replace setminus

2021_Algebra_I.tex

@@ -248,10 +248,10 @@ This generalizes some facts about matrices to matrices with elements from commut
 \end{fact}
 }
 \begin{proof}
-    Let $B$ be a field and $a \in A \sm \{0\} $. Then $a^{-1} \in  B$ is integral over $A$, hence $a^{-d} = \sum_{i=0}^{d-1} \alpha_i a^{-i}$ for some $\alpha_i \in A$. Multiplication by $a^{d-1}$ yields
+    Let $B$ be a field and $a \in A \setminus \{0\} $. Then $a^{-1} \in  B$ is integral over $A$, hence $a^{-d} = \sum_{i=0}^{d-1} \alpha_i a^{-i}$ for some $\alpha_i \in A$. Multiplication by $a^{d-1}$ yields
     $a^{-1} = \sum_{i=0}^{d-1} \alpha_i a^{d-1-i} \in A$.
 
-    On the other hand, let $B$ be integral over the field $A$. Let $b \in  B \sm \{0\}$. As $B$ is integral over $A$, there is a sub-$A$-algebra $\tilde{B} \subseteq B, b \in \tilde{B}$ finitely generated as an $A$-module, i.e. a finite-dimensional  $A$-vector space.  Since $B$ is a domain, $\tilde{B} \xrightarrow{b\cdot } \tilde{B}$ is injective, hence surjective, thus $\exists x \in \tilde{B} : b \cdot  x \cdot  1$.
+    On the other hand, let $B$ be integral over the field $A$. Let $b \in  B \setminus \{0\}$. As $B$ is integral over $A$, there is a sub-$A$-algebra $\tilde{B} \subseteq B, b \in \tilde{B}$ finitely generated as an $A$-module, i.e. a finite-dimensional  $A$-vector space.  Since $B$ is a domain, $\tilde{B} \xrightarrow{b\cdot } \tilde{B}$ is injective, hence surjective, thus $\exists x \in \tilde{B} : b \cdot  x \cdot  1$.
 \end{proof}
 \subsection{Noether normalization theorem}
 \begin{lemma}\label{nntechlemma}
@@ -282,7 +282,7 @@ is injective. $n$ and the $a_i$ can be chosen such that $A$ is finite over the i
     If suffices to show that the $a_i$ are algebraically independent.
     Since $A$ is of finite type  over $K$ and thus over $\tilde{A}$, by fact \ref{ftaiimplf} (integral and finite type $\implies$ finite) $A$ is finite over $\tilde{A}$.
     Thus we only need to show that the  $a_i$ are algebraically independent over $K$.
-    Assume there is $P \in  K[X_1,\ldots,X_n] \sm \{0\} $ such that $P(a_1,\ldots,a_n) = 0$. Let $P = \sum_{\alpha \in \N^n} p_\alpha X^{\alpha}$ and $S = \{ \alpha \in \N^n | p_\alpha \neq 0\}$. For $\vec{k} = (k_i)_{i=1}^{n} \in  \N^n$ and $\alpha \in \N^n$ we define $w_{\vec{k}}(\alpha) \coloneqq \sum_{i=1}^{n} k_i\alpha_i$.
+    Assume there is $P \in  K[X_1,\ldots,X_n] \setminus \{0\} $ such that $P(a_1,\ldots,a_n) = 0$. Let $P = \sum_{\alpha \in \N^n} p_\alpha X^{\alpha}$ and $S = \{ \alpha \in \N^n | p_\alpha \neq 0\}$. For $\vec{k} = (k_i)_{i=1}^{n} \in  \N^n$ and $\alpha \in \N^n$ we define $w_{\vec{k}}(\alpha) \coloneqq \sum_{i=1}^{n} k_i\alpha_i$.
 
     By \ref{nntechlemma} it is possible to choose $\vec{k} \in \N^n$ such that
     $k_1 = 1$ and for $\alpha \neq  \beta \in  S$ we have $w_{\vec{k}}(\alpha) \neq  w_{\vec{k}}(\beta)$.
@@ -393,7 +393,7 @@ The following proof of the Nullstellensatz only works for uncountable fields, bu
     \]
     Let $d \coloneqq \prod_{\kappa \in  I} (T - \kappa) $. Then for $\lambda \in I$ we have
     \[
-        0 = (dg)(\lambda) = x_\lambda \prod_{\kappa \in  I \sm \{\lambda\} } (\lambda - \kappa)
+        0 = (dg)(\lambda) = x_\lambda \prod_{\kappa \in  I \setminus \{\lambda\} } (\lambda - \kappa)
     \] 
     This is a contradiction as $x_\lambda \neq 0$.
 \end{proof}
@@ -509,10 +509,10 @@ Let $X$ be a topological space.
 \end{definition}
 \begin{proof}\,
     \begin{enumerate}
-        \item[A $\implies$ B] Let $A_j$ be a descending chain of closed subsets. Define $A \coloneqq \bigcap_{j = 0}^{\infty} A_j$. If A holds, the covering $X \sm A = \bigcup_{j = 0}^{\infty} (X \sm A_j)$ has a finite subcovering.
+        \item[A $\implies$ B] Let $A_j$ be a descending chain of closed subsets. Define $A \coloneqq \bigcap_{j = 0}^{\infty} A_j$. If A holds, the covering $X \setminus A = \bigcup_{j = 0}^{\infty} (X \setminus A_j)$ has a finite subcovering.
         \item[B  $\implies$ C] Suppose $\cM$ does not have a $\subseteq$-minimal element. Using DC, one can construct a counterexample $A_1 \subsetneq A_2 \supsetneq \ldots$ to B.
         \item[C $\implies$ A] Let $\bigcup_{i \in  I} V_i$ be an open covering of an open subset $U \subseteq X$.
-            By C, the set $\cM \coloneqq \{X \sm \bigcup_{i \in F} V_i | F \subseteq I \text{ finite} \}$ has a $\subseteq$-minimal element.
+            By C, the set $\cM \coloneqq \{X \setminus \bigcup_{i \in F} V_i | F \subseteq I \text{ finite} \}$ has a $\subseteq$-minimal element.
     \end{enumerate}
 \end{proof}
 
@@ -579,7 +579,7 @@ Let $X$ be a topological space.
 \begin{proof}\,
     \begin{itemize}
         \item[$A \iff B$] by definition of denseness.
-        \item[B $\iff$ C] Let $U \coloneqq X \sm A, V \coloneqq X \sm B$.
+        \item[B $\iff$ C] Let $U \coloneqq X \setminus A, V \coloneqq X \setminus B$.
         \item[B $\implies$ D] Suppose $W$  is a non-connected open subset. Then there exists a decomposition $W = U \cup V$ into disjoint open subsets.
         \item[D $\implies$ B] If $U,V \neq \emptyset$ are disjoint open subsets, then $U \cup V$ is non-connected.
     \end{itemize}
@@ -663,7 +663,7 @@ Let $X$ be a topological space.
 \end{proposition}
 \begin{proof}
     By the Nullstellensatz (\ref{hns1}), $A = \emptyset \iff I = R$. Suppose $A = B \cup C$ is a decomposition into proper closed subsets $A = V(J), B = V(K)$ where  $J = \sqrt{J}. K = \sqrt{K}$.
-    Since $A \neq B$ and $A \neq C$, there are $f \in  J \sm I, g \in K \sm I$. $fg$ vanishes on $A = B \cup C$. By the Nullstellensatz (\ref{hns3}) $fg \in \sqrt{I}  = I$ and $I$ fails to be prime.
+    Since $A \neq B$ and $A \neq C$, there are $f \in  J \setminus I, g \in K \setminus I$. $fg$ vanishes on $A = B \cup C$. By the Nullstellensatz (\ref{hns3}) $fg \in \sqrt{I}  = I$ and $I$ fails to be prime.
 
     On the other hand suppose that $fg \in I, f \notin I, g \not\in I$. By the Nullstellensatz (\ref{hns3}) and $I = \sqrt{I} $ neither $f$ nor $g$ vanishes on all of $A$. Thus $(A \cap V(f)) \cup (A \cap V(g))$ is a decomposition and $A$ fails to be irreducible.
 
@@ -752,7 +752,7 @@ In general, these inequalities may be strict.
     Every non-zero prime ideal $\fp$ of a UFD $R$ contains a prime element.
 \end{lemma}
 \begin{proof}
-    Let $p \in  \fp \sm \{0\} $ with the minimal number of prime factors, counted by multiplicity.
+    Let $p \in  \fp \setminus \{0\} $ with the minimal number of prime factors, counted by multiplicity.
     If $p $ was a unit, then $\fp \supseteq pR = R$. If  $p = ab$ with non-units $a,b$, it follows that $a \in \fp$ or $b \in \fp$ contradicting the minimality assumption.
     Thus $p$ is a prime element of $R$.
 \end{proof}
@@ -785,10 +785,10 @@ In general, these inequalities may be strict.
 
