Is this plane curve irreducible?












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I want to define a plane curve in $mathbb{A}^2(mathbb{C})$ by the polynomial $f(x,y)=x(x-1)^2-(y-1)^2=0$ where $(x,y)inmathbb{A}^2(mathbb{C})$, but my goal is for the plane curve to be irreducible.



How do I tell/prove that this curve is irreducible?



Thanks so much!










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$endgroup$

















    1












    $begingroup$


    I want to define a plane curve in $mathbb{A}^2(mathbb{C})$ by the polynomial $f(x,y)=x(x-1)^2-(y-1)^2=0$ where $(x,y)inmathbb{A}^2(mathbb{C})$, but my goal is for the plane curve to be irreducible.



    How do I tell/prove that this curve is irreducible?



    Thanks so much!










    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I want to define a plane curve in $mathbb{A}^2(mathbb{C})$ by the polynomial $f(x,y)=x(x-1)^2-(y-1)^2=0$ where $(x,y)inmathbb{A}^2(mathbb{C})$, but my goal is for the plane curve to be irreducible.



      How do I tell/prove that this curve is irreducible?



      Thanks so much!










      share|cite|improve this question











      $endgroup$




      I want to define a plane curve in $mathbb{A}^2(mathbb{C})$ by the polynomial $f(x,y)=x(x-1)^2-(y-1)^2=0$ where $(x,y)inmathbb{A}^2(mathbb{C})$, but my goal is for the plane curve to be irreducible.



      How do I tell/prove that this curve is irreducible?



      Thanks so much!







      algebraic-curves






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      edited Dec 13 '18 at 15:04







      mathgeen

















      asked Dec 13 '18 at 12:09









      mathgeenmathgeen

      224




      224






















          1 Answer
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          I start from a change of cordinates:
          begin{equation*}
          begin{cases}
          x^{prime}=x-1\
          y^{prime}=y-1
          end{cases}
          end{equation*}

          so the plane curve in the new coordinates has equation $left(x^{prime}+1right)left(x^{prime}right)^2-left(y^{prime}right)^2=0$; for simplicity, the new equation can be rewrite as $y^2=x^2(x+1)$.



          The best and simpler easy reasoning to prove the irreducibility of $X={(x,y)inmathbb{A}^2_{mathbb{C}}mid y^2=x^2(x+1)}$, in my knowldge, is the following: because $x^2(x+1)$ must be a square of a polynomial then $x+1=t^2$, so $x=t^2-1$ and $y^2=(t^2-1)^2t^2Rightarrow y=pm t(t^2-1)$. From all this, $X$ is the image of $mathbb{A}^1_{mathbb{C}}$ via the continuous map (with respect to Zariski topology)
          begin{equation*}
          varphi:tinmathbb{A}^1_{mathbb{C}}to(t^2-1,t(t^2-1))inmathbb{A}^2_{mathbb{C}};
          end{equation*}

          because the image of irreducible sets via continuous map is irreducible as well, $X$ is an irreducible subset of $mathbb{A}^2_{mathbb{C}}$.






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            1 Answer
            1






            active

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            oldest

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            active

            oldest

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            3












            $begingroup$

            I start from a change of cordinates:
            begin{equation*}
            begin{cases}
            x^{prime}=x-1\
            y^{prime}=y-1
            end{cases}
            end{equation*}

            so the plane curve in the new coordinates has equation $left(x^{prime}+1right)left(x^{prime}right)^2-left(y^{prime}right)^2=0$; for simplicity, the new equation can be rewrite as $y^2=x^2(x+1)$.



            The best and simpler easy reasoning to prove the irreducibility of $X={(x,y)inmathbb{A}^2_{mathbb{C}}mid y^2=x^2(x+1)}$, in my knowldge, is the following: because $x^2(x+1)$ must be a square of a polynomial then $x+1=t^2$, so $x=t^2-1$ and $y^2=(t^2-1)^2t^2Rightarrow y=pm t(t^2-1)$. From all this, $X$ is the image of $mathbb{A}^1_{mathbb{C}}$ via the continuous map (with respect to Zariski topology)
            begin{equation*}
            varphi:tinmathbb{A}^1_{mathbb{C}}to(t^2-1,t(t^2-1))inmathbb{A}^2_{mathbb{C}};
            end{equation*}

            because the image of irreducible sets via continuous map is irreducible as well, $X$ is an irreducible subset of $mathbb{A}^2_{mathbb{C}}$.






            share|cite|improve this answer











            $endgroup$


















              3












              $begingroup$

              I start from a change of cordinates:
              begin{equation*}
              begin{cases}
              x^{prime}=x-1\
              y^{prime}=y-1
              end{cases}
              end{equation*}

              so the plane curve in the new coordinates has equation $left(x^{prime}+1right)left(x^{prime}right)^2-left(y^{prime}right)^2=0$; for simplicity, the new equation can be rewrite as $y^2=x^2(x+1)$.



