In $mathbb{C}[x,y]$: If $langle u,v rangle$ is a maximal ideal, then $langle u-lambda,v-mu rangle$ is a...
$begingroup$
Let $u=u(x,y), v=v(x,y) in mathbb{C}[x,y]$, with $deg(u) geq 2$ and $deg(v) geq 2$.
Let $lambda, mu in mathbb{C}$.
Assume that the ideal generated by $u$ and $v$, $langle u,v rangle$, is a maximal ideal of $mathbb{C}[x,y]$.
Is it true that $langle u-lambda, v-mu rangle$ is a maximal ideal of $mathbb{C}[x,y]$?
My attempts to answer my question are:
(1) By Hilbert's Nullstellensatz, $langle u,v rangle= langle x-a,y-b rangle$, for some $a,b in mathbb{C}$, so
$x-a=F_1u+G_1v$ and $y-b=F_2u+G_2v$, for some $F_1,G_1,F_2,G_2 in mathbb{C}[x,y]$.
Then, $x=F_1u+G_1v+a$ and $y=F_2u+G_2v+b$.
(2) $frac{mathbb{C}[x,y]}{langle u,v rangle}$ is a field (since $langle u,v rangle$ is maximal); actually, $frac{mathbb{C}[x,y]}{langle u,v rangle}$ is isomorphic to $mathbb{C}$. Is it true that $frac{mathbb{C}[x,y]}{langle u,v rangle}$ is isomorphic to $frac{mathbb{C}[x,y]}{langle u-lambda,v-mu rangle}$? In other words, is it true that $frac{mathbb{C}[x,y]}{langle u-lambda,v-mu rangle}$ is isomorphic to $mathbb{C}$?
See this question.
(3) If $langle u-lambda,v-mu rangle$ is not maximal, then it is contained in some maximal ideal: $langle u-lambda,v-mu rangle subsetneq langle x-c,y-d rangle$, $c,d in mathbb{C}$.
It is not difficult to see that $(u-lambda)(c,d)=0$ and $(v-mu)(c,d)=0$,
so $u(c,d)-lambda=0$ and $v(c,d)-mu=0$, namely,
$u(c,d)=lambda$ and $v(c,d)=mu$.
Remark: Is it possible that $langle u-lambda,v-mu rangle = mathbb{C}[x,y]$. If so, then there exist $F,G in mathbb{C}[x,y]$ such that
$F(u-lambda)+G(v-mu)=1$. Then at $(a,b)$ we get:
$F(a,b)(-lambda)+G(a,b)(-mu)=1$ (since, by (1), $u(a,b)=0$ and $v(a,b)=0$).
Thank you very much!
algebraic-geometry ring-theory commutative-algebra maximal-and-prime-ideals
asked Dec 16 '18 at 0:21
user237522user237522
2,1451617
$endgroup$
|
show 2 more comments
$begingroup$
Let $u=u(x,y), v=v(x,y) in mathbb{C}[x,y]$, with $deg(u) geq 2$ and $deg(v) geq 2$.
Let $lambda, mu in mathbb{C}$.
Assume that the ideal generated by $u$ and $v$, $langle u,v rangle$, is a maximal ideal of $mathbb{C}[x,y]$.
Is it true that $langle u-lambda, v-mu rangle$ is a maximal ideal of $mathbb{C}[x,y]$?
My attempts to answer my question are:
(1) By Hilbert's Nullstellensatz, $langle u,v rangle= langle x-a,y-b rangle$, for some $a,b in mathbb{C}$, so
$x-a=F_1u+G_1v$ and $y-b=F_2u+G_2v$, for some $F_1,G_1,F_2,G_2 in mathbb{C}[x,y]$.
Then, $x=F_1u+G_1v+a$ and $y=F_2u+G_2v+b$.
(2) $frac{mathbb{C}[x,y]}{langle u,v rangle}$ is a field (since $langle u,v rangle$ is maximal); actually, $frac{mathbb{C}[x,y]}{langle u,v rangle}$ is isomorphic to $mathbb{C}$. Is it true that $frac{mathbb{C}[x,y]}{langle u,v rangle}$ is isomorphic to $frac{mathbb{C}[x,y]}{langle u-lambda,v-mu rangle}$? In other words, is it true that $frac{mathbb{C}[x,y]}{langle u-lambda,v-mu rangle}$ is isomorphic to $mathbb{C}$?
See this question.
(3) If $langle u-lambda,v-mu rangle$ is not maximal, then it is contained in some maximal ideal: $langle u-lambda,v-mu rangle subsetneq langle x-c,y-d rangle$, $c,d in mathbb{C}$.
It is not difficult to see that $(u-lambda)(c,d)=0$ and $(v-mu)(c,d)=0$,
so $u(c,d)-lambda=0$ and $v(c,d)-mu=0$, namely,
$u(c,d)=lambda$ and $v(c,d)=mu$.
Remark: Is it possible that $langle u-lambda,v-mu rangle = mathbb{C}[x,y]$. If so, then there exist $F,G in mathbb{C}[x,y]$ such that
$F(u-lambda)+G(v-mu)=1$. Then at $(a,b)$ we get:
$F(a,b)(-lambda)+G(a,b)(-mu)=1$ (since, by (1), $u(a,b)=0$ and $v(a,b)=0$).