     We call $\cH$ \vocab{matroidal}  if in addition the following conditions hold: 
     \begin{enumerate}
-        \item[M] If $m,n \in X$ and $A \subseteq X$ then $m \in \cH( \{n\}  \cup  A) \sm \cH(A) \iff n \in \cH(\{m\}  \cup A) \sm \cH(A).$ 
+        \item[M] If $m,n \in X$ and $A \subseteq X$ then $m \in \cH( \{n\}  \cup  A) \setminus \cH(A) \iff n \in \cH(\{m\}  \cup A) \setminus \cH(A).$ 
             \item[F] $\cH(A) = \bigcup_{F \subseteq A \text{ finite}} \cH(F)$.
     \end{enumerate}
-    In this case, $S \subseteq X$ is called \vocab{Independent subset}, if $s \not\in  \cH(S \sm \{s\})$ for all $s \in S$ and
+    In this case, $S \subseteq X$ is called \vocab{Independent subset}, if $s \not\in  \cH(S \setminus \{s\})$ for all $s \in S$ and
     \vocab[Generating subset]{generating}  if $X = \cH(S)$.
     $S$ is called a \vocab{base}, if it is both generating and independent.
 \end{definition}
@@ -811,8 +811,8 @@ In general, these inequalities may be strict.
 \begin{proof}
     H1, H2 and F are trivial. For an algebraically closed subfield $K \subseteq M \subseteq L$ we have $\cH(M) = M$. Thus $\cH(\cH(T)) = \cH(T)$ (H3).
 
-    Let $x,y \in L$, $T \subseteq L$ and $x \in  \cH(T \cup \{y\}) \sm \cH(T)$. We have to show that $y \in  \cH(T \cup \{x\}) \sm \cH(T)$.
+    Let $x,y \in L$, $T \subseteq L$ and $x \in  \cH(T \cup \{y\}) \setminus \cH(T)$. We have to show that $y \in  \cH(T \cup \{x\}) \setminus \cH(T)$.
-    If $y \in  \cH(T)$ we have $\cH(T \cup \{y\}) \subseteq \cH(\cH(T)) = \cH(T) \implies x \in  \cH(T) \sm \cH(T) \lightning$.
+    If $y \in  \cH(T)$ we have $\cH(T \cup \{y\}) \subseteq \cH(\cH(T)) = \cH(T) \implies x \in  \cH(T) \setminus \cH(T) \lightning$.
     Hence it is sufficient to show $y \in  \cH(T \cup  \{x\})$. \Wlog $T = \emptyset$ (replace $K$ be the subfield generated by $K \cup T$).
     Then $x$ is algebraic over the subfield $M$ of $L$ generated by $K \cup \{y\}$. Thus there exists $0 \neq P \in  M[T]$ with $P(x) = 0$.
     The coefficients $p_i$ of $P$ belong to the field of quotients of the $K$-subalgebra of $L$ generated by $y$. There are thus polynomials $Q_i, R \in K[Y]$ such that $p_i = \frac{Q_i(y)}{R(y)}$, $R(y) \neq 0$.
@@ -821,7 +821,7 @@ In general, these inequalities may be strict.
         Q(X,Y) \coloneqq \sum_{i=0}^{\infty} X^i Q_i(Y) = \sum_{i,j=0}^{\infty} q_{i,j}X^i Y^j = \sum_{j=0}^{\infty} Y^j \hat{Q_j}(X) \in  K[X,Y]
     \].
     Then $Q(x,y) = 0$.
-    Let $\hat{p_j} \coloneqq \hat{Q_j}(x)$. Then $\hat{P}(y) = 0$. As $Q \neq 0$ there is $(i,j) \in  \N^2$ such that $q_{i,j} \neq 0$ and then $\hat{p_j} \neq 0$ as $x \not\in \cH(\emptyset)$. Thus $\hat{P} \in  \hat{M}[X] \sm \{0\} $, where $\hat{M}$ is the subfield of $L$ generated by $K$ and $x$. Thus $y$ is algebraic over $\hat{M}$ and $y \in \cH(\{x\})$,
+    Let $\hat{p_j} \coloneqq \hat{Q_j}(x)$. Then $\hat{P}(y) = 0$. As $Q \neq 0$ there is $(i,j) \in  \N^2$ such that $q_{i,j} \neq 0$ and then $\hat{p_j} \neq 0$ as $x \not\in \cH(\emptyset)$. Thus $\hat{P} \in  \hat{M}[X] \setminus \{0\} $, where $\hat{M}$ is the subfield of $L$ generated by $K$ and $x$. Thus $y$ is algebraic over $\hat{M}$ and $y \in \cH(\{x\})$,
 \end{proof}
 \begin{definition}[Transcendence Base]
     Let $L / K$ be a field extension and $\cH(T)$ the algebraic closure in $L$ of the subfield  generated by $K$ and $T$. A base for $(L, \cH)$ is called a \vocab{transcendence base} and the \vocab{transcendence degree}  $\trdeg(L / K)$ is defined as the cardinality of any transcendence base of  $L / K$. 
@@ -892,7 +892,7 @@ Let $K$ be a field and $R = K[X_1,\ldots,X_n]$.
     If $n > 0$, then $K(X_1,\ldots,X_n) / K$ is not of finite type.
 \end{lemma}
 \begin{proof}
-    Suppose $(f_i)_{i=1}^m$ generate $K(X_1,\ldots,X_n)$ as a $K$-algebra. Let $f_i = \frac{a_i}{b}, a_i \in R, b \in R \sm \{0\}$.
+    Suppose $(f_i)_{i=1}^m$ generate $K(X_1,\ldots,X_n)$ as a $K$-algebra. Let $f_i = \frac{a_i}{b}, a_i \in R, b \in R \setminus \{0\}$.
     Then $bf_i \in R$, and as the $f_i$ generate $K(X_1,\ldots,X_n)$ as a $K$-algebra, for every $g \in K(X_1,\ldots,X_n)$ there is $N \in \N$ with
     \[
         b^Ng \in R \tag{+} \label{bNginR}
@@ -1055,8 +1055,8 @@ Because the elements of $S$ become units in $R_S$, $J \sqcap R$ is an $S$-satura
 % Some more remarks on localization
 
 \begin{remark}\label{locandquot}
-    Let $R$ be a domain. If $S = R \sm \{0\}$, then $R_S$ is the field of quotients $Q(R)$.
+    Let $R$ be a domain. If $S = R \setminus \{0\}$, then $R_S$ is the field of quotients $Q(R)$.
-    If $S \subseteq R \sm \{0\} $, then
+    If $S \subseteq R \setminus \{0\} $, then
      \[
     R_S \cong \left\{ \frac{a}{s} \in  K | a \in R, s \in S\right\} 
     \] 
@@ -1128,7 +1128,7 @@ Then \[
     Let $\fq$ be any prime ideal.
     Take $a_1,\ldots,a_n \in A$ as in the lemma. As the $a_i \mod \fq$ are $\mathfrak{l}$-algebraically independent, the same holds for the $a_i$ themselves.
     Thus $\trdeg(Q(A) / \mathfrak{l}) \ge n$ and the inequality is strict, if it can be shown that the $a_i$ fail to be a transcendence base of $Q(A) / \mathfrak{l}$.
-    Let $R \subseteq A$ denote the $\mathfrak{l}$-subalgebra generated by $a_1,\ldots,a_n$ and $S \coloneqq R \sm \{0\} $.
+    Let $R \subseteq A$ denote the $\mathfrak{l}$-subalgebra generated by $a_1,\ldots,a_n$ and $S \coloneqq R \setminus \{0\} $.
     We must show, that $Q(A)$ fails to be algebraic over $\mathfrak{l}_1 \coloneqq R_S = Q(R)$.
     Let $A_1 \coloneqq A_S$ and $\fq_S$ the prime ideal corresponding to $\fq$ as in \ref{idealslocbij}.
     We have $\fq_S \neq  \{0\} $ as $\{0_{A}\}_S = \{0_{A_S}\}$.
@@ -1165,15 +1165,15 @@ Then \[
     Let $R$ be a ring.  $R$ is called a \vocab{local ring}, if the following equivalent conditions hold:
     \begin{itemize}
         \item $\#\mSpec R = 1$
-        \item  $R \sm R^{\times }$ is an ideal.
+        \item  $R \setminus R^{\times }$ is an ideal.
     \end{itemize}
-    If this holds, $\mathfrak{m}_R \coloneqq R \sm R^{\times }$ is the unique maximal ideal of $R$.
+    If this holds, $\mathfrak{m}_R \coloneqq R \setminus R^{\times }$ is the unique maximal ideal of $R$.
 \end{definition}
 \begin{proof}
     Suppose $\mSpec R = \{\mathfrak{m}\}$. If $x \in  \mathfrak{m}$, then $x \not\in R^{\times }$ as otherwise $xR = R \implies \mathfrak{m} = R$.
     If $x \not\in  R^{\times }$ then $xR$ is a proper ideal, hence contained in some maximal ideal. Thus $x \in \mathfrak{m}$.
 
