Final fixes

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

iii/commalg/04_integrality_finiteness_finite_generation.tex

@@ -609,8 +609,8 @@ \subsection{Cohen--Seidenberg theorems}
     \[ \mathfrak q B_{\mathfrak p} \cap A_{\mathfrak p} = S^{-1} \mathfrak q \cap S^{-1} A = S^{-1} (\mathfrak q \cap A) = S^{-1} \mathfrak p = \mathfrak p A_{\mathfrak p} \]
     Similarly, \( \mathfrak q' B_{\mathfrak p} \cap A_{\mathfrak p} = \mathfrak p A_{\mathfrak p} \), which is a maximal ideal of \( A_{\mathfrak p} \).
     As \( A \subseteq B \) is an integral extension, \( A_{\mathfrak p} \subseteq B_{\mathfrak p} \) is also an integral extension.
-    Recall that the contraction of a maximal ideal is maximal in such an extension.
-    Now, \( \mathfrak q B_{\mathfrak p} \subseteq \mathfrak q' B_{\mathfrak p} \) are maximal ideals of \( B_{\mathfrak p} \), so they must coincide.
+    Recall that the if the contraction of a prime ideal is maximal in an integral extension, the prime ideal was maximal.
+    Thus, \( \mathfrak q B_{\mathfrak p} \subseteq \mathfrak q' B_{\mathfrak p} \) are maximal ideals of \( B_{\mathfrak p} \), so they must coincide.
 \end{proof}
 \begin{proposition}[lying over]
     Let \( A \subseteq B \) be an integral extension of rings, and let \( \mathfrak p \in \Spec A \).