I.2.13

3/1/2021

Guess who? Who's the sour loser? The sore fool? The 4-day math skipper? and by that I mean I skip through math,
like a maiden frolicking through a field. "More like mathlessly marching into March." OHHHHHHH. Listen, Shakespeare: The incident last time? We are putting that behind us. Not because
I have figured out the issue, but because I haven't. I took my break. I still periodically run through the proof in
my head and keep verifying my own absurdity like a left winger right winger being endowed with
conscoiusness. But I am effectively over it. Being over it doesn't mean you've erased it from your brain, it
means that it doesn't dominate your head anymore. The arms of the twisted cubic curve, pointing on
either end to that same infinity, have loosened from my neck. And here we are, at the next exercise.

....which wasn't easy. It took me hours + some online help, but at least I didn't prove something that shouldn't be
provable. From the mere mention of "hypersurface", we know that 2.8 is likely going to be helpful (at least for the
fact that "hypersurface" is defined there, lol). And of course, we're gonna be using 2.12 (a, b, and c. NOT d, lol).
Let's take a look:

(btw I'll denoting the coordinate ring of P^{2} to be k[r,s,t] and the coordinate ring of P^{5} to be k[y_{0},…,y_{5}])

By 2.12c, we know that Y is isomorphic to P^{2} via the 2-uple map ρ. Thus a dimension 1 curve Z ⊂ Y is
isomorphic to a dimension 1 curve D ⊂ P^{2}. Now here's the first fun trick: A dimension 1 curve in P^{2}
is a hypersurface. Thus, thanks to 2.8, I know that D = Z(f) for some irreducible f ∈ k[r,s,t].

Now let's refer to my work in 2.12c. We want to write Z = Z(g) ∩ Y for some irreducible g. I provided a
construction of something sort of close: I can write Z = Z(δ) ∩Y , for an ideal δ. What does this δ look like? Well,
I provided two cases. Let's start with the easier case:

CASE 1: deg f is a multiple of 2 (i.e. deg f is even)

Then I just let g be any substitution of the variables of f with y_{0},y_{1},…,y_{5}. Actually, I just realized that a good
way to say this is "Let g be any element in the preimage of θ" (again, refer back to 2.12 for the definition of θ). So
g ∈ θ^{-1}(f), and I claimed that Z = Z(g) ∩ Y .

Well, that's almost exactly what we wanted, except I have to show that g is irreducible. Is it? Well, suppose you can
reduce g = h ⋅ l where h,l are nonunits (positive degree polynomials). Then taking θ of both sides yields
f = θ(h) ⋅ θ(l).

You can see where we're going here. Since f is irreducible, either θ(h) or θ(l) has to be a unit. Suppose it's θ(h)
without loss of generality. Here's a little extra fact: Since g is homogenous, both h and l have to be
homogenous (EXERCISE LEFT TO THE READER... it's easy to see, I swear). Then remember my
LEMMA 1, from 2.12? h has to get mapped to a nonunit or to 0. Either way, that's a contradiction.

So that's case 1 done. Now let's look at case 2:

CASE 2: deg f is odd

What did I make δ in this case? Well, I did something clever (...plz give me some credit here, reader).
I said, let's "force" the degree to be even by considering the three polynomials rf,sf,tf and let
g_{0} ∈ θ^{-1}(rf),g_{1} ∈ θ^{-1}sf,g_{2} ∈ θ^{-1}(tf). Then I showed that Z = Z(g_{0},g_{1},g_{2}) ∩ Y . That worked fine for the
purposes of solving part (c), but there's clearly a problem here. For this exercise, I want my δ to be generated by a
single irreducible polynomial f, whereas here I have three polynomials, which I don't even necessarily know are
irreducible.

Am I screwed? The answer is yes. I played around with these 3 polynomials for hours, and convinced myself that
there's no guarantee that Z(g_{0},g_{1},g_{2}) is a hypersurface (I couldn't even show that it was a variety!!!). If I
intersect it with Y , I do get back Z. But I want to intersect a hypersurface with Y , not whatever this
Z(g_{0},g_{1},g_{2}) is.

How do I get a hypersurface? Well, here's where I looked online for help. YES. I FUCKING ADMIT IT. POINT
AND LAUGH, READER.

Instead of doing my """clever""" trick, there is a cleverer trick you can do. Instead of splitting f into rf,sf,tf, you
can use f^{2} instead (more generally, for a d-tuple embedding, you can use f^{d})... Yeah, I probably
wouldn't have thought of that on my own. So we can just derive g from f^{2}: g ∈ θ^{-1}(f^{2}) (clearly, a
point ρ(P) satisfies g iff P satisfies f^{2} iff P satisfies f, thanks to integral domain business). Hence
Z = Z(g) ∩ Y , which is almost exactly what we need, except, again, we have to verify that g is irreducible. What I saw online did not justify this at all, so we're gonna have to do it ourselves. Damn!

Now, here's where the "2" in 2-uple comes in handy. We derived g from f^{2}. That 2 in the exponent is going to be
very helpful. Suppose, again, that g = h ⋅ l for nonunits h,l. Then taking θ of both sides this time yields
f^{2} = θ(h)θ(l). From the same reasoning earlier(you know, the solid reasoning where I left the EXERCISE TO
THE READER), θ(h) and θ(l) have to each be nonunits or 0. The latter is a clear contradiciton, so we have to
assume they're nonunits. BUT: k[r,s,t] is a UFD, which means that we have to have θ(h) = f and θ(l) = f.
Both of these are contradictions. Why? They contradict the assumption that the degree of f is odd.

And that's that. Let's try to be more consistent this month (lol, jk, I just wanted try saying that to see how it
feels).