-    Assume that $\mathfrak{m} = R \sm R^{\times }$ is an ideal in $R$. As $1 \in R^{\times }$ this is a proper ideal. If $I$ is any proper ideal and $x \in I$, then $x \in \mathfrak{m}$. Hence $R = xR \subseteq I \subseteq \mathfrak{m}$. It follows that $\mathfrak{m}$ is the only maximal ideal of $R$.
+    Assume that $\mathfrak{m} = R \setminus R^{\times }$ is an ideal in $R$. As $1 \in R^{\times }$ this is a proper ideal. If $I$ is any proper ideal and $x \in I$, then $x \in \mathfrak{m}$. Hence $R = xR \subseteq I \subseteq \mathfrak{m}$. It follows that $\mathfrak{m}$ is the only maximal ideal of $R$.
 \end{proof}
 \begin{remark}
     \begin{itemize}
@@ -1185,7 +1185,7 @@ Then \[
 Many questons of commutative algebra are easier in the case of local rings. Localization at a prime ideal is a technique to reduce a problem to this case.
 
 \begin{proposition}[Localization at a prime ideal]\label{locatprime}
-    Let $A$ be a ring and $\fp \in  \Spec A$. Then $S \coloneqq A \sm \fp$ is a multiplicative subset, $A_S$ is a local ring with maximal ideal $\mathfrak{m} = \fp_S =\{\frac{p}{s}| p \in  \fp, s \in S\} $.
+    Let $A$ be a ring and $\fp \in  \Spec A$. Then $S \coloneqq A \setminus \fp$ is a multiplicative subset, $A_S$ is a local ring with maximal ideal $\mathfrak{m} = \fp_S =\{\frac{p}{s}| p \in  \fp, s \in S\} $.
 
     We have a bijection
     \begin{align}
@@ -1195,10 +1195,10 @@ Many questons of commutative algebra are easier in the case of local rings. Loca
     \end{align}
 \end{proposition}
 \begin{proof}
-    It is clear that $S$ is a multiplicative subset and that $\fp_S$ is an ideal. By \ref{ssatiis} $\frac{a}{s} \in \fp_S \iff a \in  \fp \iff a \in A \sm S$ for all $a \in A, s \in S$.
+    It is clear that $S$ is a multiplicative subset and that $\fp_S$ is an ideal. By \ref{ssatiis} $\frac{a}{s} \in \fp_S \iff a \in  \fp \iff a \in A \setminus S$ for all $a \in A, s \in S$.
     Thus, if $\frac{a}{s} \not\in  \fp_S$ then it is a unit in $A_S$ with inverse $\frac{s}{a}$. Hence $A_S$ is a local ring with maximal ideal $\fp_S$.
 
-    The claim about $\Spec A_S$ follows from \ref{idealslocbij} using the fact (\ref{primeidealssat}) that a prime ideal $\fr \in \Spec A$ is $S$-saturated iff it is disjoint from $S = A \sm \fp$ iff $\fr \subseteq \fp$.
+    The claim about $\Spec A_S$ follows from \ref{idealslocbij} using the fact (\ref{primeidealssat}) that a prime ideal $\fr \in \Spec A$ is $S$-saturated iff it is disjoint from $S = A \setminus \fp$ iff $\fr \subseteq \fp$.
 \end{proof}
 \begin{definition}
     The ring $A_S$ as in \ref{locatprime} is called the \vocab[Localization]{localization of $A$ at the prime ideal $\fp$} and denoted $A_\fp$. 
@@ -1210,7 +1210,7 @@ Many questons of commutative algebra are easier in the case of local rings. Loca
 
 \begin{remark}
     Let $B = \mathfrak{k}[X_1,\ldots,X_n]$, $x \in \mathfrak{k}^n$ and $\mathfrak{m}$ the maximal ideal such that $V(\mathfrak{m}) = \{x\}$.
-    The elements of $B_\mathfrak{m}$ are the fractions $\frac{b}{s}, b \in B, s \in B \sm \mathfrak{m}$, i.e. $s(x) \neq 0$.
+    The elements of $B_\mathfrak{m}$ are the fractions $\frac{b}{s}, b \in B, s \in B \setminus \mathfrak{m}$, i.e. $s(x) \neq 0$.
     These are precisely the rational functions which are well-defined in some neighbourhood of $x$.
     This will be rigorously formulated in \ref{proplocalring}.
     %Hence the name localization.
@@ -1276,7 +1276,7 @@ Many questons of commutative algebra are easier in the case of local rings. Loca
     \begin{enumerate}
          \item[D] Consider the ring extension $A / \fq$ of $R / \fp$. Both rings are domains and the extension is integral.
             By the fact about integrality and fields (\ref{fintaf}) $A / \fq$ is a field iff $R / \fp$ is a field. Thus $\fq \in \mSpec A \iff \fp \in \mSpec R$.
-        \item[A] Suppose $\fp \in \Spec R$ and let $S \coloneqq R \sm \fp$. Then $S$ is a multiplicative subset of both $R$ and $A$, and we may consider the localizations $R \xrightarrow{\rho} R_\fp, A \xrightarrow{\alpha} A_\fp$ with respect to $S$. By the universal property of $\rho$, there exists a unique homomorphism $R_\fp \xrightarrow{i} A_\fp$ such that $i\rho = \alpha \defon{R}$.
+        \item[A] Suppose $\fp \in \Spec R$ and let $S \coloneqq R \setminus \fp$. Then $S$ is a multiplicative subset of both $R$ and $A$, and we may consider the localizations $R \xrightarrow{\rho} R_\fp, A \xrightarrow{\alpha} A_\fp$ with respect to $S$. By the universal property of $\rho$, there exists a unique homomorphism $R_\fp \xrightarrow{i} A_\fp$ such that $i\rho = \alpha \defon{R}$.
             We have $j(\frac{r}{s}) = \frac{r}{s}$ and $j$ is easily seen to be injective.
 
             \begin{figure}[H]
@@ -1290,7 +1290,7 @@ Many questons of commutative algebra are easier in the case of local rings. Loca
                $A_\fp$ is integral over $R_\fp$.
            \end{claim} 
            \begin{subproof}
-               An element $x \in  A_\fp$ has the form $x = \frac{a}{s}$ for some $s \in  R \sm \fp$ and where $a \in  A$ is integral over $R$.
+               An element $x \in  A_\fp$ has the form $x = \frac{a}{s}$ for some $s \in  R \setminus \fp$ and where $a \in  A$ is integral over $R$.
                Hence $a^n = \sum_{i=0}^{n-1} r_ia^i$ for some $r_i \in R$. Thus $x^n = \sum_{i=0}^{n-1} \rho_i x^i$ with $\rho_i \coloneqq s^{i-n} r_i \in R_\fp$.
            \end{subproof}
            As $i$ is injective and $R_\fp \neq  \{0\} $ ($R_\fp$ is local!) $A_\fp \neq  \{0\}$, there is $\mathfrak{m} \in \mSpec A_\fp$.
@@ -1320,10 +1320,10 @@ This is part of the proof of \ref{trdegandkdim}.
 %TODO: relate to  \ref{htandcodim}
 \begin{proof}
     Let $B = \mathfrak{k}[X_1,\ldots,X_n]$ and let $X \subseteq Y \subseteq \mathfrak{k}^n$ be irreducible closed subsets of $\mathfrak{k}^n$.
-    We have to show $\codim(X,Y) = \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) \sm \mathfrak{k})$.
+    We have to show $\codim(X,Y) = \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) \setminus \mathfrak{k})$.
     The inequality
      \[
-         \codim(X,Y) \le  \trdeg(\fK(Y) \sm \mathfrak{k}) - \trdeg(\fK(X) \sm \mathfrak{k})
+         \codim(X,Y) \le  \trdeg(\fK(Y) \setminus \mathfrak{k}) - \trdeg(\fK(X) \setminus \mathfrak{k})
     \] 
     has been shown in \ref{upperboundcodim}.
     In the case of $X = \{0\} , Y = \mathfrak{k}^n$, equality holds because the chain of irreducible subsets $\{0\}  \subsetneq \{0\}  \times \mathfrak{k} \subsetneq \ldots \subsetneq \{0\} \times \mathfrak{k}^n\subsetneq \mathfrak{k}^n$
@@ -1346,7 +1346,7 @@ This is part of the proof of \ref{trdegandkdim}.
 
     Hence $\dim(Y) \ge \trdeg(\fK(Y) / \mathfrak{k})$.
 
