fix tikzcd in figures

2021_Algebra_I.tex

@@ -971,19 +971,20 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$.
 \begin{definition}[Multiplicative subset]
     A \vocab{multiplicative subset} of a ring $R$ is a subset $S \subseteq R$ such that $\prod_{i=1}^{n} f_i \in S $ when $n \in \N$ and all $f_i \in S$. 
 \end{definition}
+
 \begin{proposition}
     Let $S \subseteq R$  be a multiplicative subset. Then there is a ring homomorphism $R \xrightarrow{i} R_S$ such that $i(S) \subseteq R_S^{\times }$ and $i$ has the \vocab{universal property}  for such ring homomorphisms:
     If $R \xrightarrow{j} T$ is a ring homomorphism with $j(S) \subseteq T^{\times }$, then there is a unique ring homomorphism $R_S \xrightarrow{\iota} T$ with $j = \iota i$.
 
-    \begin{figure}[H]
-        \centering
+    \[
         \begin{tikzcd}
             R \arrow{r}{i}\arrow{d}{j}& R_S \arrow[dotted]{ld}{\existsone \iota}\\
             T 
         \end{tikzcd}
-    \end{figure}
+        \]
     
 \end{proposition}
+
 \begin{proof}
     The construction is similar to the construction of the field of quotients:
 
@@ -1078,14 +1079,13 @@ Because the elements of $S$ become units in $R_S$, $J \sqcap R$ is an $S$-satura
     $\tau_3(I_{S}) = 0$, hence a unique $R_S / I_S \xrightarrow{\tau_4} T$ whose composition with $\pi_{R_S, I_S}$ equals $\tau_3$ exists.
     This is the only one for which the composition $R \to  R_S \to  R_S / I_S \xrightarrow{\tau_4} T$ equals $\tau$. 
 
-\begin{figure}[H]
-    \centering
+    \[
     \begin{tikzcd}
         R \arrow{r}{\tau}\arrow[swap]{d}{\pi_{R,I}}& T &  R\arrow[swap]{l}{\tau}\arrow{d}{}\\
         R / I \arrow[dotted]{ru}{\existsone \tau_1}\arrow{d}{} & & R_S \arrow[dotted, swap]{lu}{\existsone \tau_3}\arrow{d}{\pi_{R_S, I_S}}\\
         (R / I)_{\overline{S}} \arrow[dotted,bend right]{ruu}{\existsone \tau_2} & & R_S / I_S \arrow[dotted, bend left, swap]{luu}{\existsone \tau_4}\\
     \end{tikzcd}
-\end{figure}
+  \]
 
 
 \end{proof}
@@ -1233,26 +1233,25 @@ Many questons of commutative algebra are easier in the case of local rings. Loca
 
     (We are given $\fp \subseteq \tilde \fp$ and  $\fq$ such that  $\fp = \fq \sqcap R$ and must make $\fq$ larger).
 
-    \begin{figure}[H]
-        \centering
+    \[
         \begin{tikzcd}
             \fq \arrow[mapsto]{d}{\cdot \sqcap R}& \subseteq &{\color{blue}\tilde\fq}\arrow[mapsto]{d}{\cdot \sqcap R}  & \in \Spec A\\
             \fq \sqcap R = \fp & \subseteq & \tilde \fp & \in \Spec R
         \end{tikzcd}
-    \end{figure}
+      \]
 
 
 \vocab{Going-down} holds for $A / R$ if for arbitrary $\tilde \fq \in  \Spec A$ and arbitrary $\fp \in \Spec R$ with $\fp \subseteq \tilde \fq \sqcap R$, there exists $\fq \in \Spec A$ with $\fq \subseteq \tilde \fq$ and $\fp = \fq \sqcap R$.
 
