I.2.2

2/2/2021

This post is mostly straight math. Feel free to skip this post if you only like squiggly math.

SOME NOTES TO MYSELF BEFORE BEGINNING:

I covered some of the cases of Z(α) = ∅ in the previous post. Note that, in (ii), the = S
case
DOESN'T break the homogenous Nullstellensatz, whereas the = S_{
+} COULD break the homogenous
Nullstellensatz.

Also, in the previous post, I supposed that the degree of 0 in S = k[x_{0},x_{1},…,x_{n}] ought to be 0. Oddly, it appears
that this isn't the case. It appears that 0 transcends degree. See condition (ii) in this exercise. The ideal
can be
equal to S_{+}, which contains NO zero degree elements. And yet, since
is an ideal, we must have 0 ∈
i.e.
0 ∈ S_{+}. So 0 is in the positive degree part of S. However. S_{0} itself is the same as k, and certainly 0 ∈ k. so
0 ∈ S_{0}. So it's in the zero degree part of S as well. SO, basically, you can't put a degree on 0. I'll keep that in
mind, I guess.

OKAY NOTES FIN. LET'S BEGIN.

(i) (ii):

Let f ∈ S be homogenous positive degree. Then vacuously ∀P ∈ Z(α) : f(P) = 0 (since (i) says Z(α) = ∅).
By the homgenous Nullstellensatz (2.1), f^{q} ∈ α for some q > 0 i.e. f ∈
.

Thus, we've established that S_{+} ⊂
.

CASE 1: 1 ∈ α

Then α = S.

CASE 2: 1α

Then certainly 1
(else 1^{q} = 1 ∈ α, a contradiction). But if 1
then c
for any nonzero constant
c ∈ k (otherwise we'd have c⋅ = 1 ∈ α, a contradiction). I.e. ∀c ∈ S_{0} : c
. So
must be contained in
S_{+}. I.e. ⊂ S_{
+}.

DONE.

(ii) (iii):

CASE 1: = S

Then 1 ∈
, which means that 1^{q} ∈ α for some q, but this just means that 1 ∈ alpha i.e. α = S. But
S ⊃ S_{d}. CASE 1 DONE.

CASE 2: = S_{
+}:

Suppose THIS were true:

| (1) |

(we're working in P^{n} by the way lol)

(also, shouldn't have used k cause that usually refers to the field but fuck it too late lol)

Then for d = k(n + 1), consider S_{d}.

Let cx_{0}^{k0}x_{1}^{k1}x_{n}^{kn} be an arbitrary term of S_{d}.

We have to have k_{0} + k_{1} + + k_{n} = d, which means that we have to have k_{i} >=k for at least one i. Assume
without loss of generality that i = 0. Then we can rewrite that arbitrary term by factoring out an
x_{0}^{k}:

x_{0}^{k}(cx_{0}^{k0-k}x_{1}^{k1}x_{n}^{kn})

But this guy is actually in α, since x_{0}^{k} ∈ α (by (1)) and x_{0}^{k} ⋅ f ∈ α for any f ∈ S (since α is an ideal). So we
basically showed that an arbitrary element of S_{0} is in α. I.e. we showed S_{0} ⊂ α.

To finish off case 2, we just need to show that (1) is true. Since = S
by assumption, we must have
x_{0},…,x_{n} ∈
. I.e. for each i = 0,…,n, we have x_{i}^{mi} ∈ α for some m_{i} > 0. But then certainly
x_{i}^{m0m1mimn} ∈ α. (e.g. (x_{0}^{m0})^{m1mn} ∈ α, etc.).

CASE 2 DONE.

(iii) (i):

We're given α ⊃ S_{d}. Taking Z(⋅) of both sides, we reverse the inclusion, yielding

| (2) |

But since x_{0}^{d},x_{1}^{d},…,x_{n}^{d} ∈ S_{d}, a point P would have to be all 0 in order to be 0 at all of these (since k is an
integral domain!) and this can't happen in P^{n}, so we actually have Z(S_{d}) = ∅. So by (2), Z(α) ⊂∅ i.e.
Z(α) = ∅.