When $langle F(x,y)y+u(x)(x-c), G(x,y)y+v(x)(x-c) rangle$ is a maximal ideal of $mathbb{C}[x,y]$?
$begingroup$
Let $F=F(x,y),G=G(x,y) in mathbb{C}[x,y]$, $u=u(x),v=v(x) in mathbb{C}[x]$, $c in mathbb{C}$.
Denote:
$A=F(x,y)y+u(x)(x-c)=Fy+u(x-c)$,
$B=G(x,y)y+v(x)(x-c)=Gy+v(x-c)$.
When $I_{A,B}:=langle A,B rangle$ is a maximal ideal of $mathbb{C}[x,y]$?
It is well-known that maximal ideals of $mathbb{C}[x,y]$ are of the form: $langle x-a,y-b rangle$, $a,b in mathbb{C}$, see this question.
Observe that, for example, $langle x-a, y-b+(x-a)^5 rangle$ is maximal, since $langle x-a, y-b+(x-a)^5 rangle = langle x-a,y-b rangle$.
Remarks:
1. Notice that $I_{A,B} subseteq langle x-c, y rangle$.
- What if we further assume that $gcd(u,v)=gcd(u(x),v(x))=1$?
Any hints and comments are welcome!
algebraic-geometry ring-theory commutative-algebra ideals
asked Dec 12 '18 at 19:34
user237522user237522
2,1601617
$endgroup$
add a comment |
$begingroup$
Let $F=F(x,y),G=G(x,y) in mathbb{C}[x,y]$, $u=u(x),v=v(x) in mathbb{C}[x]$, $c in mathbb{C}$.
Denote:
$A=F(x,y)y+u(x)(x-c)=Fy+u(x-c)$,
$B=G(x,y)y+v(x)(x-c)=Gy+v(x-c)$.
When $I_{A,B}:=langle A,B rangle$ is a maximal ideal of $mathbb{C}[x,y]$?
It is well-known that maximal ideals of $mathbb{C}[x,y]$ are of the form: $langle x-a,y-b rangle$, $a,b in mathbb{C}$, see this question.
Observe that, for example, $langle x-a, y-b+(x-a)^5 rangle$ is maximal, since $langle x-a, y-b+(x-a)^5 rangle = langle x-a,y-b rangle$.
Remarks:
1. Notice that $I_{A,B} subseteq langle x-c, y rangle$.
- What if we further assume that $gcd(u,v)=gcd(u(x),v(x))=1$?
Any hints and comments are welcome!
algebraic-geometry ring-theory commutative-algebra ideals
asked Dec 12 '18 at 19:34
user237522user237522
2,1601617
$endgroup$
add a comment |
$begingroup$
Let $F=F(x,y),G=G(x,y) in mathbb{C}[x,y]$, $u=u(x),v=v(x) in mathbb{C}[x]$, $c in mathbb{C}$.
Denote:
$A=F(x,y)y+u(x)(x-c)=Fy+u(x-c)$,
$B=G(x,y)y+v(x)(x-c)=Gy+v(x-c)$.
When $I_{A,B}:=langle A,B rangle$ is a maximal ideal of $mathbb{C}[x,y]$?
It is well-known that maximal ideals of $mathbb{C}[x,y]$ are of the form: $langle x-a,y-b rangle$, $a,b in mathbb{C}$, see this question.
Observe that, for example, $langle x-a, y-b+(x-a)^5 rangle$ is maximal, since $langle x-a, y-b+(x-a)^5 rangle = langle x-a,y-b rangle$.
Remarks:
1. Notice that $I_{A,B} subseteq langle x-c, y rangle$.
- What if we further assume that $gcd(u,v)=gcd(u(x),v(x))=1$?
Any hints and comments are welcome!
algebraic-geometry ring-theory commutative-algebra ideals
asked Dec 12 '18 at 19:34
user237522user237522
2,1601617
$endgroup$
Let $F=F(x,y),G=G(x,y) in mathbb{C}[x,y]$, $u=u(x),v=v(x) in mathbb{C}[x]$, $c in mathbb{C}$.
