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 \[ dim(\overline{Y}) = dim(Y) \]