How to prove $ left{ t^2,t^3 right}$ equals the vanishing set of $y^2-x^3$?











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Exercise 3.2 in Hartshorne is about proving that morphisms of varieties may be underlain by homeomorphisms without being isomorphisms of varieties. The morphism in consideration is $varphi:tmapsto (t^2,t^3)$ where the image is to be thought of as the curve $y^2-x^3$.



I don't understand over which fields $Bbbk$ the set $ left{ t^2,t^3 right}$ equals the vanishing set of $y^2-x^3in Bbbk[x,y]$ - only why it's contained in the vanishing set. Indeed if $(a,b)$ satisfies $a^3=b^2$ we at least need the existence of square/cube roots, since we want $t$ such that $t^2=a,t^3=b$. Suppose for convenience the field is algebraically closed. Then we have some creature worthy of the name $sqrt a$ that satisfies $sqrt a^2=a$ and by assumption $sqrt a^6=b^2$. But why should we have $sqrt a^3=b$? (more accurately, why can we choose $sqrt a$ to have this property?)



Maybe this is elementary field/Galois theory, but better late than never.










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    Exercise 3.2 in Hartshorne is about proving that morphisms of varieties may be underlain by homeomorphisms without being isomorphisms of varieties. The morphism in consideration is $varphi:tmapsto (t^2,t^3)$ where the image is to be thought of as the curve $y^2-x^3$.



    I don't understand over which fields $Bbbk$ the set $ left{ t^2,t^3 right}$ equals the vanishing set of $y^2-x^3in Bbbk[x,y]$ - only why it's contained in the vanishing set. Indeed if $(a,b)$ satisfies $a^3=b^2$ we at least need the existence of square/cube roots, since we want $t$ such that $t^2=a,t^3=b$. Suppose for convenience the field is algebraically closed. Then we have some creature worthy of the name $sqrt a$ that satisfies $sqrt a^2=a$ and by assumption $sqrt a^6=b^2$. But why should we have $sqrt a^3=b$? (more accurately, why can we choose $sqrt a$ to have this property?)



    Maybe this is elementary field/Galois theory, but better late than never.










    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Exercise 3.2 in Hartshorne is about proving that morphisms of varieties may be underlain by homeomorphisms without being isomorphisms of varieties. The morphism in consideration is $varphi:tmapsto (t^2,t^3)$ where the image is to be thought of as the curve $y^2-x^3$.



      I don't understand over which fields $Bbbk$ the set $ left{ t^2,t^3 right}$ equals the vanishing set of $y^2-x^3in Bbbk[x,y]$ - only why it's contained in the vanishing set. Indeed if $(a,b)$ satisfies $a^3=b^2$ we at least need the existence of square/cube roots, since we want $t$ such that $t^2=a,t^3=b$. Suppose for convenience the field is algebraically closed. Then we have some creature worthy of the name $sqrt a$ that satisfies $sqrt a^2=a$ and by assumption $sqrt a^6=b^2$. But why should we have $sqrt a^3=b$? (more accurately, why can we choose $sqrt a$ to have this property?)



      Maybe this is elementary field/Galois theory, but better late than never.










      share|cite|improve this question













      Exercise 3.2 in Hartshorne is about proving that morphisms of varieties may be underlain by homeomorphisms without being isomorphisms of varieties. The morphism in consideration is $varphi:tmapsto (t^2,t^3)$ where the image is to be thought of as the curve $y^2-x^3$.



      I don't understand over which fields $Bbbk$ the set $ left{ t^2,t^3 right}$ equals the vanishing set of $y^2-x^3in Bbbk[x,y]$ - only why it's contained in the vanishing set. Indeed if $(a,b)$ satisfies $a^3=b^2$ we at least need the existence of square/cube roots, since we want $t$ such that $t^2=a,t^3=b$. Suppose for convenience the field is algebraically closed. Then we have some creature worthy of the name $sqrt a$ that satisfies $sqrt a^2=a$ and by assumption $sqrt a^6=b^2$. But why should we have $sqrt a^3=b$? (more accurately, why can we choose $sqrt a$ to have this property?)



      Maybe this is elementary field/Galois theory, but better late than never.







      field-theory galois-theory






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      asked Nov 22 at 22:04









      Arrow

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          Suppose $x^3=y^2$. If $x=0$ then $y=0$ so that $(x,y)=(t^2,t^3)$ for $t=0$.



          Otherwise $xne0$. We can then define $t=y/x$. Then $t^2=y^2/x^2=x^3/x^2=x$
          and $t^3=t^2t=x(y/x)=y$. So $(x,y)=(t^2,t^3)$.