-    The general case of $\codim(X,Y)  \ge \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) \sm \mathfrak{k})$ is shown in \ref{proofcodimletrdeg}.
+    The general case of $\codim(X,Y)  \ge \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) \setminus \mathfrak{k})$ is shown in \ref{proofcodimletrdeg}.
     % TODO: reorder
     % TODO: Motivation: "Morphism" (AlGeo) and Lift of {0} x k \subseteq \ldots
 
@@ -1367,10 +1367,10 @@ This is part of the proof of \ref{trdegandkdim}.
     Induction on $n$. The case of $n < 2$ is trivial.
     Let $n \ge 2$ and the assertion be shown for a list of $n-1$ ideals one wants to avoid.
     If $n \ge 3$ we may, by reordering the $\fp_i$ assume that $\fp_1$ is a prime ideal.
-    By the induction assumption, there is $f_k \in I \sm \bigcup_{j \neq k} \fp_j$. If there is $k$ with $1 \le k\le n$ and $f_k \not\in \fp_k$, then the proof is finished.
+    By the induction assumption, there is $f_k \in I \setminus \bigcup_{j \neq k} \fp_j$. If there is $k$ with $1 \le k\le n$ and $f_k \not\in \fp_k$, then the proof is finished.
     Otherwise 
     \[
-    f_1 + \prod_{j=2}^{n} f_j \in  I \sm \bigcup_{j=1}^n \fp_j 
+    f_1 + \prod_{j=2}^{n} f_j \in  I \setminus \bigcup_{j=1}^n \fp_j 
     \] 
 \end{proof}
 
@@ -1481,11 +1481,11 @@ Recall the definition of a normal field extension in the case of finite field ex
     We must show that there exists $\sigma \in G$ such that $\fq = \sigma(\fr)$.
     This is equivalent to $\fq \subseteq \sigma(\fr)$, since the Krull going-up theorem (\ref{cohenseidenberg}) applies to the integral ring extension $B / A$, showing that there are no inclusions between different elements of $\Spec B$ lying above $\fp \in \Spec A$.
 
-    If $L / K$ is finite and there is no such $\sigma$, then by prime avoidance (\ref{primeavoidance}) there is $ x \in \fq \sm \bigcup_{\sigma \in G}  \sigma(\fr)$.
+    If $L / K$ is finite and there is no such $\sigma$, then by prime avoidance (\ref{primeavoidance}) there is $ x \in \fq \setminus \bigcup_{\sigma \in G}  \sigma(\fr)$.
-    As $\fr$ is a prime ideal, $y = \prod_{\sigma \in  G} \sigma(x) \in  \fq \sm \fr$.\footnote{$\prod_{\sigma \in G} \sigma(x) = \prod_{\sigma \in G} \sigma^{-1}(x)$}
+    As $\fr$ is a prime ideal, $y = \prod_{\sigma \in  G} \sigma(x) \in  \fq \setminus \fr$.\footnote{$\prod_{\sigma \in G} \sigma(x) = \prod_{\sigma \in G} \sigma^{-1}(x)$}
     By the characterization of $L^G$ for normal field extensions (\ref{characfixnormalfe}), there is a positive integer $k$ with $y^k \in K$. 
     As $A$ is normal, we have $y^k \in  K \cap B = A$.
-    Thus $y^k \in (A \cap \fq) \sm (A \cap \fr) = \fp \sm \fp = \emptyset \lightning$.
+    Thus $y^k \in (A \cap \fq) \setminus (A \cap \fr) = \fp \setminus \fp = \emptyset \lightning$.
 
     If $L / K$ is not finite, one applies Zorn's lemma to the poset of pairs $(M, \sigma)$ where $M$ is an intermediate field and $\sigma \in \Aut(M / K)$ such that $\sigma(\fr \cap M) = \fq \cap M$.
 
@@ -1555,7 +1555,7 @@ This is part of the proof of \ref{trdegandkdim}. %TODO: reorder
     We then used going-up to lift  a chain of prime ideals corresponding to $\mathfrak{k}^d \supsetneq \{0\}  \times \mathfrak{k}^{n-1} \supsetneq \ldots \supsetneq \{0\}$ under $Y \xrightarrow{F = (f_1,\ldots,f_d)} \mathfrak{k}^d$ to a chain of prime ideals in $A$.
     This was done left-to-right as going-up was used to make prime ideals larger. In particular, when $\{0\} \in \mathfrak{k}^d$ has several preimages under $F$ we cannot control to which of them the maximal ideal terminating the lifted chain belongs. Thus, we can show that in the inequality
     \[
-        \codim(\{y\}, Y) \le  d = \trdeg(\fK(Y) \sm \mathfrak{k})
+        \codim(\{y\}, Y) \le  d = \trdeg(\fK(Y) \setminus \mathfrak{k})
     \] 
     (see \ref{upperboundcodim})
     equality holds for at least one pint $y \in F^{-1}(\{0\})$ but cannot rule out that there are other $y \in F^{-1}(\{0\})$ for which the inequality becomes strict.
@@ -1680,7 +1680,7 @@ By lemma \ref{ltrdegresfieldtrbase} there are $a_1,\ldots,a_n \in A$ whose image
 As these images are $\mathfrak{l}$-algebraically independent, the same holds for the $a_i$ themselves.
 
 By lemma \ref{extendtotrbase} we can extend $(a_{i})_{i=1}^n$ to a transcendence base $(a_i)_{i=1}^m \in  A^m$ of  $K / \mathfrak{l}$.
-Let  $R \subseteq A$ denote the $\mathfrak{l}$-subalgebra generated by $a_1,\ldots,a_n$ and let $S \coloneqq R \sm \{0\}$.
+Let  $R \subseteq A$ denote the $\mathfrak{l}$-subalgebra generated by $a_1,\ldots,a_n$ and let $S \coloneqq R \setminus \{0\}$.
 Let $A_1 \coloneqq A_S$ and $\fp_S$ the prime ideal corresponding to $\fp$ under $\Spec(A_1) \cong \{\fr \in \Spec A | \fr \cap S = \emptyset\}$ (\ref{idealslocbij}).
 As in \ref{locandquot}, $A_1$ is a domain with $Q(A_1) \cong K = Q(A)$ and by \ref{locandfactor} $A_1 / \fp_S \cong (A / \fp)_{\overline{S}}$, where $\overline{S}$ denotes the image of $S$ in $A / \fp$.
 As in \ref{trdegresfield}, $\mathfrak{k}(\fp_S) \cong \mathfrak{k}(\fp)$ is integral over $A_1 / \fp_S$.
@@ -1692,7 +1692,7 @@ From the fact about integrality and fields (\ref{fintaf}), it follows that $A_1
 \end{remark}
 \begin{dexample}[Noetherian ring with infinite dimension]\footnote{\url{https://math.stackexchange.com/questions/1109732/noetherian-ring-with-infinite-krull-dimension-nagatas-example}}
     Let $A = \mathfrak{k}[X_i | i \in \N]$ and $m_1, m_2, \ldots \in \N$ an increasing sequence such that $m_{i+1}-m_i > m_i - m_{i-1}$.
-    Let $\fp_i \coloneqq (X_{m_{i}+1},\ldots,X_{m_{i+1}})$ and $S \coloneqq A \sm \bigcup_{i \in \N}  \fp_i$.
+    Let $\fp_i \coloneqq (X_{m_{i}+1},\ldots,X_{m_{i+1}})$ and $S \coloneqq A \setminus \bigcup_{i \in \N}  \fp_i$.
     $S$ is multiplicatively closed.
     $A_S$ is Noetherian but $\hght((\fp_i)_S) = m_{i+1}- m_{i}$ hence $\dim(A_S) = \infty$.
 \end{dexample}
@@ -1734,7 +1734,7 @@ From the fact about integrality and fields (\ref{fintaf}), it follows that $A_1
     This is called the \vocab{nil radical} of $A$. 
 \end{proposition}
 \begin{proof}
-    It is clear that elements of $\sqrt{\{0\} } $ must belong to all prime ideals. Conversely, let $a \in  A \sm \sqrt{\{0\} }$. Then  $S = a^{\N}$ is a multiplicative subset of $A$ not containing $0$.
+    It is clear that elements of $\sqrt{\{0\} } $ must belong to all prime ideals. Conversely, let $a \in  A \setminus \sqrt{\{0\} }$. Then  $S = a^{\N}$ is a multiplicative subset of $A$ not containing $0$.
     The localisation $A_S$ of $A$ is thus not the null ring. Hence $\Spec A_S \neq \emptyset$. If $\fq \in  \Spec A_S$, then by the description of $\Spec A_S$ (\ref{idealslocbij}), $\fp \coloneqq \fq \sqcap A$ is a prime ideal of $A$ disjoint from $S$, hence $a \not\in \fp$.
 \end{proof}
 
@@ -1758,7 +1758,7 @@ From the fact about integrality and fields (\ref{fintaf}), it follows that $A_1
     If $A = \Vspec(I)$, then by \ref{sqandvspec} $\sqrt{I}  = \bigcap_{\fp \in A}  \fp$. Thus, an ideal with $\sqrt{I}  = I$ can be recovered from $\Vspec( I)$. Since $\Vspec(J) = \Vspec(\sqrt{J})$, the map from ideals with $\sqrt{I}  = I$ to closed subsets is surjective.
 
