get rid of wlog

algebra.sty

@@ -29,7 +29,6 @@
 \DeclareMathOperator{\codim}{codim}
 \DeclareMathOperator{\trdeg}{trdeg}
 \DeclareMathOperator{\hght}{ht}
-\newcommand{\Wlog}{W.l.o.g. }
 
 \newcommand{\Vspec}{\ensuremath V_{\mathbb{S}}}%\Spec}}
 \newcommand{\Vs}{\ensuremath V_{\mathbb{S}}}%\Spec}}

inputs/nullstellensatz_and_zariski_topology.tex

@@ -497,7 +497,7 @@ In general, these inequalities may be strict.
 
     Let $x,y \in L$, $T \subseteq L$ and $x \in  \mathcal{H}(T \cup \{y\}) \setminus \mathcal{H}(T)$. We have to show that $y \in  \mathcal{H}(T \cup \{x\}) \setminus \mathcal{H}(T)$.
     If $y \in  \mathcal{H}(T)$ we have $\mathcal{H}(T \cup \{y\}) \subseteq \mathcal{H}(\mathcal{H}(T)) = \mathcal{H}(T) \implies x \in  \mathcal{H}(T) \setminus \mathcal{H}(T) \lightning$.
-    Hence it is sufficient to show $y \in  \mathcal{H}(T \cup  \{x\})$. \Wlog $T = \emptyset$ (replace $K$ be the subfield generated by $K \cup T$).
+    Hence it is sufficient to show $y \in  \mathcal{H}(T \cup  \{x\})$. Without loss of generality loss of generality $T = \emptyset$ (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 $p_i = \frac{Q_i(y)}{R(y)}$, $R(y) \neq 0$.
     Let
@@ -1133,7 +1133,7 @@ Recall the definition of a normal field extension in the case of finite field ex
     
 
     Alternative proof\footnote{\url{http://www.math.lsa.umich.edu/~tfylam/Math221/2.pdf}}:
-    Let $x = \frac{a}{b} \in Q(A)$ be integral over $A$. \Wlog $\gcd(a,b) = 1$.  Then $x^n + c_{n-1} x^{n-1} + \ldots + c_0 = 0$ for some $c_i \in A$.
+    Let $x = \frac{a}{b} \in Q(A)$ be integral over $A$. Without loss of generality loss of generality $\gcd(a,b) = 1$.  Then $x^n + c_{n-1} x^{n-1} + \ldots + c_0 = 0$ for some $c_i \in A$.
     Multiplication with $b^n$ yields $a^n + c_{n-1} b a^{n-1} + \ldots +c_0 b^n = 0$. Thus $b | a^n$. Since $\gcd(a,b) = 1$ it follows that $b$ is a unit, hence $x \in  A$.
 \end{proof}
 

inputs/projective_spaces.tex

@@ -211,7 +211,7 @@ Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.
 \end{proposition}
 \begin{proof}
     From the projective form of the Nullstellensatz it follows that $f$ is injective and that $f^{-1}(\Vp\left( I \right)) = \sqrt{I} = I$.
-    If $X \subseteq \mathbb{P}^n$ is closed, then $X = \Vp(J)$ for some homogeneous ideal $J \subseteq A$. \Wlog $J = \sqrt{J}$. If $J \not\subseteq A_+$, then $J = A$ (\ref{proja}), hence $X = \Vp(J) = \emptyset = \Vp(A_+)$.
+    If $X \subseteq \mathbb{P}^n$ is closed, then $X = \Vp(J)$ for some homogeneous ideal $J \subseteq A$. Without loss of generality loss of generality $J = \sqrt{J}$. If $J \not\subseteq A_+$, then $J = A$ (\ref{proja}), hence $X = \Vp(J) = \emptyset = \Vp(A_+)$.
     Thus we may assume $J \subseteq A_+$, and $f$ is surjective.
 
 
@@ -268,7 +268,7 @@ Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.
 \end{proposition}
 \begin{proof}
     Let $X \subseteq \mathbb{P}^n$ be irreducible. If $x \in X$, there is an integer $0 \le i \le n$ and $X \in  U_i = \mathbb{P}^n \setminus \Vp(X_i)$.
-    \Wlog $i = 0$. Then $\codim(X, \mathbb{P}^n) = \codim(X \cap \mathbb{A}^n, \mathbb{A}^n)$ by the locality of Krull codimension (\ref{lockrullcodim}).
+    Without loss of generality loss of generality $i = 0$. Then $\codim(X, \mathbb{P}^n) = \codim(X \cap \mathbb{A}^n, \mathbb{A}^n)$ 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 $\codim(X,Y) = \codim(X \cap \mathbb{A}^n, Y \cap \mathbb{A}^n)$, $\codim(X,Z) = \codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$ and $\codim(Y,Z) = \codim(Y \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$.
     Thus

inputs/uebersicht.tex

@@ -57,7 +57,7 @@ Thus $Q(T) = p_\alpha T^{w_k(\alpha)} + \ldots$ where $\alpha \in S$ such that $
 % 
 A first result of dimension theory:
 $A \mathfrak{l}$-algebra of finite type, $\fp, \fq \in \Spec A, \fp \subsetneq \fq$. Then $\trdeg(\mathfrak{k}(\fp) /\mathfrak{l}) > \trdeg(\mathfrak{k}(\fq) / \mathfrak{l})$:
-\Wlog  $\fp = \{0\}$ and $A$ a domain ($A' \coloneqq A / \fp$).
+Without loss of generality loss of generality  $\fp = \{0\}$ and $A$ a domain ($A' \coloneqq A / \fp$).
 For $\fq \in \MaxSpec A$, $\mathfrak{k}(\fq) = A / \fq$ finite type, hence finite (HNS)  $\implies \trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) = 0$.
 $\trdeg(Q(A) / \mathfrak{l}) = 0 \implies A$ integral over $\mathfrak{k}$ $\implies$ $A$ a field $\implies \fp = \fq \lightning$.
 

inputs/varieties.tex

@@ -92,7 +92,7 @@ is an isomorphism.
 
     Let $\phi \in  \mathcal{O}_X(X)$. 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 $U_x = X \setminus V(g_x)$.
+        Without loss of generality loss of generality we can assume $U_x = X \setminus V(g_x)$.
     \end{claim}
     \begin{subproof}
         The closed subsets $(X \setminus U_x) \subseteq \mathfrak{k}^n$ has the form $X\setminus U_x = V(J_x)$ for some ideal $J_x \subseteq R$.
@@ -100,7 +100,7 @@ is an isomorphism.
         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 $V(g_x)  \subseteq V(f_x)$.
+        Without loss of generality loss of generality we can assume $V(g_x)  \subseteq V(f_x)$.
     \end{claim}
     \begin{subproof}
         Replace $f_x$ by $f_xg_x$ and $g_x$ by $g_x^2$.