Download Lecture 20 1 Point Set Topology

Lecture 20
October, 23
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Point Set Topology
Let X be a topological space, then Z is locally closed if it is an open set in the
induced topology on Z. In other words, Z = U ∩ C where U is open in X and
C ≡ Z is closed in X.
Assume that X has dcc on closed sets (Noetherian on ideals) then a subset of
X, Z is constructible if it is a finite union of locally closed sets.
Proposition 1.
• Z contains an open subset of Z.
• Z is a finite union of disjoint locally closed sets.
Proposition 2. The set of constructible sets of X is closed under operations
finite union ∪, finite intersection ∩, and complement X − Z.
Proof of Proposition 1. Let Z be locally closed, the Z = S
C1 ∪ · · · ∪ Cn is a
union of a finite number of irreducible sets. It follows that Z = (Z ∩ Ci ), where
Z ∩ Ci is an open subset in Ci . So Z is a union of disjoint locally closed sets.
Assume that Z = Z1 ∪ · · · ∪ Zn is the union of locally closed sets, then Zi = Ci
are irreducible. If C1 ⊂ C2 ∪ · · · ∪ Cr , then D = C2 ∪ · · · ∪ Cr is a closed set
and U = Z − (Z ∩ D) is an open set in Z, as desired. So we may assume that
C1 6⊂ C2 ∪ · · · ∪ Cr .
Let Y1 = C1 −(D∩C1 ) be an open set of C1 . Then A1 = C1 −Y1 and B1 = C1 −Z1
are closed sets in C1 . But since C1 is closed in Z, it follows that A1 and B1 are
closed in Z. Then A1 ∪ B1 is a proper closed set in C1 , otherwise it would not be
irreducible. Hence Z − (A1 ∪ B1 ) is an open non-empty set of Z and we are done.
Definition. A quasi-projective variety X is a locally closed subset of Pn . Examples of them are: projective variety, affine variety. If we consider A2 − {0}, then
it is a quasiprojective variety that is neither affine nor projective.
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Main Theorem
Theorem. Let φ : A → B be a homomorphism of finitely generated algebras and
let f : Y → X be a morphism of their varieties. Then the image of f (Y ) is a
constructible set of X.
Proof. We will use Noetherian (acc) induction on B or dcc on varieties of Y ,
this will be the same as induction on dimension. We may assume that Y is an
irreducible affine variety. It follows from the fact that a finite union of constructible
sets is a constructible set and every variety is a union of finite number of irreducible
varieties. Thus we may assume that B is a domain. The base of induction is clear,
the image of an irreducible set of dimension zero is the image of a point, which must
be a point in X.
Let A = A/ker(φ), then the corresponding variety C is a closed set of X, and Y
via f factors through C. If f (Y ) is constructible in C then it is constructible in X.
So we can replace A by A, that is start with φ : A → B being injective.
Suppose s ∈ B, s 6= 0. Then the variety of Bs = B[s−1 ] = Y − V (s), since
we localized B. Note that the variety of B = B/sB is V (s). Thus Y can be
written as the union of varieties of Bs and B/sB. We would like to show that f (Y )
is constructible. But the image of V (s) is constructible by Noetherian induction.
Thus we are allowed to localize B at s ∈ B in order to simplify the problem.
If s ∈ A, s 6= 0, consider a map φs : As → Bs , then the corresponding variety
Xs is an open set because Xs = X − (set of zeros ofs in X) and the set of zeros of s
in X is a closed set. So if the image is constructible in Xs , then it is constructible
in X. This follows from the observation that if a set is open in Xs , then it is open
in X. If a set is constructible in Xs , say Z is an intersection of an open set and a
closed set Us ∩ Cs . Then Us and Xs − Cs are open in X and Z is constructible in
X, since Z = Us ∩ (X − (Xs − Cs )) that is an intersection of an open and a closed
set. Thus we are allowed to localize A in order to simplify the problem.
So far these were preliminary observations. Our plan is to choose s ∈ A appropriately and localize such that Ys → Xs will be surjective.
Let K be a fraction field of A. We extend the map φ : A → B between domains
to a map from K → BK , where BK is a fraction ring consisting of
{u−1 b | b ∈ B, u ∈ A, u 6= 0}.
We simplified notation s−1 φ(b) and for the rest of the proof we think of the image
of A as an inclusion of A in B.
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Note that B is a finite type algebra, so BK will be also a finite type algebra.
Using Noether Normalization Theorem: there exist independent y1 , . . . , yd ∈ BK
such that K[y1 , . . . , yd ] ⊂ BK and B
K is a finite module over K[Y ]. So any element
P
in BK can written as a finite sum
Pi (Y )vi , where Pi (Y ) ∈ K[Y ] and {vi ∈ BK }
is the generating set.
Since BK is a finite module over K[Y ], it follows that BK is integral over K[Y ].
Let zi be one of generators of B, then there exists a monic polynomial with coefficients in K[Y ],
X n + P1 (Y )X n−1 + · · · + Pn (Y ) = 0
such that zi is the solution of it. There is a finite number of polynomials Pi (Y ) and
all they have finite number of coefficients in K. We localize at all u ∈ A that appear
in all denominators of all coefficients. This yields that zi will be integral over A[Y ]
after localizing A.
Applying same procedure for all of generators of B, we get that B after localization is integral over A[Y ] (A is also considered after localization). Hence B is a
finitely generated module over A[Y ], so f is a surjective map. Since f is surjective,
Y entirely maps to X and we conclude the proof.
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