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.
field-theory galois-theory
asked Nov 22 at 22:04
Arrow
5,11211445
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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.
field-theory galois-theory
asked Nov 22 at 22:04
Arrow
5,11211445
add a comment |
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.
field-theory galois-theory
asked Nov 22 at 22:04
Arrow
5,11211445
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
field-theory galois-theory
asked Nov 22 at 22:04
Arrow
5,11211445
asked Nov 22 at 22:04
Arrow
5,11211445
asked Nov 22 at 22:04
Arrow
5,11211445
asked Nov 22 at 22:04
Arrow
5,11211445
asked Nov 22 at 22:04
Arrow
5,11211445
5,11211445
add a comment |
add a comment |
1 Answer
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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!
answered Nov 22 at 22:08
Lord Shark the Unknown
99.1k958131
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
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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!
answered Nov 22 at 22:08
Lord Shark the Unknown
99.1k958131
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
add a comment |
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!
answered Nov 22 at 22:08
Lord Shark the Unknown
99.1k958131
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
add a comment |
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!
answered Nov 22 at 22:08
Lord Shark the Unknown
99.1k958131
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!
answered Nov 22 at 22:08
Lord Shark the Unknown
99.1k958131
answered Nov 22 at 22:08
Lord Shark the Unknown
99.1k958131
answered Nov 22 at 22:08
Lord Shark the Unknown
99.1k958131
answered Nov 22 at 22:08
Lord Shark the Unknown
99.1k958131
99.1k958131
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
add a comment |
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
add a comment |
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