From 0171af5cd59b39491e001c910af12d9438fabe38 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Maximilian=20Ke=C3=9Fler?= Date: Wed, 16 Feb 2022 02:57:31 +0100 Subject: [PATCH] get rid of wlog --- algebra.sty | 1 - inputs/nullstellensatz_and_zariski_topology.tex | 4 ++-- inputs/projective_spaces.tex | 4 ++-- inputs/uebersicht.tex | 2 +- inputs/varieties.tex | 4 ++-- 5 files changed, 7 insertions(+), 8 deletions(-) diff --git a/algebra.sty b/algebra.sty index 6d24916..9fb13bf 100644 --- a/algebra.sty +++ b/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}} diff --git a/inputs/nullstellensatz_and_zariski_topology.tex b/inputs/nullstellensatz_and_zariski_topology.tex index e243c80..7b0220a 100644 --- a/inputs/nullstellensatz_and_zariski_topology.tex +++ b/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} diff --git a/inputs/projective_spaces.tex b/inputs/projective_spaces.tex index 993d4df..12250f6 100644 --- a/inputs/projective_spaces.tex +++ b/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 diff --git a/inputs/uebersicht.tex b/inputs/uebersicht.tex index fa705ea..7128c6b 100644 --- a/inputs/uebersicht.tex +++ b/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 $\implies \fp = \fq \lightning$. diff --git a/inputs/varieties.tex b/inputs/varieties.tex index 1af3933..535d310 100644 --- a/inputs/varieties.tex +++ b/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 $g_x^2$.