Mixing
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!
polynomials commutative-algebra substitution resultant
asked Jul 23 at 23:52
user237522
1,7941617
 |Â
show 5 more comments
up vote
2
down vote
favorite
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!
polynomials commutative-algebra substitution resultant
asked Jul 23 at 23:52
user237522
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
 |Â
show 5 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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!
polynomials commutative-algebra substitution resultant
asked Jul 23 at 23:52
user237522
1,7941617
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!
polynomials commutative-algebra substitution resultant
asked Jul 23 at 23:52
user237522
1,7941617
edited Jul 24 at 0:23
asked Jul 23 at 23:52
user237522
1,7941617
asked Jul 23 at 23:52
user237522
1,7941617
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
 |Â
show 5 more comments
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
 |Â
show 5 more comments
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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