Thank you very much!
algebraic-geometry ring-theory commutative-algebra maximal-and-prime-ideals
asked Dec 16 '18 at 0:21
user237522user237522
2,1451617
$endgroup$
$begingroup$
The isomorphism desired in (2) does not hold in general. consider, e.g. $u equiv v equiv 1$ and $lambda = mu = 1$.
$endgroup$
– mathworker21
Dec 16 '18 at 0:44
$begingroup$
Thank you. It was assumed that $deg(u) geq 2$ and $deg(v) geq 2$; perhaps the desired isomorphism in (2) may not hold also in higher degrees?
$endgroup$
– user237522
Dec 16 '18 at 0:50
1
$begingroup$
Standard algebraic geometry tells us that maximal ideals containing $(f,g)$ correspond to common zeroes of $f$ and $g$. With this in mind, $(f,g)$ is maximal if and only if $f$ and $g$ have exactly one common zero and they're not tangent when they intersect. E.g. $y = x^2$ and $y = -x^2$ only have $(0,0)$ as a common zero but it has multiplicity two and we have $(y - x^2, y + x^2) = (y,x^2)$.
$endgroup$
– Trevor Gunn
Dec 16 '18 at 1:36
$begingroup$
One thing that gets in the way of constructing examples where $(f,g)$ is maximal is Bézout's theorem which says there will always be $deg f cdot deg g$ common zeroes with multiplicity in $mathbb{CP}^2$. Somehow we want to have all but one of these zeroes to be "at infinity." The only example I've come up with so far is something like $(y - x^2, y - (x + 1)^2) = (2x + 1, 4y - 1)$.
$endgroup$
– Trevor Gunn
Dec 16 '18 at 1:39
$begingroup$
@TrevorGunn, thank you for your interesting comments (unfortunately, I am not sure if I fully understand them). Could you please elaborate your example?
$endgroup$
– user237522
Dec 16 '18 at 1:54
|
show 2 more comments
$begingroup$
Let $u=u(x,y), v=v(x,y) in mathbb{C}[x,y]$, with $deg(u) geq 2$ and $deg(v) geq 2$.
Let $lambda, mu in mathbb{C}$.
Assume that the ideal generated by $u$ and $v$, $langle u,v rangle$, is a maximal ideal of $mathbb{C}[x,y]$.
Is it true that $langle u-lambda, v-mu rangle$ is a maximal ideal of $mathbb{C}[x,y]$?
My attempts to answer my question are:
(1) By Hilbert's Nullstellensatz, $langle u,v rangle= langle x-a,y-b rangle$, for some $a,b in mathbb{C}$, so
$x-a=F_1u+G_1v$ and $y-b=F_2u+G_2v$, for some $F_1,G_1,F_2,G_2 in mathbb{C}[x,y]$.
Then, $x=F_1u+G_1v+a$ and $y=F_2u+G_2v+b$.
(2) $frac{mathbb{C}[x,y]}{langle u,v rangle}$ is a field (since $langle u,v rangle$ is maximal); actually, $frac{mathbb{C}[x,y]}{langle u,v rangle}$ is isomorphic to $mathbb{C}$. Is it true that $frac{mathbb{C}[x,y]}{langle u,v rangle}$ is isomorphic to $frac{mathbb{C}[x,y]}{langle u-lambda,v-mu rangle}$? In other words, is it true that $frac{mathbb{C}[x,y]}{langle u-lambda,v-mu rangle}$ is isomorphic to $mathbb{C}$?
See this question.
(3) If $langle u-lambda,v-mu rangle$ is not maximal, then it is contained in some maximal ideal: $langle u-lambda,v-mu rangle subsetneq langle x-c,y-d rangle$, $c,d in mathbb{C}$.
It is not difficult to see that $(u-lambda)(c,d)=0$ and $(v-mu)(c,d)=0$,
so $u(c,d)-lambda=0$ and $v(c,d)-mu=0$, namely,
$u(c,d)=lambda$ and $v(c,d)=mu$.
Remark: Is it possible that $langle u-lambda,v-mu rangle = mathbb{C}[x,y]$. If so, then there exist $F,G in mathbb{C}[x,y]$ such that
$F(u-lambda)+G(v-mu)=1$. Then at $(a,b)$ we get:
$F(a,b)(-lambda)+G(a,b)(-mu)=1$ (since, by (1), $u(a,b)=0$ and $v(a,b)=0$).
Thank you very much!
algebraic-geometry ring-theory commutative-algebra maximal-and-prime-ideals
asked Dec 16 '18 at 0:21
user237522user237522
2,1451617
$endgroup$
Let $u=u(x,y), v=v(x,y) in mathbb{C}[x,y]$, with $deg(u) geq 2$ and $deg(v) geq 2$.
Let $lambda, mu in mathbb{C}$.
Assume that the ideal generated by $u$ and $v$, $langle u,v rangle$, is a maximal ideal of $mathbb{C}[x,y]$.
Is it true that $langle u-lambda, v-mu rangle$ is a maximal ideal of $mathbb{C}[x,y]$?
