I.2.9a - Errata

2/11/2021

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

| (1) |

and perform a series of steps that got me to

| (2) |

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 | = clP^{n}(Z(β(T′)) ∩ U) | |||||

⊂ clP^{n}(Z(β(T′))) | ||||||

= Z(β(T′)) | (closure of closed set) |

I(Y ) | ⊃ I(Z(β(T′))) | (I reverses inclusion) | ||||

= | (Nullstellensatz) | |||||

⊃ Sβ(T′) |

And there's (2) again.

Phew, my proof just barely hangs together. We are SAFE!

One more thing:

I asked you to take for granted that

| (3) |

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

| (4) |

Let me justify that:

Grab some element f from T ⊂ S^{h}, like let's say (working in P^{3})...

f(w,x,y,z) | = wxy + w^{2}x + w^{3} + xyz | |||||

α(f) | = xy + x + 1 + xyz | definition of α | ||||

β(α(f)) | = w^{3}(
+
+ 1 +
) | definition of β | ||||

= wxy + w^{2}x + w^{3} + xyz | ||||||

= f |

f | = wxy + w^{2}x + w^{3} + wyz | ||

α(f) | = xy + x + 1 + yz | ||

β(α(f)) | = xy + wx + w^{2} + xz | ||

= f∕w |

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

w^{r}β(α(f)) | = w^{r}
| ||

= f |