*Loads barrel*
*Cocks gun*
*Aims*
Please complete the security check to access–
BANG BANG BANG BANG BANG BANG
BANG BANG BANG BANG BANG BANG
BANG BANG BANG BANG BANG BANG
BANG BANG BANG BANG BANG BANG
BANG BANG BANG BANG BANG BANG
BANG BANG BANG BANG BANG BANG
BANG BANG BANG BANG BANG BANG
BANG BANG BANG Whew. Sorry, was just letting off some steam. Anyway, today's exercise asks us to
prove 3 separate things. Let's start with the first:
PART 1:
_{Y,X} is a local ring
I.e. we need to show that it has a
unique maximal ideal.
Now, if P ∈ X were a
point, then the maximal ideal of
_{P,X} are all the elements that vanish at P.
Hence, let's guess that the maximal ideal of
_{Y,X} are all the elements that
vanish on Y . Define:
M  = {< U,f >∈
_{Y,X}f vanishes on Y }   
One thing I have to check is that this set is well defined. I.e., if < U,f >=< V,g >,
then if f vanishes on Y , does g also vanish on Y ? The answer is yes, thanks to
remark 3.1.1 (f = g = 0 on U ∩V ∩Y which is an open set of Y ). Also, M is
clearly an ideal.
So is it maximal? If J ⊋ M, is an ideal, then it would have to contain some
< U,f > where f does not vanish on Y . So f(Q)≠0 for some Q ∈ Y . Our
goal is to invert f, so we can produce 1. Let V be an open set of X containing Q
such that f = g∕h for polynomials g,h. Then Z(g) ∩ X is closed in X, so
W = X  (Z(g) ∩ X) is a nonempty open set in X (it's nonempty
because Q ∈ W!). V ∩ W is a nonempty open set that intersects Y
(since it contains Q), and note that f = g∕h on that set, with g never
vanishing on it. Hence we can define < V ∩ W,h∕g > as en element
of
_{Y,X}. Multiplying this by < U,f > produces a 1 in the ideal so
J = (1). Hence, every ideal is contained in M (unless its (1)). DONE.
PART 2: The residue field of
_{Y,X} is K(Y ).
Oh, btw, here's the definition of a function field:
(dw: this is, effectively, the first time I've seen it too)
So we want to show
_{Y,X}∕M ≃ K(Y ), so the typical way of doing this on this
blog is to make a morphism
ϕ :
_{Y,X}  → K(Y )   
whose kernel is M. And the "natural" choice of a map would be
< U,f > 
< U ∩ Y,f >   
So suppose < U,f >∈ M. Then f vanishes on Y , so
< U,f > 
< U ∩ Y,f >  

 =< U ∩ Y, 0 >   
hence M ⊂ ker ϕ
For the reverse inclusion,
< U,f >  ∈ ker ϕ    

ϕ(< U,f >)  =< W, 0 >    

< U ∩ Y,f >  =< W, 0 >    

f  = 0 on U ∩ W ∩ Y    

f  = 0 on Y      
Hence ker ϕ ⊂ M. PART 2 DONE.
PART 3: dim
_{Y,X} = dim X  dim Y
I have no idea how to do this lol. I am looking up a solution, brb.
Okay, I got it (...still took me a while since the solution skipped most of the
details). You have to do the "reduce to affine" thing we did last time, plus
theorem 1.8A.
Let's start by showing that reducing everything to affines works. I.e. assume that
LEMMA 1: If X is an affine variety and Y is a subvariety, then
dim
_{Y,X} = dim X  dim Y
I'll prove this at the end, but let's assume we have it for now. Like last time, we
need to show that everything works for quasiaffines, projective varieties,
quasiprojectives, etc. First, this lemma will help us:
LEMMA 2: If X is any variety and Y is a subvariety and U is an open subset of
X, then dim
_{Y ∩U,X∩U} = dim
_{Y,X}
The proof is obvious.
Now, let's go through them:
LEMMA 3: If X is a quasiaffine variety and Y is a subvariety, then
dim
_{Y,X} = dim X  dim Y
PROOF:
X is an open subset of some affine variety Z (so X = Z), so
dim
_{Y,X}  = dim
_{Y ∩X,Z∩X}    

 = dim
_{Y,Z}  (LEMMA 2)    

 = dim Z  dim Y  (LEMMA 1)    

 = dim X  dim Y    

 = dim X  dim Y      
Done with LEMMA 3. Now,
LEMMA 4: If X is an projective variety and Y is a subvariety, then
dim
_{Y,X} = dim X  dim Y
PROOF:
Using the usual isomorphism, we have an open set U such that U ∩ X is an
affine variety. Hence
dim
_{Y,X}  = dim
_{Y ∩U,X∩U}    

 = dim Y ∩ U  dim X ∩ U  (LEMMA 1)    

 = dim Y  dim X      
LEMMA 5: If X is an quasiprojective variety and Y is a subvariety, then
dim
_{Y,X} = dim X  dim Y
PROOF: Same idea as LEMMA 3.
Okay, so basically if we show LEMMA 1, we're done (moral of the story: reducing
to affine helps a lot, especially for dimension arguments)
They define α = {f ∈
(X)f vanishes on Y }, and (since this is apparently
prime), they apply 1.8A:
htα + dim
(X)∕α  = dim
(X)  (1)

htα  = dim
(X)  dim
(X)∕α  (2) 
Now, since we're working with affines, we know that
dim
(X)  = dim A(X)     

 = dim X      
So we can rewrite (2) as
htα  = dim X  dim
(X)∕α  (3) 
And here's something else: dim
(X)∕α ≃ dim Y .... Errrrm... EXERCISE
LEFT TO READER (hint, if you think about it, the left hand side is just the
coordinate ring of Y .... I think).
So now, we can rewrite (3) as
Yeah, you can guess where this is going. Since
_{Y,X} is a local ring,
dim
_{Y,X} = htM. So if we show that htM = htα, we're done.
In fact, it is very easy to produce a bijection
and that finishes off LEMMA 1, and thus this exercise.
I'M BAD @ MATH.