Can this argument about homogenous polynomials over $mathbb{R}$ in two variables having graphs that are...
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If we have some homogeneous polynomial $f(x, y)$ (for example $f(x, y) = x^3y + x^2y^2 - 17y^4$) than it's solution set is a union of straight lines (or a single point/null set). I've seen arguments for this, but I was wondering if this 'geometric' argument could be made formal.
Suppose we have a polynomial $p(x, y).$ We can look at its low degree part (all terms of highest degree) to get another polynomial $q(x, y)$ such that the graph of $q(x, y) = 0$ models the behavior of $p(x, y)$ in a neighborhood of $0.$ Hence, it's graph needs to be a union of straight lines (or degenerate point/null set) since the algebraic curve, by the implicit function theorem, describes a bunch of functions that are differentiable at $0$ and thus have linear approximations at $0$ and thus look like lines locally; hence the graph of $q(x, y) = 0$ should be a bunch of lines. As every homogeneous polynomial is clearly the low degree part of some curve, this property is true of all homogeneous polynomials.
(This argument is much more dubious...) Similarly, we could approach this problem from the opposite perspective, looking at the high degree part of the polynomial. Then, place the polynomial in the projective plane and, like before, find its derivatives 'at infinity' and therefore approximate it locally as a bunch of lines.
Can either of these two arguments be made rigorous? Is there even a reasonable way to talk about the derivative 'at infinity' in a projective plane?
algebraic-geometry
asked Dec 31 '18 at 2:10
Michael BarzMichael Barz
915
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add a comment |
$begingroup$
If we have some homogeneous polynomial $f(x, y)$ (for example $f(x, y) = x^3y + x^2y^2 - 17y^4$) than it's solution set is a union of straight lines (or a single point/null set). I've seen arguments for this, but I was wondering if this 'geometric' argument could be made formal.
Suppose we have a polynomial $p(x, y).$ We can look at its low degree part (all terms of highest degree) to get another polynomial $q(x, y)$ such that the graph of $q(x, y) = 0$ models the behavior of $p(x, y)$ in a neighborhood of $0.$ Hence, it's graph needs to be a union of straight lines (or degenerate point/null set) since the algebraic curve, by the implicit function theorem, describes a bunch of functions that are differentiable at $0$ and thus have linear approximations at $0$ and thus look like lines locally; hence the graph of $q(x, y) = 0$ should be a bunch of lines. As every homogeneous polynomial is clearly the low degree part of some curve, this property is true of all homogeneous polynomials.
(This argument is much more dubious...) Similarly, we could approach this problem from the opposite perspective, looking at the high degree part of the polynomial. Then, place the polynomial in the projective plane and, like before, find its derivatives 'at infinity' and therefore approximate it locally as a bunch of lines.
Can either of these two arguments be made rigorous? Is there even a reasonable way to talk about the derivative 'at infinity' in a projective plane?
algebraic-geometry
asked Dec 31 '18 at 2:10
Michael BarzMichael Barz
915
$endgroup$
add a comment |
$begingroup$
If we have some homogeneous polynomial $f(x, y)$ (for example $f(x, y) = x^3y + x^2y^2 - 17y^4$) than it's solution set is a union of straight lines (or a single point/null set). I've seen arguments for this, but I was wondering if this 'geometric' argument could be made formal.
Suppose we have a polynomial $p(x, y).$ We can look at its low degree part (all terms of highest degree) to get another polynomial $q(x, y)$ such that the graph of $q(x, y) = 0$ models the behavior of $p(x, y)$ in a neighborhood of $0.$ Hence, it's graph needs to be a union of straight lines (or degenerate point/null set) since the algebraic curve, by the implicit function theorem, describes a bunch of functions that are differentiable at $0$ and thus have linear approximations at $0$ and thus look like lines locally; hence the graph of $q(x, y) = 0$ should be a bunch of lines. As every homogeneous polynomial is clearly the low degree part of some curve, this property is true of all homogeneous polynomials.
(This argument is much more dubious...) Similarly, we could approach this problem from the opposite perspective, looking at the high degree part of the polynomial. Then, place the polynomial in the projective plane and, like before, find its derivatives 'at infinity' and therefore approximate it locally as a bunch of lines.
Can either of these two arguments be made rigorous? Is there even a reasonable way to talk about the derivative 'at infinity' in a projective plane?
algebraic-geometry
asked Dec 31 '18 at 2:10
Michael BarzMichael Barz
915
$endgroup$
If we have some homogeneous polynomial $f(x, y)$ (for example $f(x, y) = x^3y + x^2y^2 - 17y^4$) than it's solution set is a union of straight lines (or a single point/null set). I've seen arguments for this, but I was wondering if this 'geometric' argument could be made formal.
