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

iii/commalg/04_integrality_finiteness_finite_generation.tex

@@ -651,7 +651,7 @@ \subsection{Cohen--Seidenberg theorems}
 \end{tikzcd}\]
 \end{theorem}
 \begin{proof}
-    We have an injection \( \faktor{A}{\mathfrak p_1} \to \faktor{B}{\mathfrak q_1} \) given by \( a + \mathfrak p_1 \mapsto q + \mathfrak q_1 \).
+    We have an injection \( \faktor{A}{\mathfrak p_1} \to \faktor{B}{\mathfrak q_1} \) given by \( a + \mathfrak p_1 \mapsto a + \mathfrak q_1 \).
     This is an integral extension, so by lying over, there is a prime ideal \( \faktor{\mathfrak q_2}{\mathfrak q_1} \) of \( \faktor{B}{\mathfrak q_1} \) that contracts to \( \faktor{\mathfrak p_2}{\mathfrak p_1} \) in \( \faktor{A}{\mathfrak p_1} \).
     We claim that \( \mathfrak q_2 \cap A = \mathfrak p_2 \).
     In the diagram

iii/forcing/01_set_theory.tex

@@ -6,8 +6,8 @@ \subsection{Introduction to independence results}
     In the 19th century, it was shown that the other four postulates are satisfied by hyperbolic geometry, but this postulate is not satisfied.
     This shows that the other four axioms are insufficient to prove the parallel postulate.
     \item Let \( \varphi \) be the statement in the language of fields describing the existence of a square root of 2.
-    We know that \( \mathbb Q \) is a field satisfying \( \neg\varphi \), and \( \mathbb Q[\sqrt{2}] \) satisfies \( \varphi \).
-    The fields \( \mathbb Q \) and \( \mathbb Q[\sqrt{2}] \) are models of the theory of fields, one of which satisfies \( \varphi \), and one of which satisfies \( \neg\varphi \).
+    We know that \( \mathbb Q \) is a field satisfying \( \neg\varphi \), and \( \mathbb Q\qty[\sqrt{2}] \) satisfies \( \varphi \).
+    The fields \( \mathbb Q \) and \( \mathbb Q\qty[\sqrt{2}] \) are models of the theory of fields, one of which satisfies \( \varphi \), and one of which satisfies \( \neg\varphi \).
     This shows that the theory of fields does not prove \( \varphi \) or \( \neg\varphi \).
     A similar result holds for the statement \( \varphi \) that says that there are no roots of \( x^4 = -1 \).
     \item G\"odel's incompleteness theorem implies that there must always be an independence result in a sufficiently powerful consistent set theory.

iii/forcing/03_forcing.tex

@@ -855,7 +855,7 @@ \subsection{Verifying the axioms: part two}
     Fix \( a \in M[G] \) with name \( \dot x \in M^{\mathbb P} \), and define
     \[ S = \qty{\dot x \in M^{\mathbb P} \mid \ran \dot x \subseteq \ran \dot a} = \mathcal P(\mathbb P \times \ran \dot a)^M \]
     and let
-    \[ \dot b = \qty{\langle \Bbbone, \dot x \rangle \mid x \in S} \in M^{\mathbb P} \]
+    \[ \dot b = \qty{\langle \Bbbone, \dot x \rangle \mid \dot x \in S} \in M^{\mathbb P} \]
     Let \( c \in \mathcal P(a) \cap M[G] \); we must show that \( c \in \dot b^G \).
     Let \( \dot c \in M^{\mathbb P} \) be a name for \( c \), and let
     \[ \dot x = \qty{\langle p, \dot z \rangle \mid \dot z \in \ran \dot a \wedge (p \Vdash \dot z \in \dot c)^M} \in S \]

iii/forcing/04_forcing_and_independence.tex

@@ -285,7 +285,7 @@ \subsection{Possible sizes of the continuum}
     This can only happen if there is some \( q_1 \) with \( \langle q_1, \check n \rangle \in \dot z_{\dot x} \) such that \( r \mathrel\| q_1 \).
     Therefore, by definition, \( q_1 \in A_{\dot x, n} \).
     Since \( A_{\dot x, n} \) is an antichain containing \( q_0 \) and \( q_1 \) which are both compatible with \( r \), we must have \( q_0 = q_1 \).
-    Hence, \( \langle q_0, \check n \rangle \in \dot x_{\dot x} \).
+    Hence, \( \langle q_0, \check n \rangle \in \dot x \).
     Thus \( q_0 \Vdash \check n \in \dot x \) by definition, so since \( r \leq q_0 \), we have \( r \Vdash \check n \in \dot x \).
     Therefore \( D_{\dot x, n} \) is dense as required.
 
@@ -355,7 +355,7 @@ \subsection{Larger chain conditions}
         \kappa^+ & \text{if } \cf \kappa \leq \lambda
     \end{cases} \]
 \end{proposition}
-There is a natural bijection between \( \omega_3 \times \omega \) and \( \omega_3 \times \omega_1 \), and from this it will follow that \( 2^{\aleph_0} = 2^{\aleph_1} = \aleph_3 \).
+There is a natural bijection between \( \omega_3 \times \omega \) and \( \omega_3 \times \omega_1 \), and from this it follows that \( 2^{\aleph_0} = 2^{\aleph_1} = \aleph_3 \).
 \begin{definition}
     Let \( I, J \) be sets and let \( \kappa \) be a regular cardinal.
     Define \( \Fn_\kappa(I, J) \) to be the partial functions \( I \to J \) of size less than \( \kappa \).