              The best and simpler easy reasoning to prove the irreducibility of $X={(x,y)inmathbb{A}^2_{mathbb{C}}mid y^2=x^2(x+1)}$, in my knowldge, is the following: because $x^2(x+1)$ must be a square of a polynomial then $x+1=t^2$, so $x=t^2-1$ and $y^2=(t^2-1)^2t^2Rightarrow y=pm t(t^2-1)$. From all this, $X$ is the image of $mathbb{A}^1_{mathbb{C}}$ via the continuous map (with respect to Zariski topology)
              begin{equation*}
              varphi:tinmathbb{A}^1_{mathbb{C}}to(t^2-1,t(t^2-1))inmathbb{A}^2_{mathbb{C}};
              end{equation*}

              because the image of irreducible sets via continuous map is irreducible as well, $X$ is an irreducible subset of $mathbb{A}^2_{mathbb{C}}$.






              share|cite|improve this answer











              $endgroup$
















                3












                3








                3





                $begingroup$

                I start from a change of cordinates:
                begin{equation*}
                begin{cases}
                x^{prime}=x-1\
                y^{prime}=y-1
                end{cases}
                end{equation*}

                so the plane curve in the new coordinates has equation $left(x^{prime}+1right)left(x^{prime}right)^2-left(y^{prime}right)^2=0$; for simplicity, the new equation can be rewrite as $y^2=x^2(x+1)$.



                The best and simpler easy reasoning to prove the irreducibility of $X={(x,y)inmathbb{A}^2_{mathbb{C}}mid y^2=x^2(x+1)}$, in my knowldge, is the following: because $x^2(x+1)$ must be a square of a polynomial then $x+1=t^2$, so $x=t^2-1$ and $y^2=(t^2-1)^2t^2Rightarrow y=pm t(t^2-1)$. From all this, $X$ is the image of $mathbb{A}^1_{mathbb{C}}$ via the continuous map (with respect to Zariski topology)
                begin{equation*}
                varphi:tinmathbb{A}^1_{mathbb{C}}to(t^2-1,t(t^2-1))inmathbb{A}^2_{mathbb{C}};
                end{equation*}

                because the image of irreducible sets via continuous map is irreducible as well, $X$ is an irreducible subset of $mathbb{A}^2_{mathbb{C}}$.






                share|cite|improve this answer











                $endgroup$



                I start from a change of cordinates:
                begin{equation*}
                begin{cases}
                x^{prime}=x-1\
                y^{prime}=y-1
                end{cases}
                end{equation*}

                so the plane curve in the new coordinates has equation $left(x^{prime}+1right)left(x^{prime}right)^2-left(y^{prime}right)^2=0$; for simplicity, the new equation can be rewrite as $y^2=x^2(x+1)$.



                The best and simpler easy reasoning to prove the irreducibility of $X={(x,y)inmathbb{A}^2_{mathbb{C}}mid y^2=x^2(x+1)}$, in my knowldge, is the following: because $x^2(x+1)$ must be a square of a polynomial then $x+1=t^2$, so $x=t^2-1$ and $y^2=(t^2-1)^2t^2Rightarrow y=pm t(t^2-1)$. From all this, $X$ is the image of $mathbb{A}^1_{mathbb{C}}$ via the continuous map (with respect to Zariski topology)
                begin{equation*}
                varphi:tinmathbb{A}^1_{mathbb{C}}to(t^2-1,t(t^2-1))inmathbb{A}^2_{mathbb{C}};
                end{equation*}

                because the image of irreducible sets via continuous map is irreducible as well, $X$ is an irreducible subset of $mathbb{A}^2_{mathbb{C}}$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 13 '18 at 17:33

























                answered Dec 13 '18 at 14:10









                Armando j18eosArmando j18eos

                2,64511328




                2,64511328






























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