          Not a square or cube root in sight!






          share|cite|improve this answer

















          • 2




            A more complicated version of this argument will continue to work with $(2, 3)$ replaced by any pair of coprime exponents, e.g. using Bezout's lemma.
            – Qiaochu Yuan
            Nov 22 at 22:10










          • Not sure whether to laugh or cry. Probably cry. Thanks for the answer!
            – Arrow
            Nov 22 at 22:16











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          1 Answer
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          1 Answer
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          active

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          active

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          up vote
          4
          down vote



          accepted










          Suppose $x^3=y^2$. If $x=0$ then $y=0$ so that $(x,y)=(t^2,t^3)$ for $t=0$.



          Otherwise $xne0$. We can then define $t=y/x$. Then $t^2=y^2/x^2=x^3/x^2=x$
          and $t^3=t^2t=x(y/x)=y$. So $(x,y)=(t^2,t^3)$.



          Not a square or cube root in sight!






          share|cite|improve this answer

















          • 2




            A more complicated version of this argument will continue to work with $(2, 3)$ replaced by any pair of coprime exponents, e.g. using Bezout's lemma.
            – Qiaochu Yuan
            Nov 22 at 22:10










          • Not sure whether to laugh or cry. Probably cry. Thanks for the answer!
            – Arrow
            Nov 22 at 22:16















          up vote
          4
          down vote



          accepted










          Suppose $x^3=y^2$. If $x=0$ then $y=0$ so that $(x,y)=(t^2,t^3)$ for $t=0$.



          Otherwise $xne0$. We can then define $t=y/x$. Then $t^2=y^2/x^2=x^3/x^2=x$
          and $t^3=t^2t=x(y/x)=y$. So $(x,y)=(t^2,t^3)$.



          Not a square or cube root in sight!






          share|cite|improve this answer

















          • 2




            A more complicated version of this argument will continue to work with $(2, 3)$ replaced by any pair of coprime exponents, e.g. using Bezout's lemma.
            – Qiaochu Yuan
            Nov 22 at 22:10










          • Not sure whether to laugh or cry. Probably cry. Thanks for the answer!
            – Arrow
            Nov 22 at 22:16













          up vote
          4
          down vote



          accepted







          up vote
          4
          down vote



          accepted






          Suppose $x^3=y^2$. If $x=0$ then $y=0$ so that $(x,y)=(t^2,t^3)$ for $t=0$.



          Otherwise $xne0$. We can then define $t=y/x$. Then $t^2=y^2/x^2=x^3/x^2=x$
          and $t^3=t^2t=x(y/x)=y$. So $(x,y)=(t^2,t^3)$.



          Not a square or cube root in sight!






          share|cite|improve this answer












          Suppose $x^3=y^2$. If $x=0$ then $y=0$ so that $(x,y)=(t^2,t^3)$ for $t=0$.



          Otherwise $xne0$. We can then define $t=y/x$. Then $t^2=y^2/x^2=x^3/x^2=x$
          and $t^3=t^2t=x(y/x)=y$. So $(x,y)=(t^2,t^3)$.



          Not a square or cube root in sight!







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 at 22:08









          Lord Shark the Unknown

          97.8k958129




          97.8k958129








          • 2




            A more complicated version of this argument will continue to work with $(2, 3)$ replaced by any pair of coprime exponents, e.g. using Bezout's lemma.
            – Qiaochu Yuan
            Nov 22 at 22:10










          • Not sure whether to laugh or cry. Probably cry. Thanks for the answer!
            – Arrow
            Nov 22 at 22:16














          • 2




            A more complicated version of this argument will continue to work with $(2, 3)$ replaced by any pair of coprime exponents, e.g. using Bezout's lemma.
            – Qiaochu Yuan
            Nov 22 at 22:10










          • Not sure whether to laugh or cry. Probably cry. Thanks for the answer!
            – Arrow
            Nov 22 at 22:16








          2




          2




          A more complicated version of this argument will continue to work with $(2, 3)$ replaced by any pair of coprime exponents, e.g. using Bezout's lemma.
          – Qiaochu Yuan
          Nov 22 at 22:10




          A more complicated version of this argument will continue to work with $(2, 3)$ replaced by any pair of coprime exponents, e.g. using Bezout's lemma.
          – Qiaochu Yuan
          Nov 22 at 22:10












          Not sure whether to laugh or cry. Probably cry. Thanks for the answer!
          – Arrow
          Nov 22 at 22:16




          Not sure whether to laugh or cry. Probably cry. Thanks for the answer!
          – Arrow
          Nov 22 at 22:16


















           

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