     Sine $R$ corresponds to $\emptyset$, the proper ideals correspond to non-empty subsets of $\Spec R$. Assume that $\Vspec(I) = \Vspec(J_1) \cup \Vspec(J_2)$, where the decomposition is proper and the ideals coincide with their radicals.
-    Let $g = f_1f_2$ with $f_k \in  J_k \sm I$. Since $\Vspec(g) \supseteq \Vspec(f_k) \supseteq \Vspec(I_k), \Vspec(I) \subseteq \Vspec(g)$. Hence $g \in \sqrt{I}  = I$.
+    Let $g = f_1f_2$ with $f_k \in  J_k \setminus I$. Since $\Vspec(g) \supseteq \Vspec(f_k) \supseteq \Vspec(I_k), \Vspec(I) \subseteq \Vspec(g)$. Hence $g \in \sqrt{I}  = I$.
     As $f_k \not\in I$, $I$ fails to be a prime ideal.
     Conversely, assume that $f_1f_2 \in I$ while the factors are not in $I$. Since $I = \sqrt{I}, \Vspec(f_k) \not\supseteq \Vspec(I)$. But $\Vspec(f_1) \cup \Vspec(f_2) = \Vspec(f_1f_2) \supseteq \Vspec(I)$.
     The proper decomposition $\Vspec(I) = \left( \Vspec(I) \cap \Vspec(f_1) \right) \cup \left( \Vspec(I) \cap \Vspec(f_2) \right) $ now shows that $\Vspec(I)$ fails to be irreducible.
@@ -1880,7 +1880,7 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ and $I \subseteq R$ an ideal.
 
 
     Assume that this is the case. Let $\fp \in \Spec R, \hght(\fp) = 1$.
-    Let $p \in  \fp \sm \{0\}$. Replacing $p$ by a prime factor of $p$, we may assume $p$ to be prime. Thus $\{0\}  \subsetneq pR \subseteq \fp$ is a chain of prime ideals and since $\hght(\fp) = 1$ it follows that $\fp = pR$.
+    Let $p \in  \fp \setminus \{0\}$. Replacing $p$ by a prime factor of $p$, we may assume $p$ to be prime. Thus $\{0\}  \subsetneq pR \subseteq \fp$ is a chain of prime ideals and since $\hght(\fp) = 1$ it follows that $\fp = pR$.
 
     Conversely, assume that every $\fp \in \Spec R$ with $\hght(\fp)=1$ is a principal ideal. Let $f \in R$ be irreducible.
     Let $\fp \in \Spec R$ be a $\subseteq$-minimal element  of $V(f)$. By the principal ideal theorem (\ref{pitheorem}), $\hght(\fp)=1$.
@@ -1892,7 +1892,7 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ and $I \subseteq R$ an ideal.
     For a ring $A, \bigcap_{\mathfrak{m} \in \mSpec A} \mathfrak{m} = \{a \in A | \A x \in  A ~ 1 - ax \in A^{\times }\} \text{\reflectbox{$\coloneqq$}} \rad(A)$, the \vocab{Jacobson radical}  of $A$.
 \end{proposition}
 \begin{proof}
-    Suppose $\mathfrak{m} \in \mSpec A$ and $a \in  A \sm \mathfrak{m}$. Then $a \mod \mathfrak{m} \neq 0$ and $A / \mathfrak{m}$ is a field. Hence $a \mod \mathfrak{m}$ has an inverse $x \mod \mathfrak{m}$.
+    Suppose $\mathfrak{m} \in \mSpec A$ and $a \in  A \setminus \mathfrak{m}$. Then $a \mod \mathfrak{m} \neq 0$ and $A / \mathfrak{m}$ is a field. Hence $a \mod \mathfrak{m}$ has an inverse $x \mod \mathfrak{m}$.
     $1 - ax \in \mathfrak{m}$, hence $1 - ax \not\in A^{\times}$ and $a $ is not al element of the RHS.
 
     Conversely, let $a \in A$ belong to all $\mathfrak{m} \in \mSpec A$. If there exists $x \in  A$ such that $1 - ax \not\in  A^{\times }$ then $(1-ax) A$ was a proper ideal in $A$, hence contained in a maximal ideal $\mathfrak{m}$. As $a \in \mathfrak{m}, 1 = (1-ax) + ax \in  \mathfrak{m}$, a  contradiction.
@@ -1929,7 +1929,7 @@ Let $\mathfrak{l}$ be any field.
     
     When dealing with $\bP^n$, the usual convention is to use $0$ as the index of the first coordinate.
     
-    We denote the one-dimensional subspace generated by $(x_0,\ldots,x_n) \in \mathfrak{k}^{n+1} \sm \{0\}$ by $[x_0,\ldots,x_n] \in \bP^n$.
+    We denote the one-dimensional subspace generated by $(x_0,\ldots,x_n) \in \mathfrak{k}^{n+1} \setminus \{0\}$ by $[x_0,\ldots,x_n] \in \bP^n$.
     If $x = [x_0,\ldots,x_n] \in \bP^n$, the $(x_{i})_{i=0}^n$ are called \vocab{homogeneous coordinates}  of $x$.
     At least one of the $x_{i}$ must be $\neq 0$.
 \end{definition}
@@ -1939,7 +1939,7 @@ Let $\mathfrak{l}$ be any field.
 \begin{definition}[Infinite hyperplane]
     For $0 \le  i \le n$ let $U_i \subseteq \bP^n$ denote the set of $[x_0,\ldots,x_{n}]$ with $x_{i}\neq 0$.
     This is a correct definition since two different sets $[x_0,\ldots,x_{n}]$ and $[\xi_0,\ldots,\xi_n]$ of homogeneous coordinates for the same point $x \in \bP^n$ differ by scaling with a $\lambda \in  \mathfrak{l}^{\times}$, $x_i = \lambda \xi_i$. Since not all $x_i$ may be $0$, $\bP^n = \bigcup_{i=0}^n U_i$. We identify $\bA^n = \bA^n(\mathfrak{l}) = \mathfrak{l}^n$ with $U_0$ by identifying $(x_1,\ldots,x_n) \in \bA^n$ with $[1,x_1,\ldots,x_n] \in \bP^n$.
-    Then $\bP^1 = \bA^1 \cup \{\infty\} $ where $\infty=[0,1]$. More generally, when $n > 0$ $\bP^n \sm \bA^n$ can be identified with $\bP^{n-1}$ identifying $[0,x_1,\ldots,x_n] \in \bP^n \sm \bA^n$ with $[x_1,\ldots,x_n] \in \bP^{n-1}$.
+    Then $\bP^1 = \bA^1 \cup \{\infty\} $ where $\infty=[0,1]$. More generally, when $n > 0$ $\bP^n \setminus \bA^n$ can be identified with $\bP^{n-1}$ identifying $[0,x_1,\ldots,x_n] \in \bP^n \setminus \bA^n$ with $[x_1,\ldots,x_n] \in \bP^{n-1}$.
 
     Thus $\bP^n$ is $\bA^n \cong \mathfrak{l}^n$ with a copy of $\bP^{n-1}$ added as an \vocab{infinite hyperplane} .
 \end{definition}
@@ -2060,8 +2060,8 @@ Let $\mathfrak{l}$ be any field.
 \end{proof}
 \begin{corollary}
     The Zariski topology on $\bP^n$ is indeed a topology.
-    The induced topology on the open set $\bA^n = \bP^n \sm \Vp(X_0) \cong \mathfrak{k}^n$ is the Zariski topology on $\mathfrak{k}^n$.
+    The induced topology on the open set $\bA^n = \bP^n \setminus \Vp(X_0) \cong \mathfrak{k}^n$ is the Zariski topology on $\mathfrak{k}^n$.
-    The same holds for all $U_i = \bP^n \sm \Vp(X_i) \cong \mathfrak{k}^n$.
+    The same holds for all $U_i = \bP^n \setminus \Vp(X_i) \cong \mathfrak{k}^n$.
     Moreover, the topological space $\bP^n$ is Noetherian.
 \end{corollary}
 
@@ -2119,8 +2119,8 @@ Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.
     For a graded ring $R_\bullet$, let $\Proj(R_\bullet)$ be the set of  $\fp \in \Spec R$ such that $\fp$ is a homogeneous ideal and $\fp \not\supseteq R_+$.
 \end{definition}
 \begin{remark}\label{proja}
-    As the elements of $A_0 \sm \{0\}$ are units in $A$ it follows that for every homogeneous ideal $I$ we have $I \subseteq A_+$ or $I = A$.
+    As the elements of $A_0 \setminus \{0\}$ are units in $A$ it follows that for every homogeneous ideal $I$ we have $I \subseteq A_+$ or $I = A$.
-    In particular, $\Proj(A_\bullet) = \{\fp \in  \Spec A \sm A_+ |  \fp \text{ is homogeneous}\} $.
+    In particular, $\Proj(A_\bullet) = \{\fp \in  \Spec A \setminus A_+ |  \fp \text{ is homogeneous}\} $.
 \end{remark}
 \begin{proposition}\label{bijproj}
     There is a bijection 
@@ -2138,10 +2138,10 @@ Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.
 
