I.7.1

9/16/2021

Welcome to the new section, reader! Spread your arms! Let's do math! First off,
here's a spam of definitions n stuff:

whew. Yea, I actually read most of that, including the proofs (..most of it.).
ANYWAY... part A wants us to calculate the degree of the d-uple embedding
from 2.12 (holy shit that exercise was a lot of work lol).

ALSO: Hartshorne is generous enough to give us the answer: d^{n}. So our job is to
prove it.

Here's an interesting thing tho: degree is NOT preserved over isomorphism. If you
look at Prop 7.6(c) above, you'll see that deg P^{n} = 1. But P^{n} is isomorphic to
its d-uple embedding in P^{N} (2.12c), YET, the degree of the embedding is
certainly not 1. Kinda weird to talk about stuff that isn't isomorphism
independent, tbh, but w.e. i guess well roll w it.

OK: so let's bring back some of the notation from 2.12. The d-uple
embedding of P^{n} into P^{N} is given by Z(α), where α = ker θ, and
θ : k[y_{0},…,y_{N}] → k[x_{0},…,x_{n}] sends the y_{i}s to d-degree monomials over
x_{0},…,x_{n}. Hence, to calculate the degree of Y = Z(α), we want to calculate the
Hilbert polynomial of S(Y ) = k[y_{0},…,y_{N}]∕α, and multiply the leading
coefficient by n! (the d-uple embedding of P^{n} is isomorphic to P^{n}, hence the
dimension is n)...... except we don't. Not quite. ysee, I struggled my balls off
trying to calculate the Hilbert polynomial of S(Y ) as given above.... and so I
looked at the solution. SORRY. Here's the trick: use the isomorphism theorem,
and realize that S(Y ) ≃ imθ.... Yea... it's WAY easier to analyze the image of θ
in k[x_{0},…,x_{n}], lol. Use the image, not the quotient. Got it. let's do that.

BTW: I'm inherting the grading of M = imθ via the isomorphism. So M_{1} is all
the d-degree elements in k[x_{0},…,x_{n}], M_{2} are all the 2d-degree elements, etc.
Then, obviously, M_{l} is generated by the l-degree monomials over the
indeterminates x_{0},…,x_{n}. So we need to COUNT (up to constant multiple). Get
in here combinatorics fellas. How many ld degree monomials can you make out of
n + 1 indeterminates. FOLKS: from x_{0},…,x_{n}, you need to pick ld of them.
Order doesn't matter, and you can reuse elements. So it must be.... a
COMBINATION WITH REPLACEMENT. I.e. the one I have to look up the
formula for. HENCE, we can write the Hilbert function (polynomial) as

ϕ(l) | = | ||

= | |||

= | |||

= | |||

As per the definition of degree, let's multiply by n!, yielding just the
numerator

(n + ld)(n + ld - 1) (ld + 1) |

Of course, we care about the leading coefficient of this term (seen as
a polynomial over l). And would you look at that? There's exactly n
terms here, so the leading term is d^{n}l^{n}, and the leading coefficient is d^{n}.
DONE. NOTE THAT n here is NOT the dimension of the ambient space of
the embedding. That is N. n is the dimension of the preimage of the
embedding. ← i know this is bad gibberish but it makes sense to me lol

Okay, same deal, except this time we care about the thing from 2.14:

θ : k[{z_{ij}] | → k[x_{0},…,x_{r},y_{0},…,y_{s}] | ||

z_{ij} |
x_{i}y_{j} |

OK: Same trick as last time (USE THE IMAGE, NOT THE QUOTIENT).
Let's look at M = imθ instead of the quotient. GOT IT? Alright. And this
time the grading is: M_{1} is generated by the 2-degree monomials equally
split between x, y. M_{2} is generated by the 4-degree monomials equally
split between x, y. Etc. COUNTING TIME AGAIN. To count all the
monomials (up to constant multiple) in M_{l}, we're combining l elements from
x_{0},…,x_{r} (with replacement), then combining l elements from y_{0},…,x_{s}
(with replacement). To pick the xs, we use the same formula from part A:

= |

Similarly, to pick the ys, it's

So, using the rule of product (BY THE WAY: IF YOU'VE EVER
STRUGGLED WITH COMBINATORICS, STUDY THE RULE OF
PRODUCT (AND THE RULE OF SUM). I SUCKED COMPLETE
BALLS AT COMBINATORICS FOR YEARS AND... WELL, I STILL
SUCK, BUT ONCE I UNDERSTOOD THAT EVERYTHING YOU
DO IN COMBINATORICS BASICALLY COMES FROM THESE
TWO RULES, I FINALLY STARTED TO GET IT) the total number of
ways to pick the l xs, then the l ys is

(since order doesn't matter, this already covers the case of picking them in the
reverse direction). Okay, phew, now Imma just go ahead and multiply by the
(r + s)! and unwrap this baby (yes i'm assuming the dimension is r + s F U)

(r + s)! | = (r + s)! ⋅ |

OK: So, remember, we only care about the leading term (polynomial over l).
The leading term in the numerator of the first fraction is l^{r}, and the leading term
in the numerator of the second fraction is l^{s}. Hence, in total, the leading term is

l^{r+s} |

And that first term is indeed just = , which matches our Queen's answer, and we are DUN.