Common zeros of two univariate polynomials over $mathbbC(y)$

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Let $f=f(x), g=g(x) in mathbbC[y][x] subset mathbbC(y)[x]$, where $x,y$ are variables over $mathbbC$, and each of $f,g$ is of $x$-degree $geq 1$, namely, $f$ and $g$ are non-constant polynomials over $mathbbC(y)$ in one variable $x$.



Write $f=a_nx^n+cdots+a_1x+a_0$, $g=b_mx^m+cdots+b_1x+b_0$, where $a_i=a_i(y),b_j=b_j(y) in mathbbC[y]$, $1 leq i leq n, 1 leq j leq m$.



Fix $c in mathbbC$, and denote $tildea_i:=a_i(c)$ and $tildeb_j:=b_j(c)$,

$tildef:=tildea_nx^n+cdots+tildea_1x+tildea_0$ and
$tildeg=tildeb_mx^m+cdots+tildeb_1x+tildeb_0$,
$1 leq i leq n, 1 leq j leq m$.



(Clearly, $tildea_i,tildeb_j in mathbbC$ so $tildef,tildeg in mathbbC[x]$).



Assume that $f$ and $g$ do not have a common zero in $mathbbC(y)$ (= their resultant is non-zero).




Is it true that $tildef$ and $tildeg$ do not have a common zero in $mathbbC$?




Remarks: (1) Of course, if $f$ and $g$ do not have a common zero in $mathbbC(y)$, then (in particular) $f$ and $g$ do not have a common zero in $mathbbC$, but here I am asking about $tildef$ and $tildeg$, not $f$ and $g$.



(2) Perhaps my question in trivial and has a positive answer? I only know that if $tildef$ and $tildeg$ have a common zero $d in mathbbC$, then $f=f(x,y)|(x=d,y=c)=tildef(d)=0$ and $g=g(x,y)|(x=d,y=c)=tildeg(d)=0$, but I do not see if this should imply that $f$ and $g$ have a common zero in $mathbbC(y)$. (Perhaps a universal property is relevant here).



Any comments and hints are welcome!







share|cite|improve this question

















  • 1




    Hmm, am I misunderstanding this? It seems trivially false. Take $f = x - y, g = x - y^2$, $c=0$. Then $f,g$ have no common roots in $mathbbC(y)$, since the unique roots are $y$ and $y^2$, but $tildef = tildeg = x$.
    – Jair Taylor
    Jul 24 at 0:44










  • Thank you! It seems that you are right.. What about the converse implication? (true?)
    – user237522
    Jul 24 at 1:25







  • 1




    $tildef$ and $tildeg$ have a common zero if $c$ is a zero of the resultant of $f$ and $g$.
    – random
    Jul 24 at 1:29






  • 1




    For the converse, let $f=g = yx-1$ and $c = 0$. Then $tildef, tildeg = -1$ have no roots, but $f,g$ have a common root $1/y$ in $mathbbC(y)$.
    – Jair Taylor
    Jul 24 at 1:30










  • @random, your comment sounds interesting.
    – user237522
    Jul 24 at 1:31














up vote
2
down vote

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Let $f=f(x), g=g(x) in mathbbC[y][x] subset mathbbC(y)[x]$, where $x,y$ are variables over $mathbbC$, and each of $f,g$ is of $x$-degree $geq 1$, namely, $f$ and $g$ are non-constant polynomials over $mathbbC(y)$ in one variable $x$.



Write $f=a_nx^n+cdots+a_1x+a_0$, $g=b_mx^m+cdots+b_1x+b_0$, where $a_i=a_i(y),b_j=b_j(y) in mathbbC[y]$, $1 leq i leq n, 1 leq j leq m$.



Fix $c in mathbbC$, and denote $tildea_i:=a_i(c)$ and $tildeb_j:=b_j(c)$,

$tildef:=tildea_nx^n+cdots+tildea_1x+tildea_0$ and
$tildeg=tildeb_mx^m+cdots+tildeb_1x+tildeb_0$,
$1 leq i leq n, 1 leq j leq m$.



(Clearly, $tildea_i,tildeb_j in mathbbC$ so $tildef,tildeg in mathbbC[x]$).



Assume that $f$ and $g$ do not have a common zero in $mathbbC(y)$ (= their resultant is non-zero).




Is it true that $tildef$ and $tildeg$ do not have a common zero in $mathbbC$?




Remarks: (1) Of course, if $f$ and $g$ do not have a common zero in $mathbbC(y)$, then (in particular) $f$ and $g$ do not have a common zero in $mathbbC$, but here I am asking about $tildef$ and $tildeg$, not $f$ and $g$.



