Affine Varieties: Notes&Proofs

Clearly if \mathfrak{a} is the ideal of A generated by T, then Z(T) = Z(\mathfrak{a}).

P2

Since T \subset \mathfrak{a}, Z(\mathfrak{a}) \subset Z(T). On the other side, T generates \mathfrak{a}, so Z(T) \subset \mathfrak{a}. Hence the assertion.

Any nonempty open subset of an irreducible space is irreducible and dense.

Example 1.1.3

Say X is an irreducible space and Y its nonempty open subset.

  • Dense: If not, X =  \overline{Y} \cup Y^C, hence reducible.
  • Irreducible: If not, $Y = A \cup B$, where A, B are nonempty subset of Y. Take closure in X, then we have Y = \overline{A} \cup \overline{B}, hence Y = \overline{A} say. So the closure of A in Y equals to \overline{A}\cap Y = A, thus A = Y, contradiction.

If Y is an irreducible subset of X, then its closure \overline{Y} in X is also irreducible.

Example 1.1.4

If not, \overline{Y} = A \cup B. Then (A\cap Y)\cup (B\cap Y) = (A\cup B)\cap Y = Y, which forces Y = A\cap Y. Now $Y\subset A\subset \overline{Y}$, by definition $A = \overline{Y}$, contradiction.

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Basic Properties of  Cofinite Space and Cocountable Space

 (R,  \tau_f)

  • T1: For any 2 distinct elements x, y, there is a neighborhood of x which doesn’t contain y, and a neighborhood of y which doesn’t contain x.

Since {x} is finite hence close, (R,  \tau_f) satisfies T1.

  • T2~T4:

Since (R,  \tau_f) satisfies T1, if it also saitisfies T4, then it must satisfies T3, hence T2.

Therefore, we only need to show that (R,  \tau_f) does not satisfies T2. Say x, y are 2 distinct points, and U is a neighborhood of x, then U^c is a close set contaning y. Now any open neighborhood of y is cannot be finite, so non of them is contained in U^c. That is, (R,  \tau_f) does not saitisfies T2, hence T3 or T4.

  • C1:

If (R,  \tau_f) satisfies C1, then for any element x in (R,  \tau_f), there is a countable neighborhood base of it, say U = {U_i}. \cup (U_i)^c is also countable. R is uncountable, so there exists one element y, y\neq x, y\not\in \cup (U_i)^c. That is, y is contained in any U_i in U.

Now we consider R\{y}.  It is a open neighborhood of x, and does not contain any item of U. So (R,  \tau_f) does not satisfy C1.

  • C2: Any C2 space is C1 space.

So (R,  \tau_f) does not satisfy C2.

(R,  \tau_c)

The situation about (R,  \tau_c) is similarly.

  • T1:

Let x and y be 2 distinct points. {x} is countable, so it is close. So y\in {x}^c. That is,(R,  \tau_c) satisfies T1.

  • T2~T4:

We need only to show that (R,  \tau_c) does not satisfy T2. Say U is an arbitrary open neiborhood of x, then U^c is a neighborhood of y, and it’s countable. So any subset of U^c is countable, too. However, any open neighborhood of y must be uncountable. So (R,  \tau_c) does not satisfies T2.

  • C1 & C2

We need only to show that (R,  \tau_c) does not satisfies C1.

If (R,  \tau_c) satisfies C1, then for any element x in v, there is a countable neighborhood base of it, say U = {U_i}. \cup (U_i)^c is also countable. R is uncountable, so there exists one element y, y\neq x, y\not\in \cup (U_i)^c. That is, y is contained in any U_i in U.

Now we consider R\{y}.  It is a open neighborhood of x, and does not contain any item of U. So (R,  \tau_c) does not satisfy C1.

Other

Notice that the same proof is valid when R is substituted with any uncountable space.