $d(x)$ by definition is in fact given by



                $d(x) = gcd(f(x), g(x)); tag 9$



                that is, $d(x)$ is the greatest common divisor of $f(x)$, $g(x)$ in $Bbb R[x]$; this fact gives us a tool by which $d(x)$ may be had: calculating the irreducible, hence prime, factors of $f(x)$ and $g(x)$; we find that



                $f(x) = (x^4 - 1)(x + 2) = (x^2 + 1)(x + 1)(x - 1)(x + 2), tag{10}$



                $g(x) = (x^2 - 4)(x + 1) = (x - 2)(x + 2)( x + 1); tag{11}$



                inspection of (10), (11) reveals that every factor on the right-hand sides of these equations is a prime (irreducible) in $Bbb R[x]$, being either linear or the quadratic $x^2 + 1$ which has no roots in $Bbb R$; the factors they have in common are $x + 1$ and $x + 2$; therefore, since $Bbb R[x]$ is a unique factorization domain (as is any principal ideal domain), the unique prime factors of $d(x)$ must be these two; that is



                $d(x) = (x + 1)(x + 2). tag{12}$



                $OEDelta$.






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  It is well-known that $Bbb R[x]$ is a principal ideal domain; thus the ideal



                  $(f(x), g(x)) = (d(x)) tag 1$



                  for some



                  $d(x) in Bbb R[x]; tag 2$



                  it immediately follows that



                  $d(x) = r(x)f(x) + s(x)g(x) tag 3$



                  for some



                  $r(x), s(x) in Bbb R[x]; tag 4$



                  furthermore, it is clear from (1) that



                  $d(x) mid f(x), ; d(x) mid g(x); tag 5$



                  it is equally clear via (3) that any



                  $p(x) in Bbb R[x] tag 6$



                  such that



                  $p(x) mid f(x), ; p(x) mid g(x) tag 7$



                  will also satisfy



                  $p(x) mid d(x); tag 8$



                  (5) and (8) together show that $d(x)$ by definition is in fact given by



                  $d(x) = gcd(f(x), g(x)); tag 9$



                  that is, $d(x)$ is the greatest common divisor of $f(x)$, $g(x)$ in $Bbb R[x]$; this fact gives us a tool by which $d(x)$ may be had: calculating the irreducible, hence prime, factors of $f(x)$ and $g(x)$; we find that



                  $f(x) = (x^4 - 1)(x + 2) = (x^2 + 1)(x + 1)(x - 1)(x + 2), tag{10}$



                  $g(x) = (x^2 - 4)(x + 1) = (x - 2)(x + 2)( x + 1); tag{11}$



                  inspection of (10), (11) reveals that every factor on the right-hand sides of these equations is a prime (irreducible) in $Bbb R[x]$, being either linear or the quadratic $x^2 + 1$ which has no roots in $Bbb R$; the factors they have in common are $x + 1$ and $x + 2$; therefore, since $Bbb R[x]$ is a unique factorization domain (as is any principal ideal domain), the unique prime factors of $d(x)$ must be these two; that is



                  $d(x) = (x + 1)(x + 2). tag{12}$



                  $OEDelta$.






                  share|cite|improve this answer











                  $endgroup$



                  It is well-known that $Bbb R[x]$ is a principal ideal domain; thus the ideal



                  $(f(x), g(x)) = (d(x)) tag 1$



                  for some



                  $d(x) in Bbb R[x]; tag 2$



                  it immediately follows that



                  $d(x) = r(x)f(x) + s(x)g(x) tag 3$



                  for some



                  $r(x), s(x) in Bbb R[x]; tag 4$



                  furthermore, it is clear from (1) that



                  $d(x) mid f(x), ; d(x) mid g(x); tag 5$



                  it is equally clear via (3) that any



                  $p(x) in Bbb R[x] tag 6$



                  such that



                  $p(x) mid f(x), ; p(x) mid g(x) tag 7$



                  will also satisfy



                  $p(x) mid d(x); tag 8$



                  (5) and (8) together show that $d(x)$ by definition is in fact given by



                  $d(x) = gcd(f(x), g(x)); tag 9$



                  that is, $d(x)$ is the greatest common divisor of $f(x)$, $g(x)$ in $Bbb R[x]$; this fact gives us a tool by which $d(x)$ may be had: calculating the irreducible, hence prime, factors of $f(x)$ and $g(x)$; we find that



                  $f(x) = (x^4 - 1)(x + 2) = (x^2 + 1)(x + 1)(x - 1)(x + 2), tag{10}$



                  $g(x) = (x^2 - 4)(x + 1) = (x - 2)(x + 2)( x + 1); tag{11}$



                  inspection of (10), (11) reveals that every factor on the right-hand sides of these equations is a prime (irreducible) in $Bbb R[x]$, being either linear or the quadratic $x^2 + 1$ which has no roots in $Bbb R$; the factors they have in common are $x + 1$ and $x + 2$; therefore, since $Bbb R[x]$ is a unique factorization domain (as is any principal ideal domain), the unique prime factors of $d(x)$ must be these two; that is



                  $d(x) = (x + 1)(x + 2). tag{12}$



                  $OEDelta$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 15 at 6:49

























                  answered Jan 15 at 5:57









                  Robert LewisRobert Lewis

                  45.8k23065




                  45.8k23065























                      3












                      $begingroup$

                      No: you have shown that $(f,g)$ is contained in $(p)$ (or $(q)$), but you need to show $(f,g)$ is equal to the ideal generated by some single polynomial. In other words, you would need to show the reverse inclusion as well, that $(f,g)$ contains $(p)$. You have not proved that, and in fact it is not actually true for the $p$ you have chosen.






                      share|cite|improve this answer









                      $endgroup$


















                        3












                        $begingroup$

                        No: you have shown that $(f,g)$ is contained in $(p)$ (or $(q)$), but you need to show $(f,g)$ is equal to the ideal generated by some single polynomial. In other words, you would need to show the reverse inclusion as well, that $(f,g)$ contains $(p)$. You have not proved that, and in fact it is not actually true for the $p$ you have chosen.






                        share|cite|improve this answer









                        $endgroup$
















                          3












                          3








                          3





                          $begingroup$

                          No: you have shown that $(f,g)$ is contained in $(p)$ (or $(q)$), but you need to show $(f,g)$ is equal to the ideal generated by some single polynomial. In other words, you would need to show the reverse inclusion as well, that $(f,g)$ contains $(p)$. You have not proved that, and in fact it is not actually true for the $p$ you have chosen.






                          share|cite|improve this answer









                          $endgroup$



                          No: you have shown that $(f,g)$ is contained in $(p)$ (or $(q)$), but you need to show $(f,g)$ is equal to the ideal generated by some single polynomial. In other words, you would need to show the reverse inclusion as well, that $(f,g)$ contains $(p)$. You have not proved that, and in fact it is not actually true for the $p$ you have chosen.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 15 at 5:02









                          Eric WofseyEric Wofsey

                          185k13213339




                          185k13213339






























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