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

  1. Jun 18, 2012 #1
    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. jcsd
  3. Jun 18, 2012 #2

    haruspex

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    No. f_y is never zero at x = 1/2.
     
  4. Jun 18, 2012 #3
    Applying this method generates the POSSIBLE singular points. You should check if they really are singular points by other means.
     
  5. Jun 18, 2012 #4
    Thank you haruspex and Millennial. That clarifies things!
     
  6. Jun 18, 2012 #5

    HallsofIvy

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    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).
     
  7. Jun 19, 2012 #6
    Ah, that's a good argument. Thanks HallsofIvy!
     
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