Denote:
$A=F(x,y)y+u(x)(x-c)=Fy+u(x-c)$,
$B=G(x,y)y+v(x)(x-c)=Gy+v(x-c)$.
When $I_{A,B}:=langle A,B rangle$ is a maximal ideal of $mathbb{C}[x,y]$?
It is well-known that maximal ideals of $mathbb{C}[x,y]$ are of the form: $langle x-a,y-b rangle$, $a,b in mathbb{C}$, see this question.
Observe that, for example, $langle x-a, y-b+(x-a)^5 rangle$ is maximal, since $langle x-a, y-b+(x-a)^5 rangle = langle x-a,y-b rangle$.
Remarks:
1. Notice that $I_{A,B} subseteq langle x-c, y rangle$.
- What if we further assume that $gcd(u,v)=gcd(u(x),v(x))=1$?
Any hints and comments are welcome!
algebraic-geometry ring-theory commutative-algebra ideals
algebraic-geometry ring-theory commutative-algebra ideals
asked Dec 12 '18 at 19:34
user237522user237522
2,1601617
asked Dec 12 '18 at 19:34
user237522user237522
2,1601617
edited Dec 12 '18 at 19:40
user237522
asked Dec 12 '18 at 19:34
user237522user237522
2,1601617
asked Dec 12 '18 at 19:34
user237522user237522
2,1601617
asked Dec 12 '18 at 19:34
user237522user237522
2,1601617
2,1601617
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is true if and only if $Fv-Guin Bbb{C}$.
To see why, let
$$ M = newcommandbmat{begin{pmatrix}}newcommandemat{end{pmatrix}}bmat F & u\G
& vemat.$$
Then $$bmat A \ B emat = M bmat y \ x-c emat.$$
Then if $(A,B)=(y,x-c)$, we have that $y=e_{11}A + e_{12}B$ and $(x-c)=e_{21}A + e_{22}B$.
Then the matrix
$$ E = bmat e_{11} & e_{12} \ e_{21} & e_{22} emat $$ satisfies
$$ Ebmat A \ B emat = bmat y \ x-c emat.$$
Thus $$ EMbmat y \ x-c emat = bmat y \ x-c emat.$$
Expanding this gives
$$ bmat e_{11} F +e_{12} G & e_{11}u+e_{12}v \ e_{21}F+e_{22}G & e_{21}u+e_{22}vematbmat y \ x-cemat = bmat y \ x-c emat.$$
Then
$$(e_{11}F+e_{12}G)y+(e_{11}u+e_{12}v)(x-c) = y,$$
which tells us immediately that $e_{11}u+e_{12}v=0$, and thus that
$e_{11}F+e_{12}G=1$.
Similarly, the fact that $$(e_{21}F+e_{22}G)y + (e_{21}u+e_{22}v)(x-c) = (x-c)$$
tells us that $e_{21}F+e_{22}G=0$ and $e_{21}u+e_{22}v=1$.
Hence $F$ and $G$ are relatively prime (and not just relatively prime, they generate the unit ideal), and $u$ and $v$ also generate the unit ideal.
Moreover, since $e_{11}u+e_{12}v=0$, we have $e_{11}u=-e_{12}v$, so since $u$ and $v$ are relatively prime, $e_{11} = h_1v$ and $e_{12}=-h_1u$. Then since
$e_{11}F+e_{12}G=1$, we have $h_1vF-h_1uG=1$. Therefore $Fv-Gu=1/h_1in Bbb{C}$.
Conversely if $Fv-GuinBbb{C}$, then $M$ is invertible, so $A$ and $B$ generate $(y,x-c)$.
answered Dec 12 '18 at 19:58
jgonjgon
14.8k22042
$endgroup$
$begingroup$
Thank you very much! Your answer is very nice.
$endgroup$
– user237522
Dec 12 '18 at 21:44
$begingroup$
Please, do you think that it is easy to find a (sufficient and necessary) condition for $I_{A,B}$ to be a radical ideal? where $I$ is a radical ideal if $a^n in I$ implies that $a in I$.
