Foundations of Algebraic Geometry Class 27
نویسنده
چکیده
1.2. Proposition. — π : PnA → SpecA is a closed morphism. Proof. Suppose Z ↪→ PnA is a closed subset. We wish to show that π(Z) is closed. Suppose y / ∈ π(Z) is a closed point of SpecA. We’ll check that there is a distinguished open neighborhood D(f) of y in SpecA such that D(f) doesn’t meet π(Z). (If we could show this for all points of π(Z), we would be done. But I prefer to concentrate on closed points for now.) Suppose y corresponds to the maximal ideal m of A. We seek f ∈ A − m such that πf vanishes on Z. A picture helps here, but I haven’t put it in the notes.
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