The singular points on f = x^2 y - x y on a plane

  • Thread starter naturemath
  • Start date
  • #1
31
0
Let f(x,y) = x^2 y - xy = x(x-1)y be a polynomial in k[x,y].

I am looking for the singular subset of this function.

Taking the partials, we obtain

f_x = 2xy - y

f_y = x^2 - x.

In order to find the singular subset, both partials (with respect to x and with respect to y) must vanish. So we obtain that

f_x = 2xy - y = 0

which implies x = 1/2 or y = 0,

while

f_y = x^2 - x = 0

implies

x = 0 or x =1.

Drawing a picture of f, it is clear that the two points (0,0) and (1,0) are singular points, but what does x = 1/2 tell us? Is this point supposed to be a singular point as well?
 
Last edited:

Answers and Replies

  • #2
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
34,601
5,991
f_y = x^2 - x = 0
:
what does x = 1/2 tell us? Is this point supposed to be a singular point as well?
No. f_y is never zero at x = 1/2.
 
  • #3
296
0
Applying this method generates the POSSIBLE singular points. You should check if they really are singular points by other means.
 
  • #4
31
0
Thank you haruspex and Millennial. That clarifies things!
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,833
956
A some what better method would be to look at [itex]f_x= 2xy- y= (2x- 1)y= 0[/itex] and say, as you do, that either x= 1/2 or y= 0.

Now look at [itex]f_y= x^2- x= x(x- 1)= 0[/itex] and use the values you already have. IF x= 1/2, that is impossible but IF y= 0, x can be 0 or 1. The singular points are (0, 0) and (1, 0).
 
  • #6
31
0
Ah, that's a good argument. Thanks HallsofIvy!
 

Related Threads on The singular points on f = x^2 y - x y on a plane

  • Last Post
Replies
9
Views
10K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
2K
Replies
1
Views
2K
Replies
3
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
20
Views
4K
Replies
5
Views
707
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
1K
Top