replace \fK

2021_Algebra_I.tex

@@ -742,7 +742,7 @@ In general, these inequalities may be strict.
     \]
     it is clear that $\codim(\{0\}, \mathfrak{k}^n) \ge n$.Translation by $x \in \mathfrak{k}^n$ gives us $\codim(\{x\} , \mathfrak{k}^n) \ge n$.
 
-    The opposite inequality follows from \ref{upperbounddim} ($Z = \mathfrak{k}^n$ $\dim \mathfrak{k}^n \le \trdeg(\fK(Z) / \mathfrak{k}) = \trdeg(Q(\mathfrak{k}[X_1,\ldots,X_n]) / \mathfrak{k}) = n$).
+    The opposite inequality follows from \ref{upperbounddim} ($Z = \mathfrak{k}^n$ $\dim \mathfrak{k}^n \le \trdeg(\mathfrak{K}(Z) / \mathfrak{k}) = \trdeg(Q(\mathfrak{k}[X_1,\ldots,X_n]) / \mathfrak{k}) = n$).
 
     The theorem is a special case of \ref{htandtrdeg}.
      % DIMT
@@ -919,19 +919,19 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$.
 %i = ic
 \begin{notation}
     Let $X \subseteq \mathfrak{k}^n$ be an irreducible closed subset. Then $X = V(\fp)$ for a unique prime ideal $\fp \subseteq R$.
-    Let $\fK(X) \coloneqq Q(R / \fp)$ denote the field of quotients of $R / \fp$.
+    Let $\mathfrak{K}(X) \coloneqq Q(R / \fp)$ denote the field of quotients of $R / \fp$.
 \end{notation}
 \begin{remark}
-    As the elements of $\fp$ vanish on $X$, $R / \fp$ may be viewed as the ring of polynomials and $\fK(X)$  as the field of rational functions on $X$.
+    As the elements of $\fp$ vanish on $X$, $R / \fp$ may be viewed as the ring of polynomials and $\mathfrak{K}(X)$  as the field of rational functions on $X$.
 \end{remark}
 \begin{theorem}\label{trdegandkdim}
-    If $X \subseteq \mathfrak{k}^n$ is irreducible, then $\dim X = \trdeg (\mathfrak{k}(X) / \mathfrak{k})$ and $\codim(X, \mathfrak{k}^n) = n - \trdeg(\fK(X) / \mathfrak{k})$.
+    If $X \subseteq \mathfrak{k}^n$ is irreducible, then $\dim X = \trdeg (\mathfrak{k}(X) / \mathfrak{k})$ and $\codim(X, \mathfrak{k}^n) = n - \trdeg(\mathfrak{K}(X) / \mathfrak{k})$.
-    More generally if $Y \subseteq \mathfrak{k}^n$ is irreducible and $X \subseteq Y$, then $\codim(X,Y) = \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) / \mathfrak{k})$.
+    More generally if $Y \subseteq \mathfrak{k}^n$ is irreducible and $X \subseteq Y$, then $\codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - \trdeg(\mathfrak{K}(X) / \mathfrak{k})$.
 \end{theorem}
 \begin{proof}
     % DIMT
     One part will be shown in "A first result on dimension theory" (\ref{upperboundcodim}) 
-    and other one in "Aplication to dimension theory: Proof of $\dim Y = \trdeg(\fK(Y) / \mathfrak{k})$" (\ref{lowerbounddimy}).
+    and other one in "Aplication to dimension theory: Proof of $\dim Y = \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$" (\ref{lowerbounddimy}).
     The theorem is a special case of \ref{htandtrdeg}.
 \end{proof}
 \begin{remark}
@@ -1139,21 +1139,21 @@ Then \[
 \end{proof}
 
