Mixing
A special case of the JC where the degrees are equal
$begingroup$
Let $p,q in mathbb{C}[x,y]$ with $operatorname{Jac}(p,q):=p_xq_y-p_yq_x in mathbb{C}^{times}$.
Assume that $deg(p)=deg(q)$ (the total degree, also called the $(1,1)$-degree).
Claim: In this special case (equal degrees) it follows that $f: (x,y) mapsto (p,q)$ is an automorphism of $mathbb{C}[x,y]$.
Please, I would like to make sure that I am not missing something,
and the above claim (which I have not seen in articles) is indeed true.
If $f: (x,y) mapsto (p,q)$ is not an automorphism of $mathbb{C}[x,y]$, then (by a result that I think I have seen in van den Essen's book, perhaps due to a result of Nagata and Lang) there exists an automorphism $g$ of $mathbb{C}[x,y]$ such that:
$deg(p)=deg(g(p))$.
$deg(q)=deg(g(q))$.- $deg(g(p))=(a+b)n > (a+b)m=deg(g(q))$
(since the $(1,1)$-leading term of $g(p)$ is $lambda x^{an}y^{bn}$ and the $(1,1)$-leading term of $g(q)$ is $mu x^{am}y^{bm}$, $lambda,mu in k^{times}$).
Then $deg(p)>deg(q)$, contrary to our assumption.
Remark: Now I see that there is a problem with my claim, since every endomorphism we can bring to a form with equal degrees (just multiply with an appropriate automorphism such as $(x,y) mapsto (x+y,x-y)$), and then it would follow that the JC is true.. this seems too easy to be a valid proof for the two-dimensional JC. So the quoted result (concerning the existence of such $g$ etc.) should be wrong? (Probably I misunderstood the quoted result).
Any hints and comments are welcome!
polynomials commutative-algebra
asked Jan 29 at 21:40
user237522user237522
2,1851617
$endgroup$
add a comment |
$begingroup$
Let $p,q in mathbb{C}[x,y]$ with $operatorname{Jac}(p,q):=p_xq_y-p_yq_x in mathbb{C}^{times}$.
Assume that $deg(p)=deg(q)$ (the total degree, also called the $(1,1)$-degree).
Claim: In this special case (equal degrees) it follows that $f: (x,y) mapsto (p,q)$ is an automorphism of $mathbb{C}[x,y]$.
Please, I would like to make sure that I am not missing something,
and the above claim (which I have not seen in articles) is indeed true.
If $f: (x,y) mapsto (p,q)$ is not an automorphism of $mathbb{C}[x,y]$, then (by a result that I think I have seen in van den Essen's book, perhaps due to a result of Nagata and Lang) there exists an automorphism $g$ of $mathbb{C}[x,y]$ such that:
$deg(p)=deg(g(p))$.
$deg(q)=deg(g(q))$.- $deg(g(p))=(a+b)n > (a+b)m=deg(g(q))$
(since the $(1,1)$-leading term of $g(p)$ is $lambda x^{an}y^{bn}$ and the $(1,1)$-leading term of $g(q)$ is $mu x^{am}y^{bm}$, $lambda,mu in k^{times}$).
Then $deg(p)>deg(q)$, contrary to our assumption.
Remark: Now I see that there is a problem with my claim, since every endomorphism we can bring to a form with equal degrees (just multiply with an appropriate automorphism such as $(x,y) mapsto (x+y,x-y)$), and then it would follow that the JC is true.. this seems too easy to be a valid proof for the two-dimensional JC. So the quoted result (concerning the existence of such $g$ etc.) should be wrong? (Probably I misunderstood the quoted result).
Any hints and comments are welcome!
polynomials commutative-algebra
asked Jan 29 at 21:40
user237522user237522
2,1851617
$endgroup$
$begingroup$
Actually, there is no contradiction, since the result I was talking about is Theorem 5.11 of arxiv.org/pdf/1111.6100v2.pdf, which deals with an irreducible pair $(p,q)$, and as such it cannot have $deg(p)=deg(q)$, by Proposition 3.8.
$endgroup$
– user237522
Feb 11 at 19:02
add a comment |
$begingroup$
Let $p,q in mathbb{C}[x,y]$ with $operatorname{Jac}(p,q):=p_xq_y-p_yq_x in mathbb{C}^{times}$.
Assume that $deg(p)=deg(q)$ (the total degree, also called the $(1,1)$-degree).
Claim: In this special case (equal degrees) it follows that $f: (x,y) mapsto (p,q)$ is an automorphism of $mathbb{C}[x,y]$.
Please, I would like to make sure that I am not missing something,
and the above claim (which I have not seen in articles) is indeed true.
If $f: (x,y) mapsto (p,q)$ is not an automorphism of $mathbb{C}[x,y]$, then (by a result that I think I have seen in van den Essen's book, perhaps due to a result of Nagata and Lang) there exists an automorphism $g$ of $mathbb{C}[x,y]$ such that:
$deg(p)=deg(g(p))$.
