replace \fK
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1 changed files with 25 additions and 25 deletions
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@ -742,7 +742,7 @@ In general, these inequalities may be strict.
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\]
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it is clear that $\codim(\{0\}, \mathfrak{k}^n) \ge n$.Translation by $x \in \mathfrak{k}^n$ gives us $\codim(\{x\} , \mathfrak{k}^n) \ge n$.
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The opposite inequality follows from \ref{upperbounddim} ($Z = \mathfrak{k}^n$ $\dim \mathfrak{k}^n \le \trdeg(\fK(Z) / \mathfrak{k}) = \trdeg(Q(\mathfrak{k}[X_1,\ldots,X_n]) / \mathfrak{k}) = n$).
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The opposite inequality follows from \ref{upperbounddim} ($Z = \mathfrak{k}^n$ $\dim \mathfrak{k}^n \le \trdeg(\mathfrak{K}(Z) / \mathfrak{k}) = \trdeg(Q(\mathfrak{k}[X_1,\ldots,X_n]) / \mathfrak{k}) = n$).
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The theorem is a special case of \ref{htandtrdeg}.
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% DIMT
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@ -919,19 +919,19 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$.
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%i = ic
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\begin{notation}
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Let $X \subseteq \mathfrak{k}^n$ be an irreducible closed subset. Then $X = V(\fp)$ for a unique prime ideal $\fp \subseteq R$.
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Let $\fK(X) \coloneqq Q(R / \fp)$ denote the field of quotients of $R / \fp$.
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Let $\mathfrak{K}(X) \coloneqq Q(R / \fp)$ denote the field of quotients of $R / \fp$.
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\end{notation}
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\begin{remark}
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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 $X$.
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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 $X$.
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\end{remark}
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\begin{theorem}\label{trdegandkdim}
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If $X \subseteq \mathfrak{k}^n$ is irreducible, then $\dim X = \trdeg (\mathfrak{k}(X) / \mathfrak{k})$ and $\codim(X, \mathfrak{k}^n) = n - \trdeg(\fK(X) / \mathfrak{k})$.
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More generally if $Y \subseteq \mathfrak{k}^n$ is irreducible and $X \subseteq Y$, then $\codim(X,Y) = \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) / \mathfrak{k})$.
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If $X \subseteq \mathfrak{k}^n$ is irreducible, then $\dim X = \trdeg (\mathfrak{k}(X) / \mathfrak{k})$ and $\codim(X, \mathfrak{k}^n) = n - \trdeg(\mathfrak{K}(X) / \mathfrak{k})$.
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More generally if $Y \subseteq \mathfrak{k}^n$ is irreducible and $X \subseteq Y$, then $\codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - \trdeg(\mathfrak{K}(X) / \mathfrak{k})$.
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\end{theorem}
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\begin{proof}
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% DIMT
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One part will be shown in "A first result on dimension theory" (\ref{upperboundcodim})
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and other one in "Aplication to dimension theory: Proof of $\dim Y = \trdeg(\fK(Y) / \mathfrak{k})$" (\ref{lowerbounddimy}).
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and other one in "Aplication to dimension theory: Proof of $\dim Y = \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$" (\ref{lowerbounddimy}).
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The theorem is a special case of \ref{htandtrdeg}.
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\end{proof}
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\begin{remark}
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@ -1139,21 +1139,21 @@ Then \[
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\end{proof}
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\begin{corollary}\label{upperboundcodim}
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Let $X, Y \subseteq \mathfrak{k}^n$ be irreducible and closed. Then $\codim(X,Y) \le \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) / \mathfrak{k})$.
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Let $X, Y \subseteq \mathfrak{k}^n$ be irreducible and closed. Then $\codim(X,Y) \le \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - \trdeg(\mathfrak{K}(X) / \mathfrak{k})$.
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\end{corollary}
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\begin{proof}
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Let $X = X_0 \subsetneq X_1 \subsetneq \ldots \subsetneq X_c = Y$ be a chain of irreducible closed subsets between $X$ and $Y$.
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Then $X_i = V(\fp_i)$ for prime ideals $\fp_0 \supsetneq \fp_1 \supsetneq \ldots \supsetneq \fp_c$ in $R = \mathfrak{k}[X_1,\ldots,X_n]$.