 (We are given $\fp \subseteq \tilde \fp$ and  $\tilde \fq$ such that $\tilde \fp = \tilde \fq \sqcap R$ and must make $\tilde \fq$ smaller).
-    \begin{figure}[H]
-        \centering
+    \[
         \begin{tikzcd}
             {\color{blue}\fq} \arrow[mapsto]{d}{\cdot \sqcap R}& \subseteq &\tilde\fq\arrow[mapsto]{d}{\cdot \sqcap R}  & \in \Spec A\\
             \fp & \subseteq & \tilde \fp = \tilde \fq \sqcap R & \in \Spec R
         \end{tikzcd}
-    \end{figure}
+      \]
 \end{definition}
+
 \begin{remark}
     In the situation of \ref{goupgodown}, we say $\fq \in \Spec A$ \vocab[Primeideal!lies above]{lies above} $\fp \in \Spec R$ if $\fq \sqcap R = \fp$.
 \end{remark}
@@ -1276,13 +1275,12 @@ Many questons of commutative algebra are easier in the case of local rings. Loca
         \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]
-                \centering
+            \[
                 \begin{tikzcd}
             R \arrow{r}{\rho}\arrow[hookrightarrow]{d}{\subseteq}&  R_\fp \arrow[hookrightarrow, dotted]{d}{\existsone i}\\
                     A \arrow{r}{\alpha} & A_\fp
                 \end{tikzcd}
-            \end{figure}
+            \]
            \begin{claim}
                $A_\fp$ is integral over $R_\fp$.
            \end{claim} 
@@ -1500,13 +1498,12 @@ Recall the definition of a normal field extension in the case of finite field ex
     It follows from the assumptions that the field of quotients $Q(B)$ is an algebraic field extension of $Q(A)$.
     There is an algebraic extension $L$ of $Q(B)$ such that $L / Q(A)$ is normal (for instance an algebraic closure of $Q(B)$).
     Let  $C$ be the integral closure of $A$ in $L$. Then $B \subseteq C$ and $C / B$ is integral.
-\begin{figure}[H]
-    \centering
+  \[
     \begin{tikzcd}
         Q(A) \arrow[hookrightarrow]{r}{} &    Q(B) \arrow[hookrightarrow]{r}{} & L \coloneqq \overline{Q(B)} \\
         A \arrow[hookrightarrow]{u}{}\arrow[hookrightarrow]{r}{} & B \arrow[hookrightarrow]{r}{}\arrow[hookrightarrow]{u}{}& C \arrow[hookrightarrow]{u}{}\\	
     \end{tikzcd}
-\end{figure}
+  \]
 
     \begin{claim}
         Going-down holds for $C / A$.
@@ -2655,18 +2652,18 @@ If $X$ is a set, then $\cB \subseteq \mathcal{P}(X)$ is a base for some topology
 \begin{proof}
     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
+    \[
         \begin{tikzcd}
             \cO_X(X) \arrow{d}{\cdot |_U}\arrow{r}{x \mapsto \frac{x}{1}} & \cO_X(X)_\lambda \arrow[dotted, bend left]{dl}{\existsone \phi} \\
             \cO_X(U) &
         \end{tikzcd}
-        \hspace{50pt}
+      \]
+      \[
         \begin{tikzcd}
             &Y \arrow[bend right, swap]{ld}{\pi_0} \arrow[bend right, swap]{d}{\pi}&\cO_Y(Y) \cong A_\lambda \arrow{d}{\mathfrak{s}}& \\
             X \arrow[hookrightarrow]{r}{}& U \arrow[swap]{u}{\sigma} & \cO_X(U)
         \end{tikzcd}
-    \end{figure}
+      \]
     For the rest of the proof, we may assume $X = V(I) \subseteq \mathfrak{k}^n$ where $I = \sqrt{I} \subseteq R \coloneqq\mathfrak{k}[X_1,\ldots,X_n]$ is an ideal.
     Then $A \coloneqq \cO_X(X) \cong R / I$ and there is $\ell \in R$ such that $\ell\defon{X} = \lambda$.
     Let $Y = V(J) \subseteq \mathfrak{k}^{n+1}$ where $J \subseteq \mathfrak{k}[Z,X_1,\ldots,X_n]$ is generated by the elements of $I$ and $1 - Z\ell(X_1,\ldots,X_n)$.
@@ -2763,13 +2760,12 @@ If $X$ is a set, then $\cB \subseteq \mathcal{P}(X)$ is a base for some topology
     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]
-        \centering
+    \[
         \begin{tikzcd}
             A \arrow{r}{} \arrow{d}{\lambda \mapsto \lambda_x} & A_{\mathfrak{m}_x} \arrow[dotted, bend left]{ld}{\existsone \iota} \\
             \cO_{X,x}
         \end{tikzcd}
-    \end{figure}
+      \]
     commutes.
    
     The morphism $A_{\mathfrak{m}_x}\xrightarrow{\iota} \cO_{X,x}$ is an isomorphism.