My attempts to answer my question are:
(1) By Hilbert's Nullstellensatz, $langle u,v rangle= langle x-a,y-b rangle$, for some $a,b in mathbb{C}$, so
$x-a=F_1u+G_1v$ and $y-b=F_2u+G_2v$, for some $F_1,G_1,F_2,G_2 in mathbb{C}[x,y]$.
Then, $x=F_1u+G_1v+a$ and $y=F_2u+G_2v+b$.
(2) $frac{mathbb{C}[x,y]}{langle u,v rangle}$ is a field (since $langle u,v rangle$ is maximal); actually, $frac{mathbb{C}[x,y]}{langle u,v rangle}$ is isomorphic to $mathbb{C}$. Is it true that $frac{mathbb{C}[x,y]}{langle u,v rangle}$ is isomorphic to $frac{mathbb{C}[x,y]}{langle u-lambda,v-mu rangle}$? In other words, is it true that $frac{mathbb{C}[x,y]}{langle u-lambda,v-mu rangle}$ is isomorphic to $mathbb{C}$?
See this question.
(3) If $langle u-lambda,v-mu rangle$ is not maximal, then it is contained in some maximal ideal: $langle u-lambda,v-mu rangle subsetneq langle x-c,y-d rangle$, $c,d in mathbb{C}$.
It is not difficult to see that $(u-lambda)(c,d)=0$ and $(v-mu)(c,d)=0$,
so $u(c,d)-lambda=0$ and $v(c,d)-mu=0$, namely,
$u(c,d)=lambda$ and $v(c,d)=mu$.
Remark: Is it possible that $langle u-lambda,v-mu rangle = mathbb{C}[x,y]$. If so, then there exist $F,G in mathbb{C}[x,y]$ such that
$F(u-lambda)+G(v-mu)=1$. Then at $(a,b)$ we get:
$F(a,b)(-lambda)+G(a,b)(-mu)=1$ (since, by (1), $u(a,b)=0$ and $v(a,b)=0$).
Thank you very much!
algebraic-geometry ring-theory commutative-algebra maximal-and-prime-ideals
algebraic-geometry ring-theory commutative-algebra maximal-and-prime-ideals
asked Dec 16 '18 at 0:21
user237522user237522
2,1451617
asked Dec 16 '18 at 0:21
user237522user237522
2,1451617
edited Dec 16 '18 at 1:29
user237522
asked Dec 16 '18 at 0:21
user237522user237522
2,1451617
asked Dec 16 '18 at 0:21
user237522user237522
2,1451617
asked Dec 16 '18 at 0:21
user237522user237522
2,1451617
2,1451617
$begingroup$
The isomorphism desired in (2) does not hold in general. consider, e.g. $u equiv v equiv 1$ and $lambda = mu = 1$.
$endgroup$
– mathworker21
Dec 16 '18 at 0:44
$begingroup$
Thank you. It was assumed that $deg(u) geq 2$ and $deg(v) geq 2$; perhaps the desired isomorphism in (2) may not hold also in higher degrees?
$endgroup$
– user237522
Dec 16 '18 at 0:50
1
$begingroup$
Standard algebraic geometry tells us that maximal ideals containing $(f,g)$ correspond to common zeroes of $f$ and $g$. With this in mind, $(f,g)$ is maximal if and only if $f$ and $g$ have exactly one common zero and they're not tangent when they intersect. E.g. $y = x^2$ and $y = -x^2$ only have $(0,0)$ as a common zero but it has multiplicity two and we have $(y - x^2, y + x^2) = (y,x^2)$.
$endgroup$
– Trevor Gunn
Dec 16 '18 at 1:36
$begingroup$
One thing that gets in the way of constructing examples where $(f,g)$ is maximal is Bézout's theorem which says there will always be $deg f cdot deg g$ common zeroes with multiplicity in $mathbb{CP}^2$. Somehow we want to have all but one of these zeroes to be "at infinity." The only example I've come up with so far is something like $(y - x^2, y - (x + 1)^2) = (2x + 1, 4y - 1)$.
$endgroup$
– Trevor Gunn
Dec 16 '18 at 1:39
$begingroup$
@TrevorGunn, thank you for your interesting comments (unfortunately, I am not sure if I fully understand them). Could you please elaborate your example?
$endgroup$
– user237522
Dec 16 '18 at 1:54
|
show 2 more comments
$begingroup$
The isomorphism desired in (2) does not hold in general. consider, e.g. $u equiv v equiv 1$ and $lambda = mu = 1$.
$endgroup$
– mathworker21
Dec 16 '18 at 0:44
$begingroup$
Thank you. It was assumed that $deg(u) geq 2$ and $deg(v) geq 2$; perhaps the desired isomorphism in (2) may not hold also in higher degrees?
$endgroup$
– user237522
Dec 16 '18 at 0:50
1
$begingroup$
Standard algebraic geometry tells us that maximal ideals containing $(f,g)$ correspond to common zeroes of $f$ and $g$. With this in mind, $(f,g)$ is maximal if and only if $f$ and $g$ have exactly one common zero and they're not tangent when they intersect. E.g. $y = x^2$ and $y = -x^2$ only have $(0,0)$ as a common zero but it has multiplicity two and we have $(y - x^2, y + x^2) = (y,x^2)$.