$endgroup$
– user237522
Dec 13 '18 at 14:38
add a comment |
Your Answer
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$begingroup$
This is true if and only if $Fv-Guin Bbb{C}$.
To see why, let
$$ M = newcommandbmat{begin{pmatrix}}newcommandemat{end{pmatrix}}bmat F & u\G
& vemat.$$
Then $$bmat A \ B emat = M bmat y \ x-c emat.$$
Then if $(A,B)=(y,x-c)$, we have that $y=e_{11}A + e_{12}B$ and $(x-c)=e_{21}A + e_{22}B$.
Then the matrix
$$ E = bmat e_{11} & e_{12} \ e_{21} & e_{22} emat $$ satisfies
$$ Ebmat A \ B emat = bmat y \ x-c emat.$$
Thus $$ EMbmat y \ x-c emat = bmat y \ x-c emat.$$
Expanding this gives
$$ bmat e_{11} F +e_{12} G & e_{11}u+e_{12}v \ e_{21}F+e_{22}G & e_{21}u+e_{22}vematbmat y \ x-cemat = bmat y \ x-c emat.$$
Then
$$(e_{11}F+e_{12}G)y+(e_{11}u+e_{12}v)(x-c) = y,$$
which tells us immediately that $e_{11}u+e_{12}v=0$, and thus that
$e_{11}F+e_{12}G=1$.
Similarly, the fact that $$(e_{21}F+e_{22}G)y + (e_{21}u+e_{22}v)(x-c) = (x-c)$$
tells us that $e_{21}F+e_{22}G=0$ and $e_{21}u+e_{22}v=1$.
Hence $F$ and $G$ are relatively prime (and not just relatively prime, they generate the unit ideal), and $u$ and $v$ also generate the unit ideal.
Moreover, since $e_{11}u+e_{12}v=0$, we have $e_{11}u=-e_{12}v$, so since $u$ and $v$ are relatively prime, $e_{11} = h_1v$ and $e_{12}=-h_1u$. Then since
$e_{11}F+e_{12}G=1$, we have $h_1vF-h_1uG=1$. Therefore $Fv-Gu=1/h_1in Bbb{C}$.
Conversely if $Fv-GuinBbb{C}$, then $M$ is invertible, so $A$ and $B$ generate $(y,x-c)$.
answered Dec 12 '18 at 19:58
jgonjgon
14.8k22042
$endgroup$
$begingroup$
Thank you very much! Your answer is very nice.
$endgroup$
– user237522
Dec 12 '18 at 21:44
$begingroup$
Please, do you think that it is easy to find a (sufficient and necessary) condition for $I_{A,B}$ to be a radical ideal? where $I$ is a radical ideal if $a^n in I$ implies that $a in I$.
$endgroup$
– user237522
Dec 13 '18 at 14:38
add a comment |
$begingroup$
This is true if and only if $Fv-Guin Bbb{C}$.
To see why, let
$$ M = newcommandbmat{begin{pmatrix}}newcommandemat{end{pmatrix}}bmat F & u\G
& vemat.$$
Then $$bmat A \ B emat = M bmat y \ x-c emat.$$
Then if $(A,B)=(y,x-c)$, we have that $y=e_{11}A + e_{12}B$ and $(x-c)=e_{21}A + e_{22}B$.
Then the matrix
$$ E = bmat e_{11} & e_{12} \ e_{21} & e_{22} emat $$ satisfies
$$ Ebmat A \ B emat = bmat y \ x-c emat.$$
Thus $$ EMbmat y \ x-c emat = bmat y \ x-c emat.$$
Expanding this gives
$$ bmat e_{11} F +e_{12} G & e_{11}u+e_{12}v \ e_{21}F+e_{22}G & e_{21}u+e_{22}vematbmat y \ x-cemat = bmat y \ x-c emat.$$
Then
$$(e_{11}F+e_{12}G)y+(e_{11}u+e_{12}v)(x-c) = y,$$
which tells us immediately that $e_{11}u+e_{12}v=0$, and thus that
$e_{11}F+e_{12}G=1$.