 
     Suppose $\fp \in \Proj(A_\bullet)$. Then $\fp \neq  A_+$ hence $X = \Vp(\fp) \neq \emptyset$ by the proven part of the proposition.
-    Assume $X = X_1 \cup X_2$ is a decomposition into proper closed subsets, where $X_k = \Vp(I_k)$ for some $I_k \subseteq A_+, I_k = \sqrt{I_k}$. Since $X_k$ is a proper subset of $X$, there is $f_k \in I_k \sm \fp$.
+    Assume $X = X_1 \cup X_2$ is a decomposition into proper closed subsets, where $X_k = \Vp(I_k)$ for some $I_k \subseteq A_+, I_k = \sqrt{I_k}$. Since $X_k$ is a proper subset of $X$, there is $f_k \in I_k \setminus \fp$.
     We have $\Vp(f_1f_2) \supseteq \Vp(f_k) \supseteq \Vp(I_k)$ hence $\Vp(f_1f_2) \supseteq \Vp(I_1) \cup \Vp(I_2) = X = \Vp(\fp)$ and it follows that $f_1f_2\in  \sqrt{\fp} = \fp \lightning$.
 
-    Assume $X = \Vp(\fp)$ is irreducible, where $\fp = \sqrt{\fp}  \in A_+$ is homogeneous. The $\fp \neq  A_+$ as $X = \emptyset$ otherwise. Assume that $f_1f_2 \in  \fp$ but $f_i \not\in  A_{d_i} \sm \fp$.
+    Assume $X = \Vp(\fp)$ is irreducible, where $\fp = \sqrt{\fp}  \in A_+$ is homogeneous. The $\fp \neq  A_+$ as $X = \emptyset$ otherwise. Assume that $f_1f_2 \in  \fp$ but $f_i \not\in  A_{d_i} \setminus \fp$.
     Then $X \not \subseteq \Vp(f_i)$ by the projective Nullstellensatz when $d_i > 0$ and because $\Vp(1) = \emptyset$ when $d_i = 0$.
     Thus $X = (X \cap \Vp\left( f_1 \right)) \cup (X \cap \Vp(f_2))$ is a proper decomposition $\lightning$.
     By lemma \ref{homprime}, $\fp$ is a prime ideal.
@@ -2189,7 +2189,7 @@ Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.
     \end{itemize}
 \end{proposition}
 \begin{proof}
-    Let $X \subseteq \bP^n$ be irreducible. If $x \in X$, there is an integer $0 \le i \le n$ and $X \in  U_i = \bP^n \sm \Vp(X_i)$.
+    Let $X \subseteq \bP^n$ be irreducible. If $x \in X$, there is an integer $0 \le i \le n$ and $X \in  U_i = \bP^n \setminus \Vp(X_i)$.
     \Wlog $i = 0$. Then $\codim(X, \bP^n) = \codim(X \cap \bA^n, \bA^n)$ by the locality of Krull codimension (\ref{lockrullcodim}).
     Applying this with $X = \{x\}$ and our results about the affine case gives the second assertion.
     If $Y$ and $Z$ are also irreducible with $X \subseteq Y \subseteq Z$, then $\codim(X,Y) = \codim(X \cap \bA^n, Y \cap \bA^n)$, $\codim(X,Z) = \codim(X \cap \bA^n, Z \cap \bA^n)$ and $\codim(Y,Z) = \codim(Y \cap \bA^n, Z \cap \bA^n)$.
@@ -2207,7 +2207,7 @@ Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.
 \begin{definition}
     If $X \subseteq \bP^n$ is closed, we define the \vocab{affine cone over $X$}
     \[
-        C(X) = \{0\} \cup \{(x_0,\ldots,x_n) \in  \mathfrak{k}^{n+1} \sm \{0\}  | [x_0,\ldots,x_n] \in X\} 
+        C(X) = \{0\} \cup \{(x_0,\ldots,x_n) \in  \mathfrak{k}^{n+1} \setminus \{0\}  | [x_0,\ldots,x_n] \in X\} 
     \] 
     If $X = \Vp(I)$ where $I \subseteq A_+ = \mathfrak{k}[X_0,\ldots,X_n]_+$ is homogeneous, then $C(X) = \Va(I)$.
 \end{definition}
@@ -2380,12 +2380,12 @@ is an isomorphism.
 
     Let $\phi \in  \cO_X(X)$. for $x \in X$, there are an open subset $U_x \subseteq X$ and $f_x, g_x \in R$ such that $\phi = \frac{f_x}{g_x}$ on $U_x$.
     \begin{claim}
-        \Wlog we can assume $U_x = X \sm V(g_x)$.
+        \Wlog we can assume $U_x = X \setminus V(g_x)$.
     \end{claim}
     \begin{subproof}
-        The closed subsets $(X \sm U_x) \subseteq \mathfrak{k}^n$ has the form $X\sm U_x = V(J_x)$ for some ideal $J_x \subseteq R$.
+        The closed subsets $(X \setminus U_x) \subseteq \mathfrak{k}^n$ has the form $X\setminus U_x = V(J_x)$ for some ideal $J_x \subseteq R$.
-        As $x \not\in  X \sm V_x$ there is $h_x \in J_x$ with $h_x(x) \neq 0$.
+        As $x \not\in  X \setminus V_x$ there is $h_x \in J_x$ with $h_x(x) \neq 0$.
-        Replacing $U_x$ by $X \sm V(h_x)$, $f_x$ by $f_xh_x$ and $g_x$ by $g_xh_x$, we may assume $U_x = X \sm V(g_x)$.
+        Replacing $U_x$ by $X \setminus V(h_x)$, $f_x$ by $f_xh_x$ and $g_x$ by $g_xh_x$, we may assume $U_x = X \setminus V(g_x)$.
     \end{subproof}
     \begin{claim}
         \Wlog we can assume $V(g_x)  \subseteq V(f_x)$.
@@ -2396,7 +2396,7 @@ is an isomorphism.
     As $X$ is quasi-compact, there are finitely many points $(x_i)_{i=1}^m$ such that the $U_{x_i}$ cover $X$.
     Let $U_i \coloneqq U_{x_i}, f_i \coloneqq f_{x_i}, g_i \coloneqq g_{x_i}$.
 
-    As the $U_i = X \sm V(g_i)$ cover $X$, $V(I) \cap \bigcap_{i=1}^m V(g_i) = X \cap \bigcap_{i=1}^m V(g_i) = \emptyset$.
+    As the $U_i = X \setminus V(g_i)$ cover $X$, $V(I) \cap \bigcap_{i=1}^m V(g_i) = X \cap \bigcap_{i=1}^m V(g_i) = \emptyset$.
     By the Nullstellensatz (\ref{hns1}) the ideal of $R$ generated by $I$ and the $a_i$ equals $R$. 
     There are thus $n \ge m \in \N$ and elements $(g_i)_{i = m+1}^n$ of $I$ and $(a_i)_{i=1}^n \in R^n$ such that $1 = \sum_{i=1}^{n} a_ig_i$.
     Let for $i > m$ $f_i \coloneqq 0$, $F = \sum_{i=1}^{n} a_if_i = \sum_{i=1}^{m} a_if_i \in R$.
@@ -2406,7 +2406,7 @@ is an isomorphism.
     \end{claim}
     \begin{subproof}
         If $x \in V_i$ this follows by our choice of $f_i$ and $g_i$.
-        If $x \in X \sm V_i$ or $i > m$ both sides are zero.
+        If $x \in X \setminus V_i$ or $i > m$ both sides are zero.
     \end{subproof}
     It follows that
     \[
@@ -2423,14 +2423,14 @@ Let $X \subseteq \bP^n$ be closed and $R_\bullet = \mathfrak{k}[X_0,\ldots,X_n]$
 
 \begin{remark}
     This is a subsheaf of rings of the sheaf of $\mathfrak{k}$-valued functions on $X$.
-Under the identification $\bA^n =\mathfrak{k}^n$ with $\bP^n \sm \Vp(X_0)$, one has $\cO_X \defon{X \sm \Vp(X_0)} = \cO_{X \cap \bA^n}$ as subsheaves of the sheaf of $\mathfrak{k}$-valued functions, where the second sheaf is a sheaf on a closed subset of $\mathfrak{k}^n$:
+Under the identification $\bA^n =\mathfrak{k}^n$ with $\bP^n \setminus \Vp(X_0)$, one has $\cO_X \defon{X \setminus \Vp(X_0)} = \cO_{X \cap \bA^n}$ as subsheaves of the sheaf of $\mathfrak{k}$-valued functions, where the second sheaf is a sheaf on a closed subset of $\mathfrak{k}^n$:
 