Suppose we have a polynomial $p(x, y).$ We can look at its low degree part (all terms of highest degree) to get another polynomial $q(x, y)$ such that the graph of $q(x, y) = 0$ models the behavior of $p(x, y)$ in a neighborhood of $0.$ Hence, it's graph needs to be a union of straight lines (or degenerate point/null set) since the algebraic curve, by the implicit function theorem, describes a bunch of functions that are differentiable at $0$ and thus have linear approximations at $0$ and thus look like lines locally; hence the graph of $q(x, y) = 0$ should be a bunch of lines. As every homogeneous polynomial is clearly the low degree part of some curve, this property is true of all homogeneous polynomials.
(This argument is much more dubious...) Similarly, we could approach this problem from the opposite perspective, looking at the high degree part of the polynomial. Then, place the polynomial in the projective plane and, like before, find its derivatives 'at infinity' and therefore approximate it locally as a bunch of lines.
Can either of these two arguments be made rigorous? Is there even a reasonable way to talk about the derivative 'at infinity' in a projective plane?
algebraic-geometry
algebraic-geometry
asked Dec 31 '18 at 2:10
Michael BarzMichael Barz
915
asked Dec 31 '18 at 2:10
Michael BarzMichael Barz
915
edited Dec 31 '18 at 2:18
Michael Barz
asked Dec 31 '18 at 2:10
Michael BarzMichael Barz
915
asked Dec 31 '18 at 2:10
Michael BarzMichael Barz
915
asked Dec 31 '18 at 2:10
Michael BarzMichael Barz
915
915
add a comment |
add a comment |
1 Answer
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It's easier: if $f(x,y)$ is homogeneous of degree $d$, then $f(lambda x,lambda y)=lambda^d f(x,y)$. So if $f(x,y)=0$ for a point $(x,y)neq (0,0)$, then $f$ also vanishes on all the points on the line through the origin passing through $(x,y)$.
answered Dec 31 '18 at 2:21
KReiserKReiser
9,94021435
$endgroup$
$begingroup$
I've seen that argument (that's the argument I had originally), but I was wondering if the geometric argument I had could be made rigorous--I saw this appearing again with algebraic curves and was trying to look for a geometric explanation of why the low and high degree parts were unions of lines.
$endgroup$
– Michael Barz
Dec 31 '18 at 2:26
3
$begingroup$
I suppose the rigorous way to do this for the low-degree terms is to look at the tangent cone. What's really going on with the high-degree stuff "at infinity" is that you want to take the projective completion of your curve and look at what happens at some point on the completed bit, and the high degree terms get turned in to low-degree terms when you specialize to the affine patch containing the point you picked which was originally not in your affine patch.
$endgroup$
– KReiser
Dec 31 '18 at 2:34
add a comment |
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1 Answer
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1 Answer
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$begingroup$
It's easier: if $f(x,y)$ is homogeneous of degree $d$, then $f(lambda x,lambda y)=lambda^d f(x,y)$. So if $f(x,y)=0$ for a point $(x,y)neq (0,0)$, then $f$ also vanishes on all the points on the line through the origin passing through $(x,y)$.
answered Dec 31 '18 at 2:21
KReiserKReiser
9,94021435
$endgroup$
$begingroup$
I've seen that argument (that's the argument I had originally), but I was wondering if the geometric argument I had could be made rigorous--I saw this appearing again with algebraic curves and was trying to look for a geometric explanation of why the low and high degree parts were unions of lines.
$endgroup$
– Michael Barz
Dec 31 '18 at 2:26
3
$begingroup$
I suppose the rigorous way to do this for the low-degree terms is to look at the tangent cone. What's really going on with the high-degree stuff "at infinity" is that you want to take the projective completion of your curve and look at what happens at some point on the completed bit, and the high degree terms get turned in to low-degree terms when you specialize to the affine patch containing the point you picked which was originally not in your affine patch.
$endgroup$
– KReiser
Dec 31 '18 at 2:34
add a comment |
$begingroup$
It's easier: if $f(x,y)$ is homogeneous of degree $d$, then $f(lambda x,lambda y)=lambda^d f(x,y)$. So if $f(x,y)=0$ for a point $(x,y)neq (0,0)$, then $f$ also vanishes on all the points on the line through the origin passing through $(x,y)$.
answered Dec 31 '18 at 2:21
KReiserKReiser
9,94021435
$endgroup$
$begingroup$
I've seen that argument (that's the argument I had originally), but I was wondering if the geometric argument I had could be made rigorous--I saw this appearing again with algebraic curves and was trying to look for a geometric explanation of why the low and high degree parts were unions of lines.
