Rank of derivative polynomial map equals dimension image?
I've been told that given a polynomial map $f:Xto Y$ in characteristic zero, there exists an open dense subset $U$ of $X$ such that for all points $x$ in $U$, the rank of the derivative of $f$ in $x$ equals the dimension of the image of $f$.
Could someone please explain to me why this is true, or direct me to some reference where this is explained?
algebraic-geometry
asked Jun 20 '14 at 14:20
Jasmine
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reputation from Dániel G. ending in 4 days.
Looking for an answer drawing from credible and/or official sources.
add a comment |
I've been told that given a polynomial map $f:Xto Y$ in characteristic zero, there exists an open dense subset $U$ of $X$ such that for all points $x$ in $U$, the rank of the derivative of $f$ in $x$ equals the dimension of the image of $f$.
Could someone please explain to me why this is true, or direct me to some reference where this is explained?
algebraic-geometry
asked Jun 20 '14 at 14:20
Jasmine
161
This question has an open bounty worth +50
reputation from Dániel G. ending in 4 days.
Looking for an answer drawing from credible and/or official sources.
A reference is Shafarevich, Basic Algebraic Geometry Volume 1, II.6.2 Theorem 2. (To apply the statement as written in your case, you need to remove the singular locus of $X$, but that doesn't cause any problems.)
– user64687
Jun 20 '14 at 16:05
Could someone elaborate on the reference given in the previous comment? The theorem in question seems much more general and I'm having trouble applying it to the specific case in the question.
– Dániel G.
Jan 3 at 22:22
add a comment |
I've been told that given a polynomial map $f:Xto Y$ in characteristic zero, there exists an open dense subset $U$ of $X$ such that for all points $x$ in $U$, the rank of the derivative of $f$ in $x$ equals the dimension of the image of $f$.
Could someone please explain to me why this is true, or direct me to some reference where this is explained?
algebraic-geometry
asked Jun 20 '14 at 14:20
Jasmine
161
I've been told that given a polynomial map $f:Xto Y$ in characteristic zero, there exists an open dense subset $U$ of $X$ such that for all points $x$ in $U$, the rank of the derivative of $f$ in $x$ equals the dimension of the image of $f$.
Could someone please explain to me why this is true, or direct me to some reference where this is explained?
algebraic-geometry
algebraic-geometry
asked Jun 20 '14 at 14:20
Jasmine
161
asked Jun 20 '14 at 14:20
Jasmine
161
asked Jun 20 '14 at 14:20
Jasmine
161
asked Jun 20 '14 at 14:20
Jasmine
161
asked Jun 20 '14 at 14:20
Jasmine
161
161
This question has an open bounty worth +50
reputation from Dániel G. ending in 4 days.
Looking for an answer drawing from credible and/or official sources.
This question has an open bounty worth +50
reputation from Dániel G. ending in 4 days.
Looking for an answer drawing from credible and/or official sources.
A reference is Shafarevich, Basic Algebraic Geometry Volume 1, II.6.2 Theorem 2. (To apply the statement as written in your case, you need to remove the singular locus of $X$, but that doesn't cause any problems.)
– user64687
Jun 20 '14 at 16:05
Could someone elaborate on the reference given in the previous comment? The theorem in question seems much more general and I'm having trouble applying it to the specific case in the question.
– Dániel G.
Jan 3 at 22:22
add a comment |
A reference is Shafarevich, Basic Algebraic Geometry Volume 1, II.6.2 Theorem 2. (To apply the statement as written in your case, you need to remove the singular locus of $X$, but that doesn't cause any problems.)
– user64687
Jun 20 '14 at 16:05
Could someone elaborate on the reference given in the previous comment? The theorem in question seems much more general and I'm having trouble applying it to the specific case in the question.
– Dániel G.
Jan 3 at 22:22
A reference is Shafarevich, Basic Algebraic Geometry Volume 1, II.6.2 Theorem 2. (To apply the statement as written in your case, you need to remove the singular locus of $X$, but that doesn't cause any problems.)
– user64687
Jun 20 '14 at 16:05
A reference is Shafarevich, Basic Algebraic Geometry Volume 1, II.6.2 Theorem 2. (To apply the statement as written in your case, you need to remove the singular locus of $X$, but that doesn't cause any problems.)
– user64687
Jun 20 '14 at 16:05
Could someone elaborate on the reference given in the previous comment? The theorem in question seems much more general and I'm having trouble applying it to the specific case in the question.
– Dániel G.
Jan 3 at 22:22
Could someone elaborate on the reference given in the previous comment? The theorem in question seems much more general and I'm having trouble applying it to the specific case in the question.
– Dániel G.
Jan 3 at 22:22
add a comment |
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A reference is Shafarevich, Basic Algebraic Geometry Volume 1, II.6.2 Theorem 2. (To apply the statement as written in your case, you need to remove the singular locus of $X$, but that doesn't cause any problems.)
– user64687
Jun 20 '14 at 16:05
Could someone elaborate on the reference given in the previous comment? The theorem in question seems much more general and I'm having trouble applying it to the specific case in the question.
– Dániel G.
Jan 3 at 22:22