 Indeed, if $W$ is as in the definition then $\phi([1,y_1,\ldots,y_n]) = \frac{f(1,y_1,\ldots,y_n)}{g(1,y_1,\ldots,y_n)}$ for $[1,y_1,\ldots,y_n] \in  W$.
 Conversely  if $\phi([1,y_1,\ldots,y_n]) = \frac{f(y_1,\ldots,y_n)}{g(y_1,\ldots,y_n)}$ on an open subset $W $ of $X \cap \bA^n$ then 
 $\phi([y_0,\ldots,y_n]) = \frac{F(y_0,\ldots,y_n)}{G(y_0,\ldots,y_n)}$ on $W$ where $F(X_0,\ldots,X_n) \coloneqq X_0^d f(\frac{X_1}{X_0}, \ldots, \frac{X_n}{X_0})$ and $G(X_0,\ldots,X_n) = X_0^d g(\frac{X_1}{X_0},\ldots, \frac{X_n}{X_0})$ with a sufficiently large $d \in \N$.
 \end{remark}
 \begin{remark}
-    It follows from the previous remark and the similar result in the affine case that the elements of $\cO_X(U)$ are continuous on $U \sm V(X_0)$.
+    It follows from the previous remark and the similar result in the affine case that the elements of $\cO_X(U)$ are continuous on $U \setminus V(X_0)$.
     Since the situation is symmetric in the homogeneous coordinates, they are continuous on all of $U$.
 \end{remark}
 The following is somewhat harder than in the affine case:
@@ -2557,7 +2557,7 @@ The following is somewhat harder than in the affine case:
             The $V_x$ cover $U$. By the sheaf axiom for $\cO_X$ there is $\ell \in \cO_X(U)$ with $\ell\defon{V_x} =\lambda_x$. It follows that  $\ell=\lambda$.
         \item By the definition of variety, every $x \in U$ has a quasi-affine neighbourhood $V \subseteq U$. We can assume  $U$ to be quasi-affine and $X = V(I) \subseteq \mathfrak{k}^n$, as the general assertion follows by an application of ii).
             If $x \in U$ there are a neighbourhood $x \in W \subseteq U$ and $a,b \in R = \mathfrak{k}[X_1,\ldots,X_n]$ such that $\vartheta(y) = \frac{a(y)}{b(y)}$ for $y \in W$, with $b(y) \neq 0$.
-            Then $a(x) \neq 0$ as $\vartheta(x) \neq 0$. Replacing $W$ by $W \sm V(a)$, we may assume that $a$ has no zeroes on $W$.
+            Then $a(x) \neq 0$ as $\vartheta(x) \neq 0$. Replacing $W$ by $W \setminus V(a)$, we may assume that $a$ has no zeroes on $W$.
             Then $\lambda(y) = \frac{b(y)}{a(y)}$ for $y \in W$ has a non-vanishing denominator and $\lambda \in \cO_X(U)$.
             We have $\lambda \cdot \vartheta = 1$, thus $\vartheta \in  \cO_X(U)^{\times}$.
     \end{enumerate}
@@ -2612,7 +2612,7 @@ The following is somewhat harder than in the affine case:
         $f$ is a morphism in $\Var_\mathfrak{k}$
     \end{claim}
     \begin{subproof}
-        For open $\Omega \subseteq Y, U = f^{-1}(\Omega) = \{x \in X | \A \lambda \in J ~ (\phi(\lambda))(x) \neq 0\}$ is open in $X$, where $Y \sm \Omega = V(J)$.
+        For open $\Omega \subseteq Y, U = f^{-1}(\Omega) = \{x \in X | \A \lambda \in J ~ (\phi(\lambda))(x) \neq 0\}$ is open in $X$, where $Y \setminus \Omega = V(J)$.
         If $\lambda \in \cO_Y(\Omega)$ and $x \in U$, then $f(x)$ has a neighbourhood $V$ such that there are $a,b \in R$ with $\lambda(v) = \frac{a(v)}{b(v)}$ and $b(v) \neq 0$ for all $v \in V$.
         Let $W \coloneqq f^{-1}(V)$. Then $\alpha \coloneqq \phi(a)\defon{W} \in \cO_X(W)$,  $\beta \coloneqq \phi(b)\defon{W} \in \cO_X(W)$.
         By the second part of \ref{localinverse} $\beta \in \cO_X(W)^{\times}$ and $f\st(\lambda)\defon{W}  = \frac{\alpha}{\beta} \in \cO_X(W)$.
@@ -2652,11 +2652,11 @@ If $X$ is a set, then $\cB \subseteq \cP(X)$ is a base for some topology on $X$
     
 \end{definition}
 \begin{proposition}\label{oxulocaf}
-    Let $X$ be an affine variety over $\mathfrak{k}$, $\lambda \in \cO_X(X)$ and $U = X \sm V(\lambda)$.
+    Let $X$ be an affine variety over $\mathfrak{k}$, $\lambda \in \cO_X(X)$ and $U = X \setminus V(\lambda)$.
     Then  $U$ is an affine variety and the morphism $\phi: \cO_X(X)_\lambda \to \cO_X(U)$ defined by the restriction $\cO_X(X) \xrightarrow{\cdot |_U } \cO_X(U)$ and the universal property of the localization is an isomorphism.
 \end{proposition}
 \begin{proof}
-    Let $X$ be an affine variety over $\mathfrak{k}, \lambda \in  \cO_X(X)$ and $U = X \sm V(\lambda)$. The fact that $\lambda\defon{U}  \in \cO_x(U)^{\times}$ follows from \ref{localinverse}.
+    Let $X$ be an affine variety over $\mathfrak{k}, \lambda \in  \cO_X(X)$ and $U = X \setminus V(\lambda)$. The fact that $\lambda\defon{U}  \in \cO_x(U)^{\times}$ follows from \ref{localinverse}.
     Thus the universal property of the localization $\cO_X(X)_\lambda$ can be applied to $\cO_X(X) \xrightarrow{\cdot |_U} \cO_X(U)$.
     \begin{figure}[H]
         \centering
@@ -2689,7 +2689,7 @@ If $X$ is a set, then $\cB \subseteq \cP(X)$ is a base for some topology on $X$
     The affine open subsets of a variety $X$ are a topology base on $X$.
 \end{corollary}
 \begin{proof}
-    Let $X = V(I) \subseteq \mathfrak{k}^n$ with $I = \sqrt{I}$. If $U \subseteq X$ is open then $X \sm U = V(J)$ with $J \supseteq I$ and $U = \bigcup_{f \in J} (X \sm V(f))$.
+    Let $X = V(I) \subseteq \mathfrak{k}^n$ with $I = \sqrt{I}$. If $U \subseteq X$ is open then $X \setminus U = V(J)$ with $J \supseteq I$ and $U = \bigcup_{f \in J} (X \setminus V(f))$.
     Thus $U$ is a union of affine open subsets. The same then holds for arbitrary quasi-affine varieties.
     
