← I.2.12d I.2.14 →



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 P2 to be k[r,s,t] and the coordinate ring of P5 to be k[y0,,y5])

By 2.12c, we know that Y is isomorphic to P2 via the 2-uple map ρ. Thus a dimension 1 curve Z Y is isomorphic to a dimension 1 curve D P2. Now here's the first fun trick: A dimension 1 curve in P2 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 y0,y1,,y5. 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 g0 θ-1(rf),g1 θ-1sf,g2 θ-1(tf). Then I showed that Z = Z(g0,g1,g2) 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(g0,g1,g2) 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(g0,g1,g2) 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 f2 instead (more generally, for a d-tuple embedding, you can use fd)... Yeah, I probably wouldn't have thought of that on my own. So we can just derive g from f2: g θ-1(f2) (clearly, a point ρ(P) satisfies g iff P satisfies f2 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 f2. 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 f2 = 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).