I.5.5

6/16/2021

Now, reader, shall I? that fossil sitting behind your cheekbone teases a hallway of
intrigue within. Indeed, it is beckoning me to secrete my secrets into its corridors.
Shall I run my tongue across that golden helix? Shall I nibble that lobule, dribble
my tongue against that sensitive antitragus, or leave the whole artifact soaked in
an indiscriminate slobber? Would it admit my silly trifles, my somber troubles?
Would it submit to greedily brandished incisors? Would it forgive a lie; "Don't
worry, I don't bite x3"? Would you be so kind as to lend it to me, today? I will
give it back, I swear. Pretend my lips are the lips of a seashell, and my
warm breath the sound of the ocean. For today I only have one little
amulet to donate. One secret. One confession.... I fuked your mom lolololo jk but
actually what I was gonna say is i looked at the solution for this exercise. *breathes sex noises into your ear*

I'm sorry, reader. Also, I'm skipping 5.4 because MY ABSTRACT ALGEBRA
KNOWLEDGE IS SO FUCKING DEFICIENT LOL.

So, anyway, you may recall that we had a definition for nonsingularity for affine
varieties:

In this exercise however, we're working with P^{2}, so here's a generalization of the
definition for more arbitrary varieties:

Now, this definition is, as you can see, confusing as fuck. HOWEVER: Note that
nonsingularity only depends on the local ring. So we can use open sets in P^{2}
instead of P^{2} itself. I.e. we can cover P^{2} with A^{2}s and just default to the more
convenient definition with the Jacobian. So this problem is back to checking for
nonzero Jacobians.

The solution says that

x^{d} + y^{d} + z^{d} | = 0 | (1) |

works in almost all cases. Let's take the open set z≠0 (analogous argument for x≠0,y≠0. Then our curve equation turns into

x^{d} + y^{d} + 1 | = 0 |

in A

The Jacobian is then:

J | = |

Now the issue is that if chark divides d, then both components are 0, which is bad! So those are the cases for which this doesn't work, and we'll handle those later. However, in the cases that chark does not divide d, note that we would need x = 0 and y = 0 to make the Jacobian 0. But note that this isn't a point on the original curve, so the Jacobian is always nonzero.

Now we handle the chark|d case. The solution gives

xy^{d-1} + yz^{d-1} + zx^{d-1} | = 0 |

(yea, I wouldn't have thought of this on my own). So taking z≠0, we get

xy^{d-1} + y + x^{d-1} | = 0 | (2) |

And the Jacobian is

J | = | ||

=
(since chark|d) | |||

So we need

y^{d-1} | = x^{d-2} | (3) |

xy^{d-2} | = 1 | (4) |

Multiplying both sides of the (4) by y and substuting (3) yields

xy^{d-1} | = y | (5) |

x^{d-1} | = y | (6) |

Now let's plug in (5) and (6) into the curve equation (2):

xy^{d-1} + y + x^{d-1} | = 0 | ||

y + y + y | = 0 | ||

3y | = 0 | ||

y | = 0 |

and thus (6) tells us that x = 0 as well. But this doesn't satisfy (4). Done.