$endgroup$
– Trevor Gunn
Dec 16 '18 at 1:36
$begingroup$
One thing that gets in the way of constructing examples where $(f,g)$ is maximal is Bézout's theorem which says there will always be $deg f cdot deg g$ common zeroes with multiplicity in $mathbb{CP}^2$. Somehow we want to have all but one of these zeroes to be "at infinity." The only example I've come up with so far is something like $(y - x^2, y - (x + 1)^2) = (2x + 1, 4y - 1)$.
$endgroup$
– Trevor Gunn
Dec 16 '18 at 1:39
$begingroup$
@TrevorGunn, thank you for your interesting comments (unfortunately, I am not sure if I fully understand them). Could you please elaborate your example?
$endgroup$
– user237522
Dec 16 '18 at 1:54
$begingroup$
The isomorphism desired in (2) does not hold in general. consider, e.g. $u equiv v equiv 1$ and $lambda = mu = 1$.
$endgroup$
– mathworker21
Dec 16 '18 at 0:44
$begingroup$
The isomorphism desired in (2) does not hold in general. consider, e.g. $u equiv v equiv 1$ and $lambda = mu = 1$.
$endgroup$
– mathworker21
Dec 16 '18 at 0:44
$begingroup$
Thank you. It was assumed that $deg(u) geq 2$ and $deg(v) geq 2$; perhaps the desired isomorphism in (2) may not hold also in higher degrees?
$endgroup$
– user237522
Dec 16 '18 at 0:50
$begingroup$
Thank you. It was assumed that $deg(u) geq 2$ and $deg(v) geq 2$; perhaps the desired isomorphism in (2) may not hold also in higher degrees?
$endgroup$
– user237522
Dec 16 '18 at 0:50
1
1
$begingroup$
Standard algebraic geometry tells us that maximal ideals containing $(f,g)$ correspond to common zeroes of $f$ and $g$. With this in mind, $(f,g)$ is maximal if and only if $f$ and $g$ have exactly one common zero and they're not tangent when they intersect. E.g. $y = x^2$ and $y = -x^2$ only have $(0,0)$ as a common zero but it has multiplicity two and we have $(y - x^2, y + x^2) = (y,x^2)$.
$endgroup$
– Trevor Gunn
Dec 16 '18 at 1:36
$begingroup$
Standard algebraic geometry tells us that maximal ideals containing $(f,g)$ correspond to common zeroes of $f$ and $g$. With this in mind, $(f,g)$ is maximal if and only if $f$ and $g$ have exactly one common zero and they're not tangent when they intersect. E.g. $y = x^2$ and $y = -x^2$ only have $(0,0)$ as a common zero but it has multiplicity two and we have $(y - x^2, y + x^2) = (y,x^2)$.
$endgroup$
– Trevor Gunn
Dec 16 '18 at 1:36
$begingroup$
One thing that gets in the way of constructing examples where $(f,g)$ is maximal is Bézout's theorem which says there will always be $deg f cdot deg g$ common zeroes with multiplicity in $mathbb{CP}^2$. Somehow we want to have all but one of these zeroes to be "at infinity." The only example I've come up with so far is something like $(y - x^2, y - (x + 1)^2) = (2x + 1, 4y - 1)$.
$endgroup$
– Trevor Gunn
Dec 16 '18 at 1:39
$begingroup$
One thing that gets in the way of constructing examples where $(f,g)$ is maximal is Bézout's theorem which says there will always be $deg f cdot deg g$ common zeroes with multiplicity in $mathbb{CP}^2$. Somehow we want to have all but one of these zeroes to be "at infinity." The only example I've come up with so far is something like $(y - x^2, y - (x + 1)^2) = (2x + 1, 4y - 1)$.
$endgroup$
– Trevor Gunn
Dec 16 '18 at 1:39
$begingroup$
@TrevorGunn, thank you for your interesting comments (unfortunately, I am not sure if I fully understand them). Could you please elaborate your example?
$endgroup$
– user237522
Dec 16 '18 at 1:54
$begingroup$
@TrevorGunn, thank you for your interesting comments (unfortunately, I am not sure if I fully understand them). Could you please elaborate your example?
$endgroup$
– user237522
Dec 16 '18 at 1:54
|
show 2 more comments
1 Answer
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I think this should work:
By the Nullstellensatz $(u, v)$ is maximal if and only if $V(u,v)$ is a single (closed) point.
Why is this? First suppose $(u,v)$ is maximal. Then you're comfortable with saying that $(u,v) = (x - a, y - b)$ by the (weak) Nullstellensatz. Thus $$V(u,v) = V(x - a, y - b) = {(a,b)}$$
is a single point.
Conversely, if $V(u,v) = {(a,b)}$ is a single point, then the (strong) Nullstellensatz tells us that $$operatorname{rad}(u,v) = I(V(u,v)) = I({(a,b)}) = (x - a, y - b).$$
The condition that $(u,v) = operatorname{rad}(u,v)$ is the same as saying $(a,b)$ doesn't have extra multiplicity as a common zero.
