I.1.9
12/20/20
A frightening exercise if you’re clueless, but if you follow the thread of the previous exercise, then you’ll get
there.
I’m going to write α = 𝔞 for this post since it looks nicer when sent through htlatex. Now, letting Y = Z(α) break
up Y into irreducible components

Like before, letting d = dimS_{1}, our goal is to show that d ≥ n  r
Suppose α = (f_{1},…,f_{r}).
Like in the last exercise, we know that J is a minimal prime containing α. Now let β = (f_{2},…f_{r}), so that
α = (f_{1}) + β. Now, unlike the last exercise, β isn’t prime, so let’s consider a minimal prime ideal P containing β.
Then, like in the last exercise,
dimA∕P  = ht(J∕P) + dim(A∕P)∕(J∕P)  
= ht(J∕P) + dimA∕J  
= ht(J∕P) + d  
d  = dimA∕P  ht(J∕P)  
LEMMA: π(f) is a nonzerodivisor:
Well, since P is prime, A∕P is an integral domain, so it has no nonzero zerodivisors. So if π(f) is nonzero, it’s not a
zerodivisor. And [f] = [0]f ∈ P(f) ⊂ PZ(f) ⊃ Z(P)H ⊃ Y contradicting the Y ⊈ H assumption
from the exercise. So we’re done.
LEMMA: π(f) is a nonunit:
[f][g] = [1][fg  1] = [0]fg  1 ∈ P
Now, err, I got worried here for a while. How does this give you a contradiction? Well, 1 is a special term in a ring,
so let’s isolate it:
fg  1 ∈ Pfg  1 = p (where p ∈ P), so fg p = 1 But this means that (f) + P = (1)Y ∩H = A^{n}H = A^{n}
a contradiction since H is a hypersurface. DUN.
∏ _{i=1}^{s}((f_{ 1}) + P_{i})  = (1)  
⋃ _{i=1}^{s}Z(f_{ 1}) ∩ Z(P_{i})  = ∅  Taking Z of both sides  
Z(f_{1}) ∩⋃ _{i=1}^{s}Z(P_{ i})  = ∅  
Z(f_{1}) ∩ Z(β)  = ∅  
Z(f_{1},f_{2},…,f_{r})  = ∅  
Y  = ∅  
 (1) 
Now what? Well. P is the a minimal prime ideal containing β = (f_{2},…,f_{r}), so we can just repeat the same process
that we did for J and α. I.e. Let δ = (f_{3},…,f_{r}) and let P′ be the smallest prime containing δ. Then we’ll get
 (2) 
So plugging it into the earlier equation,
 (3) 
Just keep repeating the process, until you’re left with
 (4) 
where Q is a minimal prime containing (f_{r}). It’s easy to show that ht(Q) ≤ 1 (Set P to be the minimal prime
containing (0), i.e. P = (0) and repeat the same fucky argument as above). So
dimA  = dimA∕Q + ht(Q)  
≤ dimA∕Q + 1  
dimA∕Q  ≥ dimA  1  
= n  1  
d  ≥ n  1  (r  1)  
= n  r  