Random Note 1

Commutative Algebra

Let $A$ be a commutative ring with identity. If $I \subset A$ is an ideal, then

$$dim(A/I) + ht(I) \leq dim(A)$$

In case any of $dim(A/I), ht(I), dim(A)$ is maximal, the inequality trivially holds. So we may assume that they are finite. Let $dim(A/I) = k$, then there exists a chain of distinct prime ideals 

$$\mathfrak{p}_0 < \cdots < \mathfrak{p}_k$$  

where every $\mathfrak{p}_i$ contains $I$. This follows from the fact that every prime ideals in $A/I$ has a order-preserving bijection with prime ideals in $A$ containing $I$. 

Now let $ht(I) = n$, then by definition, we have

$$ht(I) = \inf \{ ht(\mathfrak{p}) ~|~ \mathfrak{p} \supset I \} = ht(\mathfrak{p}')$$

for some $\mathfrak{p}'$. In particular, for any $\mathfrak{p} \supset I, n \leq ht(\mathfrak{p})$. This implies that there exists a chain of distinct primes

$$\mathfrak{q}_0 < \cdots < \mathfrak{q}_n = \mathfrak{p}_0$$

Then 

$$\mathfrak{q}_0 < \cdots < \mathfrak{q}_n = \mathfrak{p}_0 < \cdots < \mathfrak{p}_k$$

has $n+k+1$ many distinct prime ideals. This shows that $dim(A) \geq n+k$. This completes the proof.

Topology

Let $Y \subset X$ be a subspace. $A \subset Y$ and let $B$ be a closure of $A$ in $Y$, then

$$B = \overline{A} \cap Y$$

Clearly, $\overline{A} \cap Y$ is closed in $Y$. $A \subset \overline{A} \cap Y$. By definition of closure, $B \subset \overline{A} \cap Y$. 

Conversely, $B$ closed in $Y$ implies that $B = C \cap Y$ for some closed set $C$ in $X$. In particular, $A \subset B = C \cap Y \subset C$. As $C$ is closed in $X$, it follows that $\overline{A} \subset C \Rightarrow \overline{A} \cap Y = C \cap Y = B$.

In particular, if $A$ is closed in $Y$, we obtain 

$$A = \overline{A} \cap Y$$

Algebraic Geometry (Hartshorne Proposition 1.10)

Let $Y$ be quasi-affine. We wish to prove that $dim~Y =  dim~\overline{Y}$. Consider a chain of distinct closed irreducible subsets of Y

$$Y_1 < \cdots < Y_n$$

We claim that

$$\overline{Y_1} < \cdots < \overline{Y_n}$$

is a chain of distinct cllosed irreducible subsets of $Y_i$.

Closure of an irreducible subset is irreducible.

$$A \subset Y \subset \overline{Y} \Rightarrow \overline{A} \subset \overline{Y}$$

Here we have that $\overline{A}$ is closed in $X$, hence $\overline{A} = \overline{A} \cap \overline{Y} \subset \overline{Y}$ shows that $\overline{A}$ is closed in $\overline{Y}$ 

In particular, all $\overline{Y_i}$ are closed in $\overline{Y}$

If $\overline{Y_i} = \overline{Y_j}$ for some $i \neq j$, then taking the intersection with $Y$, we obtain $Y_i = Y_j$ which is a contradiction.

This shows that $dim~Y \leq dim~\overline{Y}$

$$dim(\overline{Y}) = dim(k[x_1, \ldots, x_m]/I(\overline{Y})) \leq dim(k[x_1, \ldots, x_m]) =m$$

so we have that $dim(Y)$ is finite. Say $n$. Given a chain of maximal length,

$$Y_0 < \cdots < Y_n$$

$Y_0$ has to be a point. This follows from the fact that $\mathbb{A}^n$ is $T_1$, so single points are closed irreducible subset. We have that 

$$\overline{Y_0} < \cdots < \overline{Y_n}$$

is also a maximal chain starting with $\overline{Y_0}$. If not, then there exists $W$ closed in $\overline{Y}$ among the chain. Take the intersection of $Y$, then we have $Y_i$ with extra $W$ as $\overline{Y_i} \cap Y = Y_i$ as shown above. 

$Y$ is quasi-affine, meaning that $Y$ is open in some closed set $V$, i.e. $Y = U \cap V$ for some open set $U$ in $X$. We have that $W \subset \overline{Y} \subset V \Rightarrow Y \cap W = (U \cap V) \cap W = U \cap W$, hence open in $W$. Hence dense in $W$ and irreducible. $W$ is closed in $\overline{Y}$, hence taking the intersection, it is $W \cap Y$ is closed in $Y$.  There are two cases,

1) $\overline{Y_k} \subset W$

2) $W \subset \overline{Y_k}$

for some $k$.

In either case, by the maximality, we get $Y_k = W \cap Y$. By this implies that $\overline{Y_k} = W$ by the denseness of $W \cap Y$. $\overline{Y_0}$ corresponds to a maximal ideal $\mathfrak{m}$, and by the maximality we just proved, $ht(\mathfrak{m}) = n$. 

Because a ideal corresponding to a point is $(x_1 - a_1, \ldots, x_m - a_m)$, we have that $A(\overline{Y})/\mathfrak{m} \cong k$. We have the theorem that

$$ht(\mathfrak{p}) + dim(B/\mathfrak{p}) = dim(B)$$

for $B$ integral domain that is finitely generated $k$-algebra with $k$ a field and $\mathfrak{p}$ a prime ideal. Substituting $\mathfrak{m}$ for $\mathfrak{p}$ and $A(\overline{Y})$ for $B$, we obtain that

$$n + 0 = dim(A(\overline{Y}))$$

This proves that $$dim(\overline{Y}) = dim(Y)$$