Let $bar{u}(x,y,z)$ and $bar{v}(x,y,z)$ be the homogenizations of $u, v$. By Bézout's theorem, $V(bar{u},bar{v}) subseteq mathbb{CP}^2$ consists of $deg(bar{u})deg(bar{v}) = deg(u)deg(v)$ points counted with multiplicity.
By the first observation, $V(u,v)$ consists of $1$ point, meaning the other $deg(u)deg(v) - 1$ points occur at infinity. I.e. they are the common zeroes of $bar{u}(x,y,0)$ and $bar{v}(x,y,0)$
If we modify $u, v$ by adding $lambda, mu$ then the new homogenizations are
$$ overline{u - lambda} = bar{u} + lambda z^{deg u}, quad overline{v - mu} = bar{u} + mu z^{deg u} $$
Since $overline{u - lambda}(x,y,0) = bar{u}(x,y,0)$ and $overline{v - mu}(x,y,0) = bar{v}(x,y,0)$ it is also true that $overline{u - lambda}$ and $overline{v - mu}$ have $deg(u)deg(v) - 1$ common zeroes at infinity.
Thus $u - lambda, v - mu$ have exactly one common zero not at infinity, so by the Nullstellensatz, $(u - lambda, v - mu)$ is maximal.
answered Dec 16 '18 at 1:53
Trevor GunnTrevor Gunn
14.3k32046
$endgroup$
$begingroup$
Thank you very much! Unfortunately, I am still not familiar with algebraic geometry (I had a feeling that algebraic geometry may help. Several pure algebraic questions I have asked in the past got excellent answers relying on algebraic geometry). I hope that your answer is correct.
$endgroup$
– user237522
Dec 16 '18 at 2:05
$begingroup$
@user237522 I've added some explanation of why we are looking at common zeroes with multiplicity 1. Unfortunately, it's hard to give a good explaination of the multiplicity 1 condition without more algebraic geometry but I hope that modulo that, you are able to follow my reasoning.
$endgroup$
– Trevor Gunn
Dec 16 '18 at 2:07
$begingroup$
Thank you very much again! I have some gaps in understanding your answer, but hopefully it is correct. (I have upvoted it, since it 'seems' correct).
$endgroup$
– user237522
Dec 16 '18 at 2:10
$begingroup$
There is a paper I wish to write, in which your answer will be helpful. Please, could you email me user237522mse@gmail.com I will give you more details about the paper, and you will let me know if you want to be a co-author or just being quoted+acknowledged. (I will probably email you back tomorrow, since now it is very late at night here).
$endgroup$
– user237522
Dec 16 '18 at 2:28
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I think this should work:
By the Nullstellensatz $(u, v)$ is maximal if and only if $V(u,v)$ is a single (closed) point.
Why is this? First suppose $(u,v)$ is maximal. Then you're comfortable with saying that $(u,v) = (x - a, y - b)$ by the (weak) Nullstellensatz. Thus $$V(u,v) = V(x - a, y - b) = {(a,b)}$$
is a single point.
Conversely, if $V(u,v) = {(a,b)}$ is a single point, then the (strong) Nullstellensatz tells us that $$operatorname{rad}(u,v) = I(V(u,v)) = I({(a,b)}) = (x - a, y - b).$$
The condition that $(u,v) = operatorname{rad}(u,v)$ is the same as saying $(a,b)$ doesn't have extra multiplicity as a common zero.
Let $bar{u}(x,y,z)$ and $bar{v}(x,y,z)$ be the homogenizations of $u, v$. By Bézout's theorem, $V(bar{u},bar{v}) subseteq mathbb{CP}^2$ consists of $deg(bar{u})deg(bar{v}) = deg(u)deg(v)$ points counted with multiplicity.
By the first observation, $V(u,v)$ consists of $1$ point, meaning the other $deg(u)deg(v) - 1$ points occur at infinity. I.e. they are the common zeroes of $bar{u}(x,y,0)$ and $bar{v}(x,y,0)$
If we modify $u, v$ by adding $lambda, mu$ then the new homogenizations are
$$ overline{u - lambda} = bar{u} + lambda z^{deg u}, quad overline{v - mu} = bar{u} + mu z^{deg u} $$
Since $overline{u - lambda}(x,y,0) = bar{u}(x,y,0)$ and $overline{v - mu}(x,y,0) = bar{v}(x,y,0)$ it is also true that $overline{u - lambda}$ and $overline{v - mu}$ have $deg(u)deg(v) - 1$ common zeroes at infinity.
Thus $u - lambda, v - mu$ have exactly one common zero not at infinity, so by the Nullstellensatz, $(u - lambda, v - mu)$ is maximal.
answered Dec 16 '18 at 1:53
Trevor GunnTrevor Gunn
14.3k32046
$endgroup$
$begingroup$
Thank you very much! Unfortunately, I am still not familiar with algebraic geometry (I had a feeling that algebraic geometry may help. Several pure algebraic questions I have asked in the past got excellent answers relying on algebraic geometry). I hope that your answer is correct.
$endgroup$
– user237522
Dec 16 '18 at 2:05
$begingroup$
@user237522 I've added some explanation of why we are looking at common zeroes with multiplicity 1. Unfortunately, it's hard to give a good explaination of the multiplicity 1 condition without more algebraic geometry but I hope that modulo that, you are able to follow my reasoning.