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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
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\[
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c + \trdeg(\fK(X) / \mathfrak{k}) = c + \trdeg(\mathfrak{k}(\fp_0) / \mathfrak{k}) \le \trdeg(\mathfrak{k}(\fp_c) / \mathfrak{k}) = \trdeg(\fK(Y) / \mathfrak{k})
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c + \trdeg(\mathfrak{K}(X) / \mathfrak{k}) = c + \trdeg(\mathfrak{k}(\fp_0) / \mathfrak{k}) \le \trdeg(\mathfrak{k}(\fp_c) / \mathfrak{k}) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k})
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\]
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As $\codim(X,Y) = \sup \{c \in \N | \exists X = X_0 \subsetneq \ldots \subsetneq X_c = Y \text{ irreducible, closed}\}$ it follows that
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$$\codim(X,Y) \le \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) / \mathfrak{k})$$
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$$\codim(X,Y) \le \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - \trdeg(\mathfrak{K}(X) / \mathfrak{k})$$
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\end{proof}
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\begin{corollary}\label{upperbounddim}
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Let $Z \subseteq \mathfrak{k}^n$ be irreducible and closed.
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Then \[\dim Z \le \trdeg(\fK(Z) / \mathfrak{k})\] and \[\codim(Z, \mathfrak{k}^n) \le n - \trdeg(\fK(Z) / \mathfrak{k}\]
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Then \[\dim Z \le \trdeg(\mathfrak{K}(Z) / \mathfrak{k})\] and \[\codim(Z, \mathfrak{k}^n) \le n - \trdeg(\mathfrak{K}(Z) / \mathfrak{k}\]
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\end{corollary}
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\begin{proof}
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Take $X = \{z\} $ and $Y = Z$ or $X = Z$ and $Y = \mathfrak{k}^n$ in \ref{upperboundcodim}.
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@ -1313,17 +1313,17 @@ Many questons of commutative algebra are easier in the case of local rings. Loca
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\begin{remark}
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The proof of \ref{cohenseidenberg} does not use Noetherianness, as this is not an assumption.
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\end{remark}
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\subsubsection{Application to dimension theory: Proof of \texorpdfstring{$\dim Y = \trdeg(\fK(Y) / \mathfrak{k})$}{dim Y = trdeg(K(Y) / k)}}
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\subsubsection{Application to dimension theory: Proof of \texorpdfstring{$\dim Y = \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$}{dim Y = trdeg(K(Y) / k)}}
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\label{lowerbounddimy}
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This is part of the proof of \ref{trdegandkdim}.
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%It uses going-up.
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%TODO: relate to \ref{htandcodim}
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\begin{proof}
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Let $B = \mathfrak{k}[X_1,\ldots,X_n]$ and let $X \subseteq Y \subseteq \mathfrak{k}^n$ be irreducible closed subsets of $\mathfrak{k}^n$.
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We have to show $\codim(X,Y) = \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) \setminus \mathfrak{k})$.
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We have to show $\codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - \trdeg(\mathfrak{K}(X) \setminus \mathfrak{k})$.
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The inequality
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\[
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\codim(X,Y) \le \trdeg(\fK(Y) \setminus \mathfrak{k}) - \trdeg(\fK(X) \setminus \mathfrak{k})
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\codim(X,Y) \le \trdeg(\mathfrak{K}(Y) \setminus \mathfrak{k}) - \trdeg(\mathfrak{K}(X) \setminus \mathfrak{k})
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\]
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has been shown in \ref{upperboundcodim}.
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In the case of $X = \{0\} , Y = \mathfrak{k}^n$, equality holds because the chain of irreducible subsets $\{0\} \subsetneq \{0\} \times \mathfrak{k} \subsetneq \ldots \subsetneq \{0\} \times \mathfrak{k}^n\subsetneq \mathfrak{k}^n$
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@ -1331,7 +1331,7 @@ This is part of the proof of \ref{trdegandkdim}.
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We have $Y = V(\fp)$ for a unique $\fp \in \Spec B$. Let $A = B / \fp$ be the ring of polynomials on $Y$.
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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 $f_i$.
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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 $d = \trdeg \fK(Y) / \mathfrak{k}$.
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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 $d = \trdeg \mathfrak{K}(Y) / \mathfrak{k}$.
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\begin{align}
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@ -1344,9 +1344,9 @@ This is part of the proof of \ref{trdegandkdim}.
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Let $\tilde \fq_i \coloneqq \pi_{B,\fp}^{-1}(\fq_i), Y_i \coloneqq V(\tilde \fq_i)$.
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This is a chain $Y = Y_0 \supsetneq Y_1 \supsetneq \ldots \supsetneq Y_d$ of irreducible subsets of $\mathfrak{k}^n$.
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Hence $\dim(Y) \ge \trdeg(\fK(Y) / \mathfrak{k})$.
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Hence $\dim(Y) \ge \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$.
