Hello, my loves. Welcome to a new section. Let's start by getting the
matching out of the way (just graph them on wolframalpha or something):
The rest has to do with identifying singularities! And this exercise is based on the
preliminary definition of (non)singularity:
Now, as the exercise says, each of the varieties given are curves, which means they
each have dimension 1. Hence, for each of these, a point is nonsingular iff the rank
of the Jacobian is 2  1 = 1. I.e. a point is singular iff the rank of the Jacobian is
0. In particular, the Jacobian in each of these examples is gonna look like
J  =
  
(so really, it's just the gradient vector of
f) The rank of the matrix is the row
span, which is just a single vector, and that is therefore
1 unless both
and
are zero. Hence, for each of these problems we are solving
 = 0  

 = 0  

  
in addition to
f = 0, so our solutions actually lie on the curve. Hence this whole
exercise is basically solving systems of linear equations (over an algebraically
closed field)
"Errr, can't I just look at the graphs and tell from there?" NO. Those graphs are
sketches on "real" axes (i.e. non algebraically closed field). Those graphs could be
missing some "imaginary"/"complex" singular points, so we do have to
go through the algebra (although we
do know by looking at the graphs
that at least
(0, 0) is a singularity for each). See part (c) as an example.
Part a
Here, f = x^{4} + y^{4}  x^{2}. The Jacobian matrix is
J  =
  
Hence, we need
4x^{3}  2x  = 0  

4y^{3}  = 0   
The second equation yields
y = 0. And the first equation yields
x = 0 or
x = ±
. However,
(0,
) isn't a point on the curve, so the only singular
point is
(0, 0)
Part b
Here, f = x^{6} + y^{6}  xy. The Jacobian matrix is
J  =
  
Hence, we need
6x^{5}  y  = 0  

6y^{5}  x  = 0   
Plugging the second into the first yields
6(6y^{5})^{5}  y  = 0  

6^{6}y^{25}  y  = 0  

y(6^{6}y^{24}  1)  = 0  

  
So
y = 0 or
y =
Similarly,
x = 0 or
x =
Plugging back each into the curve equation, clearly only
(0, 0) satisfies it
Part c
Here, f = x^{4} + y^{4} + y^{2}  x^{3}. So we need
4x^{3}  3x^{2}  = 0  

4y^{3} + 2y  = 0   
Which factor into
x^{2}(4x  3)  = 0  

2y(y^{2} + 1)  = 0  

  
By "i" I mean
, which is defined because we're in an algebraically closed
field.
Plugging in all the possibilities back into the curve equation, (0, 0) obviously
works and (
, 0) obviously doesn't. Note that
(i)^{4} + (i)^{2}  = i^{4} + i^{2}  

 = (i^{2})^{2} + i^{2}  

 = (1)^{2} + (1)  

 = 1  1  

 = 0   
Hence,
(0,i) and
(0,i) work and
(
,±i) doesn't. So the answers are
(0, 0), (0,±i). Note that
(0,±i) are points that aren't visible on the "real"
graphs on Figure 4.
I cannot fucking figure out how to do part d. The system of equations is
x^{4} + y^{4}  x^{2}y  xy^{2}  = 0 
4x^{3}  2xy  y^{2}  = 0 
  

4y^{3}  2xy  x^{2}  = 0    

    
Reader: If you can solve this system of equations nicely, let me know. Otherwise,
I'm skipping it.
(BTW: Some
3s and
6s showed up in this post. I was assuming that the
characteristic of
k wasn't
3, but if you look back you can see that the
case of char
k = 3 actually gets us to the answers even more quickly)