Can two different branches of Puiseux expansion have the same tangent?
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Let $C$ be a curve in complex affine plane defined by an irreducible polynomial $f in mathbbC[x,y]$. If we create a Puiseux expansion of $C$ in the neighbourhhod of $(0,0)$, $C$ can be split into several branches (ie local parametrizations $(x,y)=(t^m,sum_1^infty a_r t^r)$).
My question is about tangents: is it possible for two different branches to have the same tangent at $(0,0)$? Can you give me an example of such irreducible $f$?
geometry algebraic-geometry parametrization
edited Sep 4 at 8:37
Bernard
112k635104
asked Sep 4 at 7:15
antizaba
253
add a comment |Â
up vote
0
down vote
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Let $C$ be a curve in complex affine plane defined by an irreducible polynomial $f in mathbbC[x,y]$. If we create a Puiseux expansion of $C$ in the neighbourhhod of $(0,0)$, $C$ can be split into several branches (ie local parametrizations $(x,y)=(t^m,sum_1^infty a_r t^r)$).
My question is about tangents: is it possible for two different branches to have the same tangent at $(0,0)$? Can you give me an example of such irreducible $f$?
geometry algebraic-geometry parametrization
edited Sep 4 at 8:37
Bernard
112k635104
asked Sep 4 at 7:15
antizaba
253
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $C$ be a curve in complex affine plane defined by an irreducible polynomial $f in mathbbC[x,y]$. If we create a Puiseux expansion of $C$ in the neighbourhhod of $(0,0)$, $C$ can be split into several branches (ie local parametrizations $(x,y)=(t^m,sum_1^infty a_r t^r)$).
My question is about tangents: is it possible for two different branches to have the same tangent at $(0,0)$? Can you give me an example of such irreducible $f$?
geometry algebraic-geometry parametrization
edited Sep 4 at 8:37
Bernard
112k635104
asked Sep 4 at 7:15
antizaba
253
Let $C$ be a curve in complex affine plane defined by an irreducible polynomial $f in mathbbC[x,y]$. If we create a Puiseux expansion of $C$ in the neighbourhhod of $(0,0)$, $C$ can be split into several branches (ie local parametrizations $(x,y)=(t^m,sum_1^infty a_r t^r)$).
My question is about tangents: is it possible for two different branches to have the same tangent at $(0,0)$? Can you give me an example of such irreducible $f$?
geometry algebraic-geometry parametrization
geometry algebraic-geometry parametrization
edited Sep 4 at 8:37
Bernard
112k635104
asked Sep 4 at 7:15
antizaba
253
edited Sep 4 at 8:37
Bernard
112k635104
asked Sep 4 at 7:15
antizaba
253
edited Sep 4 at 8:37
Bernard
112k635104
edited Sep 4 at 8:37
Bernard
112k635104
edited Sep 4 at 8:37
Bernard
112k635104
112k635104
asked Sep 4 at 7:15
antizaba
253
asked Sep 4 at 7:15
antizaba
253
asked Sep 4 at 7:15
antizaba
253
253
add a comment |Â
add a comment |Â
1 Answer
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$f(x,y) = (x-y^2)(x+y^2) + x^3$.
More explanation : It's easy to check that it's irreducible. Now, for a polynomial $f = f_r + f_r+1 + dots $ where $f_d$ is the homogenous part of degree $d$, the branches of $f$ and $f_r$ are in bijection and have the same tangent direction as an easy calculation shows. In particular the branches of $f$ at $0$ are $x-y^2 = 0$ and $x+y^2 = 0$ which are indeed tangent.
answered Sep 4 at 13:11
Nicolas Hemelsoet
5,215417
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$f(x,y) = (x-y^2)(x+y^2) + x^3$.
More explanation : It's easy to check that it's irreducible. Now, for a polynomial $f = f_r + f_r+1 + dots $ where $f_d$ is the homogenous part of degree $d$, the branches of $f$ and $f_r$ are in bijection and have the same tangent direction as an easy calculation shows. In particular the branches of $f$ at $0$ are $x-y^2 = 0$ and $x+y^2 = 0$ which are indeed tangent.
answered Sep 4 at 13:11
Nicolas Hemelsoet
5,215417
add a comment |Â
up vote
1
down vote
accepted
$f(x,y) = (x-y^2)(x+y^2) + x^3$.
More explanation : It's easy to check that it's irreducible. Now, for a polynomial $f = f_r + f_r+1 + dots $ where $f_d$ is the homogenous part of degree $d$, the branches of $f$ and $f_r$ are in bijection and have the same tangent direction as an easy calculation shows. In particular the branches of $f$ at $0$ are $x-y^2 = 0$ and $x+y^2 = 0$ which are indeed tangent.
answered Sep 4 at 13:11
Nicolas Hemelsoet
5,215417
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$f(x,y) = (x-y^2)(x+y^2) + x^3$.
More explanation : It's easy to check that it's irreducible. Now, for a polynomial $f = f_r + f_r+1 + dots $ where $f_d$ is the homogenous part of degree $d$, the branches of $f$ and $f_r$ are in bijection and have the same tangent direction as an easy calculation shows. In particular the branches of $f$ at $0$ are $x-y^2 = 0$ and $x+y^2 = 0$ which are indeed tangent.
answered Sep 4 at 13:11
Nicolas Hemelsoet
5,215417
$f(x,y) = (x-y^2)(x+y^2) + x^3$.
More explanation : It's easy to check that it's irreducible. Now, for a polynomial $f = f_r + f_r+1 + dots $ where $f_d$ is the homogenous part of degree $d$, the branches of $f$ and $f_r$ are in bijection and have the same tangent direction as an easy calculation shows. In particular the branches of $f$ at $0$ are $x-y^2 = 0$ and $x+y^2 = 0$ which are indeed tangent.
answered Sep 4 at 13:11
Nicolas Hemelsoet
5,215417
edited Sep 5 at 6:55
answered Sep 4 at 13:11
Nicolas Hemelsoet
5,215417
answered Sep 4 at 13:11
Nicolas Hemelsoet
5,215417
answered Sep 4 at 13:11
Nicolas Hemelsoet
5,215417
5,215417
add a comment |Â
add a comment |Â
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