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The general case of $\codim(X,Y) \ge \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) \setminus \mathfrak{k})$ is shown in \ref{proofcodimletrdeg}.
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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}.
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% TODO: reorder
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% TODO: Motivation: "Morphism" (AlGeo) and Lift of {0} x k \subseteq \ldots
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@ -1540,30 +1540,30 @@ Recall the definition of a normal field extension in the case of finite field ex
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\end{dexample}
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\subsubsection{Proof of \texorpdfstring{$\codim(\{y\},Y) = \trdeg(\fK(Y) / \mathfrak{k})$}{codim(\{y\},Y) = trdeg(K(Y) /k)}}
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\subsubsection{Proof of \texorpdfstring{$\codim(\{y\},Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$}{codim(\{y\},Y) = trdeg(K(Y) /k)}}
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\label{proofcodimletrdeg}
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This is part of the proof of \ref{trdegandkdim}. %TODO: reorder
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\begin{proof}
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% DIMT
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Let $B = \mathfrak{k}[X_1,\ldots,X_n]$ and $X \subseteq Y = V(\fp) \subseteq \mathfrak{k}^n$ irreducible closed subsets of $\mathfrak{k}^n$.
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We want to show that $\codim(X,Y) = \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) / \mathfrak{k})$.
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We want to show that $\codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - \trdeg(\mathfrak{K}(X) / \mathfrak{k})$.
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$\le $ was shown in \ref{upperboundcodim}.
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$\dim Y \ge \trdeg(\fK(Y) / \mathfrak{k})$ was shown in \ref{lowerbounddimy} by
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$\dim Y \ge \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$ was shown in \ref{lowerbounddimy} by
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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.
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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 $A$.
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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
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\[
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\codim(\{y\}, Y) \le d = \trdeg(\fK(Y) \setminus \mathfrak{k})
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\codim(\{y\}, Y) \le d = \trdeg(\mathfrak{K}(Y) \setminus \mathfrak{k})
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\]
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(see \ref{upperboundcodim})
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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.
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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.
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From this $\codim(X,Y) \ge \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) / \mathfrak{k})$ can be derived similarly to \ref{upperboundcodim}.
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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}.
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Thus
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\[
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\codim(X,Y) = \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) / \mathfrak{k})
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\codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - \trdeg(\mathfrak{K}(X) / \mathfrak{k})
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\]
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follows (see \ref{htandcodim} and \ref{htandtrdeg}).
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\end{proof}
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@ -1578,7 +1578,7 @@ This is part of the proof of \ref{trdegandkdim}. %TODO: reorder
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% i = ic
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\subsection{The height of a prime ideal}
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In order to complete the proof of \ref{proofcodimletrdeg} and show $\codim(X,Y) = \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) / \mathfrak{k})$,
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In order to complete the proof of \ref{proofcodimletrdeg} and show $\codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - \trdeg(\mathfrak{K}(X) / \mathfrak{k})$,
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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.
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To formulate a result which still applies in this context, we need the following:
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\begin{definition}[Height of a prime ideal]
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@ -2889,7 +2889,7 @@ Let $R$ be the ring generated by $\mathfrak{l}$ and the $a_i$. Localize with res
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% Dim k^n
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$\dim(\mathfrak{k}^n)$
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$ \ge n$ build chian
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$\le n$ a first result in dim T ($\fp \subsetneq \fq \implies \trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) < \trdeg(\mathfrak{k}(\fp) / \mathfrak{l})$. Thus $\codim(X,Y) \le \trdeg(\fK(Y) / \mathfrak{l}) - \trdeg(\fK(X) / \mathfrak{l})$.
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$\le n$ a first result in dim T ($\fp \subsetneq \fq \implies \trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) < \trdeg(\mathfrak{k}(\fp) / \mathfrak{l})$. Thus $\codim(X,Y) \le \trdeg(\mathfrak{K}(Y) / \mathfrak{l}) - \trdeg(\mathfrak{K}(X) / \mathfrak{l})$.
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TODO
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% List of proofs of HNS
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@ -2926,7 +2926,7 @@ The ht p and trdeg
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% Definitions
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Zariski-Topology, Spec, $\mathfrak{k}^n$
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Residue field $\mathfrak{k}(\fp) \coloneqq Q(A / \fp), \fK(V(\fp)) \coloneqq \mathfrak{k}(\fp)$. TODO?
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Residue field $\mathfrak{k}(\fp) \coloneqq Q(A / \fp), \mathfrak{K}(V(\fp)) \coloneqq \mathfrak{k}(\fp)$. TODO?
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% Counterexamples
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no going-up
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% list of definitions of codim, dim, trdeg, ht
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