 \begin{corollary}\label{upperboundcodim}
-    Let $X, Y \subseteq \mathfrak{k}^n$ be irreducible and closed. Then $\codim(X,Y) \le \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) / \mathfrak{k})$.
+    Let $X, Y \subseteq \mathfrak{k}^n$ be irreducible and closed. Then $\codim(X,Y) \le \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - \trdeg(\mathfrak{K}(X) / \mathfrak{k})$.
 \end{corollary}
 \begin{proof}
     Let $X = X_0 \subsetneq X_1 \subsetneq \ldots \subsetneq X_c = Y$ be a chain of irreducible closed subsets between $X$ and $Y$.
     Then $X_i = V(\fp_i)$ for prime ideals $\fp_0 \supsetneq \fp_1 \supsetneq \ldots \supsetneq \fp_c$ in $R = \mathfrak{k}[X_1,\ldots,X_n]$.
     By \ref{trdegresfield} we have $\trdeg(\mathfrak{k}(\fp_i) / \mathfrak{k}) < \trdeg(\mathfrak{k}(\fp_{i+1}) / \mathfrak{k})$ for all $0 \le i < c$. Thus
     \[
-        c + \trdeg(\fK(X) / \mathfrak{k}) = c + \trdeg(\mathfrak{k}(\fp_0) / \mathfrak{k}) \le  \trdeg(\mathfrak{k}(\fp_c) / \mathfrak{k}) = \trdeg(\fK(Y) / \mathfrak{k})
+        c + \trdeg(\mathfrak{K}(X) / \mathfrak{k}) = c + \trdeg(\mathfrak{k}(\fp_0) / \mathfrak{k}) \le  \trdeg(\mathfrak{k}(\fp_c) / \mathfrak{k}) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k})
     \]
     As $\codim(X,Y) = \sup \{c \in \N | \exists X = X_0 \subsetneq \ldots \subsetneq X_c = Y \text{ irreducible, closed}\}$ it follows that
-    $$\codim(X,Y) \le \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) / \mathfrak{k})$$
+    $$\codim(X,Y) \le \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - \trdeg(\mathfrak{K}(X) / \mathfrak{k})$$
 \end{proof}
 \begin{corollary}\label{upperbounddim}
     Let $Z \subseteq \mathfrak{k}^n$ be irreducible and closed. 
-    Then \[\dim Z \le  \trdeg(\fK(Z) / \mathfrak{k})\] and \[\codim(Z, \mathfrak{k}^n) \le  n - \trdeg(\fK(Z) / \mathfrak{k}\]
+    Then \[\dim Z \le  \trdeg(\mathfrak{K}(Z) / \mathfrak{k})\] and \[\codim(Z, \mathfrak{k}^n) \le  n - \trdeg(\mathfrak{K}(Z) / \mathfrak{k}\]
 \end{corollary}
 \begin{proof}
     Take $X = \{z\} $ and $Y = Z$ or $X = Z$ and $Y = \mathfrak{k}^n$ in \ref{upperboundcodim}.
@@ -1313,17 +1313,17 @@ Many questons of commutative algebra are easier in the case of local rings. Loca
 \begin{remark}
     The proof of \ref{cohenseidenberg} does not use Noetherianness, as this is not an assumption.
 \end{remark}
-\subsubsection{Application to dimension theory: Proof of \texorpdfstring{$\dim Y = \trdeg(\fK(Y) / \mathfrak{k})$}{dim Y = trdeg(K(Y) / k)}}
+\subsubsection{Application to dimension theory: Proof of \texorpdfstring{$\dim Y = \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$}{dim Y = trdeg(K(Y) / k)}}
 \label{lowerbounddimy}
 This is part of the proof of \ref{trdegandkdim}. 
 %It uses going-up.
 %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 $\codim(X,Y) = \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) \setminus \mathfrak{k})$.
+    We have to show $\codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - \trdeg(\mathfrak{K}(X) \setminus \mathfrak{k})$.
     The inequality
      \[
-         \codim(X,Y) \le  \trdeg(\fK(Y) \setminus \mathfrak{k}) - \trdeg(\fK(X) \setminus \mathfrak{k})
+         \codim(X,Y) \le  \trdeg(\mathfrak{K}(Y) \setminus \mathfrak{k}) - \trdeg(\mathfrak{K}(X) \setminus \mathfrak{k})
     \] 
     has been shown in \ref{upperboundcodim}.
     In the case of $X = \{0\} , Y = \mathfrak{k}^n$, equality holds because the chain of irreducible subsets $\{0\}  \subsetneq \{0\}  \times \mathfrak{k} \subsetneq \ldots \subsetneq \{0\} \times \mathfrak{k}^n\subsetneq \mathfrak{k}^n$
@@ -1331,7 +1331,7 @@ This is part of the proof of \ref{trdegandkdim}.
 
