← I.2.9a I.2.9b →

I.2.9a - Errata



Remember that C is irreducible

Hole in my heart? More like hole in my proof. Yes, reader, just remember.. You remember, right? C is irreducible.... C is irreducible.... You're being hypnotized.... Yess..... Now....... lick my dick..........

Talk about an asspull. Yes, I pulled that one out of the hole in my ass, persay. For whatever reason, since Z(β(T)) U was irreducible, I assumed that Z(β(T)) was irreducible. NOT necessarily true. Z(β(T)) is merely a closed set. There's not much reason to assume it's irreducible right off the bat.

Well, what do then? It's thanks to that asspull that I was able to say

--         ′
Y =  Z(β(T  ))

and perform a series of steps that got me to

I(Y-) ⊃ S (β (T′))

HOLD ON. The first equation is an equality, but second equation is an inclusion. There might be some wiggle room here. Check it out:

Y = clPn(Z(β(T)) U)
= Z(β(T)) (closure of closed set)
Okay, okay, steady... steady... Now, let me follow yesterday's work except paying attention to the inclusion.:
I(Y ) I(Z(β(T))) (I reverses inclusion)
= ∘ ------′
  S β(T ) (Nullstellensatz)

And there's (2) again.

Phew, my proof just barely hangs together. We are SAFE!
PIC One more thing:
I asked you to take for granted that
β (α (T)) ⊃ T

claiming that "it's in my notes" (lol), then proceeded to conclude.

Sβ (α (T))S ⊃  T

Let me justify that:
Grab some element f from T Sh, like let's say (working in P3)...

f(w,x,y,z) = wxy + w2x + w3 + xyz
=⇒α(f) = xy + x + 1 + xyz definition of α
=⇒β(α(f)) = w3( x-
w y-
w + x-
w + 1 + x-
w -y
w z-
w) definition of β
= wxy + w2x + w3 + xyz
= f
Hmph, interesting... β(α(f)) = f. Let's try an example where all terms have a w:
f = wxy + w2x + w3 + wyz
= ⇒α(f) = xy + x + 1 + yz
= ⇒β(α(f)) = xy + wx + w2 + xz
= f∕w
The pattern you'll see is that β(α(f)) = f∕(wr) for some int r 0. "Wait, doesn't this lay doubt on equation (3)??" Indeed it does, because it's PROBABLY WRONG. *Applause*. Ahh, thank you, thank you *Bow*. I'd like to thank my family, my friends, and all my loyal fans...

Okay, so we don't exactly recover f the way we want, but if β(α(T)) were an ideal, then we could say that
wrβ(α(f)) = wr f
= f
thereby recovering f. Which is why, DUN DUN DUN, (4) is actually CORRECT. So we are, you know...