$endgroup$
– Trevor Gunn
Dec 16 '18 at 2:07
$begingroup$
Thank you very much again! I have some gaps in understanding your answer, but hopefully it is correct. (I have upvoted it, since it 'seems' correct).
$endgroup$
– user237522
Dec 16 '18 at 2:10
$begingroup$
There is a paper I wish to write, in which your answer will be helpful. Please, could you email me user237522mse@gmail.com I will give you more details about the paper, and you will let me know if you want to be a co-author or just being quoted+acknowledged. (I will probably email you back tomorrow, since now it is very late at night here).
$endgroup$
– user237522
Dec 16 '18 at 2:28
add a comment |
$begingroup$
I think this should work:
By the Nullstellensatz $(u, v)$ is maximal if and only if $V(u,v)$ is a single (closed) point.
Why is this? First suppose $(u,v)$ is maximal. Then you're comfortable with saying that $(u,v) = (x - a, y - b)$ by the (weak) Nullstellensatz. Thus $$V(u,v) = V(x - a, y - b) = {(a,b)}$$
is a single point.
Conversely, if $V(u,v) = {(a,b)}$ is a single point, then the (strong) Nullstellensatz tells us that $$operatorname{rad}(u,v) = I(V(u,v)) = I({(a,b)}) = (x - a, y - b).$$
The condition that $(u,v) = operatorname{rad}(u,v)$ is the same as saying $(a,b)$ doesn't have extra multiplicity as a common zero.
Let $bar{u}(x,y,z)$ and $bar{v}(x,y,z)$ be the homogenizations of $u, v$. By Bézout's theorem, $V(bar{u},bar{v}) subseteq mathbb{CP}^2$ consists of $deg(bar{u})deg(bar{v}) = deg(u)deg(v)$ points counted with multiplicity.
By the first observation, $V(u,v)$ consists of $1$ point, meaning the other $deg(u)deg(v) - 1$ points occur at infinity. I.e. they are the common zeroes of $bar{u}(x,y,0)$ and $bar{v}(x,y,0)$
If we modify $u, v$ by adding $lambda, mu$ then the new homogenizations are
$$ overline{u - lambda} = bar{u} + lambda z^{deg u}, quad overline{v - mu} = bar{u} + mu z^{deg u} $$
Since $overline{u - lambda}(x,y,0) = bar{u}(x,y,0)$ and $overline{v - mu}(x,y,0) = bar{v}(x,y,0)$ it is also true that $overline{u - lambda}$ and $overline{v - mu}$ have $deg(u)deg(v) - 1$ common zeroes at infinity.
Thus $u - lambda, v - mu$ have exactly one common zero not at infinity, so by the Nullstellensatz, $(u - lambda, v - mu)$ is maximal.
answered Dec 16 '18 at 1:53
Trevor GunnTrevor Gunn
14.3k32046
$endgroup$
$begingroup$
Thank you very much! Unfortunately, I am still not familiar with algebraic geometry (I had a feeling that algebraic geometry may help. Several pure algebraic questions I have asked in the past got excellent answers relying on algebraic geometry). I hope that your answer is correct.
$endgroup$
– user237522
Dec 16 '18 at 2:05
$begingroup$
@user237522 I've added some explanation of why we are looking at common zeroes with multiplicity 1. Unfortunately, it's hard to give a good explaination of the multiplicity 1 condition without more algebraic geometry but I hope that modulo that, you are able to follow my reasoning.
$endgroup$
– Trevor Gunn
Dec 16 '18 at 2:07
$begingroup$
Thank you very much again! I have some gaps in understanding your answer, but hopefully it is correct. (I have upvoted it, since it 'seems' correct).
$endgroup$
– user237522
Dec 16 '18 at 2:10
$begingroup$
There is a paper I wish to write, in which your answer will be helpful. Please, could you email me user237522mse@gmail.com I will give you more details about the paper, and you will let me know if you want to be a co-author or just being quoted+acknowledged. (I will probably email you back tomorrow, since now it is very late at night here).
$endgroup$
– user237522
Dec 16 '18 at 2:28
add a comment |
$begingroup$
I think this should work:
By the Nullstellensatz $(u, v)$ is maximal if and only if $V(u,v)$ is a single (closed) point.
Why is this? First suppose $(u,v)$ is maximal. Then you're comfortable with saying that $(u,v) = (x - a, y - b)$ by the (weak) Nullstellensatz. Thus $$V(u,v) = V(x - a, y - b) = {(a,b)}$$
is a single point.
Conversely, if $V(u,v) = {(a,b)}$ is a single point, then the (strong) Nullstellensatz tells us that $$operatorname{rad}(u,v) = I(V(u,v)) = I({(a,b)}) = (x - a, y - b).$$
The condition that $(u,v) = operatorname{rad}(u,v)$ is the same as saying $(a,b)$ doesn't have extra multiplicity as a common zero.