(2) Perhaps my question in trivial and has a positive answer? I only know that if $tildef$ and $tildeg$ have a common zero $d in mathbbC$, then $f=f(x,y)|(x=d,y=c)=tildef(d)=0$ and $g=g(x,y)|(x=d,y=c)=tildeg(d)=0$, but I do not see if this should imply that $f$ and $g$ have a common zero in $mathbbC(y)$. (Perhaps a universal property is relevant here).



Any comments and hints are welcome!







share|cite|improve this question

















  • 1




    Hmm, am I misunderstanding this? It seems trivially false. Take $f = x - y, g = x - y^2$, $c=0$. Then $f,g$ have no common roots in $mathbbC(y)$, since the unique roots are $y$ and $y^2$, but $tildef = tildeg = x$.
    – Jair Taylor
    Jul 24 at 0:44










  • Thank you! It seems that you are right.. What about the converse implication? (true?)
    – user237522
    Jul 24 at 1:25







  • 1




    $tildef$ and $tildeg$ have a common zero if $c$ is a zero of the resultant of $f$ and $g$.
    – random
    Jul 24 at 1:29






  • 1




    For the converse, let $f=g = yx-1$ and $c = 0$. Then $tildef, tildeg = -1$ have no roots, but $f,g$ have a common root $1/y$ in $mathbbC(y)$.
    – Jair Taylor
    Jul 24 at 1:30










  • @random, your comment sounds interesting.
    – user237522
    Jul 24 at 1:31












up vote
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down vote

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down vote

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Let $f=f(x), g=g(x) in mathbbC[y][x] subset mathbbC(y)[x]$, where $x,y$ are variables over $mathbbC$, and each of $f,g$ is of $x$-degree $geq 1$, namely, $f$ and $g$ are non-constant polynomials over $mathbbC(y)$ in one variable $x$.



Write $f=a_nx^n+cdots+a_1x+a_0$, $g=b_mx^m+cdots+b_1x+b_0$, where $a_i=a_i(y),b_j=b_j(y) in mathbbC[y]$, $1 leq i leq n, 1 leq j leq m$.



Fix $c in mathbbC$, and denote $tildea_i:=a_i(c)$ and $tildeb_j:=b_j(c)$,

$tildef:=tildea_nx^n+cdots+tildea_1x+tildea_0$ and
$tildeg=tildeb_mx^m+cdots+tildeb_1x+tildeb_0$,
$1 leq i leq n, 1 leq j leq m$.



(Clearly, $tildea_i,tildeb_j in mathbbC$ so $tildef,tildeg in mathbbC[x]$).



Assume that $f$ and $g$ do not have a common zero in $mathbbC(y)$ (= their resultant is non-zero).




Is it true that $tildef$ and $tildeg$ do not have a common zero in $mathbbC$?




Remarks: (1) Of course, if $f$ and $g$ do not have a common zero in $mathbbC(y)$, then (in particular) $f$ and $g$ do not have a common zero in $mathbbC$, but here I am asking about $tildef$ and $tildeg$, not $f$ and $g$.



(2) Perhaps my question in trivial and has a positive answer? I only know that if $tildef$ and $tildeg$ have a common zero $d in mathbbC$, then $f=f(x,y)|(x=d,y=c)=tildef(d)=0$ and $g=g(x,y)|(x=d,y=c)=tildeg(d)=0$, but I do not see if this should imply that $f$ and $g$ have a common zero in $mathbbC(y)$. (Perhaps a universal property is relevant here).



Any comments and hints are welcome!







share|cite|improve this question













Let $f=f(x), g=g(x) in mathbbC[y][x] subset mathbbC(y)[x]$, where $x,y$ are variables over $mathbbC$, and each of $f,g$ is of $x$-degree $geq 1$, namely, $f$ and $g$ are non-constant polynomials over $mathbbC(y)$ in one variable $x$.



Write $f=a_nx^n+cdots+a_1x+a_0$, $g=b_mx^m+cdots+b_1x+b_0$, where $a_i=a_i(y),b_j=b_j(y) in mathbbC[y]$, $1 leq i leq n, 1 leq j leq m$.



Fix $c in mathbbC$, and denote $tildea_i:=a_i(c)$ and $tildeb_j:=b_j(c)$,

$tildef:=tildea_nx^n+cdots+tildea_1x+tildea_0$ and
$tildeg=tildeb_mx^m+cdots+tildeb_1x+tildeb_0$,
$1 leq i leq n, 1 leq j leq m$.



(Clearly, $tildea_i,tildeb_j in mathbbC$ so $tildef,tildeg in mathbbC[x]$).



Assume that $f$ and $g$ do not have a common zero in $mathbbC(y)$ (= their resultant is non-zero).




