I.4.3b

5/29/2021

This exercise is... weirdly put. First, look at the definition of a rational map:

Hmmmm, in the case that the image is k = A^{1}, doesn't this just conicide with
the definition of a function field? Indeed. And didn't we already handle
that case in part (a)? Indeed. The only difference is that we now embed
A^{1} into P^{1}. But that shouldn't change anything about where the map
f = x_{1}∕x_{0} is defined, as f would still only be defined where x_{0}≠0, right,
so isn't the answer the exact same as part (a)? Fucking nope, lol.

You see, you might get starting a bit suspicious when you suspect that
the only thing we did is rename stuff, and the answer otherwise matches
part (a), so you go ahead and look up the solution, and you see this:

ψ : P^{2} -{(0, 0, 1)} | → P^{1} | ||

(x_{0},x_{1},x_{2}) | → (x_{0},x_{1}) |

B-b-b-b-b-b-b-b-b-b-b-but how is that possible? T-t-t-t-t-t-t-the map f, which is what this part (B) is all about is given as f = x

You have to interpret this question in a screwy way to get this to work. Recall from part (a) how I discovered the map for U

(x_{0},x_{1},x_{2}) |
x_{1}∕x_{0} |

and then we are embedding this is P

x_{1}∕x_{0} |
(1,x_{1}∕x_{0}) |

So on U

(x_{0},x_{1},x_{2}) |
(1,x_{1}∕x_{0}) |

And now, since we're in projective coordinates, we can rewrite this as

(x_{0},x_{1},x_{2}) |
(x_{0},x_{1}) |

And taking this definition in and of itself, this is, indeed, the map brought up in the solutions, and one can easily see that it's indeed defined everywhere except (0, 0, 1).

So we're restricting to an open set, dividing by x

One of the points of this exercise is to loosen up. This section in general is I think an opportunity to not be overprecise with definitions and stuff, because they're going to be very cumbersome for some of the stuff that goes on in this section. Neither f nor ϕ are actual set functions, and yet we give formulas for them. This is some abuse of notation that I wasn't expecting, but now I "get" it. Relax your muscles, lay back, stretch your slipperless feet, let your boner protrude from your sweatpants.