Ahhhh, the smell of errata in the morning. 'Tis been a while since I've casually stated a false statement and not gotten away with it. Let's take a look.

Firstly, I meant to say "no y_i terms". There is no such thing as a "Y " term, since Y is a variety, not an indeterminate in a polynomial ring. Secondly, that would still be wrong. Let's try this again.

Let f ∈ I(Z_i). Define

(i.e. analogous to before, plug in y for the y_i coordinates and leave the x_i coordinates as indeterminates, giving us a polynomial in k[{x_i}]. We do this for every y ∈ Y ). Then

x	∈ Xi
⇐⇒x × Y	⊂ Xi × Y
⇐⇒∀y ∈ Y : x × y	∈ Zi
⇐⇒∀y ∈ Y,∀f ∈ I(Zi) : f(x × y)	= 0
⇐⇒∀f ∈ I(Zi),∀y ∈ Y : f(x × y)	= 0
⇐⇒∀f ∈ I(Zi),∀g ∈ Sf : g(x)	= 0
⇐⇒∀g ∈⋃ f∈I(Zi)Sf : g(x)	= 0
⇐⇒x	∈ Z(⋃ f∈I(Zi)Sf)

That last set is closed, so X_i is closed. So that's fixed.

Also, boy, I am slowing down in updates again. The ebb and flow of Hartshorned is mysterious, dependent more on your dear admin's whim than their actual busyness. A new week is upon us, let's see how we do.