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\subsubsection{Matroids}
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\begin{definition}[Hull operator]
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Let $X$ be a set, $\cP(X)$ the power set of $X$. A \vocab{Hull operator} on $X$ is a map $\cP(X) \xrightarrow{\cH} \cP(X)$ such that
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Let $X$ be a set, $\mathcal{P}(X)$ the power set of $X$. A \vocab{Hull operator} on $X$ is a map $\mathcal{P}(X) \xrightarrow{\mathcal{H}} \mathcal{P}(X)$ such that
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\begin{enumerate}
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\item[H1] $\forall A \in \cP(X) ~ A \subseteq \cH(A)$.
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\item[H2] $A \subseteq B \subseteq X \implies \cH(A) \subseteq \cH(B)$.
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\item[H3] $\cH(\cH(X)) = \cH(X)$.
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\item[H1] $\forall A \in \mathcal{P}(X) ~ A \subseteq \mathcal{H}(A)$.
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\item[H2] $A \subseteq B \subseteq X \implies \mathcal{H}(A) \subseteq \mathcal{H}(B)$.
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\item[H3] $\mathcal{H}(\mathcal{H}(X)) = \mathcal{H}(X)$.
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\end{enumerate}
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We call $\cH$ \vocab{matroidal} if in addition the following conditions hold:
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We call $\mathcal{H}$ \vocab{matroidal} if in addition the following conditions hold:
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\begin{enumerate}
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\item[M] If $m,n \in X$ and $A \subseteq X$ then $m \in \cH( \{n\} \cup A) \setminus \cH(A) \iff n \in \cH(\{m\} \cup A) \setminus \cH(A).$
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\item[F] $\cH(A) = \bigcup_{F \subseteq A \text{ finite}} \cH(F)$.
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\item[M] If $m,n \in X$ and $A \subseteq X$ then $m \in \mathcal{H}( \{n\} \cup A) \setminus \mathcal{H}(A) \iff n \in \mathcal{H}(\{m\} \cup A) \setminus \mathcal{H}(A).$
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\item[F] $\mathcal{H}(A) = \bigcup_{F \subseteq A \text{ finite}} \mathcal{H}(F)$.
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\end{enumerate}
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In this case, $S \subseteq X$ is called \vocab{Independent subset}, if $s \not\in \cH(S \setminus \{s\})$ for all $s \in S$ and
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\vocab[Generating subset]{generating} if $X = \cH(S)$.
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In this case, $S \subseteq X$ is called \vocab{Independent subset}, if $s \not\in \mathcal{H}(S \setminus \{s\})$ for all $s \in S$ and
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\vocab[Generating subset]{generating} if $X = \mathcal{H}(S)$.
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$S$ is called a \vocab{base}, if it is both generating and independent.
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\end{definition}
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\begin{theorem}
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If $\cH$ is a matroidal hull operator on $X$, then a basis exists, every independent set is contained in a base and two arbitrary bases have the same cardinality.
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If $\mathcal{H}$ is a matroidal hull operator on $X$, then a basis exists, every independent set is contained in a base and two arbitrary bases have the same cardinality.
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\end{theorem}
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\begin{example}
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Let $K$ be a field, $V$ a $K$-vector space and $\cL(T)$ the $K$-linear hull of $T$ for $T \subseteq V$.
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Then $\cL$ is a matroidal hull operator on $V$.
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Let $K$ be a field, $V$ a $K$-vector space and $\mathcal{L}(T)$ the $K$-linear hull of $T$ for $T \subseteq V$.
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Then $\mathcal{L}$ is a matroidal hull operator on $V$.
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\end{example}
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\subsubsection{Transcendence degree}
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\begin{lemma}
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Let $L / K$ be a field extension and let $\cH(T)$ be the algebraic closure in $L$ of the subfield of $L$ generated by $K$ and $T$.\footnote{This is the intersection of all subfields of $L$ containing $K \cup T$, or the field of quotients of the sub-$K$-algebra of $L$ generated by $T$.}
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Then $\cH$ is a matroidal hull operator.
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Let $L / K$ be a field extension and let $\mathcal{H}(T)$ be the algebraic closure in $L$ of the subfield of $L$ generated by $K$ and $T$.\footnote{This is the intersection of all subfields of $L$ containing $K \cup T$, or the field of quotients of the sub-$K$-algebra of $L$ generated by $T$.}
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Then $\mathcal{H}$ is a matroidal hull operator.
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\end{lemma}
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\begin{proof}
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H1, H2 and F are trivial. For an algebraically closed subfield $K \subseteq M \subseteq L$ we have $\cH(M) = M$. Thus $\cH(\cH(T)) = \cH(T)$ (H3).
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H1, H2 and F are trivial. For an algebraically closed subfield $K \subseteq M \subseteq L$ we have $\mathcal{H}(M) = M$. Thus $\mathcal{H}(\mathcal{H}(T)) = \mathcal{H}(T)$ (H3).
