For which $f in mathbbC[x,y]$, $fy-lambda$ is irreducible in $mathbbC[x,y]$ for infinitely many $mathbbC ni lambda$'s
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Let $f=f(x,y) in mathbbC[x,y]$,
and for every $lambda in mathbbC$, denote $F_lambda:=fy-lambda$.
Is it possible to characterize all $f in mathbbC[x,y]$ such that $F_lambda$ is irreducible in $mathbbC[x,y]$ for infinitely many $mathbbC ni lambda$'s?
Example: If I am not wrong, $f=x$ is such that $F_lambda=xy-lambda$ is irreducible for every $lambda in mathbbC-0$.
See also this question concerning irreducibility of a polynomial in two variables.
Any hints and comments are welcome!
algebraic-geometry polynomials irreducible-polynomials
asked Aug 12 at 4:52
user237522
1,8141617
 |Â
show 1 more comment
up vote
2
down vote
favorite
Let $f=f(x,y) in mathbbC[x,y]$,
and for every $lambda in mathbbC$, denote $F_lambda:=fy-lambda$.
Is it possible to characterize all $f in mathbbC[x,y]$ such that $F_lambda$ is irreducible in $mathbbC[x,y]$ for infinitely many $mathbbC ni lambda$'s?
Example: If I am not wrong, $f=x$ is such that $F_lambda=xy-lambda$ is irreducible for every $lambda in mathbbC-0$.
See also this question concerning irreducibility of a polynomial in two variables.
Any hints and comments are welcome!
algebraic-geometry polynomials irreducible-polynomials
asked Aug 12 at 4:52
user237522
1,8141617
To go just a little further with your example, I think any $f$ which does not depend on $y$ qualifies. And any non-constant $f$ that does not depend on $x$ does not qualify, since $F_lambda$ is then a polynomial in $y$ only with degree at least 2, and therefore always factors.
– aschepler
Aug 12 at 10:13
Thanks for your comment. Please, by '$f$ which does not depend on $y$' did you mean that $f$ and $y$ are algebraically independent over $mathbbC$? (which is equivalent to saying that $f_x in mathbbC[x,y]-0$).
– user237522
Aug 12 at 10:21
I mean $f_y = 0$, or given $f=sum a_ij x^i y^j$, we have $j>0 rightarrow a_ij=0$.
– aschepler
Aug 12 at 10:24
I am not sure I understand; $f_y=0$ implies that $f in mathbbC[x]$, and your second condition also says that $f in mathbbC[x]$?
– user237522
Aug 12 at 10:34
Yes, I just put it two different ways to try to be clear.
– aschepler
Aug 12 at 14:50
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $f=f(x,y) in mathbbC[x,y]$,
and for every $lambda in mathbbC$, denote $F_lambda:=fy-lambda$.
Is it possible to characterize all $f in mathbbC[x,y]$ such that $F_lambda$ is irreducible in $mathbbC[x,y]$ for infinitely many $mathbbC ni lambda$'s?
Example: If I am not wrong, $f=x$ is such that $F_lambda=xy-lambda$ is irreducible for every $lambda in mathbbC-0$.
See also this question concerning irreducibility of a polynomial in two variables.
Any hints and comments are welcome!
algebraic-geometry polynomials irreducible-polynomials
asked Aug 12 at 4:52
user237522
1,8141617
Let $f=f(x,y) in mathbbC[x,y]$,
and for every $lambda in mathbbC$, denote $F_lambda:=fy-lambda$.
Is it possible to characterize all $f in mathbbC[x,y]$ such that $F_lambda$ is irreducible in $mathbbC[x,y]$ for infinitely many $mathbbC ni lambda$'s?
Example: If I am not wrong, $f=x$ is such that $F_lambda=xy-lambda$ is irreducible for every $lambda in mathbbC-0$.
See also this question concerning irreducibility of a polynomial in two variables.
Any hints and comments are welcome!
algebraic-geometry polynomials irreducible-polynomials
asked Aug 12 at 4:52
user237522
1,8141617
edited Aug 12 at 12:26
asked Aug 12 at 4:52
user237522
1,8141617
asked Aug 12 at 4:52
user237522
1,8141617
asked Aug 12 at 4:52
user237522
1,8141617
1,8141617
To go just a little further with your example, I think any $f$ which does not depend on $y$ qualifies. And any non-constant $f$ that does not depend on $x$ does not qualify, since $F_lambda$ is then a polynomial in $y$ only with degree at least 2, and therefore always factors.
– aschepler
Aug 12 at 10:13
Thanks for your comment. Please, by '$f$ which does not depend on $y$' did you mean that $f$ and $y$ are algebraically independent over $mathbbC$? (which is equivalent to saying that $f_x in mathbbC[x,y]-0$).