Similarly, the fact that $$(e_{21}F+e_{22}G)y + (e_{21}u+e_{22}v)(x-c) = (x-c)$$
tells us that $e_{21}F+e_{22}G=0$ and $e_{21}u+e_{22}v=1$.
Hence $F$ and $G$ are relatively prime (and not just relatively prime, they generate the unit ideal), and $u$ and $v$ also generate the unit ideal.
Moreover, since $e_{11}u+e_{12}v=0$, we have $e_{11}u=-e_{12}v$, so since $u$ and $v$ are relatively prime, $e_{11} = h_1v$ and $e_{12}=-h_1u$. Then since
$e_{11}F+e_{12}G=1$, we have $h_1vF-h_1uG=1$. Therefore $Fv-Gu=1/h_1in Bbb{C}$.
Conversely if $Fv-GuinBbb{C}$, then $M$ is invertible, so $A$ and $B$ generate $(y,x-c)$.
answered Dec 12 '18 at 19:58
jgonjgon
14.8k22042
$endgroup$
$begingroup$
Thank you very much! Your answer is very nice.
$endgroup$
– user237522
Dec 12 '18 at 21:44
$begingroup$
Please, do you think that it is easy to find a (sufficient and necessary) condition for $I_{A,B}$ to be a radical ideal? where $I$ is a radical ideal if $a^n in I$ implies that $a in I$.
$endgroup$
– user237522
Dec 13 '18 at 14:38
add a comment |
$begingroup$
This is true if and only if $Fv-Guin Bbb{C}$.
To see why, let
$$ M = newcommandbmat{begin{pmatrix}}newcommandemat{end{pmatrix}}bmat F & u\G
& vemat.$$
Then $$bmat A \ B emat = M bmat y \ x-c emat.$$
Then if $(A,B)=(y,x-c)$, we have that $y=e_{11}A + e_{12}B$ and $(x-c)=e_{21}A + e_{22}B$.
Then the matrix
$$ E = bmat e_{11} & e_{12} \ e_{21} & e_{22} emat $$ satisfies
$$ Ebmat A \ B emat = bmat y \ x-c emat.$$
Thus $$ EMbmat y \ x-c emat = bmat y \ x-c emat.$$
Expanding this gives
$$ bmat e_{11} F +e_{12} G & e_{11}u+e_{12}v \ e_{21}F+e_{22}G & e_{21}u+e_{22}vematbmat y \ x-cemat = bmat y \ x-c emat.$$
Then
$$(e_{11}F+e_{12}G)y+(e_{11}u+e_{12}v)(x-c) = y,$$
which tells us immediately that $e_{11}u+e_{12}v=0$, and thus that
$e_{11}F+e_{12}G=1$.
Similarly, the fact that $$(e_{21}F+e_{22}G)y + (e_{21}u+e_{22}v)(x-c) = (x-c)$$
tells us that $e_{21}F+e_{22}G=0$ and $e_{21}u+e_{22}v=1$.
Hence $F$ and $G$ are relatively prime (and not just relatively prime, they generate the unit ideal), and $u$ and $v$ also generate the unit ideal.
Moreover, since $e_{11}u+e_{12}v=0$, we have $e_{11}u=-e_{12}v$, so since $u$ and $v$ are relatively prime, $e_{11} = h_1v$ and $e_{12}=-h_1u$. Then since
$e_{11}F+e_{12}G=1$, we have $h_1vF-h_1uG=1$. Therefore $Fv-Gu=1/h_1in Bbb{C}$.
Conversely if $Fv-GuinBbb{C}$, then $M$ is invertible, so $A$ and $B$ generate $(y,x-c)$.
answered Dec 12 '18 at 19:58
jgonjgon
14.8k22042
$endgroup$
This is true if and only if $Fv-Guin Bbb{C}$.