Let $bar{u}(x,y,z)$ and $bar{v}(x,y,z)$ be the homogenizations of $u, v$. By Bézout's theorem, $V(bar{u},bar{v}) subseteq mathbb{CP}^2$ consists of $deg(bar{u})deg(bar{v}) = deg(u)deg(v)$ points counted with multiplicity.
By the first observation, $V(u,v)$ consists of $1$ point, meaning the other $deg(u)deg(v) - 1$ points occur at infinity. I.e. they are the common zeroes of $bar{u}(x,y,0)$ and $bar{v}(x,y,0)$
If we modify $u, v$ by adding $lambda, mu$ then the new homogenizations are
$$ overline{u - lambda} = bar{u} + lambda z^{deg u}, quad overline{v - mu} = bar{u} + mu z^{deg u} $$
Since $overline{u - lambda}(x,y,0) = bar{u}(x,y,0)$ and $overline{v - mu}(x,y,0) = bar{v}(x,y,0)$ it is also true that $overline{u - lambda}$ and $overline{v - mu}$ have $deg(u)deg(v) - 1$ common zeroes at infinity.
Thus $u - lambda, v - mu$ have exactly one common zero not at infinity, so by the Nullstellensatz, $(u - lambda, v - mu)$ is maximal.
answered Dec 16 '18 at 1:53
Trevor GunnTrevor Gunn
14.3k32046
$endgroup$
I think this should work:
By the Nullstellensatz $(u, v)$ is maximal if and only if $V(u,v)$ is a single (closed) point.
Why is this? First suppose $(u,v)$ is maximal. Then you're comfortable with saying that $(u,v) = (x - a, y - b)$ by the (weak) Nullstellensatz. Thus $$V(u,v) = V(x - a, y - b) = {(a,b)}$$
is a single point.
Conversely, if $V(u,v) = {(a,b)}$ is a single point, then the (strong) Nullstellensatz tells us that $$operatorname{rad}(u,v) = I(V(u,v)) = I({(a,b)}) = (x - a, y - b).$$
The condition that $(u,v) = operatorname{rad}(u,v)$ is the same as saying $(a,b)$ doesn't have extra multiplicity as a common zero.
Let $bar{u}(x,y,z)$ and $bar{v}(x,y,z)$ be the homogenizations of $u, v$. By Bézout's theorem, $V(bar{u},bar{v}) subseteq mathbb{CP}^2$ consists of $deg(bar{u})deg(bar{v}) = deg(u)deg(v)$ points counted with multiplicity.
By the first observation, $V(u,v)$ consists of $1$ point, meaning the other $deg(u)deg(v) - 1$ points occur at infinity. I.e. they are the common zeroes of $bar{u}(x,y,0)$ and $bar{v}(x,y,0)$
If we modify $u, v$ by adding $lambda, mu$ then the new homogenizations are
$$ overline{u - lambda} = bar{u} + lambda z^{deg u}, quad overline{v - mu} = bar{u} + mu z^{deg u} $$
Since $overline{u - lambda}(x,y,0) = bar{u}(x,y,0)$ and $overline{v - mu}(x,y,0) = bar{v}(x,y,0)$ it is also true that $overline{u - lambda}$ and $overline{v - mu}$ have $deg(u)deg(v) - 1$ common zeroes at infinity.
Thus $u - lambda, v - mu$ have exactly one common zero not at infinity, so by the Nullstellensatz, $(u - lambda, v - mu)$ is maximal.
answered Dec 16 '18 at 1:53
Trevor GunnTrevor Gunn
14.3k32046
edited Dec 16 '18 at 2:04
answered Dec 16 '18 at 1:53
Trevor GunnTrevor Gunn
14.3k32046
answered Dec 16 '18 at 1:53
Trevor GunnTrevor Gunn
14.3k32046
answered Dec 16 '18 at 1:53
Trevor GunnTrevor Gunn
14.3k32046
14.3k32046
$begingroup$
Thank you very much! Unfortunately, I am still not familiar with algebraic geometry (I had a feeling that algebraic geometry may help. Several pure algebraic questions I have asked in the past got excellent answers relying on algebraic geometry). I hope that your answer is correct.
$endgroup$
– user237522
Dec 16 '18 at 2:05
$begingroup$
@user237522 I've added some explanation of why we are looking at common zeroes with multiplicity 1. Unfortunately, it's hard to give a good explaination of the multiplicity 1 condition without more algebraic geometry but I hope that modulo that, you are able to follow my reasoning.
$endgroup$
– Trevor Gunn
Dec 16 '18 at 2:07
$begingroup$
Thank you very much again! I have some gaps in understanding your answer, but hopefully it is correct. (I have upvoted it, since it 'seems' correct).
$endgroup$
– user237522
Dec 16 '18 at 2:10
$begingroup$
There is a paper I wish to write, in which your answer will be helpful. Please, could you email me user237522mse@gmail.com I will give you more details about the paper, and you will let me know if you want to be a co-author or just being quoted+acknowledged. (I will probably email you back tomorrow, since now it is very late at night here).
$endgroup$
– user237522
Dec 16 '18 at 2:28
add a comment |
$begingroup$
Thank you very much! Unfortunately, I am still not familiar with algebraic geometry (I had a feeling that algebraic geometry may help. Several pure algebraic questions I have asked in the past got excellent answers relying on algebraic geometry). I hope that your answer is correct.