Is it true that $tildef$ and $tildeg$ do not have a common zero in $mathbbC$?




Remarks: (1) Of course, if $f$ and $g$ do not have a common zero in $mathbbC(y)$, then (in particular) $f$ and $g$ do not have a common zero in $mathbbC$, but here I am asking about $tildef$ and $tildeg$, not $f$ and $g$.



(2) Perhaps my question in trivial and has a positive answer? I only know that if $tildef$ and $tildeg$ have a common zero $d in mathbbC$, then $f=f(x,y)|(x=d,y=c)=tildef(d)=0$ and $g=g(x,y)|(x=d,y=c)=tildeg(d)=0$, but I do not see if this should imply that $f$ and $g$ have a common zero in $mathbbC(y)$. (Perhaps a universal property is relevant here).



Any comments and hints are welcome!









share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 24 at 0:23
























asked Jul 23 at 23:52









user237522

1,7941617




1,7941617







  • 1




    Hmm, am I misunderstanding this? It seems trivially false. Take $f = x - y, g = x - y^2$, $c=0$. Then $f,g$ have no common roots in $mathbbC(y)$, since the unique roots are $y$ and $y^2$, but $tildef = tildeg = x$.
    – Jair Taylor
    Jul 24 at 0:44










  • Thank you! It seems that you are right.. What about the converse implication? (true?)
    – user237522
    Jul 24 at 1:25







  • 1




    $tildef$ and $tildeg$ have a common zero if $c$ is a zero of the resultant of $f$ and $g$.
    – random
    Jul 24 at 1:29






  • 1




    For the converse, let $f=g = yx-1$ and $c = 0$. Then $tildef, tildeg = -1$ have no roots, but $f,g$ have a common root $1/y$ in $mathbbC(y)$.
    – Jair Taylor
    Jul 24 at 1:30










  • @random, your comment sounds interesting.
    – user237522
    Jul 24 at 1:31












  • 1




    Hmm, am I misunderstanding this? It seems trivially false. Take $f = x - y, g = x - y^2$, $c=0$. Then $f,g$ have no common roots in $mathbbC(y)$, since the unique roots are $y$ and $y^2$, but $tildef = tildeg = x$.
    – Jair Taylor
    Jul 24 at 0:44










  • Thank you! It seems that you are right.. What about the converse implication? (true?)
    – user237522
    Jul 24 at 1:25







  • 1




    $tildef$ and $tildeg$ have a common zero if $c$ is a zero of the resultant of $f$ and $g$.
    – random
    Jul 24 at 1:29






  • 1




    For the converse, let $f=g = yx-1$ and $c = 0$. Then $tildef, tildeg = -1$ have no roots, but $f,g$ have a common root $1/y$ in $mathbbC(y)$.
    – Jair Taylor
    Jul 24 at 1:30










  • @random, your comment sounds interesting.
    – user237522
    Jul 24 at 1:31







1




1




Hmm, am I misunderstanding this? It seems trivially false. Take $f = x - y, g = x - y^2$, $c=0$. Then $f,g$ have no common roots in $mathbbC(y)$, since the unique roots are $y$ and $y^2$, but $tildef = tildeg = x$.
– Jair Taylor
Jul 24 at 0:44




Hmm, am I misunderstanding this? It seems trivially false. Take $f = x - y, g = x - y^2$, $c=0$. Then $f,g$ have no common roots in $mathbbC(y)$, since the unique roots are $y$ and $y^2$, but $tildef = tildeg = x$.
– Jair Taylor
Jul 24 at 0:44












Thank you! It seems that you are right.. What about the converse implication? (true?)
– user237522
Jul 24 at 1:25





Thank you! It seems that you are right.. What about the converse implication? (true?)
– user237522
Jul 24 at 1:25





1




1




$tildef$ and $tildeg$ have a common zero if $c$ is a zero of the resultant of $f$ and $g$.
– random
Jul 24 at 1:29




$tildef$ and $tildeg$ have a common zero if $c$ is a zero of the resultant of $f$ and $g$.
– random
Jul 24 at 1:29




1




1




For the converse, let $f=g = yx-1$ and $c = 0$. Then $tildef, tildeg = -1$ have no roots, but $f,g$ have a common root $1/y$ in $mathbbC(y)$.
– Jair Taylor
Jul 24 at 1:30




For the converse, let $f=g = yx-1$ and $c = 0$. Then $tildef, tildeg = -1$ have no roots, but $f,g$ have a common root $1/y$ in $mathbbC(y)$.
– Jair Taylor
Jul 24 at 1:30












@random, your comment sounds interesting.
– user237522
Jul 24 at 1:31




@random, your comment sounds interesting.
– user237522
Jul 24 at 1:31















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