To see why, let
$$ M = newcommandbmat{begin{pmatrix}}newcommandemat{end{pmatrix}}bmat F & u\G
& vemat.$$
Then $$bmat A \ B emat = M bmat y \ x-c emat.$$
Then if $(A,B)=(y,x-c)$, we have that $y=e_{11}A + e_{12}B$ and $(x-c)=e_{21}A + e_{22}B$.
Then the matrix
$$ E = bmat e_{11} & e_{12} \ e_{21} & e_{22} emat $$ satisfies
$$ Ebmat A \ B emat = bmat y \ x-c emat.$$
Thus $$ EMbmat y \ x-c emat = bmat y \ x-c emat.$$
Expanding this gives
$$ bmat e_{11} F +e_{12} G & e_{11}u+e_{12}v \ e_{21}F+e_{22}G & e_{21}u+e_{22}vematbmat y \ x-cemat = bmat y \ x-c emat.$$
Then
$$(e_{11}F+e_{12}G)y+(e_{11}u+e_{12}v)(x-c) = y,$$
which tells us immediately that $e_{11}u+e_{12}v=0$, and thus that
$e_{11}F+e_{12}G=1$.
Similarly, the fact that $$(e_{21}F+e_{22}G)y + (e_{21}u+e_{22}v)(x-c) = (x-c)$$
tells us that $e_{21}F+e_{22}G=0$ and $e_{21}u+e_{22}v=1$.
Hence $F$ and $G$ are relatively prime (and not just relatively prime, they generate the unit ideal), and $u$ and $v$ also generate the unit ideal.
Moreover, since $e_{11}u+e_{12}v=0$, we have $e_{11}u=-e_{12}v$, so since $u$ and $v$ are relatively prime, $e_{11} = h_1v$ and $e_{12}=-h_1u$. Then since
$e_{11}F+e_{12}G=1$, we have $h_1vF-h_1uG=1$. Therefore $Fv-Gu=1/h_1in Bbb{C}$.
Conversely if $Fv-GuinBbb{C}$, then $M$ is invertible, so $A$ and $B$ generate $(y,x-c)$.
answered Dec 12 '18 at 19:58
jgonjgon
14.8k22042
answered Dec 12 '18 at 19:58
jgonjgon
14.8k22042
answered Dec 12 '18 at 19:58
jgonjgon
14.8k22042
answered Dec 12 '18 at 19:58
jgonjgon
14.8k22042
14.8k22042
$begingroup$
Thank you very much! Your answer is very nice.
$endgroup$
– user237522
Dec 12 '18 at 21:44
$begingroup$
Please, do you think that it is easy to find a (sufficient and necessary) condition for $I_{A,B}$ to be a radical ideal? where $I$ is a radical ideal if $a^n in I$ implies that $a in I$.
$endgroup$
– user237522
Dec 13 '18 at 14:38
add a comment |
$begingroup$
Thank you very much! Your answer is very nice.
$endgroup$
– user237522
Dec 12 '18 at 21:44
$begingroup$
Please, do you think that it is easy to find a (sufficient and necessary) condition for $I_{A,B}$ to be a radical ideal? where $I$ is a radical ideal if $a^n in I$ implies that $a in I$.
$endgroup$
– user237522
Dec 13 '18 at 14:38
$begingroup$
Thank you very much! Your answer is very nice.
$endgroup$
– user237522
Dec 12 '18 at 21:44
$begingroup$
Thank you very much! Your answer is very nice.
$endgroup$
– user237522
Dec 12 '18 at 21:44
$begingroup$
Please, do you think that it is easy to find a (sufficient and necessary) condition for $I_{A,B}$ to be a radical ideal? where $I$ is a radical ideal if $a^n in I$ implies that $a in I$.
$endgroup$
– user237522
Dec 13 '18 at 14:38
$begingroup$
Please, do you think that it is easy to find a (sufficient and necessary) condition for $I_{A,B}$ to be a radical ideal? where $I$ is a radical ideal if $a^n in I$ implies that $a in I$.
$endgroup$
– user237522
Dec 13 '18 at 14:38
add a comment |
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