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