     Let $X$ be any variety, $U \subseteq X$ open and $x \in U$.
@@ -2753,9 +2753,9 @@ If $X$ is a set, then $\cB \subseteq \cP(X)$ is a base for some topology on $X$
     This is indeed a local ring, with maximal ideal $\mathfrak{m}_x = \{f \in \cO_{X,x} | f(x) = 0\}$.
 \end{definition}
 \begin{proof}
-    By \ref{localring} it suffices to show that $\mathfrak{m}_x$ is a proper ideal, which is trivial, and that the elements of $\cO_{X,x} \sm \mathfrak{m}_x$ are units in $\cO_{X,x}$.
+    By \ref{localring} it suffices to show that $\mathfrak{m}_x$ is a proper ideal, which is trivial, and that the elements of $\cO_{X,x} \setminus \mathfrak{m}_x$ are units in $\cO_{X,x}$.
     Let $g = (U, \gamma)/\sim \in \cO_{X,x}$ and $g(x) \neq 0$.
-    $\gamma$ is  Zariski continuous (first point of \ref{localinverse}). Thus $V(\gamma)$ is closed. By replacing $U$ by $U \sm V(\gamma)$ we may assume that $\gamma$ vanishes nowhere on $U$.
+    $\gamma$ is  Zariski continuous (first point of \ref{localinverse}). Thus $V(\gamma)$ is closed. By replacing $U$ by $U \setminus V(\gamma)$ we may assume that $\gamma$ vanishes nowhere on $U$.
     By the third point of \ref{localinverse} we have $\gamma \in \cO_X(U)^{\times}$.
     $(\gamma^{-1})_x$ is an inverse to $g$.
 \end{proof}
@@ -2763,7 +2763,7 @@ If $X$ is a set, then $\cB \subseteq \cP(X)$ is a base for some topology on $X$
 \begin{proposition}\label{proplocalring}
     Let $X = \Va(I) \subseteq \mathfrak{k}^n$ be equipped with its usual structure sheaf, where $I = \sqrt{I} \subseteq R = \mathfrak{k}[X_1,\ldots,X_n]$ . Let $x \in X$ and $A = \cO_X(X) \cong R / I$.
     $\{P \in  R | P(x) = 0\} \text{\reflectbox{$\coloneqq$}} \fn_x \subseteq R$ is maximal, $I \subseteq \fn_x$ and  $\mathfrak{m}_x \coloneqq \fn_x / I$ is the maximal ideal of elements of $A$ vanishing at $x$.
-    If $\lambda \in  A \sm \mathfrak{m}_x$, we have $\lambda_x \in  \cO_{X,x}^{\times}$, where $\lambda_x$ denotes the image under $A \cong \cO_X(X) \to  \cO_{X,x}$.
+    If $\lambda \in  A \setminus \mathfrak{m}_x$, we have $\lambda_x \in  \cO_{X,x}^{\times}$, where $\lambda_x$ denotes the image under $A \cong \cO_X(X) \to  \cO_{X,x}$.
     By the universal property of the localization, there exists a unique ring homomorphism $A_{\mathfrak{m}_x} \xrightarrow{\iota} \cO_{X,x}$ 
     such that 
     \begin{figure}[H]
@@ -2780,22 +2780,22 @@ If $X$ is a set, then $\cB \subseteq \cP(X)$ is a base for some topology on $X$
 \end{proposition}
 \begin{proof}
     To show surjectivity, let $\ell = (U, \lambda) / \sim  \in  \cO_{X,x}$, where $U$ is an open neighbourhood of $x$ in $X$. 
-    We have $X \sm U = V(J)$ where $J \subseteq A$ is an ideal. As $x \in U$ there is $f \in J$ with $f(x) \neq 0$. Replacing $U $ by $X \sm V(f)$ we may assume $U = X \sm V(f)$.
+    We have $X \setminus U = V(J)$ where $J \subseteq A$ is an ideal. As $x \in U$ there is $f \in J$ with $f(x) \neq 0$. Replacing $U $ by $X \setminus V(f)$ we may assume $U = X \setminus V(f)$.
     By \ref{oxulocaf}, $\cO_X(U) \cong A_f$, and $\lambda = f^{-n}\vartheta$ for some $n \in \N$ and $\vartheta \in A$.
     Then $\ell = \iota(f^{-n} \vartheta)$ where the last fraction is taken in $A_{\mathfrak{m}_x}$.
 
 
     Let $\lambda = \frac{\vartheta}{g} \in  A_{\mathfrak{m}_x}$ with $\iota(\lambda) = 0$.
-    It is easy to see that $\iota(\lambda) = (X \sm V(g), \frac{\vartheta}{g}) / \sim $.
+    It is easy to see that $\iota(\lambda) = (X \setminus V(g), \frac{\vartheta}{g}) / \sim $.
-    Thus there is an open neighbourhood $U$ of $x$ in $X \sm V(g)$ such that $\vartheta$ vanishes on $U$.
+    Thus there is an open neighbourhood $U$ of $x$ in $X \setminus V(g)$ such that $\vartheta$ vanishes on $U$.
-    Similar as before there is $h \in A$ with $h(x) \neq 0$ and $W = X \sm V(h) \subseteq U$.
+    Similar as before there is $h \in A$ with $h(x) \neq 0$ and $W = X \setminus V(h) \subseteq U$.
     By the isomorphism $\cO_X(W) \cong A_h$, there is $n \in \N$ with $h^{n}\vartheta = 0$ in $A$. Since $h \not\in \mathfrak{m}_x$, $h$ is a unit and the image of $\vartheta$ in $A_{\mathfrak{m}_x}$ vanishes, implying $\lambda = 0$.
 \end{proof}
 \subsubsection{Intersection multiplicities and Bezout's theorem}
 \begin{definition}
     Let $R = \mathfrak{k}[X_0,X_1,X_2]$  equipped with its usual grading and let $x \in \bP^{2}$.
 Let $G \in R_g, H \in R_h$ be homogeneous polynomials with $x \in V(G) \cap V(h)$.
-Let $\ell\in R_1$ such that $\ell(x) \neq 0$. Then $x \in U = \bP^2 \sm V(\ell)$ and the rational functions $\gamma = \ell^{-g}G, \eta = \ell^{-h}H$ are elements of $\cO_{\bP^2}(U)$.
+Let $\ell\in R_1$ such that $\ell(x) \neq 0$. Then $x \in U = \bP^2 \setminus V(\ell)$ and the rational functions $\gamma = \ell^{-g}G, \eta = \ell^{-h}H$ are elements of $\cO_{\bP^2}(U)$.
 Let $I_x(G,H) \subseteq \cO_{\bP^2,x}$ denote the ideal generated by $\gamma_x$ and $\eta_x$.
 
 
@@ -2858,7 +2858,7 @@ For $s_1 \neq s_2$, % TODO
 
 Noether normalization: 
 $a_i \in A$ minimal  such that $A$ is integral over the subalgebra genereted by the $a_i$.
-Suppose $\exists P \in K[X_1,\ldots,X_n] \sm \{ 0\} ~ P(a_1,\ldots,a_n) = 0$. $P = \sum_{\alpha \in \N^n} p_\alpha X^\alpha, S \coloneqq \{ \alpha \in  \N^n | p_\alpha \neq 0\}$.
+Suppose $\exists P \in K[X_1,\ldots,X_n] \setminus \{ 0\} ~ P(a_1,\ldots,a_n) = 0$. $P = \sum_{\alpha \in \N^n} p_\alpha X^\alpha, S \coloneqq \{ \alpha \in  \N^n | p_\alpha \neq 0\}$.
 Choose $k$ as in the lemma.
 $b_i \coloneqq a_{i+1} - a_1^{k_{i+1}}, 1 \le i <n$. Claim: $A$ is integral ober subalgebra $B$ generated by the $b_i$ ($\lightning$ minimality)
 $Q(T) \coloneqq P(T, b_1 + T^{k_2}, \ldots, b_{n-1} + T^{k_n})$. $Q(a_1) = P(\vec a) = 0$.
@@ -2881,7 +2881,7 @@ For $\fq \in \mSpec A$, $\mathfrak{k}(\fq) = A / \fq$ finite type, hence finite
 $\trdeg(Q(A) / \mathfrak{l}) = 0 \implies A$ integral over $\mathfrak{k}$ $\implies$ $A$ a field $\implies \fp = \fq \lightning$.
 
 If $\fq \not\in \mSpec A$, let $a_1,\ldots,a_n \in A$ alg. independent such that the $\overline{a_i}$ are a transcendence base for $\mathfrak{k}(\fq) / \mathfrak{k}$
-Let $R$ be the ring generated by $\mathfrak{l}$ and the $a_i$. Localize with respect to $S \coloneqq R \sm \{0\}$.
+Let $R$ be the ring generated by $\mathfrak{l}$ and the $a_i$. Localize with respect to $S \coloneqq R \setminus \{0\}$.
 %TODO
 % TODO: LERNEN
 
@@ -2910,10 +2910,10 @@ Then $\Aut(L / K)$ transitively acts on $\{\fq \in \Spec B | \fq \cap A = \fp\}$
 \begin{itemize}
     \item $\fq,  \fr \in \Spec B$ lying over $\fp$.
     \item only need to show $\fq \subseteq \sigma(\fr)$ for some  $\sigma \in G$ (Krull going-up, no inclusions)
-    \item Suppose not. Then  $x \in \fq \sm \bigcup_{\sigma \in G}  \sigma(\fr)$ (prime aviodance)
+    \item Suppose not. Then  $x \in \fq \setminus \bigcup_{\sigma \in G}  \sigma(\fr)$ (prime aviodance)
-    \item $y = \prod_{\sigma \in G} \sigma(x) \in \fq \sm \fr$ ($\fr$ prime ideal)
+    \item $y = \prod_{\sigma \in G} \sigma(x) \in \fq \setminus \fr$ ($\fr$ prime ideal)
     \item $\exists k \in \N$ s.t. $y^k \in K$ ($y \in L^G$)
-    \item $y^k \in K \cap B = A $ ($A$ normal). Thus $y^k \in (A \cap \fq) \sm (A \cap \fr) = \fp \sm \fp$.
+    \item $y^k \in K \cap B = A $ ($A$ normal). Thus $y^k \in (A \cap \fq) \setminus (A \cap \fr) = \fp \setminus \fp$.
     \item  $L / K$ infinite: Apply Zorn to pairs $(M, \sigma)$ where $K \subseteq M \subseteq L$ and $\sigma \in \Aut(M /K)$ s.t. $\sigma(\fr \cap M) = \fq \cap M$.
 \end{itemize}