$endgroup$
– user237522
Dec 16 '18 at 2:05
$begingroup$
@user237522 I've added some explanation of why we are looking at common zeroes with multiplicity 1. Unfortunately, it's hard to give a good explaination of the multiplicity 1 condition without more algebraic geometry but I hope that modulo that, you are able to follow my reasoning.
$endgroup$
– Trevor Gunn
Dec 16 '18 at 2:07
$begingroup$
Thank you very much again! I have some gaps in understanding your answer, but hopefully it is correct. (I have upvoted it, since it 'seems' correct).
$endgroup$
– user237522
Dec 16 '18 at 2:10
$begingroup$
There is a paper I wish to write, in which your answer will be helpful. Please, could you email me user237522mse@gmail.com I will give you more details about the paper, and you will let me know if you want to be a co-author or just being quoted+acknowledged. (I will probably email you back tomorrow, since now it is very late at night here).
$endgroup$
– user237522
Dec 16 '18 at 2:28
$begingroup$
Thank you very much! Unfortunately, I am still not familiar with algebraic geometry (I had a feeling that algebraic geometry may help. Several pure algebraic questions I have asked in the past got excellent answers relying on algebraic geometry). I hope that your answer is correct.
$endgroup$
– user237522
Dec 16 '18 at 2:05
$begingroup$
Thank you very much! Unfortunately, I am still not familiar with algebraic geometry (I had a feeling that algebraic geometry may help. Several pure algebraic questions I have asked in the past got excellent answers relying on algebraic geometry). I hope that your answer is correct.
$endgroup$
– user237522
Dec 16 '18 at 2:05
$begingroup$
@user237522 I've added some explanation of why we are looking at common zeroes with multiplicity 1. Unfortunately, it's hard to give a good explaination of the multiplicity 1 condition without more algebraic geometry but I hope that modulo that, you are able to follow my reasoning.
$endgroup$
– Trevor Gunn
Dec 16 '18 at 2:07
$begingroup$
@user237522 I've added some explanation of why we are looking at common zeroes with multiplicity 1. Unfortunately, it's hard to give a good explaination of the multiplicity 1 condition without more algebraic geometry but I hope that modulo that, you are able to follow my reasoning.
$endgroup$
– Trevor Gunn
Dec 16 '18 at 2:07
$begingroup$
Thank you very much again! I have some gaps in understanding your answer, but hopefully it is correct. (I have upvoted it, since it 'seems' correct).
$endgroup$
– user237522
Dec 16 '18 at 2:10
$begingroup$
Thank you very much again! I have some gaps in understanding your answer, but hopefully it is correct. (I have upvoted it, since it 'seems' correct).
$endgroup$
– user237522
Dec 16 '18 at 2:10
$begingroup$
There is a paper I wish to write, in which your answer will be helpful. Please, could you email me user237522mse@gmail.com I will give you more details about the paper, and you will let me know if you want to be a co-author or just being quoted+acknowledged. (I will probably email you back tomorrow, since now it is very late at night here).
$endgroup$
– user237522
Dec 16 '18 at 2:28
$begingroup$
There is a paper I wish to write, in which your answer will be helpful. Please, could you email me user237522mse@gmail.com I will give you more details about the paper, and you will let me know if you want to be a co-author or just being quoted+acknowledged. (I will probably email you back tomorrow, since now it is very late at night here).
$endgroup$
– user237522
Dec 16 '18 at 2:28
add a comment |
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The isomorphism desired in (2) does not hold in general. consider, e.g. $u equiv v equiv 1$ and $lambda = mu = 1$.
$endgroup$
– mathworker21
Dec 16 '18 at 0:44
$begingroup$
Thank you. It was assumed that $deg(u) geq 2$ and $deg(v) geq 2$; perhaps the desired isomorphism in (2) may not hold also in higher degrees?
$endgroup$
– user237522
Dec 16 '18 at 0:50
1
$begingroup$
Standard algebraic geometry tells us that maximal ideals containing $(f,g)$ correspond to common zeroes of $f$ and $g$. With this in mind, $(f,g)$ is maximal if and only if $f$ and $g$ have exactly one common zero and they're not tangent when they intersect. E.g. $y = x^2$ and $y = -x^2$ only have $(0,0)$ as a common zero but it has multiplicity two and we have $(y - x^2, y + x^2) = (y,x^2)$.
$endgroup$
– Trevor Gunn
Dec 16 '18 at 1:36
$begingroup$
One thing that gets in the way of constructing examples where $(f,g)$ is maximal is Bézout's theorem which says there will always be $deg f cdot deg g$ common zeroes with multiplicity in $mathbb{CP}^2$. Somehow we want to have all but one of these zeroes to be "at infinity." The only example I've come up with so far is something like $(y - x^2, y - (x + 1)^2) = (2x + 1, 4y - 1)$.
$endgroup$
– Trevor Gunn
Dec 16 '18 at 1:39
$begingroup$
@TrevorGunn, thank you for your interesting comments (unfortunately, I am not sure if I fully understand them). Could you please elaborate your example?
$endgroup$
– user237522
Dec 16 '18 at 1:54