– user237522
Aug 12 at 10:21
I mean $f_y = 0$, or given $f=sum a_ij x^i y^j$, we have $j>0 rightarrow a_ij=0$.
– aschepler
Aug 12 at 10:24
I am not sure I understand; $f_y=0$ implies that $f in mathbbC[x]$, and your second condition also says that $f in mathbbC[x]$?
– user237522
Aug 12 at 10:34
Yes, I just put it two different ways to try to be clear.
– aschepler
Aug 12 at 14:50
 |Â
show 1 more comment
To go just a little further with your example, I think any $f$ which does not depend on $y$ qualifies. And any non-constant $f$ that does not depend on $x$ does not qualify, since $F_lambda$ is then a polynomial in $y$ only with degree at least 2, and therefore always factors.
– aschepler
Aug 12 at 10:13
Thanks for your comment. Please, by '$f$ which does not depend on $y$' did you mean that $f$ and $y$ are algebraically independent over $mathbbC$? (which is equivalent to saying that $f_x in mathbbC[x,y]-0$).
– user237522
Aug 12 at 10:21
I mean $f_y = 0$, or given $f=sum a_ij x^i y^j$, we have $j>0 rightarrow a_ij=0$.
– aschepler
Aug 12 at 10:24
I am not sure I understand; $f_y=0$ implies that $f in mathbbC[x]$, and your second condition also says that $f in mathbbC[x]$?
– user237522
Aug 12 at 10:34
Yes, I just put it two different ways to try to be clear.
– aschepler
Aug 12 at 14:50
To go just a little further with your example, I think any $f$ which does not depend on $y$ qualifies. And any non-constant $f$ that does not depend on $x$ does not qualify, since $F_lambda$ is then a polynomial in $y$ only with degree at least 2, and therefore always factors.
– aschepler
Aug 12 at 10:13
To go just a little further with your example, I think any $f$ which does not depend on $y$ qualifies. And any non-constant $f$ that does not depend on $x$ does not qualify, since $F_lambda$ is then a polynomial in $y$ only with degree at least 2, and therefore always factors.
– aschepler
Aug 12 at 10:13
Thanks for your comment. Please, by '$f$ which does not depend on $y$' did you mean that $f$ and $y$ are algebraically independent over $mathbbC$? (which is equivalent to saying that $f_x in mathbbC[x,y]-0$).
– user237522
Aug 12 at 10:21
Thanks for your comment. Please, by '$f$ which does not depend on $y$' did you mean that $f$ and $y$ are algebraically independent over $mathbbC$? (which is equivalent to saying that $f_x in mathbbC[x,y]-0$).
– user237522
Aug 12 at 10:21
I mean $f_y = 0$, or given $f=sum a_ij x^i y^j$, we have $j>0 rightarrow a_ij=0$.
– aschepler
Aug 12 at 10:24
I mean $f_y = 0$, or given $f=sum a_ij x^i y^j$, we have $j>0 rightarrow a_ij=0$.
– aschepler
Aug 12 at 10:24
I am not sure I understand; $f_y=0$ implies that $f in mathbbC[x]$, and your second condition also says that $f in mathbbC[x]$?
– user237522
Aug 12 at 10:34
I am not sure I understand; $f_y=0$ implies that $f in mathbbC[x]$, and your second condition also says that $f in mathbbC[x]$?
– user237522
Aug 12 at 10:34
Yes, I just put it two different ways to try to be clear.
– aschepler
Aug 12 at 14:50
Yes, I just put it two different ways to try to be clear.
– aschepler
Aug 12 at 14:50
 |Â
show 1 more comment
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To go just a little further with your example, I think any $f$ which does not depend on $y$ qualifies. And any non-constant $f$ that does not depend on $x$ does not qualify, since $F_lambda$ is then a polynomial in $y$ only with degree at least 2, and therefore always factors.
– aschepler
Aug 12 at 10:13
Thanks for your comment. Please, by '$f$ which does not depend on $y$' did you mean that $f$ and $y$ are algebraically independent over $mathbbC$? (which is equivalent to saying that $f_x in mathbbC[x,y]-0$).
– user237522
Aug 12 at 10:21
I mean $f_y = 0$, or given $f=sum a_ij x^i y^j$, we have $j>0 rightarrow a_ij=0$.
– aschepler
Aug 12 at 10:24
I am not sure I understand; $f_y=0$ implies that $f in mathbbC[x]$, and your second condition also says that $f in mathbbC[x]$?
– user237522
Aug 12 at 10:34
Yes, I just put it two different ways to try to be clear.
– aschepler
Aug 12 at 14:50