Some edits

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

iii/cat/07_additive_and_abelian_categories.tex

@@ -300,7 +300,7 @@ \subsection{Abelian categories}
     \end{pmatrix} = \coker \begin{pmatrix}
         f \\ g
     \end{pmatrix} \), so the square is also a pushout.
-    We show that \( g \) is a pseudoepimorphism; this suffices as \( \mathcal A \) is additive.
+    We show that \( g \) is a pseudoepimorphism; this suffices as \( \mathcal A \) is abelian.
     Suppose we have \( \ell : C \to E \) with \( \ell g = 0 \).
     Then \( \begin{pmatrix}
         \ell & (B \xrightarrow 0 E)

iii/commalg/04_integrality_finiteness_finite_generation.tex

@@ -353,7 +353,7 @@ \subsection{Noether normalisation}
     \[ A = k[T, T^{-1}] = k[T, T^{-1} - cT] \]
     We claim that \( k[T^{-1} - cT] \subseteq A \) is a finite extension for most values of \( c \), and in particular, for at least one.
     First, note \( T^{-1} T - 1 = 0 \), and then change variables to
-    \[ ((T^{-1} - cT) + cT) T - 1 = 0 \implies \underbrace{c}_{\in k} T^2 + \underbrace{(T^{-1} - cT)}_{\in k[T^{-1} - ct]} T - \underbrace{1}_{\in k[T^{-1} - cT]} = 0 \]
+    \[ ((T^{-1} - cT) + cT) T - 1 = 0 \implies \underbrace{c}_{\in k} T^2 + \underbrace{(T^{-1} - cT)}_{\in k[T^{-1} - cT]} T - \underbrace{1}_{\in k[T^{-1} - cT]} = 0 \]
     Hence if \( c \neq 0 \), \( T \) is integral over \( k[T^{-1} - cT] \).
 \end{example}
 \begin{proof}
@@ -550,7 +550,7 @@ \subsection{Integrality over ideals}
     \[ \sqrt{\mathfrak a \overline A} = \sqrt{\sqrt{\mathfrak a}\, \overline A} \]
     The forwards inclusion is clear.
     For the other direction, it is a general fact that \( \sqrt{I}^e \subseteq \sqrt{I^e} \), so
-    \[ \sqrt{\mathfrak a} \overline A \subseteq \sqrt{\mathfrak a \overline A} \]
+    \[ \sqrt{\mathfrak a}\, \overline A \subseteq \sqrt{\mathfrak a \overline A} \]
     Taking radicals on both sides,
     \[ \sqrt{\sqrt{\mathfrak a}\, \overline A} \subseteq \sqrt{\sqrt{\mathfrak a \overline A}} = \sqrt{\mathfrak a \overline A} \]
 \end{proof}
@@ -591,7 +591,7 @@ \subsection{Integrality over ideals}
 \end{proof}
 
 \subsection{Cohen--Seidenberg theorems}
-If \( A \subseteq B \) is an extension of rings, the inclusion \( \iota : A \to B \) gives rise to \( \iota^\star : \Spec B \to \Spec A \) given by \( \iota(\mathfrak q) = \mathfrak q \cap A \).
+If \( A \subseteq B \) is an extension of rings, the inclusion \( \iota : A \to B \) gives rise to \( \iota^\star : \Spec B \to \Spec A \) given by \( \iota^\star(\mathfrak q) = \mathfrak q \cap A \).
 We will study the fibres of this induced map on spectra.
 \begin{proposition}[incomparability]
     Let \( A \subseteq B \) be an integral extension, and let \( \mathfrak q, \mathfrak q' \) be prime ideals of \( B \).

iii/commalg/06_direct_and_inverse_limits.tex

@@ -138,7 +138,7 @@ \subsection{Graded rings and modules}
     Hilbert's basis theorem shows that (ii) implies (i).
     For the converse, \( A_0 \) is Noetherian as it is isomorphic to a quotient of the Noetherian ring \( A \).
     Note that \( A_+ \) is generated by the set of homogeneous elements of positive degree.
-    By (i), \( A_+ \) an ideal in a Noetherian ring so is generated by a finite set \( \qty{x_1, \dots, x_s} \), and we can take each \( x_i \) to be homogeneous, say, \( x_i \in A_{k_i} \) where \( k_i > 0 \).
+    By (i), \( A_+ \) is an ideal in a Noetherian ring so is generated by a finite set \( \qty{x_1, \dots, x_s} \), and we can take each \( x_i \) to be homogeneous, say, \( x_i \in A_{k_i} \) where \( k_i > 0 \).
     Let \( A' \) be the \( A_0 \)-subalgebra of \( A \) generated by \( \qty{x_1, \dots, x_s} \); we want to show \( A' = A \).
     It suffices to show that \( A_n \subseteq A' \) for every \( n \geq 0 \), which we will show by induction.
     The case \( n = 0 \) is clear.

iii/commalg/07_dimension_theory.tex

@@ -29,7 +29,7 @@ \subsection{???}
 \begin{proof}
     First, we show that \( \dim A \leq \dim B \).
     Consider a chain of prime ideals \( \mathfrak p_0 \subsetneq \dots \subsetneq \mathfrak p_d \) in \( \Spec A \).
-    By the lying over theirem and the going up theorem, we obtain a chain of prime ideals \( \mathfrak q_0 \subseteq \dots \subseteq \mathfrak q_d \) in \( \Spec B \).
+    By the lying over theorem and the going up theorem, we obtain a chain of prime ideals \( \mathfrak q_0 \subseteq \dots \subseteq \mathfrak q_d \) in \( \Spec B \).
     As \( \mathfrak p_i = \mathfrak q_i \cap A \) and \( \mathfrak p_i \neq \mathfrak p_{i+1} \), we must have \( \mathfrak q_i \neq \mathfrak q_{i+1} \).
     So this produces a chain of length \( d \) in \( B \), as required.