$deg(q)=deg(g(q))$.- $deg(g(p))=(a+b)n > (a+b)m=deg(g(q))$
(since the $(1,1)$-leading term of $g(p)$ is $lambda x^{an}y^{bn}$ and the $(1,1)$-leading term of $g(q)$ is $mu x^{am}y^{bm}$, $lambda,mu in k^{times}$).
Then $deg(p)>deg(q)$, contrary to our assumption.
Remark: Now I see that there is a problem with my claim, since every endomorphism we can bring to a form with equal degrees (just multiply with an appropriate automorphism such as $(x,y) mapsto (x+y,x-y)$), and then it would follow that the JC is true.. this seems too easy to be a valid proof for the two-dimensional JC. So the quoted result (concerning the existence of such $g$ etc.) should be wrong? (Probably I misunderstood the quoted result).
Any hints and comments are welcome!
polynomials commutative-algebra
asked Jan 29 at 21:40
user237522user237522
2,1851617
$endgroup$
Let $p,q in mathbb{C}[x,y]$ with $operatorname{Jac}(p,q):=p_xq_y-p_yq_x in mathbb{C}^{times}$.
Assume that $deg(p)=deg(q)$ (the total degree, also called the $(1,1)$-degree).
Claim: In this special case (equal degrees) it follows that $f: (x,y) mapsto (p,q)$ is an automorphism of $mathbb{C}[x,y]$.
Please, I would like to make sure that I am not missing something,
and the above claim (which I have not seen in articles) is indeed true.
If $f: (x,y) mapsto (p,q)$ is not an automorphism of $mathbb{C}[x,y]$, then (by a result that I think I have seen in van den Essen's book, perhaps due to a result of Nagata and Lang) there exists an automorphism $g$ of $mathbb{C}[x,y]$ such that:
$deg(p)=deg(g(p))$.
$deg(q)=deg(g(q))$.- $deg(g(p))=(a+b)n > (a+b)m=deg(g(q))$
(since the $(1,1)$-leading term of $g(p)$ is $lambda x^{an}y^{bn}$ and the $(1,1)$-leading term of $g(q)$ is $mu x^{am}y^{bm}$, $lambda,mu in k^{times}$).
Then $deg(p)>deg(q)$, contrary to our assumption.
Remark: Now I see that there is a problem with my claim, since every endomorphism we can bring to a form with equal degrees (just multiply with an appropriate automorphism such as $(x,y) mapsto (x+y,x-y)$), and then it would follow that the JC is true.. this seems too easy to be a valid proof for the two-dimensional JC. So the quoted result (concerning the existence of such $g$ etc.) should be wrong? (Probably I misunderstood the quoted result).
Any hints and comments are welcome!
polynomials commutative-algebra
polynomials commutative-algebra
asked Jan 29 at 21:40
user237522user237522
2,1851617
asked Jan 29 at 21:40
user237522user237522
2,1851617
edited Jan 29 at 22:07
user237522
asked Jan 29 at 21:40
user237522user237522
2,1851617
asked Jan 29 at 21:40
user237522user237522
2,1851617
asked Jan 29 at 21:40
user237522user237522
2,1851617
2,1851617
$begingroup$
Actually, there is no contradiction, since the result I was talking about is Theorem 5.11 of arxiv.org/pdf/1111.6100v2.pdf, which deals with an irreducible pair $(p,q)$, and as such it cannot have $deg(p)=deg(q)$, by Proposition 3.8.
$endgroup$
– user237522
Feb 11 at 19:02
add a comment |
$begingroup$
Actually, there is no contradiction, since the result I was talking about is Theorem 5.11 of arxiv.org/pdf/1111.6100v2.pdf, which deals with an irreducible pair $(p,q)$, and as such it cannot have $deg(p)=deg(q)$, by Proposition 3.8.
$endgroup$
– user237522
Feb 11 at 19:02
$begingroup$
Actually, there is no contradiction, since the result I was talking about is Theorem 5.11 of arxiv.org/pdf/1111.6100v2.pdf, which deals with an irreducible pair $(p,q)$, and as such it cannot have $deg(p)=deg(q)$, by Proposition 3.8.
$endgroup$
– user237522
Feb 11 at 19:02
$begingroup$
Actually, there is no contradiction, since the result I was talking about is Theorem 5.11 of arxiv.org/pdf/1111.6100v2.pdf, which deals with an irreducible pair $(p,q)$, and as such it cannot have $deg(p)=deg(q)$, by Proposition 3.8.
$endgroup$
– user237522
Feb 11 at 19:02
add a comment |
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$begingroup$
Actually, there is no contradiction, since the result I was talking about is Theorem 5.11 of arxiv.org/pdf/1111.6100v2.pdf, which deals with an irreducible pair $(p,q)$, and as such it cannot have $deg(p)=deg(q)$, by Proposition 3.8.
$endgroup$
– user237522
Feb 11 at 19:02