I.3.15ac
5/11/2021
Apologies for the broken links in the last post. I have fixed them. Now I just have
to fix my broken heart </3.
"What happened to 3.14?" It broke my heart. Well, every exercise breaks my
heart. The holeinmyheart just gets bigger and bigger. Soon, it will be wide enough
for some very fancy ero guro.
Listen: 3.14 has to do with two of my worst topics on this blog. Part (a) has to do
with linear algebra. And part (b)? Here's something funny. After a day's worth of
struggling my butt off in the "easy" world of linear algebra, I decided that it
might be time to give up, and then I realized that I actually had not read part (b)
yet. I took a glance at it, and what do I see? I'll just give you the abbreviation:
TCC
If you ain't doing nothing, at least walk.
Be kind. Treat others how you would like to be treated. Just don't be an asshole. It's simple as that.
Follow your heart. Don't give a fuck about what anyone tells you. Do what you think is right, and everything will fall into place.
If you're ever in the woods, and you see someone that calls themselves holeinmyheart, run. Please, just fucking run. They'll act all nice and try to convince you that they're you're friend, but they're NOT. If you encounter this person, you need to fucking get away ASAP. Don't give into any of their advances and don't believe a word that comes out of their mouth. Don't. Let. Them. Get. Close. Make as much distance as you can, then run. Run for your fucking life. This person is no laughing matter. Even if–

and define

First job is to show that X = X_{1} ∪ X_{2}. Now clearly X ⊃ X_{1} ∪ X_{2}, so we
want to show that given x ∈ X, x is either in X_{1} or X_{2}.
Yep. I got stuck here for hours. "Little early to get stuck, eh?" As early as my
posting schedule is late, eh? Hahahahaha, *clings teacups*. Here's the thing: If x
is neither in X_{1} nor X_{2}, then that means that x×Y is contained neither in Z_{1}
nor Z_{2}. I.e.: this is the light bulb, x × Y can be split up. Can we use this
"splitting up" to split up the irreducible Y and earn a contradiction?
Let's say that x = (p_{1},…,p_{n})
Let's also say that x = (p_{1},…,p_{n}) and let's write the coordinate ring of A^{n+m}
as k[x_{1},…,x_{n},y_{1},…,y_{m}], and
Note two things:
 (1) 
(here Z is taken in A^{n+m})
 (2) 
(here I is taken in A(A^{m}) and Z is taken in A^{n+m})
Hence, intersecting (1) and (2) shows that x × Y is closed. So...
x × Y  = (x × Y ) ∩ (X × Y )  
= (x × Y ) ∩ (Z_{1} ∪ Z_{2})  
= ((x × Y ) ∩ Z_{1}) ∪ ((x × Y ) ∩ Z_{2})  
= C_{1} ∪ C_{2} 

I.e. I plug in x into f to get a polynomial in k[y_{1},…,y_{m}] (I am transferring a
polynomial function on x×A^{m} to a polynomial function on A^{m}), where clearly

Soo, let me set

And let D_{i} = Z_{m}(E_{i}) (writing Z_{m}(⋅) to denote that I'm taking the zero set in
A^{m} here). And note:
x × y  ∈ C_{i}  
⇐⇒f(x × y)  = 0(∀f ∈ I(C_{i}))  
⇐⇒f′(y)  = 0(∀f′∈ E_{i})  
⇐⇒y  ∈ Z_{m}(E_{i})  
⇐⇒y  ∈ D_{i} 
x × Y  = (x × D_{1}) ∪ (x × D_{2})  
= x × (D_{1} ∪ D_{2})  
Y  = D_{1} ∪ D_{2} 

Z_{i} is clearly generated by polynomials that have no Y terms, and these
polynomials determine X_{i}.
PART (a) done.
NOW, you might be wondering why I am skipping part (b)? Well, what's wrong
with part (b)?
Hmmm? What's that symbol over there? That's a fancy looking little symbol, eh?
Oooooh, very cute. It's like a multiplication inside a circle. And a little k nestled
up on the side. Very interesting. Would you like to know what my reaction to
seeing this symbol is? Here it is:
No.