Fulton Exercise 1.31 finding irreducible components
$begingroup$
This is Fulton, Algebraic Curves exercise 1.31:
We have to find irreducible components of $V(f) =V(y^2-xy-x^2y+x^3)$ in $mathbb{A^2({mathbb{R}})}$ and $mathbb{A^2({mathbb{C}})}$.
My attempt:
Is to write this polynomial as = $(y-x^2)(y-x)$
And so we have $V(f)= V(y-x^2) cup V(y-x)$
My guess is that it would be the same set of points in both $mathbb{A^2({mathbb{R}})}$ and $mathbb{A^2({mathbb{C}})}$.
Is my guess correct? Any alternate approach or output on my idea will be appreciated, thanks.
$V(f)$ denotes the set of all points at which $f$ is zero in $mathbb{A^2({mathbb{k}})}$, $k$ a field.
algebraic-geometry
asked Jan 14 at 6:18
clearclear
1,44121130
$endgroup$
add a comment |
$begingroup$
This is Fulton, Algebraic Curves exercise 1.31:
We have to find irreducible components of $V(f) =V(y^2-xy-x^2y+x^3)$ in $mathbb{A^2({mathbb{R}})}$ and $mathbb{A^2({mathbb{C}})}$.
My attempt:
Is to write this polynomial as = $(y-x^2)(y-x)$
And so we have $V(f)= V(y-x^2) cup V(y-x)$
My guess is that it would be the same set of points in both $mathbb{A^2({mathbb{R}})}$ and $mathbb{A^2({mathbb{C}})}$.
Is my guess correct? Any alternate approach or output on my idea will be appreciated, thanks.
$V(f)$ denotes the set of all points at which $f$ is zero in $mathbb{A^2({mathbb{k}})}$, $k$ a field.
algebraic-geometry
asked Jan 14 at 6:18
clearclear
1,44121130
$endgroup$
$begingroup$
What do you mean, the same set of points? Consider $(i,i)$, for example.
$endgroup$
– KReiser
Jan 14 at 6:44
$begingroup$
Oh, yeah you are right.
$endgroup$
– clear
Jan 14 at 6:57
$begingroup$
Any hint on how to explicitly find those sets? In case of $mathbb{R}$, it's obvious that it will be union of a parabola and a line. I don't know what I can do in complex case.
$endgroup$
– clear
Jan 14 at 6:59
1
$begingroup$
You've correctly identified them in this case, for both fields. Think about the correspondence between varieties and ideals: if some polynomial $g$ divides $f$, this means that $(f)subset (g)$, and so $V(g)subset V(f)$. Since irreducible components correspond to prime ideals, the question is asking you to find all prime factors of $f$, which you did.
$endgroup$
– KReiser
Jan 14 at 7:20
add a comment |
$begingroup$
This is Fulton, Algebraic Curves exercise 1.31:
We have to find irreducible components of $V(f) =V(y^2-xy-x^2y+x^3)$ in $mathbb{A^2({mathbb{R}})}$ and $mathbb{A^2({mathbb{C}})}$.
My attempt:
Is to write this polynomial as = $(y-x^2)(y-x)$
And so we have $V(f)= V(y-x^2) cup V(y-x)$
My guess is that it would be the same set of points in both $mathbb{A^2({mathbb{R}})}$ and $mathbb{A^2({mathbb{C}})}$.
Is my guess correct? Any alternate approach or output on my idea will be appreciated, thanks.
$V(f)$ denotes the set of all points at which $f$ is zero in $mathbb{A^2({mathbb{k}})}$, $k$ a field.
algebraic-geometry
asked Jan 14 at 6:18
clearclear
1,44121130
$endgroup$
This is Fulton, Algebraic Curves exercise 1.31:
We have to find irreducible components of $V(f) =V(y^2-xy-x^2y+x^3)$ in $mathbb{A^2({mathbb{R}})}$ and $mathbb{A^2({mathbb{C}})}$.
My attempt:
Is to write this polynomial as = $(y-x^2)(y-x)$
And so we have $V(f)= V(y-x^2) cup V(y-x)$
My guess is that it would be the same set of points in both $mathbb{A^2({mathbb{R}})}$ and $mathbb{A^2({mathbb{C}})}$.
Is my guess correct? Any alternate approach or output on my idea will be appreciated, thanks.