     We have $Y = V(\fp)$ for a unique $\fp \in \Spec B$. Let $A = B / \fp$ be the ring of polynomials on $Y$. 
     Apply the Noether normaization theorem to $A$. This yields $(f_i)_{i=1}^d \in A^d$ which are algebraically independent over $\mathfrak{k}$ and such that $A$ is finite over the subalgebra generated by the $f_i$.
-    Let $L$ be the algebraic closure in $\fK(Y)$ of the subfield of $\fK(Y)$ generated by $\mathfrak{k}$ and the $f_i$. We have  $A \subseteq L$ and since $\fK(Y) = Q(B / \fp) = Q(A)$\footnote{by definition} it follows that $\fK(Y) = L$. Hence $(f_i)_{i=1}^d$ is a transcendence base for $\fK(y) / \mathfrak{k}$ and $d = \trdeg \fK(Y) / \mathfrak{k}$. 
+    Let $L$ be the algebraic closure in $\mathfrak{K}(Y)$ of the subfield of $\mathfrak{K}(Y)$ generated by $\mathfrak{k}$ and the $f_i$. We have  $A \subseteq L$ and since $\mathfrak{K}(Y) = Q(B / \fp) = Q(A)$\footnote{by definition} it follows that $\mathfrak{K}(Y) = L$. Hence $(f_i)_{i=1}^d$ is a transcendence base for $\mathfrak{K}(y) / \mathfrak{k}$ and $d = \trdeg \mathfrak{K}(Y) / \mathfrak{k}$. 
 
     
     \begin{align}
@@ -1344,9 +1344,9 @@ This is part of the proof of \ref{trdegandkdim}.
     Let $\tilde \fq_i \coloneqq \pi_{B,\fp}^{-1}(\fq_i), Y_i \coloneqq V(\tilde \fq_i)$.
     This is a chain $Y = Y_0 \supsetneq Y_1 \supsetneq \ldots \supsetneq Y_d$ of irreducible subsets of $\mathfrak{k}^n$.
 
-    Hence $\dim(Y) \ge \trdeg(\fK(Y) / \mathfrak{k})$.
+    Hence $\dim(Y) \ge \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$.
 
-    The general case of $\codim(X,Y)  \ge \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) \setminus \mathfrak{k})$ is shown in \ref{proofcodimletrdeg}.
+    The general case of $\codim(X,Y)  \ge \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - \trdeg(\mathfrak{K}(X) \setminus \mathfrak{k})$ is shown in \ref{proofcodimletrdeg}.
     % TODO: reorder
     % TODO: Motivation: "Morphism" (AlGeo) and Lift of {0} x k \subseteq \ldots
 
@@ -1540,30 +1540,30 @@ Recall the definition of a normal field extension in the case of finite field ex
 \end{dexample}
 
 
-\subsubsection{Proof of \texorpdfstring{$\codim(\{y\},Y) = \trdeg(\fK(Y) / \mathfrak{k})$}{codim(\{y\},Y) = trdeg(K(Y) /k)}}
+\subsubsection{Proof of \texorpdfstring{$\codim(\{y\},Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$}{codim(\{y\},Y) = trdeg(K(Y) /k)}}
 \label{proofcodimletrdeg}
 This is part of the proof of \ref{trdegandkdim}. %TODO: reorder
 
 \begin{proof}
     % DIMT
     Let $B = \mathfrak{k}[X_1,\ldots,X_n]$ and $X \subseteq Y = V(\fp) \subseteq \mathfrak{k}^n$ irreducible closed subsets of  $\mathfrak{k}^n$.
-    We want to show that $\codim(X,Y) = \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) / \mathfrak{k})$.
+    We want to show that $\codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - \trdeg(\mathfrak{K}(X) / \mathfrak{k})$.
     $\le $ was shown in \ref{upperboundcodim}.
-    $\dim Y \ge  \trdeg(\fK(Y) / \mathfrak{k})$ was shown in \ref{lowerbounddimy} by 
+    $\dim Y \ge  \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$ was shown in \ref{lowerbounddimy} by 
 
