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: dⁿ. 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 Pⁿ = 1. But Pⁿ 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 Pⁿ into P^N is given by Z(α), where α = ker θ, and θ : k[y₀,…,y_N] → k[x₀,…,x_n] sends the y_is to d-degree monomials over x₀,…,x_n. Hence, to calculate the degree of Y = Z(α), we want to calculate the Hilbert polynomial of S(Y ) = k[y₀,…,y_N]∕α, and multiply the leading coefficient by n! (the d-uple embedding of Pⁿ is isomorphic to Pⁿ, 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[x₀,…,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 M₁ is all the d-degree elements in k[x₀,…,x_n], M₂ are all the 2d-degree elements, etc. Then, obviously, M_l is generated by the l-degree monomials over the indeterminates x₀,…,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 x₀,…,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 dⁿlⁿ, and the leading coefficient is dⁿ. 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

B

 


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

θ : k[{zij]	→ k[x0,…,xr,y0,…,ys]
zij	xiyj

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: M₁ is generated by the 2-degree monomials equally split between x, y. M₂ 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 x₀,…,x_r (with replacement), then combining l elements from y₀,…,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

lr+s

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