$V(f)$ denotes the set of all points at which $f$ is zero in $mathbb{A^2({mathbb{k}})}$, $k$ a field.
algebraic-geometry
algebraic-geometry
asked Jan 14 at 6:18
clearclear
1,44121130
asked Jan 14 at 6:18
clearclear
1,44121130
asked Jan 14 at 6:18
clearclear
1,44121130
asked Jan 14 at 6:18
clearclear
1,44121130
asked Jan 14 at 6:18
clearclear
1,44121130
1,44121130
$begingroup$
What do you mean, the same set of points? Consider $(i,i)$, for example.
$endgroup$
– KReiser
Jan 14 at 6:44
$begingroup$
Oh, yeah you are right.
$endgroup$
– clear
Jan 14 at 6:57
$begingroup$
Any hint on how to explicitly find those sets? In case of $mathbb{R}$, it's obvious that it will be union of a parabola and a line. I don't know what I can do in complex case.
$endgroup$
– clear
Jan 14 at 6:59
1
$begingroup$
You've correctly identified them in this case, for both fields. Think about the correspondence between varieties and ideals: if some polynomial $g$ divides $f$, this means that $(f)subset (g)$, and so $V(g)subset V(f)$. Since irreducible components correspond to prime ideals, the question is asking you to find all prime factors of $f$, which you did.
$endgroup$
– KReiser
Jan 14 at 7:20
add a comment |
$begingroup$
What do you mean, the same set of points? Consider $(i,i)$, for example.
$endgroup$
– KReiser
Jan 14 at 6:44
$begingroup$
Oh, yeah you are right.
$endgroup$
– clear
Jan 14 at 6:57
$begingroup$
Any hint on how to explicitly find those sets? In case of $mathbb{R}$, it's obvious that it will be union of a parabola and a line. I don't know what I can do in complex case.
$endgroup$
– clear
Jan 14 at 6:59
1
$begingroup$
You've correctly identified them in this case, for both fields. Think about the correspondence between varieties and ideals: if some polynomial $g$ divides $f$, this means that $(f)subset (g)$, and so $V(g)subset V(f)$. Since irreducible components correspond to prime ideals, the question is asking you to find all prime factors of $f$, which you did.
$endgroup$
– KReiser
Jan 14 at 7:20
$begingroup$
What do you mean, the same set of points? Consider $(i,i)$, for example.
$endgroup$
– KReiser
Jan 14 at 6:44
$begingroup$
What do you mean, the same set of points? Consider $(i,i)$, for example.
$endgroup$
– KReiser
Jan 14 at 6:44
$begingroup$
Oh, yeah you are right.
$endgroup$
– clear
Jan 14 at 6:57
$begingroup$
Oh, yeah you are right.
$endgroup$
– clear
Jan 14 at 6:57
$begingroup$
Any hint on how to explicitly find those sets? In case of $mathbb{R}$, it's obvious that it will be union of a parabola and a line. I don't know what I can do in complex case.
$endgroup$
– clear
Jan 14 at 6:59
$begingroup$
Any hint on how to explicitly find those sets? In case of $mathbb{R}$, it's obvious that it will be union of a parabola and a line. I don't know what I can do in complex case.
$endgroup$
– clear
Jan 14 at 6:59
1
1
$begingroup$
You've correctly identified them in this case, for both fields. Think about the correspondence between varieties and ideals: if some polynomial $g$ divides $f$, this means that $(f)subset (g)$, and so $V(g)subset V(f)$. Since irreducible components correspond to prime ideals, the question is asking you to find all prime factors of $f$, which you did.
$endgroup$
– KReiser
Jan 14 at 7:20
$begingroup$
You've correctly identified them in this case, for both fields. Think about the correspondence between varieties and ideals: if some polynomial $g$ divides $f$, this means that $(f)subset (g)$, and so $V(g)subset V(f)$. Since irreducible components correspond to prime ideals, the question is asking you to find all prime factors of $f$, which you did.
$endgroup$
– KReiser
Jan 14 at 7:20
add a comment |
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$begingroup$
What do you mean, the same set of points? Consider $(i,i)$, for example.
$endgroup$
– KReiser
Jan 14 at 6:44
$begingroup$
Oh, yeah you are right.
$endgroup$
– clear
Jan 14 at 6:57
$begingroup$
Any hint on how to explicitly find those sets? In case of $mathbb{R}$, it's obvious that it will be union of a parabola and a line. I don't know what I can do in complex case.
$endgroup$
– clear
Jan 14 at 6:59
1
$begingroup$
You've correctly identified them in this case, for both fields. Think about the correspondence between varieties and ideals: if some polynomial $g$ divides $f$, this means that $(f)subset (g)$, and so $V(g)subset V(f)$. Since irreducible components correspond to prime ideals, the question is asking you to find all prime factors of $f$, which you did.
$endgroup$
– KReiser
Jan 14 at 7:20