Show that $f(x,y,z)=x^2-y^2z$ is irreducible in $mathbb{C}[x,y,z]$.












2












$begingroup$


Let $pinmathbb{C}[x,y,z]$ be defined by $p(x,y,z)=x^2-y^2z$.



Goal: Prove that $p$ is irreducible.





Let $Isubsetmathbb{C}[x,y,z]$ be the ideal defined by



$$I:=(p).$$



My approach is to show that



$$mathbb{C}[x,y,z]/I,$$
is an integral domain and, hence, $I$ prime ideal or equivalently here, $p$ is irreducible.



Let us consider a ring homomorphism



$$varphi:mathbb{C}[x,y,z]tomathbb{C}[t_1,t_2]$$
$$begin{cases}xmapsto t_1^2t_2 \
ymapsto t_1^2 \
z mapsto t_2^2.
end{cases}$$

Notice that, since $varphi$ is a ring homomorphism,
$$varphi(p)=varphi(x^2-y^2z)=varphi(x^2)-varphi(y^2)varphi(z)=t_1^4t_2^2-t_1^4t_2^2=0.$$





Here is a claim, which I think is true (it atleast feels like it), but I don't know how to prove it. My argument relies on this:



Claim: Any polynomial $finmathbb{C}[x,y,z]$ can be written as
$$f(x,y,z)=f_0(y,z)+xf_1(y,z)+(x^2-y^2z)g(x,y,z).$$



I read on another link that, for any $f(x,y)in mathbb{C}[x,y]$, we can write



$$f(x,y)=f_0(x)+yf_1(x)+(x^3-y^2)g(x,y).$$



My approach is inspired by the above identity.



$$text{ }$$



It feels like my claim should be true, since $f_0(y,z)$ takes care of all expressions of the form $sum_{k=0}^msum_{j=0}^nx^0y^jz^k$ and $xf_1(y,z)$ takes care of everything of the form $sum_{k=0}^msum_{j=0}^nx^1y^jz^k$.



Lastly, we have $x^2g(x,y,z)$ which feels like it should take care of any polynomial



$$sum_{k=0}^msum_{j=0}^nsum_{l=2}^o x^ly^jz^kquad (o text{ was quite ugly to use in the summation, but anyway)},$$



so I don't really understand what's so special with the



$$-y^2zg(x,y,z)-text{part},$$



it feels like it doesn't really contribute, or restrict, the polynomial $f$ in any way. So it feels like you could do a similar construction for any polynomial. But probably not.



$text{ }$



My questions now is:




  • Is my claim true?

  • If it is, could you please tell me why?




Continuing on the actual problem:



Thus, applying $varphi$ on $finkervarphi$ gives us
$$varphi(f)=varphi(f_0)+varphi(x)varphi(f_1)+varphi(x^2-y^2z)varphi(g)$$
$$=f_0(t_1^2,t_2^2)+t_1^2t_2f_1(t_1^2,t_2^2)=0$$
$$Rightarrow f_0=f_1=0,$$
Where the last implication holds since $f_0(t_1^2,t_2^2)$ is of even degree while $t_1^2t_2f_1(t_1^2,t_2^2)$ is of odd degree. Hence the only case the sum can be $0$ is if $f_0$ and $f_1$ are identically $0$.



This shows us that $kervarphi=I$. By the first isomorphism theorem, we have
$$mathbb{C}[x,y,z]/Icong operatorname{im}(varphi).$$



But this shows us that $mathbb{C}[x,y,z]/I$ is a subring of the integral domain $mathbb{C}[t_1,t_2]$. Hence $I$ is a prime ideal, so $p$ is irreducible, which is what we wanted to prove.



Questions:




  • Does my approach work? In particular, is my claim true and if so, why?

  • If this approach does not work, could you please help me with a better approach and solution?


Thanks for your time!










share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Let $pinmathbb{C}[x,y,z]$ be defined by $p(x,y,z)=x^2-y^2z$.



    Goal: Prove that $p$ is irreducible.





    Let $Isubsetmathbb{C}[x,y,z]$ be the ideal defined by



    $$I:=(p).$$



    My approach is to show that



    $$mathbb{C}[x,y,z]/I,$$
    is an integral domain and, hence, $I$ prime ideal or equivalently here, $p$ is irreducible.



