Fixes

Signed-off-by: zeramorphic <[email protected]>

iii/commalg/04_integrality_finiteness_finite_generation.tex

@@ -598,7 +598,7 @@ \subsection{Cohen--Seidenberg theorems}
     Suppose that \( \mathfrak q \) and \( \mathfrak q' \) contract to the same prime ideal \( \mathfrak p = \mathfrak q \cap A = \mathfrak q' \cap A \) of \( A \), and that \( \mathfrak q \subseteq \mathfrak q' \).
     Then \( \mathfrak q = \mathfrak q' \).
 \end{proposition}
-We will write \( B_{\mathfrak p} \) for \( (A \setminus \mathfrak p)^{-1} B \), but this is not in general a ring.
+In this section, we will write \( B_{\mathfrak p} \) to abbreviate \( (A \setminus \mathfrak p)^{-1} B \).
 \begin{proof}
     Define \( S = A \setminus \mathfrak p \).
     Then \( \mathfrak q \) and \( \mathfrak q' \) are prime ideals of \( B \) not intersecting \( S \).

iii/commalg/07_dimension_theory.tex

@@ -174,31 +174,31 @@ \subsection{Dimension theory of local Noetherian rings}
 
     Let \( \mathfrak q \) be an \( \mathfrak m \)-primary ideal of \( A \), generated by \( x_1, \dots, x_s \) where \( s = \delta(\mathfrak q) \).
     Then
-    \[ G_{\mathfrak q}(A) = \faktor{A}{\mathfrak q} \oplus \faktor{\mathfrak q}{\mathfrak q^2} \oplus \oplus_{n \geq 2} \faktor{\mathfrak q^n}{\mathfrak q^{n+1}} \]
+    \[ G_{\mathfrak q}(A) = \faktor{A}{\mathfrak q} \oplus \faktor{\mathfrak q}{\mathfrak q^2} \oplus \bigoplus_{n \geq 2} \faktor{\mathfrak q^n}{\mathfrak q^{n+1}} \]
     The first factor \( \faktor{A}{\mathfrak q} \) is Artinian, and the images of \( x_1, \dots, x_s \) generate \( G_{\mathfrak q}(A) \) as an \( \faktor{A}{\mathfrak q} \)-algebra, where the \( x_i \) are of degree 1.
     Then \( \ell\qty(\faktor{\mathfrak q^n}{\mathfrak q^{n+1}}) < \infty \).
     From the theorem on Hilbert polynomials, \( \ell\qty(\faktor{\mathfrak q^n}{\mathfrak q^{n+1}}) \) is a polynomial in \( n \) of degree at most \( \delta(\mathfrak q) - 1 \), for sufficiently large \( n \).
 
     Fix some \( \mathfrak m \)-primary ideal \( \mathfrak q_0 \) such that \( \delta(\mathfrak q_0) = \delta(A) \).
     We consider two special cases: \( \mathfrak q = \mathfrak q_0 \) and \( \mathfrak q = \mathfrak m \).
-    For \( \mathfrak q \), we have
-    \[ \deg \ell\qty(\faktor{\mathfrak q_0^n}{\mathfrak q^0_{n+1}}) \leq \delta(A) - 1 \]
+    For \( \mathfrak q_0 \), we have
+    \[ \deg \ell\qty(\faktor{\mathfrak q_0^n}{\mathfrak q_0^{n+1}}) \leq \delta(A) - 1 \]
     As
     \[ \ell\qty(\faktor{A}{\mathfrak q_0^n}) = \sum_{i=0}^{n-1} \ell\qty(\faktor{\mathfrak q_0^i}{\mathfrak q_0^{i+1}}) \]
     we have
     \[ \deg \ell\qty(\faktor{A}{\mathfrak q_0^n}) \leq \delta(A) \]
     For \( \mathfrak m \),
     \[ \deg \ell\qty(\faktor{\mathfrak m^n}{\mathfrak m^{n+1}}) = d(G_{\mathfrak m}(A)) - 1 \]
     and hence
-    \[ \deg \ell\qty(\faktor{A}{\mathfrak m^n}) = d(G_{\mathfrak m})(A) \]
+    \[ \deg \ell\qty(\faktor{A}{\mathfrak m^n}) = d(G_{\mathfrak m}(A)) \]
 
     Now, there exists \( t \geq 1 \) such that \( \mathfrak m^t \subseteq \mathfrak q_0 \subseteq \mathfrak m \).
     Then
     \[ \ell\qty(\faktor{A}{\mathfrak m^n}) \leq \ell\qty(\faktor{A}{\mathfrak q_0^n}) \leq \ell\qty(\faktor{A}{\mathfrak m^{tn}}) \]
     But all of these terms are eventually polynomial, and the degrees of the left-hand and right-hand sides are the same, so we must have \( \ell\qty(\faktor{A}{\mathfrak q_0^n}) = \ell\qty(\faktor{A}{\mathfrak m^n}) \).
 
     \begin{proposition}
-        \( \delta(A) \geq d(G_{\mathfrak m})(A) \)
+        \( \delta(A) \geq d(G_{\mathfrak m}(A)) \).
     \end{proposition}
     \begin{proof}
         \[ \delta(A) = \delta(\mathfrak q_0) \geq \deg \ell\qty(\faktor{A}{\mathfrak q_0^n}) = \deg \ell\qty(\faktor{A}{\mathfrak m^n}) = d(G_{\mathfrak m}(A)) \]