     Applying Noether normalization to $A \coloneqq B / \fp$, giving us $(f_i)_{i=1}^d \in A^d$ such that the $f_i$ are algebraically independent and $A$ finite over the subalgebra generated by them.
     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
     \[
-        \codim(\{y\}, Y) \le  d = \trdeg(\fK(Y) \setminus \mathfrak{k})
+        \codim(\{y\}, Y) \le  d = \trdeg(\mathfrak{K}(Y) \setminus \mathfrak{k})
     \] 
     (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.
     However using going-down (\ref{gdkrull}) for $F$, we can use a similar argument, but start lifting of the chain at the right end for the point $y \in Y$ for which we would like to show equality.
-    From this $\codim(X,Y) \ge  \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) / \mathfrak{k})$ can be derived similarly to \ref{upperboundcodim}.
+    From this $\codim(X,Y) \ge  \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - \trdeg(\mathfrak{K}(X) / \mathfrak{k})$ can be derived similarly to \ref{upperboundcodim}.
     Thus 
      \[
-         \codim(X,Y) = \trdeg(\fK(Y) / \mathfrak{k}) -  \trdeg(\fK(X) / \mathfrak{k})
+         \codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) -  \trdeg(\mathfrak{K}(X) / \mathfrak{k})
     \] 
     follows (see \ref{htandcodim} and \ref{htandtrdeg}).
 \end{proof}
@@ -1578,7 +1578,7 @@ This is part of the proof of \ref{trdegandkdim}. %TODO: reorder
 % i = ic
 
 \subsection{The height of a prime ideal}
-In order to complete the proof of \ref{proofcodimletrdeg} and show $\codim(X,Y) = \trdeg(\fK(Y) / \mathfrak{k}) -  \trdeg(\fK(X) / \mathfrak{k})$,
+In order to complete the proof of \ref{proofcodimletrdeg} and show $\codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) -  \trdeg(\mathfrak{K}(X) / \mathfrak{k})$,
 we need to localize the $\mathfrak{k}$-algebra with respect to a multiplicative subset and replace the ground field by a larger subfield of that localization which is no longer algebraically closed.
 To formulate a result which still applies in this context, we need the following:
 \begin{definition}[Height of a prime ideal]
@@ -2889,7 +2889,7 @@ Let $R$ be the ring generated by $\mathfrak{l}$ and the $a_i$. Localize with res
 % Dim k^n
 $\dim(\mathfrak{k}^n)$
 $ \ge  n$ build chian
-$\le n$ a first result in dim  T ($\fp \subsetneq \fq \implies \trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) < \trdeg(\mathfrak{k}(\fp) / \mathfrak{l})$. Thus $\codim(X,Y) \le \trdeg(\fK(Y) / \mathfrak{l}) - \trdeg(\fK(X) / \mathfrak{l})$.
+$\le n$ a first result in dim  T ($\fp \subsetneq \fq \implies \trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) < \trdeg(\mathfrak{k}(\fp) / \mathfrak{l})$. Thus $\codim(X,Y) \le \trdeg(\mathfrak{K}(Y) / \mathfrak{l}) - \trdeg(\mathfrak{K}(X) / \mathfrak{l})$.
 
 TODO
 % List of proofs of HNS
@@ -2926,7 +2926,7 @@ The ht p and trdeg
 
 % Definitions
 Zariski-Topology, Spec, $\mathfrak{k}^n$
-Residue field $\mathfrak{k}(\fp) \coloneqq Q(A / \fp), \fK(V(\fp)) \coloneqq \mathfrak{k}(\fp)$. TODO?
+Residue field $\mathfrak{k}(\fp) \coloneqq Q(A / \fp), \mathfrak{K}(V(\fp)) \coloneqq \mathfrak{k}(\fp)$. TODO?
 % Counterexamples
  no going-up
 % list of definitions of codim, dim, trdeg, ht