    Let us consider a ring homomorphism



    $$varphi:mathbb{C}[x,y,z]tomathbb{C}[t_1,t_2]$$
    $$begin{cases}xmapsto t_1^2t_2 \
    ymapsto t_1^2 \
    z mapsto t_2^2.
    end{cases}$$

    Notice that, since $varphi$ is a ring homomorphism,
    $$varphi(p)=varphi(x^2-y^2z)=varphi(x^2)-varphi(y^2)varphi(z)=t_1^4t_2^2-t_1^4t_2^2=0.$$





    Here is a claim, which I think is true (it atleast feels like it), but I don't know how to prove it. My argument relies on this:



    Claim: Any polynomial $finmathbb{C}[x,y,z]$ can be written as
    $$f(x,y,z)=f_0(y,z)+xf_1(y,z)+(x^2-y^2z)g(x,y,z).$$



    I read on another link that, for any $f(x,y)in mathbb{C}[x,y]$, we can write



    $$f(x,y)=f_0(x)+yf_1(x)+(x^3-y^2)g(x,y).$$



    My approach is inspired by the above identity.



    $$text{ }$$



    It feels like my claim should be true, since $f_0(y,z)$ takes care of all expressions of the form $sum_{k=0}^msum_{j=0}^nx^0y^jz^k$ and $xf_1(y,z)$ takes care of everything of the form $sum_{k=0}^msum_{j=0}^nx^1y^jz^k$.



    Lastly, we have $x^2g(x,y,z)$ which feels like it should take care of any polynomial



    $$sum_{k=0}^msum_{j=0}^nsum_{l=2}^o x^ly^jz^kquad (o text{ was quite ugly to use in the summation, but anyway)},$$



    so I don't really understand what's so special with the



    $$-y^2zg(x,y,z)-text{part},$$



    it feels like it doesn't really contribute, or restrict, the polynomial $f$ in any way. So it feels like you could do a similar construction for any polynomial. But probably not.



    $text{ }$



    My questions now is:




    • Is my claim true?

    • If it is, could you please tell me why?




    Continuing on the actual problem:



    Thus, applying $varphi$ on $finkervarphi$ gives us
    $$varphi(f)=varphi(f_0)+varphi(x)varphi(f_1)+varphi(x^2-y^2z)varphi(g)$$
    $$=f_0(t_1^2,t_2^2)+t_1^2t_2f_1(t_1^2,t_2^2)=0$$
    $$Rightarrow f_0=f_1=0,$$
    Where the last implication holds since $f_0(t_1^2,t_2^2)$ is of even degree while $t_1^2t_2f_1(t_1^2,t_2^2)$ is of odd degree. Hence the only case the sum can be $0$ is if $f_0$ and $f_1$ are identically $0$.



    This shows us that $kervarphi=I$. By the first isomorphism theorem, we have
    $$mathbb{C}[x,y,z]/Icong operatorname{im}(varphi).$$



    But this shows us that $mathbb{C}[x,y,z]/I$ is a subring of the integral domain $mathbb{C}[t_1,t_2]$. Hence $I$ is a prime ideal, so $p$ is irreducible, which is what we wanted to prove.



    Questions:




    • Does my approach work? In particular, is my claim true and if so, why?

    • If this approach does not work, could you please help me with a better approach and solution?


    Thanks for your time!










    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      Let $pinmathbb{C}[x,y,z]$ be defined by $p(x,y,z)=x^2-y^2z$.



      Goal: Prove that $p$ is irreducible.





      Let $Isubsetmathbb{C}[x,y,z]$ be the ideal defined by



      $$I:=(p).$$



      My approach is to show that



      $$mathbb{C}[x,y,z]/I,$$
      is an integral domain and, hence, $I$ prime ideal or equivalently here, $p$ is irreducible.



      Let us consider a ring homomorphism



      $$varphi:mathbb{C}[x,y,z]tomathbb{C}[t_1,t_2]$$
      $$begin{cases}xmapsto t_1^2t_2 \
      ymapsto t_1^2 \
      z mapsto t_2^2.
      end{cases}$$

      Notice that, since $varphi$ is a ring homomorphism,
      $$varphi(p)=varphi(x^2-y^2z)=varphi(x^2)-varphi(y^2)varphi(z)=t_1^4t_2^2-t_1^4t_2^2=0.$$





      Here is a claim, which I think is true (it atleast feels like it), but I don't know how to prove it. My argument relies on this:



      Claim: Any polynomial $finmathbb{C}[x,y,z]$ can be written as
      $$f(x,y,z)=f_0(y,z)+xf_1(y,z)+(x^2-y^2z)g(x,y,z).$$



      I read on another link that, for any $f(x,y)in mathbb{C}[x,y]$, we can write



      $$f(x,y)=f_0(x)+yf_1(x)+(x^3-y^2)g(x,y).$$



      My approach is inspired by the above identity.



      $$text{ }$$



      It feels like my claim should be true, since $f_0(y,z)$ takes care of all expressions of the form $sum_{k=0}^msum_{j=0}^nx^0y^jz^k$ and $xf_1(y,z)$ takes care of everything of the form $sum_{k=0}^msum_{j=0}^nx^1y^jz^k$.



