Library of Mathexandria

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...

This is in Lemma 3.1 of Hartshorne's Algebraic Geometry. "An open subset of an open subset is open" is the trick. Proposition. Let \( Y \) be a topological space. \( Z \subset Y \) is an closed set if and only if there exists an open cover \( \{ U_\alpha \} \) such that \( Z \cap U_\alpha \) is closed in \( U_\alpha \) for all \( \alpha \). Proof)  \( (\Rightarrow) \): Suppose \( Z \) is closed. \( Y \) is an open cover of \( Y \) and \( Z \cap Y \) is closed in \( Y \). \( (\Leftarrow) \): If \( Z \cap U_\alpha \) is closed in \( U_\alpha \), then \( Z^c \cap U_\alpha \) is open in \( U_\alpha \). Hence \( Z^c \cap U_\alpha = Z_\alpha \cap U_\alpha \) for some \( Z_\alpha \) open in \( Y \). In particular, \( Z^c \cap U_\alpha \) is open in \( Y \). \[ Z^c = Z^c \cap Y  = Z^c \cap \left ( \bigcup U_\alpha \right ) = \bigcup ( Z^c \cap U_\alpha) \] hence \( Z^c \) is open. Hence \( Z \) is closed.