Subset of C2 such that any vanishing cubic polynomial is reducible [closed]

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I'm looking for a subset of $ mathbb C^2$ consisting of four points such that any polynomial of degree 3 vanishing on this set is reducible.

Don't really know where to start, any ideas on how to approach this are appreciated.

algebraic-geometry projective-geometry

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edited Jul 16 at 16:32

asked Jul 16 at 16:21

casimir

406

closed as off-topic by amWhy, John Ma, Cesareo, Isaac Browne, Parcly Taxel Jul 18 at 1:45

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, John Ma, Cesareo, Isaac Browne, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You mean $Bbb C^2$?
  – Kenny Lau
  Jul 16 at 16:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What do you mean by "cubic polynomial"?
  – Kenny Lau
  Jul 16 at 16:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, I did - edited accordingly.
  – casimir
  Jul 16 at 16:33
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

I'm looking for a subset of $ mathbb C^2$ consisting of four points such that any polynomial of degree 3 vanishing on this set is reducible.

Don't really know where to start, any ideas on how to approach this are appreciated.

algebraic-geometry projective-geometry

share|cite|improve this question

edited Jul 16 at 16:32

asked Jul 16 at 16:21

casimir

406

closed as off-topic by amWhy, John Ma, Cesareo, Isaac Browne, Parcly Taxel Jul 18 at 1:45

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, John Ma, Cesareo, Isaac Browne, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You mean $Bbb C^2$?
  – Kenny Lau
  Jul 16 at 16:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What do you mean by "cubic polynomial"?
  – Kenny Lau
  Jul 16 at 16:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, I did - edited accordingly.
  – casimir
  Jul 16 at 16:33
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I'm looking for a subset of $ mathbb C^2$ consisting of four points such that any polynomial of degree 3 vanishing on this set is reducible.

Don't really know where to start, any ideas on how to approach this are appreciated.

algebraic-geometry projective-geometry

share|cite|improve this question

edited Jul 16 at 16:32

asked Jul 16 at 16:21

casimir

406

I'm looking for a subset of $ mathbb C^2$ consisting of four points such that any polynomial of degree 3 vanishing on this set is reducible.

Don't really know where to start, any ideas on how to approach this are appreciated.

algebraic-geometry projective-geometry

share|cite|improve this question

edited Jul 16 at 16:32

asked Jul 16 at 16:21

casimir

406

share|cite|improve this question

share|cite|improve this question

edited Jul 16 at 16:32

edited Jul 16 at 16:32

edited Jul 16 at 16:32

asked Jul 16 at 16:21

casimir

406

asked Jul 16 at 16:21

casimir

406

asked Jul 16 at 16:21

casimir

406

406

closed as off-topic by amWhy, John Ma, Cesareo, Isaac Browne, Parcly Taxel Jul 18 at 1:45

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, John Ma, Cesareo, Isaac Browne, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by amWhy, John Ma, Cesareo, Isaac Browne, Parcly Taxel Jul 18 at 1:45

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, John Ma, Cesareo, Isaac Browne, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You mean $Bbb C^2$?
  – Kenny Lau
  Jul 16 at 16:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What do you mean by "cubic polynomial"?
  – Kenny Lau
  Jul 16 at 16:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, I did - edited accordingly.
  – casimir
  Jul 16 at 16:33
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You mean $Bbb C^2$?
  – Kenny Lau
  Jul 16 at 16:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What do you mean by "cubic polynomial"?
  – Kenny Lau
  Jul 16 at 16:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, I did - edited accordingly.
  – casimir
  Jul 16 at 16:33
  

  
  
  
  

You mean $Bbb C^2$?
– Kenny Lau
Jul 16 at 16:24

You mean $Bbb C^2$?
– Kenny Lau
Jul 16 at 16:24

What do you mean by "cubic polynomial"?
– Kenny Lau
Jul 16 at 16:25

What do you mean by "cubic polynomial"?
– Kenny Lau
Jul 16 at 16:25

Yes, I did - edited accordingly.
– casimir
Jul 16 at 16:33

Yes, I did - edited accordingly.
– casimir
Jul 16 at 16:33

add a comment | 

1 Answer
1

active

oldest

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up vote
4
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accepted

Take any four points that lie on a line will do. After a change of coordinates suppose your four points are given by $(0,0), (0,1), (0,2), (0,3)$, and let $f(x,y)$ denote your cubic polynomial.

Then by assumption, we have that $f(0,y)$ has four roots $y=0,1,2,3$. But $f(0,y)$ has degree at most $3$, so if it has four roots it must imply that $f(0,y)=0$. But this implies that $f(x,y) = xg(x,y)$ for some $g$.

share|cite|improve this answer

answered Jul 16 at 17:36

loch

1,403179

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote



accepted

Take any four points that lie on a line will do. After a change of coordinates suppose your four points are given by $(0,0), (0,1), (0,2), (0,3)$, and let $f(x,y)$ denote your cubic polynomial.

Then by assumption, we have that $f(0,y)$ has four roots $y=0,1,2,3$. But $f(0,y)$ has degree at most $3$, so if it has four roots it must imply that $f(0,y)=0$. But this implies that $f(x,y) = xg(x,y)$ for some $g$.

share|cite|improve this answer

answered Jul 16 at 17:36

loch

1,403179

add a comment | 

up vote
4
down vote



accepted

Take any four points that lie on a line will do. After a change of coordinates suppose your four points are given by $(0,0), (0,1), (0,2), (0,3)$, and let $f(x,y)$ denote your cubic polynomial.

Then by assumption, we have that $f(0,y)$ has four roots $y=0,1,2,3$. But $f(0,y)$ has degree at most $3$, so if it has four roots it must imply that $f(0,y)=0$. But this implies that $f(x,y) = xg(x,y)$ for some $g$.

share|cite|improve this answer

answered Jul 16 at 17:36

loch

1,403179

add a comment | 

up vote
4
down vote



accepted

up vote
4
down vote



accepted

Take any four points that lie on a line will do. After a change of coordinates suppose your four points are given by $(0,0), (0,1), (0,2), (0,3)$, and let $f(x,y)$ denote your cubic polynomial.

Then by assumption, we have that $f(0,y)$ has four roots $y=0,1,2,3$. But $f(0,y)$ has degree at most $3$, so if it has four roots it must imply that $f(0,y)=0$. But this implies that $f(x,y) = xg(x,y)$ for some $g$.

share|cite|improve this answer

answered Jul 16 at 17:36

loch

1,403179

Take any four points that lie on a line will do. After a change of coordinates suppose your four points are given by $(0,0), (0,1), (0,2), (0,3)$, and let $f(x,y)$ denote your cubic polynomial.

Then by assumption, we have that $f(0,y)$ has four roots $y=0,1,2,3$. But $f(0,y)$ has degree at most $3$, so if it has four roots it must imply that $f(0,y)=0$. But this implies that $f(x,y) = xg(x,y)$ for some $g$.

share|cite|improve this answer

answered Jul 16 at 17:36

loch

1,403179

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 17:36

loch

1,403179

answered Jul 16 at 17:36

loch

1,403179

answered Jul 16 at 17:36

loch

1,403179

1,403179

add a comment | 

add a comment |