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

1. Jun 18, 2012

### naturemath

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: Jun 18, 2012
2. Jun 18, 2012

### haruspex

No. f_y is never zero at x = 1/2.

3. Jun 18, 2012

### Millennial

Applying this method generates the POSSIBLE singular points. You should check if they really are singular points by other means.

4. Jun 18, 2012

### naturemath

Thank you haruspex and Millennial. That clarifies things!

5. Jun 18, 2012

### HallsofIvy

Staff Emeritus
A some what better method would be to look at $f_x= 2xy- y= (2x- 1)y= 0$ and say, as you do, that either x= 1/2 or y= 0.

Now look at $f_y= x^2- x= x(x- 1)= 0$ 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. Jun 19, 2012

### naturemath

Ah, that's a good argument. Thanks HallsofIvy!