$endgroup$
– Michael Barz
Dec 31 '18 at 2:26
3
$begingroup$
I suppose the rigorous way to do this for the low-degree terms is to look at the tangent cone. What's really going on with the high-degree stuff "at infinity" is that you want to take the projective completion of your curve and look at what happens at some point on the completed bit, and the high degree terms get turned in to low-degree terms when you specialize to the affine patch containing the point you picked which was originally not in your affine patch.
$endgroup$
– KReiser
Dec 31 '18 at 2:34
add a comment |
$begingroup$
It's easier: if $f(x,y)$ is homogeneous of degree $d$, then $f(lambda x,lambda y)=lambda^d f(x,y)$. So if $f(x,y)=0$ for a point $(x,y)neq (0,0)$, then $f$ also vanishes on all the points on the line through the origin passing through $(x,y)$.
answered Dec 31 '18 at 2:21
KReiserKReiser
9,94021435
$endgroup$
It's easier: if $f(x,y)$ is homogeneous of degree $d$, then $f(lambda x,lambda y)=lambda^d f(x,y)$. So if $f(x,y)=0$ for a point $(x,y)neq (0,0)$, then $f$ also vanishes on all the points on the line through the origin passing through $(x,y)$.
answered Dec 31 '18 at 2:21
KReiserKReiser
9,94021435
answered Dec 31 '18 at 2:21
KReiserKReiser
9,94021435
answered Dec 31 '18 at 2:21
KReiserKReiser
9,94021435
answered Dec 31 '18 at 2:21
KReiserKReiser
9,94021435
9,94021435
$begingroup$
I've seen that argument (that's the argument I had originally), but I was wondering if the geometric argument I had could be made rigorous--I saw this appearing again with algebraic curves and was trying to look for a geometric explanation of why the low and high degree parts were unions of lines.
$endgroup$
– Michael Barz
Dec 31 '18 at 2:26
3
$begingroup$
I suppose the rigorous way to do this for the low-degree terms is to look at the tangent cone. What's really going on with the high-degree stuff "at infinity" is that you want to take the projective completion of your curve and look at what happens at some point on the completed bit, and the high degree terms get turned in to low-degree terms when you specialize to the affine patch containing the point you picked which was originally not in your affine patch.
$endgroup$
– KReiser
Dec 31 '18 at 2:34
add a comment |
$begingroup$
I've seen that argument (that's the argument I had originally), but I was wondering if the geometric argument I had could be made rigorous--I saw this appearing again with algebraic curves and was trying to look for a geometric explanation of why the low and high degree parts were unions of lines.
$endgroup$
– Michael Barz
Dec 31 '18 at 2:26
3
$begingroup$
I suppose the rigorous way to do this for the low-degree terms is to look at the tangent cone. What's really going on with the high-degree stuff "at infinity" is that you want to take the projective completion of your curve and look at what happens at some point on the completed bit, and the high degree terms get turned in to low-degree terms when you specialize to the affine patch containing the point you picked which was originally not in your affine patch.
$endgroup$
– KReiser
Dec 31 '18 at 2:34
$begingroup$
I've seen that argument (that's the argument I had originally), but I was wondering if the geometric argument I had could be made rigorous--I saw this appearing again with algebraic curves and was trying to look for a geometric explanation of why the low and high degree parts were unions of lines.
$endgroup$
– Michael Barz
Dec 31 '18 at 2:26
$begingroup$
I've seen that argument (that's the argument I had originally), but I was wondering if the geometric argument I had could be made rigorous--I saw this appearing again with algebraic curves and was trying to look for a geometric explanation of why the low and high degree parts were unions of lines.
$endgroup$
– Michael Barz
Dec 31 '18 at 2:26
3
3
$begingroup$
I suppose the rigorous way to do this for the low-degree terms is to look at the tangent cone. What's really going on with the high-degree stuff "at infinity" is that you want to take the projective completion of your curve and look at what happens at some point on the completed bit, and the high degree terms get turned in to low-degree terms when you specialize to the affine patch containing the point you picked which was originally not in your affine patch.
$endgroup$
– KReiser
Dec 31 '18 at 2:34
$begingroup$
I suppose the rigorous way to do this for the low-degree terms is to look at the tangent cone. What's really going on with the high-degree stuff "at infinity" is that you want to take the projective completion of your curve and look at what happens at some point on the completed bit, and the high degree terms get turned in to low-degree terms when you specialize to the affine patch containing the point you picked which was originally not in your affine patch.
$endgroup$
– KReiser
Dec 31 '18 at 2:34
add a comment |
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