# 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.

# $(R, \tau_f)$

• 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.