Hi. This post will be quick and dry (just like sex with your wife, hahaha). Yes, an extremely scarylooking exercise. But it's actually a very similar process to what I did in 2.12b . In fact, it's even
easier.
Letting θ denote the k[{z_{ij}}] → k[x_{0},…,x_{r},y_{0},…,y_{s}] map the hint gave us, we set α = ker θ and we wanna
show that imϕ = Z(α).
"One inclusion is easier than the other," said 2.12b, and the statement holds analogously here. Give me a point
P = (a_{0},…,a_{r}) × (b_{0},…,b_{s}) ∈ P^{r} × P^{s}, so that ϕ(P) = (…,a_{i}b_{j},…). Now give me an f(…,z_{ij},…) ∈ α.
f(…,z_{ij},…)  ∈ ker θ  

θ(f(…,z_{ij},…))  = 0  

f(…,θ(z_{ij}),…)  = 0  

f(…,x_{i}y_{j},…)  = 0  

f(…,a_{i},b_{j},…)  = 0  

f(ϕ(P))  = 0  

  
Since
f ∈ α was arbitrary,
ϕ(P) ∈ Z(α). And since
P was arbitrary, im
ϕ ⊂ Z(α) Now for the reverse inclusion. Given
C = (…,c_{ij},…) ∈ Z(α), can we find
P = (a_{0},…,a_{r}) × (b_{0},…,b_{s}) ∈ P^{r} ×P^{s}
such that
ϕ(P) = C I.e. we need to satisfy
(…,a_{i}b_{j},…) = (…,c_{ij},…).
I.e. we need to satisfy
a_{i}b_{j} = c_{ij} (up to a constant multiple independent of i,j).
So what do we do? Same thing in 2.12b: Line up the requirements, "force" values for our
a_{i},b_{j}s, verify that they
actually work.
First, I'll assume that
c_{00}≠0 without loss of generality. Now, let
a_{0} be.... anything but zero. Yes. You heard me
right. Let
a_{0} be literally
fucking anything you want it to be, reader, as long as it's nonzero and from
k.
Then check out these requirements:
a_{0}b_{0}  = c_{00}  a_{0}b_{0}  = c_{00}    

a_{0}b_{1}  = c_{01}  a_{1}b_{0}  = c_{10}    

 
   

a_{0}b_{j}  = c_{0j}  a_{i}b_{0}  = c_{i0}    

 
   

a_{0}b_{s}  = c_{0s}  a_{r}b_{0}  = c_{r0}    

    
As soon as
a_{0} is fixed, so is everything else:
b_{0}  = c_{00}∕a_{0}  a_{0}  = c_{00}∕b_{0}    

b_{1}  = c_{01}∕a_{0}  a_{1}  = c_{10}∕b_{0}    

 
   

b_{j}  = c_{0j}∕a_{0}  a_{i}  = c_{i0}∕b_{0}    

 
   

b_{s}  = c_{0s}∕a_{0}  a_{r}  = c_{r0}∕b_{0}    

    
So choosing
anything for
a_{0} forces all of
P. Now we just have to check that
ϕ(P) does indeed equal
C. I.e. we need
to check if
a_{i}b_{j} = c_{ij}. Well, based on our settings,
a_{i}b_{j}  =
 

 =
 

  
Now suppose, for the sake of contradiction,
did NOT equal
c_{ij}. Then that would mean
 ≠c_{ij}  

c_{i0}c_{0j}  ≠c_{ij}c_{00}  

c_{i0}c_{0j}  c_{ij}c_{00}  ≠ 0  

  
I.e. we'd have that
C does NOT satisfy the polynomial
But, of course,
θ(g)  = x_{i}y_{0}x_{0}y_{j}  x_{i}y_{j}x_{0}y_{0}  

 = 0   
So
g ∈ ker θ i.e.
g ∈ α. By assumption,
C ∈ Z(α), yielding a contradiction.
Don't worry. The next exercise is another 3parter.