      Lastly, we have $x^2g(x,y,z)$ which feels like it should take care of any polynomial



      $$sum_{k=0}^msum_{j=0}^nsum_{l=2}^o x^ly^jz^kquad (o text{ was quite ugly to use in the summation, but anyway)},$$



      so I don't really understand what's so special with the



      $$-y^2zg(x,y,z)-text{part},$$



      it feels like it doesn't really contribute, or restrict, the polynomial $f$ in any way. So it feels like you could do a similar construction for any polynomial. But probably not.



      $text{ }$



      My questions now is:




      • Is my claim true?

      • If it is, could you please tell me why?




      Continuing on the actual problem:



      Thus, applying $varphi$ on $finkervarphi$ gives us
      $$varphi(f)=varphi(f_0)+varphi(x)varphi(f_1)+varphi(x^2-y^2z)varphi(g)$$
      $$=f_0(t_1^2,t_2^2)+t_1^2t_2f_1(t_1^2,t_2^2)=0$$
      $$Rightarrow f_0=f_1=0,$$
      Where the last implication holds since $f_0(t_1^2,t_2^2)$ is of even degree while $t_1^2t_2f_1(t_1^2,t_2^2)$ is of odd degree. Hence the only case the sum can be $0$ is if $f_0$ and $f_1$ are identically $0$.



      This shows us that $kervarphi=I$. By the first isomorphism theorem, we have
      $$mathbb{C}[x,y,z]/Icong operatorname{im}(varphi).$$



      But this shows us that $mathbb{C}[x,y,z]/I$ is a subring of the integral domain $mathbb{C}[t_1,t_2]$. Hence $I$ is a prime ideal, so $p$ is irreducible, which is what we wanted to prove.



      Questions:




      • Does my approach work? In particular, is my claim true and if so, why?

      • If this approach does not work, could you please help me with a better approach and solution?


      Thanks for your time!










      share|cite|improve this question











      $endgroup$




      Let $pinmathbb{C}[x,y,z]$ be defined by $p(x,y,z)=x^2-y^2z$.



      Goal: Prove that $p$ is irreducible.





      Let $Isubsetmathbb{C}[x,y,z]$ be the ideal defined by



      $$I:=(p).$$



      My approach is to show that



      $$mathbb{C}[x,y,z]/I,$$
      is an integral domain and, hence, $I$ prime ideal or equivalently here, $p$ is irreducible.



      Let us consider a ring homomorphism



      $$varphi:mathbb{C}[x,y,z]tomathbb{C}[t_1,t_2]$$
      $$begin{cases}xmapsto t_1^2t_2 \
      ymapsto t_1^2 \
      z mapsto t_2^2.
      end{cases}$$

      Notice that, since $varphi$ is a ring homomorphism,
      $$varphi(p)=varphi(x^2-y^2z)=varphi(x^2)-varphi(y^2)varphi(z)=t_1^4t_2^2-t_1^4t_2^2=0.$$





      Here is a claim, which I think is true (it atleast feels like it), but I don't know how to prove it. My argument relies on this:



      Claim: Any polynomial $finmathbb{C}[x,y,z]$ can be written as
      $$f(x,y,z)=f_0(y,z)+xf_1(y,z)+(x^2-y^2z)g(x,y,z).$$



      I read on another link that, for any $f(x,y)in mathbb{C}[x,y]$, we can write



      $$f(x,y)=f_0(x)+yf_1(x)+(x^3-y^2)g(x,y).$$



      My approach is inspired by the above identity.



      $$text{ }$$



      It feels like my claim should be true, since $f_0(y,z)$ takes care of all expressions of the form $sum_{k=0}^msum_{j=0}^nx^0y^jz^k$ and $xf_1(y,z)$ takes care of everything of the form $sum_{k=0}^msum_{j=0}^nx^1y^jz^k$.



      Lastly, we have $x^2g(x,y,z)$ which feels like it should take care of any polynomial



      $$sum_{k=0}^msum_{j=0}^nsum_{l=2}^o x^ly^jz^kquad (o text{ was quite ugly to use in the summation, but anyway)},$$



      so I don't really understand what's so special with the



      $$-y^2zg(x,y,z)-text{part},$$



      it feels like it doesn't really contribute, or restrict, the polynomial $f$ in any way. So it feels like you could do a similar construction for any polynomial. But probably not.



      $text{ }$



      My questions now is:




      • Is my claim true?

      • If it is, could you please tell me why?