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Let $x,y \in L$, $T \subseteq L$ and $x \in \cH(T \cup \{y\}) \setminus \cH(T)$. We have to show that $y \in \cH(T \cup \{x\}) \setminus \cH(T)$.
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If $y \in \cH(T)$ we have $\cH(T \cup \{y\}) \subseteq \cH(\cH(T)) = \cH(T) \implies x \in \cH(T) \setminus \cH(T) \lightning$.
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Hence it is sufficient to show $y \in \cH(T \cup \{x\})$. \Wlog $T = \emptyset$ (replace $K$ be the subfield generated by $K \cup T$).
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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 $y \in \mathcal{H}(T \cup \{x\}) \setminus \mathcal{H}(T)$.
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If $y \in \mathcal{H}(T)$ we have $\mathcal{H}(T \cup \{y\}) \subseteq \mathcal{H}(\mathcal{H}(T)) = \mathcal{H}(T) \implies x \in \mathcal{H}(T) \setminus \mathcal{H}(T) \lightning$.
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Hence it is sufficient to show $y \in \mathcal{H}(T \cup \{x\})$. \Wlog $T = \emptyset$ (replace $K$ be the subfield generated by $K \cup T$).
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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 $P(x) = 0$.
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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 $p_i = \frac{Q_i(y)}{R(y)}$, $R(y) \neq 0$.
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Let
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@ -821,10 +821,10 @@ In general, these inequalities may be strict.
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Q(X,Y) \coloneqq \sum_{i=0}^{\infty} X^i Q_i(Y) = \sum_{i,j=0}^{\infty} q_{i,j}X^i Y^j = \sum_{j=0}^{\infty} Y^j \hat{Q_j}(X) \in K[X,Y]
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\].
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Then $Q(x,y) = 0$.
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Let $\hat{p_j} \coloneqq \hat{Q_j}(x)$. Then $\hat{P}(y) = 0$. As $Q \neq 0$ there is $(i,j) \in \N^2$ such that $q_{i,j} \neq 0$ and then $\hat{p_j} \neq 0$ as $x \not\in \cH(\emptyset)$. Thus $\hat{P} \in \hat{M}[X] \setminus \{0\} $, where $\hat{M}$ is the subfield of $L$ generated by $K$ and $x$. Thus $y$ is algebraic over $\hat{M}$ and $y \in \cH(\{x\})$,
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Let $\hat{p_j} \coloneqq \hat{Q_j}(x)$. Then $\hat{P}(y) = 0$. As $Q \neq 0$ there is $(i,j) \in \N^2$ such that $q_{i,j} \neq 0$ and then $\hat{p_j} \neq 0$ as $x \not\in \mathcal{H}(\emptyset)$. Thus $\hat{P} \in \hat{M}[X] \setminus \{0\} $, where $\hat{M}$ is the subfield of $L$ generated by $K$ and $x$. Thus $y$ is algebraic over $\hat{M}$ and $y \in \mathcal{H}(\{x\})$,
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\end{proof}
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\begin{definition}[Transcendence Base]
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Let $L / K$ be a field extension and $\cH(T)$ the algebraic closure in $L$ of the subfield generated by $K$ and $T$. A base for $(L, \cH)$ is called a \vocab{transcendence base} and the \vocab{transcendence degree} $\trdeg(L / K)$ is defined as the cardinality of any transcendence base of $L / K$.
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Let $L / K$ be a field extension and $\mathcal{H}(T)$ the algebraic closure in $L$ of the subfield generated by $K$ and $T$. A base for $(L, \mathcal{H})$ is called a \vocab{transcendence base} and the \vocab{transcendence degree} $\trdeg(L / K)$ is defined as the cardinality of any transcendence base of $L / K$.
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\end{definition}
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\begin{remark}
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$L / K$ is algebraic iff $\trdeg(L / K) = 0$.
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@ -2644,7 +2644,7 @@ The following is somewhat harder than in the affine case:
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A set $\cB$ of open subsets of a topological space $X$ is called a \vocab{topology base} for $X$ if every open subset of $X$ can be written as a (possibly empty) union of elements of $\cB$.
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\end{definition}
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\begin{fact}
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If $X$ is a set, then $\cB \subseteq \cP(X)$ is a base for some topology on $X$ iff $X = \bigcup_{U \in \cB} U$ and for arbitrary $U, V \in \cB, U \cap V$ is a union of elements of $\cB$.
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If $X$ is a set, then $\cB \subseteq \mathcal{P}(X)$ is a base for some topology on $X$ iff $X = \bigcup_{U \in \cB} U$ and for arbitrary $U, V \in \cB, U \cap V$ is a union of elements of $\cB$.
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\end{fact}
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\begin{definition}
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Let $X$ be a variety.
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