      Continuing on the actual problem:



      Thus, applying $varphi$ on $finkervarphi$ gives us
      $$varphi(f)=varphi(f_0)+varphi(x)varphi(f_1)+varphi(x^2-y^2z)varphi(g)$$
      $$=f_0(t_1^2,t_2^2)+t_1^2t_2f_1(t_1^2,t_2^2)=0$$
      $$Rightarrow f_0=f_1=0,$$
      Where the last implication holds since $f_0(t_1^2,t_2^2)$ is of even degree while $t_1^2t_2f_1(t_1^2,t_2^2)$ is of odd degree. Hence the only case the sum can be $0$ is if $f_0$ and $f_1$ are identically $0$.



      This shows us that $kervarphi=I$. By the first isomorphism theorem, we have
      $$mathbb{C}[x,y,z]/Icong operatorname{im}(varphi).$$



      But this shows us that $mathbb{C}[x,y,z]/I$ is a subring of the integral domain $mathbb{C}[t_1,t_2]$. Hence $I$ is a prime ideal, so $p$ is irreducible, which is what we wanted to prove.



      Questions:




      • Does my approach work? In particular, is my claim true and if so, why?

      • If this approach does not work, could you please help me with a better approach and solution?


      Thanks for your time!







      algebraic-geometry irreducible-polynomials algebraic-curves integral-domain several-complex-variables






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      edited Dec 6 '18 at 21:11







      Joe

















      asked Dec 6 '18 at 16:52









      JoeJoe

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          $begingroup$

          Yes, your claim is true, for a very simple reason; considering $f(x,yz)$ as a polynomial in the integral domain $mathbf C[y,z][x]$, you always can perform in this ring a Euclidean division by a monic (in $x$) polynomial. The remainder has degree $le 1$ for a quadratic monic polynomial.



          However, this polynomial is irreducible for a shorter reason: Eisenstein criterion. Indeed $mathbf C[y,z]$ is a U.F.D. and the irreducible element $z$ divides all coefficients except the lead coefficient (of $x^2),,$ and $z^2$ does not divide the "constant" coefficient $y^2z$ (of $x^0)$






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            $begingroup$

            Yes, your claim is true, for a very simple reason; considering $f(x,yz)$ as a polynomial in the integral domain $mathbf C[y,z][x]$, you always can perform in this ring a Euclidean division by a monic (in $x$) polynomial. The remainder has degree $le 1$ for a quadratic monic polynomial.



            However, this polynomial is irreducible for a shorter reason: Eisenstein criterion. Indeed $mathbf C[y,z]$ is a U.F.D. and the irreducible element $z$ divides all coefficients except the lead coefficient (of $x^2),,$ and $z^2$ does not divide the "constant" coefficient $y^2z$ (of $x^0)$






            share|cite|improve this answer











            $endgroup$


















              3












              $begingroup$

              Yes, your claim is true, for a very simple reason; considering $f(x,yz)$ as a polynomial in the integral domain $mathbf C[y,z][x]$, you always can perform in this ring a Euclidean division by a monic (in $x$) polynomial. The remainder has degree $le 1$ for a quadratic monic polynomial.



              However, this polynomial is irreducible for a shorter reason: Eisenstein criterion. Indeed $mathbf C[y,z]$ is a U.F.D. and the irreducible element $z$ divides all coefficients except the lead coefficient (of $x^2),,$ and $z^2$ does not divide the "constant" coefficient $y^2z$ (of $x^0)$






              share|cite|improve this answer











              $endgroup$
















                3












                3








                3





                $begingroup$

                Yes, your claim is true, for a very simple reason; considering $f(x,yz)$ as a polynomial in the integral domain $mathbf C[y,z][x]$, you always can perform in this ring a Euclidean division by a monic (in $x$) polynomial. The remainder has degree $le 1$ for a quadratic monic polynomial.



                However, this polynomial is irreducible for a shorter reason: Eisenstein criterion. Indeed $mathbf C[y,z]$ is a U.F.D. and the irreducible element $z$ divides all coefficients except the lead coefficient (of $x^2),,$ and $z^2$ does not divide the "constant" coefficient $y^2z$ (of $x^0)$






                share|cite|improve this answer











                $endgroup$



                Yes, your claim is true, for a very simple reason; considering $f(x,yz)$ as a polynomial in the integral domain $mathbf C[y,z][x]$, you always can perform in this ring a Euclidean division by a monic (in $x$) polynomial. The remainder has degree $le 1$ for a quadratic monic polynomial.



                However, this polynomial is irreducible for a shorter reason: Eisenstein criterion. Indeed $mathbf C[y,z]$ is a U.F.D. and the irreducible element $z$ divides all coefficients except the lead coefficient (of $x^2),,$ and $z^2$ does not divide the "constant" coefficient $y^2z$ (of $x^0)$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 6 '18 at 21:16

























                answered Dec 6 '18 at 17:19









                BernardBernard

                119k740113




                119k740113






























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